groups 2 groups janko vol 2

de Gruyter Expositions in Mathematics 47

Editors

V. P. Maslov, Academy of Sciences, MoscowW. D. Neumann, Columbia University, New YorkR. O.Wells, Jr., International University, Bremen

Groups of Prime Power Order

Volume 2

by

Yakov Berkovich and Zvonimir Janko

≥Walter de Gruyter · Berlin · New York

Authors

Yakov Berkovich Zvonimir JankoJerusalem str. 53, apt. 15 Mathematisches InstitutAfula 18251 Ruprecht-Karls-Universität HeidelbergIsrael Im Neuenheimer Feld 288E-Mail: [email protected] 69120 Heidelberg

E-Mail: [email protected]

Mathematics Subject Classification 2000: 20-02, 20D15, 20E07

Key words: Finite p-group theory, minimal nonabelian subgroups, metacyclic subgroups, extraspe-cial subgroups, equally partitioned groups, p-groups with given maximal subgroups, 2-groups withfew cyclic subgroups of given order, Ward’s theorem on quaternion-free groups, 2-groups with

small centralizers of an involution, Blackburn’s theorem on minimal nonmetacyclic groups.

�� Printed on acid-free paper which falls within the guidelinesof the ANSI to ensure permanence and durability.

ISSN 0938-6572ISBN 978-3-11-020419-3

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The Deutsche Nationalbibliothek lists this publication in the Deutsche Nationalbibliografie;detailed bibliographic data are available in the Internet at http://dnb.d-nb.de.

� Copyright 2008 by Walter de Gruyter GmbH & Co. KG, 10785 Berlin, Germany.All rights reserved, including those of translation into foreign languages. No part of this book maybe reproduced or transmitted in any form or by any means, electronic or mechanical, includingphotocopy, recording, or any information storage or retrieval system, without permission in writing

from the publisher.

Typeset using the authors’ TeX files: Kay Dimler, Müncheberg.Printing and binding: Hubert & Co. GmbH & Co. KG, Göttingen.

Cover design: Thomas Bonnie, Hamburg.

Contents

List of definitions and notations . . . . . . . . . . . . . . . . . . . . . . . . . viii

Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xiv

�46 Degrees of irreducible characters of Suzuki p-groups . . . . . . . . . . 1

�47 On the number of metacyclic epimorphic images of finite p-groups . . . 14

�48 On 2-groups with small centralizer of an involution, I . . . . . . . . . . 19

�49 On 2-groups with small centralizer of an involution, II . . . . . . . . . . 28

�50 Janko’s theorem on 2-groups without normal elementary abelian subgroupsof order 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

�51 2-groups with self centralizing subgroup isomorphic to E8 . . . . . . . . 52

�52 2-groups with �2-subgroup of small order . . . . . . . . . . . . . . . . 75

�53 2-groups G with c2.G/ D 4 . . . . . . . . . . . . . . . . . . . . . . . . 96

�54 2-groups G with cn.G/ D 4, n > 2 . . . . . . . . . . . . . . . . . . . . 109

�55 2-groups G with small subgroup hx 2 G j o.x/ D 2ni . . . . . . . . . . 122

�56 Theorem of Ward on quaternion-free 2-groups . . . . . . . . . . . . . . 134

�57 Nonabelian 2-groups all of whose minimal nonabelian subgroups are iso-morphic and have exponent 4 . . . . . . . . . . . . . . . . . . . . . . . 140

�58 Non-Dedekindian p-groups all of whose nonnormal subgroups of the sameorder are conjugate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147

�59 p-groups with few nonnormal subgroups . . . . . . . . . . . . . . . . . 150

�60 The structure of the Burnside group of order 212 . . . . . . . . . . . . . 151

�61 Groups of exponent 4 generated by three involutions . . . . . . . . . . . 163

�62 Groups with large normal closures of nonnormal cyclic subgroups . . . . 169

�63 Groups all of whose cyclic subgroups of composite orders are normal . . 172

vi Groups of prime power order

�64 p-groups generated by elements of given order . . . . . . . . . . . . . . 179

�65 A2-groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 188

�66 A new proof of Blackburn’s theorem on minimal nonmetacyclic 2-groups 197

�67 Determination of U2-groups . . . . . . . . . . . . . . . . . . . . . . . . 202

�68 Characterization of groups of prime exponent . . . . . . . . . . . . . . . 206

�69 Elementary proofs of some Blackburn’s theorems . . . . . . . . . . . . 209

�70 Non-2-generator p-groups all of whose maximal subgroups are 2-generator 214

�71 Determination of A2-groups . . . . . . . . . . . . . . . . . . . . . . . . 233

�72 An-groups, n > 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 248

�73 Classification of modular p-groups . . . . . . . . . . . . . . . . . . . . 257

�74 p-groups with a cyclic subgroup of index p2 . . . . . . . . . . . . . . . 274

�75 Elements of order � 4 in p-groups . . . . . . . . . . . . . . . . . . . . 277

�76 p-groups with fewA1-subgroups . . . . . . . . . . . . . . . . . . . . . 282

�77 2-groups with a self-centralizing abelian subgroup of type .4; 2/ . . . . . 316

�78 Minimal nonmodular p-groups . . . . . . . . . . . . . . . . . . . . . . 323

�79 Nonmodular quaternion-free 2-groups . . . . . . . . . . . . . . . . . . 334

�80 Minimal non-quaternion-free 2-groups . . . . . . . . . . . . . . . . . . 356

�81 Maximal abelian subgroups in 2-groups . . . . . . . . . . . . . . . . . . 361

�82 A classification of 2-groups with exactly three involutions . . . . . . . . 368

�83 p-groups G with�2.G/ or ��2.G/ extraspecial . . . . . . . . . . . . . 396

�84 2-groups whose nonmetacyclic subgroups are generated by involutions . 399

�85 2-groups with a nonabelian Frattini subgroup of order 16 . . . . . . . . 402

�86 p-groups G with metacyclic��2.G/ . . . . . . . . . . . . . . . . . . . 406

�87 2-groups with exactly one nonmetacyclic maximal subgroup . . . . . . . 412

�88 Hall chains in normal subgroups of p-groups . . . . . . . . . . . . . . . 437

�89 2-groups with exactly six cyclic subgroups of order 4 . . . . . . . . . . 454

Contents vii

�90 Nonabelian 2-groups all of whose minimal nonabelian subgroups are of or-der 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 463

�91 Maximal abelian subgroups of p-groups . . . . . . . . . . . . . . . . . 467

�92 On minimal nonabelian subgroups of p-groups . . . . . . . . . . . . . . 474

Appendix

A.16 Some central products . . . . . . . . . . . . . . . . . . . . . . . . . . . 485

A.17 Alternate proofs of characterization theorems of Miller and Janko on 2-groups, and some related results . . . . . . . . . . . . . . . . . . . . . . 492

A.18 Replacement theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . 501

A.19 New proof of Ward’s theorem on quaternion-free 2-groups . . . . . . . . 506

A.20 Some remarks on automorphisms . . . . . . . . . . . . . . . . . . . . . 509

A.21 Isaacs’ examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 512

A.22 Minimal nonnilpotent groups . . . . . . . . . . . . . . . . . . . . . . . 516

A.23 Groups all of whose noncentral conjugacy classes have the same size . . 519

A.24 On modular 2-groups . . . . . . . . . . . . . . . . . . . . . . . . . . . 522

A.25 Schreier’s inequality for p-groups . . . . . . . . . . . . . . . . . . . . . 526

A.26 p-groups all of whose nonabelian maximal subgroups are either absolutelyregular or of maximal class . . . . . . . . . . . . . . . . . . . . . . . . 529

Research problems and themes II . . . . . . . . . . . . . . . . . . . . . . . . 531

Author index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 593

Subject index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 594

List of definitions and notations

Set theory

jM j is the cardinality of a setM (if G is a finite group, then jGj is called its order).

x 2 M (x 62 M ) means that x is (is not) an element of a setM . N � M (N 6� M )means that N is (is not) a subset of the set M ; moreover, if M ¤ N � M we writeN � M .

¿ is the empty set.

N is called a nontrivial subset ofM , if N ¤ ¿ and N � M . If N � M we say thatN is a proper subset ofM .

M \N is the intersection andM [N is the union of setsM and N . IfM;N are sets,then N �M D fx 2 N j x 62 M g is the difference of N andM .

Z is the set (ring) of integers: Z D f0;˙1;˙2; : : : g.

N is the set of all natural numbers.

Q is the set (field) of all rational numbers.

R is the set (field) of all real numbers.

C is the set (field) of all complex numbers.

Number theory and general algebra

p is always a prime number.

� is a set of primes; � 0 is the set of all primes not contained in � .

m, n, k, r , s are, as a rule, natural numbers.

�.m/ is the set of prime divisors of m; then m is a �-number.

np is the p-part of n, n� is the �-part of n.

.m; n/ is the greatest common divisor of m and n.

m j n should be read as: m divides n.

m − n should be read as: m does not divide n.

List of definitions and notations ix

GF.pm/ is the finite field containing pm elements.

F� is the multiplicative group of a field F.

L.G/ is the lattice of all subgroups of a group G.

If n D p˛1

1 : : : p˛k

kis the standard prime decomposition of n, then �.n/ D

PkiD1 ˛i .

Groups

We consider only finite groups which are denoted, with a pair exceptions, by uppercase Latin letters.

If G is a group, then �.G/ D �.jGj/.

G is a p-group if jGj is a power of p; G is a �-group if �.G/ � � .

G is, as a rule, a finite p-group.

H � G means thatH is a subgroup of G.

H < G means that H � G and H ¤ G (in that case H is called a proper subgroupof G). f1g denotes the group containing only one element.

H is a nontrivial subgroup of G if f1g < H < G.

H is a maximal subgroup of G if H < G and it follows from H � M < G thatH D M .

H E G means that H is a normal subgroup of G; moreover, if, in addition, H ¤ G

wewriteH GG and say thatH is a proper normal subgroup ofG. Expressions ‘normalsubgroup of G’ and ‘G-invariant subgroup’ are synonyms.

H GG is called a nontrivial normal subgroup of G provided H > f1g.

H is a minimal normal subgroup of G if (a) H E G; (b) H > f1g; (c) N G G andN < H implies N D f1g. Thus, the group f1g has no minimal normal subgroup.

G is simple if it is a minimal normal subgroup of G (so jGj > 1).

H is a maximal normal subgroup of G ifH < G and G=H is simple.

The subgroup generated by all minimal normal subgroups of G is called the socle ofG and denoted by Sc.G/. We put, by definition, Sc.f1g/ D f1g.

NG.M/ D fx 2 G j x�1Mx D M g is the normalizer of a subsetM in G.

CG.x/ is the centralizer of an element x in G W CG.x/ D fz 2 G j zx D xzg.

CG.M/ DT

x2M CG.x/ is the centralizer of a subsetM in G.

If A � B and A;B E G, then CG.B=A/ D H , whereH=A D CG=A.B=A/.

x Groups of prime power order

AwrB is the wreath product of the ‘passive’ group A and the transitive permutationgroup B (in what follows we assume that B is regular); B is called the active factor ofthe wreath product). Then the order of that group is jAjjBjjBj.

Aut.G/ is the group of automorphisms of G (the automorphism group of G).

Inn.G/ is the group of all inner automorphisms of G.

Out.G/ D Aut.G/=Inn.G/, the outer automorphism group of G.

If a; b 2 G, then ab D b�1ab.

An element x 2 G inverts a subgroup H � G if hx D h�1 for all h 2 H .

IfM � G, then hM i D hx j x 2 M i is the subgroup of G generated byM .

Mx D x�1Mx D fyx j y 2 M g for x 2 G andM � G.

Œx; y� D x�1y�1xy D x�1xy is the commutator of elements x; y ofG. IfM;N � G

then ŒM;N � D hŒx; y� j x 2 M;y 2 N i is a subgroup of G.

o.x/ is the order of an element x of G.

An element x 2 G is a �-element if �.o.x// � � .

G is a �-group, if �.G/ � � . Obviously, G is a �-group if and only if all of itselements are �-elements.

G0 is the subgroup generated by all commutators Œx; y�, x; y 2 G (i.e., G0 D ŒG;G�),G.2/ D ŒG0; G0� D G00 D .G0/0, G.3/ D ŒG00; G00� D .G00/0 and so on. G0 is calledthe commutator (or derived) subgroup of G.

Z.G/ DT

x2G CG.x/ is the center of G.

Zi.G/ is the i -th member of the upper central series of G; in particular, Z0.G/ D f1g,Z1.G/ D Z.G/.

Ki.G/ is the i -th member of the lower central series of G; in particular, K2.G/ D G0.We have Ki.G/ D ŒG; : : : ; G� (i � 1 times). We set K1.G/ D G.

If G is nonabelian, then �.G/=K3.G/ D Z.G=K3.G//.

M.G/ D hx 2 G j CG.x/ D CG.xp/ is the Mann subgroup of a p-group G.

Sylp.G/ is the set of p-Sylow subgroups of an arbitrary finite group G.

Sn is the symmetric group of degree n.

An is the alternating group of degree n

†pn is a Sylow p-subgroup of Spn .

GL.n; F / is the set of all nonsingular n � n matrices with entries in a field F , then-dimensional general linear group over F , SL.n; F / D fA 2 GL.n; F / j det.A/ D

1 2 F g, the n-dimensional special linear group over F .

List of definitions and notations xi

If H � G, then HG DT

x2G x�1Hx is the core of the subgroup H in G and

HG DT

H�N EG N is the normal closure or normal hull of H in G. Obviously,HG E G.

If G is a p-group, then pb.x/ D jG W CG.x/j; b.x/ is said to be the breadth of x 2 G,where G is a p-group; b.G/ D max fb.x/ j x 2 Gg is the breadth of G.

ˆ.G/ is the Frattini subgroup ofG (= the intersection of all maximal subgroups of G),ˆ.f1g/ D f1g, pd.G/ D jG W ˆ.G/j.

�i D fH < G j ˆ.G/ � H; jG W H j D pi g, i D 1; : : : ; d.G/, where G > f1g.

IfH < G, then �1.H/ is the set of all maximal subgroups of H .

exp.G/ is the exponent of G (the least common multiple of the orders of elements ofG). If G is a p-group, then exp.G/ D max fo.x/ j x 2 Gg.

k.G/ is the number of conjugacy classes of G (D G-classes), the class number of G.

Kx is the G-class containing an element x (sometimes we also write cclG.x/).

Cm is the cyclic group of order m.

Gm is the direct product of m copies of a group G.

A � B is the direct product of groups A and B .

A � B is a central product of groups A and B , i.e., A � B D AB with ŒA;B� D f1g.

Epm D Cmp is the elementary abelian group of order pm. G is an elementary abelian

p-group if and only if it is a p-group > f1g and G coincides with its socle. Next, f1gis elementary abelian for each prime p.

A group G is said to be homocyclic if it is a direct product of isomorphic cyclic sub-groups (obviously, elementary abelian p-groups are homocyclic).

ES.m;p/ is an extraspecial group of order p1C2m (a p-group G is said to be extraspe-cial if G0 D ˆ.G/ D Z.G/ is of order p). Note that for eachm 2 N, there are exactlytwo nonisomorphic extraspecial groups of order p2mC1.

S.p3/ is a nonabelian group of order p3 and exponent p > 2.

A special p-group is a nonabelian p-group G such that G0 D ˆ.G/ D Z.G/ iselementary abelian. Direct products of extraspecial p-groups are special.

D2m is the dihedral group of order 2m; m > 2. Some authors consider E22 as thedihedral group D4.

Q2m is the generalized quaternion group of order 2m � 23.

SD2m is the semidihedral group of order 2m � 24.

Mpm is a nonabelian p-group containing exactly p cyclic subgroups of index p.

xii Groups of prime power order

cl.G/ is the nilpotence class of a p-group G.

dl.G/ is the derived length of a p-group G.

CL.G/ is the set of all G-classes.

A p-group of maximal class is a nonabelian groupG of order pm with cl.G/ D m�1.

�m.G/ D hx 2 G j o.x/ � pmi, ��m.G/ D hx 2 G j o.x/ D pmi and Ãm.G/ D

hxpm

j x 2 Gi.

A p-group is absolutely regular if jG=Ã1.G/j < pp .

A p-group is thin if it is either absolutely regular or of maximal class.

G D A B is a semidirect product with kernel B and complement A.

A group G is an extension of N E G by a group H if G=N Š H . A group G splitsover N if G D H N withH � G andH \N D f1g (in that case, G is a semidirectproduct ofH and N with kernel N ).

H # D H � feH g, where eH is the identity element of the group H . IfM � G, thenM # D M � feGg.

An automorphism ˛ of G is regular (D fixed-point-free) if it induces a regular permu-tation on G# (a permutation is said to be regular if it has no fixed points).

An involution is an element of order 2 in a group.

A section of a group G is an epimorphic image of some subgroup of G.

If F D GF.pn/, then we write GL.m;pn/;SL.m;pn/; : : : instead of GL.m;F /,SL.m;F /; : : : .

cn.G/ is the number of cyclic subgroups of order pn in a p-group G.

sn.G/ is the number of subgroups of order pn in a p-group G.

en.G/ is the number of subgroups of order pn and exponent p in G.

An-group is a p-group G all of whose subgroups of index pn are abelian but G con-tains a nonabelian subgroup of index pn�1. In particular, A1-group is a minimalnonabelian p-group for some p.

˛n.G/ is the number of An-subgroups in a p-group G.

Characters and representations

Irr.G/ is the set of all irreducible characters of G over C.

A character of degree 1 is said to be linear.

Lin.G/ is the set of all linear characters of G (obviously, Lin.G/ � Irr.G/).

List of definitions and notations xiii

Irr1.G/ D Irr.G/ � Lin.G/ is the set of all nonlinear irreducible characters of G;n.G/ D jIrr1.G/j.

.1/ is the degree of a character of G,

H is the restriction of a character of G toH � G.

G is the character of G induced from the character of some subgroup of G.

N is a character of G defined as follows: N.x/ D .x/ (here Nw is the complex conju-gate of w 2 C).

Irr./ is the set of irreducible constituents of a character of G.

If is a character of G, then ker./ D fx 2 G j .x/ D .1/g is the kernel of acharacter .

Z./ D fx 2 G j j.x/j D .1/g is the quasikernel of .

If N E G, then Irr.G j N/ D f 2 Irr.G/ j N — ker./g.

h; i D jGj�1P

x2G .x/.x�1/ is the inner product of characters and of G.

IG.�/ D hx 2 G j �x D �i is the inertia subgroup of � 2 Irr.H/ inG, whereH GG.

1G is the principal character of G (1G.x/ D 1 for all x 2 G).

M.G/ is the Schur multiplier of G.

cd.G/ D f.1/ j 2 Irr.G/g.

mc.G/ D k.G/=jGj is the measure of commutativity of G.

T.G/ DP

�2Irr.G/ .1/; f.G/ D T.G/=jGj.

Preface

This is the second part of the book. Sections 48–57, 60, 61, 66, 67, 70, 71, 73–75, 77–87, 89–92 and Appendix 19 are written by the second author, all other sections – bythe first author. This volume contains a number of very strong results on 2-groups dueto the second author. All exercises and about all problems are due to the first author.All material of this part is appeared in the book form at the first time.Some outstanding problems of p-group theory are solved in this volume:

(i) classification of 2-groups with exactly three involutions,

(ii) classification of 2-groups containing exactly one nonmetacyclic maximal sub-group,

(iii) classification of 2-groups G containing an involution t such that CG.t/ D

hti �Q, whereQ contains only one involution,

(iv) classification of 2-groups all of whose minimal nonabelian subgroups have thesame order 8,

(v) classification of 2-groups of rank 3 all of whose maximal subgroups are ofrank 2,

(vi) classification of 2-groups containing selfcentralizing noncyclic abelian sub-groups of order 8, and so on.

There are, in this part, a number of new proofs of known important results:

(a) Blackburn’s classification of minimal nonmetacyclic groups (we presented, inSec. 66 and 69, two different proofs),

(b) classification of p-groups all of whose subgroups of index p2 are abelian,

(c) Ward’s theorem on quaternion-free 2-groups (we presented two differentproofs),

(d) classification of p-groups with cyclic subgroup of index p2,

(e) Kazarin’s classification of p-groups all of whose cyclic subgroups of order> p are normal,

(f) Iwasawa’s classification of modular p-groups, and so on.

Some results proved in this part have no analogs in existing literature:

(a) classification of 2-groups all of whose minimal nonabelian subgroups are iso-morphic and have order 16,

(b) study the 2-groups with at most 1C p C p2 minimal nonabelian subgroups,

Preface xv

(c) classification of 2-groups all of whose nonabelian two-generator subgroups areof maximal class,

(d) classification of 2-groups G with j�2.G/j � 24 and j��2.G/j D 24,

(e) classification of p-groups G with �2.G/ or ��2.G/ is metacyclic (extraspe-

cial),

(f) classification of 2-groups all of whose nonabelian subgroups are generated byinvolutions, and so on.

As the previous part, this one contains a great number of open problems posed, as arule, by the first author; some of these problems are solved and solutions are presentedbelow.The first author is indebted to Avinoam Mann for numerous useful discussions and

help. The correspondence with Martin Isaacs allowed us to acquaint the reader with anumber of his old and new important results. Moreover, Mann and Isaacs familiarizedus with a number of their papers prior of publication. Noboru Ito read a numberof sections and all appendices and made numerous useful remarks and suggestions.The help of Lev Kazarin was very important and allowed us to improve a numberof places of the book, especially, in ��46 and 63; he also acquainted the first authorwith a fragment of his PhD thesis (see ��65, 71). The first author also indebted toGregory Freiman, Marcel Herzog (both at Tel-Aviv University), Moshe Roitman andIzu Vaisman (both at University of Haifa) for help and support.The publication of the book gives us great pleasure. We are grateful to the publish-

ing house of Walter de Gruyter and all who promoted the publication, among of themProf. M. Hazewinkel, Dr. R. Plato and K. Dimler, for their support and competenthandling of the project.

�46

Degrees of irreducible charactersof Suzuki p-groups

The results of this section are due to I. A. Sagirov [Sag1, Sag2]. Throughout thissection we use the following notation: F D GF.pm/, m > 1; � is an automorphismof F of order k for some divisor k > 1 of m (recall that the group of automorphismsof F is cyclic of order m whose generator is a 7! ap for all a 2 F); n D m

k.< m/.

Let, for example, � W a 7! apn

(a 2 F). The set of fixed points of � is a subfieldF� of F satisfying apn

D 1 so containing pn elements (for example, if k D m, thenF� D F0, the prime subfield of F). Let F� be the (cyclic) multiplicative group ofF. Next, let Irr1.G/ denote the set of all nonlinear irreducible characters of G. Putcd.G/ D f.1/ j 2 Irr.G/g. Next, Irr.t/.G/ denotes the number of characters ofdegree t in Irr.G/.

Definition. The Suzuki p-group Ap.m; �/ is the set F � F with multiplication definedas follows: .a; b/.c; d/ D .a C c; b C d C a�.c//.

Let G D Ap.m; �/; then jGj D jFj2 D p2m. It follows from the definition thatelements .a; b/ and .c; d/ commute if and only if c�.a/ D a�.c/, i.e., either a D 0

or �.c=a/ D c=a so c D au for some u 2 F� . The identity element of G is .0; 0/and .a; b/�1 D .�a;�b C a�.a//. Since so defined multiplication is associative, Gindeed is a group. If .c; d/ 2 Z.G/ and .a; b/ 2 G, then c D au for u 2 F� and alla 2 F which implies c D 0 since k > 1. Thus, Z.G/ D f.0; d/ j d 2 Fg. We haveZ.G/ Š Epm . It is easy to prove, by induction that .a; b/n D .na; nb C

�

n2

�

a�.a//.Taking n D p, we get .a; b/p 2 Z.G/; moreover, if p > 2, then exp.G/ D p.

1o. Throughout this subsection p D 2 and G D A.m; �/ D A2.m; �/. Our aim isto find cd.G/ and the number of irreducible characters of every degree s 2 cd.G/.

Theorem 46.1 ([Sag1]). Suppose that p D 2, G D A.m; �/, wherem > 1 and � is anautomorphism of the field F D GF.2m/ of order k > 1 and n D m

k. Then one of the

following holds:

(a) If k is odd, then cd.G/ D f1; 212

.m�n/g.

(b) If k D 2, then cd.G/ D f1; 2m=2g D f1; 2ng.

(c) If k > 2 is even, then cd.G/ D f1; 2m=2; 2.m=2/�ng, jIrr.2m=2/.G/j D .2m�1/2n

2nC1

and jIrr.2.m=2/�n/.G/j D .2m�1/22n

2nC1 .

2 Groups of prime power order

If x 2 F then, since � is an automorphism of F of order k > 1, then �.x/ D x2s

for some nonnegative s independent of x and �k.x/ D x, x 2 F. On the other hand,�k.x/ D x2sk

so x2sk

D x, and this is true for each x 2 F. If x is a primitive elementof F, it follows that 2m � 1 divides 2sk � 1 so m.D nk/ divides sk (see Lemma46.5 below) hence n divides s, and we conclude that s D nt , where .t; k/ D 1 sinceo.�/ D k. Thus, �.x/ D x2nt

. Therefore, the number of automorphisms of F oforder k equals '.k/, where '.�/ is the Euler’s totient function. Then fa 2 F j �.a/ D

ag D F� is the set of elements a 2 F such that a2nt

D a and the cardinality of thatset is 2n. As we have shown, if a ¤ 0, then CG..a; b// D f.az; u/ j z 2 F� ; u 2 Fg

so jCG..a; b//j D 2mCn. In particular, the size of every noncentral G-class equals22m

2mCn D 2m�n. If a 2 F� f0g, then .a; b/2 D .0; a�.a// ¤ .0; 0/ so Z.G/ D �1.G/.Therefore, since exp.G/ D 4, we get�1.G/ D Ã1.G/ D ˆ.G/.

Lemma 46.2. jG0j � 2m�n.

Indeed, jG0j is at least the size of a noncentral G-class. However, all noncentralG-classes have the same size 2m�n. From the last assertion follows

Lemma 46.3. k.G/ D jZ.G/j C jGj�jZ.G/j2m�n D 2mCn C 2m � 2n.

An element � 2 F is said to be primitive if the (cyclic) multiplicative group F� D

h�i.

Lemma 46.4. A mapping '�..a; b// D .�a; ��.�/b/ is an automorphism of G oforder o.�/ for every element � 2 .F�/#.

Proof. Set ��.�/ D . Then '�..a; b// D .�a; b/, '�..c; d// D .�c; d/,

'�..a; b/.c; d// D '�..a C c; b C d C a�.c// D .�.a C c/; .b C d C a�.c///:

On the other hand,

'�..a; b//'�..c; d// D .�a; b/.�c; d/ D .�.a C c/; b C d C �a�.�c//

D .�.a C c/; .b C d C a�.c///:

It follows that '� is an endomorphism of G. Next, .�a; b/ D '�..a; b// D .0; 0/ ifand only if .a; b/ D .0; 0/ so '� is an automorphism of G since G is finite.Set o.�/ D d . Then 'd

�..a; b// D .�da; .��.�//d b/ D .a; b/ and, if a ¤ 0, then

'r�..a; b// ¤ .a; b/ for r 2 f1; 2; : : : ; d � 1g. It follows that o.'�/ D d D o.�/.

Lemma 46.5. Given a > 1, m;n, we have .am � 1; an � 1/ D a.m;n/ � 1.

Proof. Set .m:n/ D ı and d D .am � 1; an � 1/. Then there exist u; v 2 N such thatı D mu � nv. Clearly, aı � 1 divides d . On the other hand, d divides amu � 1 andanv � 1. Hence d divides the number amu � 1 � .anv � 1/ D anv.amu�nv � 1/ D

anv.aı �1/ so d divides aı �1 since .d; a/ D 1, and we conclude that d D aı �1.

�46 Degrees of irreducible characters of Suzuki p-groups 3

Lemma 46.6. Let m D nk, where n; k 2 N, and let t 2 N be coprime with k; then.m; nt/ D n.

(a) Set d D .2m � 1; 2nt C 1/. Then d D 1 if k is odd and d D 2n C 1 if k is even.

(b) Suppose that 2n C 1 divides 2m � 1. Then 2n divides m.

Proof. (a) The number d0 D .2m � 1; 22nt � 1/ equals 2.nk;2nt/ � 1 D 2n � 1 if k isodd and d0 D 22n � 1 if k is even (Lemma 46.5). Since 2nt C 1 divides 22nt � 1, thenumber d divides d0.We claim that .2nt C 1; 2n � 1/ D 1. Indeed, if a prime p divides that number, then

2n 1 .mod p/ so 2nt 1t 1 .mod p/, a contradiction since p (if it exists) mustbe odd and 2nt �1 .mod p/.Suppose that k is odd. By definition, d divides 2nt C 1 and, by the first paragraph,

d divides d0 D 2n �1 so d divides .2nt C1; 2n �1/ D 1 (see the previous paragraph),and we get d D 1. This proves the first assertion in (a).Now suppose that k is even; then t is odd. In that case, by the first paragraph, d

divides d0 D 22n � 1 D .2n � 1/.2n C 1/ and, by definition, d divides 2nt C 1.Therefore, by the second paragraph, d divides .2n C 1; 2nt C 1/ D 2n C 1 since t isodd. Since 2n C 1 divides d , we get d D 2n C 1. The proof of (a) is complete.

(b) Since .2n � 1; 2n C 1/ D 1 and 2n � 1 divides 2m � 1 D 2nk � 1, it follows that22n � 1 D .2n � 1/.2n C 1/ divides 2m � 1 and so 2n divides m (Lemma 46.5).

Write ˆ� D h'�i (� 2 F#); then jˆ�j D o.�/. If � is primitive, we write ˆ D ˆ�

Lemma 46.7. Let � 2 F be primitive; then jˆj D o.�/ D 2m � 1 and:

(a) The size of everyˆ-orbit on the set G � Z.G/ equals 2m � 1.

(b) For odd k, there is only one ˆ-orbit on the set Z.G/#. (In that case, ˆ G is aFrobenius group with minimal normal subgroup Z.G/, its kernel.)

(c) For even k, the size of everyˆ-orbit on the set Z.G/# equals 2m�12nC1

(in that case,the number of ˆ-orbits on Z.G/# equals 2n C 1/.

Proof. (a) For .a; b/ 2 G and i 2 N, we have 'i�..a; b// D .�ia; .��.�//ib/. Let

.a; b/ 2 G � Z.G/. In that case, a ¤ 0 so 'i�..a; b// D .a; b/ if and only if �i D 1,

i.e., i is divisible by 2m � 1 since � is primitive. This proves (a)(b) Now let k be odd and b ¤ 0. Then 'i

�..0; b// D .0; .��.�/ib/ D .0; b/ if and

only if .��.�//i D 1. Since �.�/ D �2nt

with .t; k/ D 1, we get 1 D .��.�//i D

�.2nt C1/i . Therefore, o.�/ D 2m � 1 divides .2nt C 1/i . By Lemma 46.6(a), if k isodd, 2m � 1 divides i since .2m � 1; 2nt C 1/ D 1, so the ˆ-orbit of .0; b/ contains2m � 1 elements. Then all elements of Z.G/# are ˆ-conjugate, and (b) is proven.(c) Let k be even. The minimal positive integer i such that �.2nt C1/i D 1 equals

2m�12nC1 , by Lemma 46.6(a), and this number is the size of the ˆ-orbit of the element.0; b/ 2 Z.G/#. It follows that the number of ˆ-orbits on Z.G/# equals 2n C 1,completing the proof of (c).


For � 2 F� � f1Fg, '� has no fixed points on .G=Z.G//# (check!) so h'�; G=Z.G/iis a Frobenius group.

Corollary 46.8. If k is odd, then G0 D Z.G/.

Proof. Indeed, Z.G/ is a minimal normal subgroup of h'�;Z.G/i, where � is a prim-itive element of F, by Lemma 46.7(b), and G0 � Z.G/ so G0 D Z.G/.

Remark. Suppose that 2 Irr1.G/, .1/ D 2a andZ0 D Z.G/\ker./, whereG D

A.m; �/. Then jZ.G/ W Z0j D 2 since the center of G= ker./ is cyclic. It followsthatG0Z0 D Z.G/, cd.G=Z0/ D f1; 2ag [Isa1, Theorem 2.31] and so jIrr1.G=Z0/j D2mC1�2m

22a D 2m�2a. A character 2 Irr1.G/ determines Z0 uniquely.

Theorem 46.9 ([Han]). If G D A.m; �/ is a Suzuki 2-group, where � is an automor-phism of F of odd order k > 1, then cd.G/ D f1; 2.m�n/=2g (here n D m

k).

Proof. Let � 2 F be primitive. Consider the action of the automorphism '� on sub-groups of index 2 in Z.G/. By Lemma 46.7(b), h'�;Z.G/i is a Frobenius group andZ.G/# is the unique '�-orbit. By Maschke’s theorem, if '

s�

¤ id, then that automor-phism has no invariant subgroups of index 2 in Z.G/, i.e., h'�i acts on the set of suchsubgroups in a fixed-point-free manner. Since the order of '� equals 2m �1, all 2m �1

subgroups of index 2 in Z.G/ are .ˆ D/h'�i-conjugate.Let 2 Irr1.G/, Z0 D Z.G/ \ ker./. Then the isomorphism type of G=Z0

independent of , since all subgroups of index 2 in Z.G/ are ˆ-conjugate. It followsthat cd.G/ D f1; dg for some d > 1, by the Remark. By Lemma 46.3 and Corollary46.8, we get 2m C .2mCn � 2n/d2 D 22m so d D 2

12

.m�n/.

Thus, Theorem 46.1(a) is proven. In what follows, k is even.

Lemma 46.10. Let � 2 F be primitive and k even. Then:

(a) If k > 2, i.e., m D kn > 2n, then G0 D Z.G/ and the number of ˆ-orbits on.G0/# equals 2n C 1, and all these orbits have the same size 2m�1

2nC1.

(b) If k D 2, i.e., m D 2n, then either G0 D Z.G/ and .G0/# has 2n C 1 distinctˆ-orbits of size 2m�1

2nC1D 2n � 1 or else jG0j D 2n.D 2

m2 / and all elements of

.G0/# are ˆ-conjugate.

Proof. The assertion on the sizes of ˆ-orbits follows from Lemma 46.7(c). Supposethat the number of ˆ-orbits on .G0/# equals t and jG0j D 2f . Then 2f � 1 D t 2m�1

2nC1

(see Lemma 46.7(c)). It follows that t � 1 D t 2m C 2n � 2f Cn � 2f . By Lemma46.2, f � m � n � n so 2n divides t � 1. It follows that either t D 1 or t � 2n C 1.On the other hand, f � m since G0 � Z.G/ and, by the previous paragraph,

t 2m�12nC1

D 2f � 1 � 2m � 1. It follows that t � 2n C 1 so t 2 f1; 2n C 1g.

Suppose that t D 1. Then, by the previous paragraph, 2f �1 D 2m�12nC1 so 2

n.2m�nC

1/ D 2f .2n C 1/. It follows that then f D n and m� n D n, i.e.,m D 2n, k D 2. Inthat case, f D n D 1

2m, and we have the second possibility in (b).

Suppose that t D 2n C 1. Then 2f � 1 D .2n C 1/2m�12nC1 D 2m � 1 so f D m.


In what follows we first assume that G0 D Z.G/. Then the number of ˆ-orbits on.G0/# equals 2n C 1 and all these orbits have the same size 2m�1

2nC1(Lemma 46.10).

Lemma 46.11. Suppose that G0 D Z.G/, � 2 F is primitive and k is even.

(a) Under ˆ, the set of subgroups of index 2 in Z.G/ is partitioned in 2n C 1 orbitsof size 2m�1

2nC1.

(b) Under ˆ, the set Irr.G/ � f1Gg is partitioned in 2n orbits of size 2m � 1 and2n C 1 orbits of size 2m�1

2nC1.

Proof. Z.G/ has 2m � 1 subgroups of index 2 and jIrr.G/j D 2mCn C 2m � 2n.

(a) By Lemma 46.7, the restriction of '� to Z.G/# is a product of 2nC1 independentcycles of size 2m�1

2nC1. This proves (a) (see the first paragraph in the proof of Lemma

46.9).

(b) Define the action of '� on the set Irr.G/ as follows. Let R be a representationaffording the character 2 Irr.G/. Set T D R ı '�1

�, i.e., T .g/ D R.g'�1

�/ for

g 2 G. If h 2 G, then since '� 2 Aut.G/, we get

T .gh/ D R..gh/'�1� / D R.g'�1

� h'�1� / D R.g'�1

� /R.h'�1� / D T .g/T .h/;

i.e., T is a representation of G. Clearly, the value of the character '�.g/ of T equals.g'�1

�/ for every g 2 G, and this character is also irreducible.

Let CL.G/ D fK1; : : : ;Krg be the set of G-classes, xi 2 Ki , all i , and letX D X.G/ D .i .xj // be the character table ofG. Then '� acts on columns and rowsofX by permutations �1 and �2, respectively, in such a way that if one considers these

permutations as elements of GL.r;C/, then �2X D .'�

i .xj // D .i .x'�1

�

j // D X�1.

Since the matrix X is nonsingular, we get �1 D X�1�2X in GL.r;C/. Then, byVishnevetsky’s lemma [BZ, Lemma 10.4(a)], �1 and �2 are conjugate in the sym-metric group Sr and so they have the same cycle structure. Now the result followsfrom Lemma 46.7. Indeed, under ˆ, there are exactly 2n C 1 orbits of size 2m�1

2nC1on

nonidentity central G-classes and 2mCn�2n

2m�1D 2n orbits of size 2m � 1 on noncentral

G-classes. It follows thatˆ has 2n C1 orbits of size 2m�12nC1 and 2

n orbits of size 2m �1

on the set Irr.G/ � f1Gg.

Lemma 46.12. If 2 Irr1.G/, k is even and .1/ D 2a, then 12m � n � a � 1

2m.

Proof. By the Remark, we have 2 Irr1.G=Z0/, whereZ0 < Z.G/ is of index 2, andcd.G=Z0/ D f1; 2ag so jIrr1.G=Z0/j D 2m�2a. It follows that m � 2a � 0 soa � 1

2m.

LetZ1; : : : ; Z2nC1 be a transversal of the set ofˆ-orbits on the setM of subgroupsof index 2 in Z.G/. By the Remark, cd.G=Zi / D f1; 2ai g, jIrr1.G=Zi /j D 2m�2ai , alli . Next, if Zi and Zi;1 belong to the same ˆ-orbit, then, obviously, G=Zi Š G=Zi;1.Since jIrr1.G/j D 2mCn �2n, we get 2mCn �2n D 2m�1

2nC1

P2nC1iD1 2m�2ai (here 2m�1

2nC1

is the common size of ˆ-orbit) which implies thatP2nC1

iD1 2m�2ai D 2n.2n C 1/.


Assume that m � 2ai > 2n for some i . Then 2m�2ai � 22nC1 > 2n.2n C 1/,contrary to the result of the previous paragraph. It follows that m � 2ai � 2n soai � 1

2m � n for all i .

Theorem 46.13. If k is even and G0 D Z.G/, then cd.G/ D f1; 2.m=2/�n; 2m=2g.Next, jIrr.2.m=2�n/.G/j D .2m�1/22n=.2nC1/, jIrr.2m=2/.G/j D .2m�1/2n=.2nC1/.

Proof. Set s D 2m�12nC1

.By Lemma 46.11(b), the set Irr.G/�f1Gg is partitioned under the action ofˆ in 2n

orbits of size 2m �1 and 2n C1 orbits of size s. Since hˆ;Gi=G0 is a Frobenius group(see the paragraph following Lemma 46.7), all jG W G0j � 1 D 2m � 1 nonprincipallinear characters of G are ˆ-conjugate. Therefore, under ˆ, there are exactly 2n � 1

orbits of size 2m � 1 on Irr1.G/.By Lemma 46.11(a), the set M of subgroups of index 2 in Z.G/ is partitioned in

exactly 2n C 1 orbits of size s under ˆ. Then, if Z0 2 M, then 's�.Z0/ D Z0 since

the order of the restriction of '� toM is s.Suppose that the size of the ˆ-orbit of 2 Irr1.G/ equals 2m � 1 and 2

Irr1.G=Z0/ for some Z0 2 M. Then ¤ 'i� 2 Irr1.G=Z0/ for 0 < i < 2n C 1 so

the number of ˆ-conjugates of equals 2n C 1.Suppose that for some Z0 2 M, Irr1.G=Z0/ has no characters belonging to a ˆ-

orbit of size s. Let cd.G=Z0/ D f1; 2ag. Then all of the 2m�2a characters of the setIrr1.G=Z0/ are partitioned in hˆi-orbits of the same odd size > 1, which is not thecase. It follows that, for each Z0 2 M, there exists in Irr1.G=Z0/ a character suchthat its ˆ-orbit has size s. Since the number of such ˆ-orbits is 2n C 1 and, under ˆ,the setM has exactly 2n C1 orbits of size s (Lemma 46.11(a)), it follows that each setIrr1.G=Z0/ (where Z0 2 M) has exactly one character belonging to a ˆ-orbit of sizes since s.2n C 1/ D 2m � 1 is the total number of nonprincipal irreducible charactersin ˆ-orbits of size s, by Lemma 46.11(b).Now suppose that the size of the ˆ-orbit of a character 2 Irr1.G/ is 2m � 1 (such

exists since jIrr1.G/j D 2n.2m � 1/ > s.2n C 1/), .1/ D 2a and 2 Irr1.G=Z0/,whereZ0 2 M. By the previous paragraph, jIrr1.G=Z0/j D 2m�2a D 1C t .2n C 1/,where t is the number of orbits of characters in Irr1.G=Z0/ belonging to ˆ-orbits ofsize¤ s (there are in Irr1.G=Z0/ exactly 2n C1 characters conjugate with underˆ).It follows that 2n C 1 divides 2m�2a � 1, and so n divides m� 2a (Lemma 46.5) andm�2a

n is even (Lemma 46.6(b)). By Lemma 46.12, m � 2a � 2n � m so m�2an � 2,

i.e., m�2an

2 f0; 2g. Since G=Z0 has at least two nonlinear irreducible characters, bythe previous paragraph, m � 2a > 0. Hence, m � 2a D 2n so a D 1

2m � n and

Irr1.G=Z0/ has exactly 2m�2a D 22n characters that lie in t D 2m�2a�12nC1

D 22n�12nC1

D

2n � 1 distinct ˆ-orbits of size 2m � 1 and one ˆ-orbit of size s.We see that there are 22ns D 22n 2m�1

2nC1 nonlinear irreducible characters of G

of degree 2.m=2/�n. The sum of squares of their degrees is 2m�2n22n 2m�12nC1

D

2m 2m�12nC1 . The sum of squares of degrees of remaining t1 D 2n.2m �1/�22n 2m�1

2nC1 D


2n 2m�12nC1 nonlinear irreducible characters of G is † D 2m.2m � 1/ � 2m 2m�1

2nC1 D

2m 2n.2m�1/2nC1

. If 2 Irr1.G/, then .1/2 � jG W Z.G/j D 2m. Since † D t1 2m, it

follows that all latter characters have the same degree 2m=2. This means that cd.G/ D

f1; 2.m=2/�n; 2m=2g, andG has exactly 22ns irreducible characters of degree 2.m=2/�n

and t1 D 2ns irreducible characters of degree 2m=2.

Corollary 46.14. If k D 2, i.e.,m D 2n, then jG0j D 2m=2.

Proof. In that case, by Lemma 46.10, either jG0j D 2m=2 or else jG0j D 2m. In thesecond case, however, by Theorem 46.13, cd.G/ D f1; 2.m=2/�n; 2m=2g, which is notthe case since m � 2n D 0.

Theorem 46.15. If k D 2, then cd.G/ D f1; 2m=2g.

Proof. By Corollary 46.14, jG0j D 2m=2, jG W G0j D 2mC.m=2/ D 2mCn D jLin.G/j.Since k.G/ D 2mCn C 2m � 2n, we get jIrr1.G/j D jIrr.G/j � 2mCn D 2m �

2n D 2m=2.2m=2 � 1/. By the remark following Corollary 46.8, if 2 Irr1.G/, then 2 Irr1.G=Z0/, where Z0 2 M, G0 6� Z0. The number of such subgroups Z0

equals 2m � 1� .2m=2 � 1/ D 2m=2.2m=2 � 1/ D jIrr1.G/j. (Here 2m � 1 D jMj and2m=2�1 is the number of subgroups of index 2 in Z.G/=G0; this explains our formula.)Thus, to every subgroup Z0 corresponds the unique nonlinear irreducible character ofG=Z0. Since jG=Z0j D 2mC1 and j.G=Z0/

0j D 2 so G=Z0 is extraspecial, we getcd.G=Z0/ D f1; 2m=2g.

Theorem 46.1 is proven.

Corollary 46.16. If k D 2 and 2 Irr1.G/, then G=Z0 is extraspecial for everysubgroup Z0 of index 2 in Z.G/ such that G0 6� Z0.

Proof. Suppose that 2 Irr1.G=Z0/; then .1/ D 212

m, by Theorem 46.15. It followsthat jZ.G=Z0/j D 2, so G=Z0 is extraspecial since its center and derived subgrouphave order 2.

2o. In this subsection, p > 2. Thus, F D GF.pm/, m 2 N with m > 1, q D pm,� is an automorphism of F of order k for some divisor k > 1 of m, m

kD n and G D

Ap.m; �/ is the generalized Suzuki p-group of order q2. Our aim in this subsection isto prove the following

Theorem 46.17 ([Sag2]). If G D Ap.m; �/ is a p-group of order q2, where p > 2,m > 1 is an integer, q D pm, � is an automorphism of the field F D GF.q/ of orderk > 1 and m

kD n, then one of the following holds:

(a) If k is odd, then cd.G/ D f1; p.m�n/=2g.

(b) If k D 2, then cd.G/ D f1; pm=2g.


(c) If k > 2 is even, then cd.G/ D f1; pm=2; p.m=2/�ng, jIrr.pm=2/.G/j D pm�1pnC1

pn

and jIrr.p.m=2/�n/.G/j D pm�1pnC1

p2n.

By the text preceding 1o, exp.G/ D p, .a; b/�1 D .�a;�b C a�.a//, Z.G/ D

f.0; b/ j b 2 Fg is of order q and the centralizer of every element from G � Z.G/ hasorder pmCn. It follows that k.G/ D pmCn C pm � pn so the number of noncentralG-classes is pmCn � pn. Next, jG0j � pm�n. We see that Lemmas 46.2, 46.3 and46.4 are also true for odd p. Argument following Lemma 46.4, is also true for odd p.Therefore, for every x 2 F, �.x/ D xpnt

for some t coprime with k.

Lemma 46.18. (a) Suppose that m D nk, .t; k/ D 1 and .pm � 1; pnt C 1/ D d . Ifk is odd, then d D 2. If k is even, then d D pn C 1.

(b) Suppose that k; n 2 N, k even and .pm � 1; pn C 1/ D d . Then d D 2 if n.m=2;n/

is even and d D p.m=2;n/ C 1 if n.m=2;n/

is odd.

Proof. (a) Since .k; t/ D 1, then d divides .pm � 1; p2nt � 1/ D .pm � 1; p2n �

1/ D p.m;2n/ � 1 (Lemma 46.5). If k is odd then .m; 2n/ D .nk; 2n/ D n so ddivides pn � 1. If k is even, then .m; 2n/ D 2n so d divides p2n � 1. Let us findd0 D .pnt C 1; pn � 1/.Let k be odd; then d divides pn � 1 so d divides d0. If r > 2 is a prime divisor

of d0, then pn 1 .mod r/ so pnt C 1 2 .mod r/, a contradiction. The sameargument also works in the case r D 4. It follows that d0 D 2 since d0 is even. Sincethe even number d divides d0 D 2, we get d D 2.Suppose that k is even; then t is odd and so pn C 1 divides .pm � 1; pnt C 1/ D d .

Next, pnt C1 D .pn C1/A and pm �1 D .pn C1/B , whereA D pn.t�1/ �pn.t�2/ C

C p2n � pn C 1 and B D pn.k�1/ � pn.k�2/ C C pn � 1. The number A isodd since t is odd, and B is even since k is even. In that case, d D .pn C 1/.A;B/

so dpnC1

D .A;B/ D u is odd. Since d D .pn C 1/u with odd u divides p2n � 1 D

.pn � 1/.pn C 1/ and .pnt C 1; pn � 1/ D 2 (see the previous paragraph!), we getd D pn C 1.

(b) Set d D .pm � 1; pn C 1/, m D 2m1, d0 D .p.m1;n/ C 1; pn C 1/. Then ddivides the number

.pm � 1; p2n � 1/ D p.m;2n/ � 1 D p2.m1;n/ � 1 D .p.m1;n/ � 1/.p.m1;n/ C 1/:

As in the proof of (a), .p.m1;n/ � 1; pn C 1/ D 2 and d0 D 2 if n.m1;n/

is even and

d0 D p.m1;n/ C 1 if n.m1;n/

is odd.Suppose that n

.m1;n/is even. Then d 2 f2; 4g since d divides

.pn C 1; p.m1;n/ � 1/.pn C 1; p.m1;n/ C 1/ D 2:

Indeed, since n is even, we get pn C 1 2 .mod 4/ so d D 2.


Suppose that s D n.m1;n/

is odd and set t D .m1; n/. Then

pn C 1 D pts C 1 D .pt C 1/.pt.s�1/ � pt.s�2/ C C p2t � pt C 1/

so pnC1pt C1

is odd.

Let �� be as in Lemma 46.4 and set ˆ D h��i provided � is a primitive element ofthe field F.

Lemma 46.19. If � is a primitive element of the field F, then:

(a) The size of ˆ-orbits on G � Z.G/ is pm � 1.

(b) The size of ˆ-orbits on Z.G/# is 12.pm � 1/ for odd k and pm�1

pnC1for even k.

Proof. By definition, �i�..a; b// D .�ia; .��.�//ib/. If a ¤ 0, the assertion follows

since � is primitive. Let a D 0. Then �i�.0; b/ D .0; .��.�//ib/ D .0; b/ if and

only if .��.�//i D 1. Since �.�/ D �pn

, we obtain �.pnC1/i D 1. It follows thato.�/ D pm �1 divides .pn C1/i , and the assertion follows from Lemma 46.18(a).

Lemma 46.20. Suppose that 2 Irr1.G/, .1/ D pa, Z0 D Z.G/ \ ker./. ThenjZ.G/ W Z0j D p, G0 6� Z0, cd.G=Z0/ D f1; pag, and jIrr1.G=Z0/j D .p �

1/pm�2a. Next, every nonlinear irreducible character of G is a character of exactlyone group G=Z, where Z is a subgroup of index p in Z.G/ not containing G0.

Proof. Since Z.G= ker.// is cyclic and Z.G/ is elementary abelian, we get jZ.G/ W

Z0j D p. Let us consider the quotient group G=Z0. Since is nonlinear andG0 6� Z0

but G0 � Z.G/, we get G0Z0 D Z.G/. Next, j.G=Z0/0j D p and exp.G=Z0/ D p.

It follows that G=Z0 D .G1=Z0/� .Z1=Z0/, whereZ1=Z0 < Z.G=Z0/ andG1=Z0

is extraspecial. Now all assertions of the lemma are obvious.

Theorem 46.21. IfG D Ap.m; �/, where � an automorphism of F of odd order k > 1,then cd.G/ D f1; p

m�n2 g, where n D m

k.

Proof. Let � be a primitive element of F. In that case, there are two ˆ-orbits ofsize 1

2.pm � 1/ on Z.G/, where ˆ is a cyclic subgroup generated by ��, by Lemma

46.19(b). If g 2 G0, then the ˆ-orbit of g lies in G0 since G0 is characteristic in G. Itfollows that jG0j � 1 � 1

2.pm � 1/ so jG0j D pm since jG0j divides pm. It follows

that G0 D Z.G/. By Lemma 46.19(b), there are at most two ˆ-orbits on the set of allsubgroups of order p in Z.G/. It follows that there are at most two ˆ-orbits on theset of maximal subgroups of Z.G/. Since G=Z0 Š G=Z

��

0 for every subgroup Z0

of index p in Z.G/ and jcd.G=Z0/j D 2, by Lemma 46.20, then G has at most twodistinct degrees of nonlinear irreducible characters.Assume that cd.G/ D f1; pa ; pbg. Then there are two ˆ-orbits on the set of sub-

groups of index p in Z.G/ and the sizes of these orbits are pm�12.p�1/

. In that case, there

are pm�12.p�1/

.p � 1/pm�2a D 12 .p

m � 1/pm�2a characters of degree pa in Irr1.G/


and, similarly, pm�12.p�1/

.p � 1/pm�2b D 12.pm � 1/pm�2b characters of degree pb in

Irr1.G/. It follows that

pmCn � pn D jIrr1.G/j D1

2.pm � 1/pm�2a C

1

2.pm � 1/pm�2b

so that pn D 12.pm�2a C pm�2b/ or, what is the same, pnC2aC2b D 1

2pm.p2a C

p2b/. Let, for definiteness, a � b. Then pnC2aC2b D 12p

mC2b.p2a�2b �1/. The lastequality is possible if and only if p2a�2b � 1 D 0. Thus a D b so cd.G/ D f1; pag.It follows that jGj D p2m D pm C .pmCn � pn/p2a so a D 1

2.m � n/.

Hence, the Theorem 46.17 is proven for odd k.

Lemma 46.22. Suppose that � 2 F is primitive and k is even. Then:

(a) If k > 2, i.e., m > 2n, then G0 D Z.G/ and the number of ˆ-orbits on .G0/#

equals pn C 1; all these orbits have size pm�1pnC1

.

(b) If k D 2, i.e., m D 2n, then either G0 D Z.G/ and the number of ˆ-orbitson .G0/# equals pn C 1 and all these orbits are of size pm�1

pnC1D pn � 1, or

jG0j D pm2 D pn and all elements of .G0/# are conjugate under ˆ.

Proof. Repeat, word for word, the proof of Lemma 46.10.

Lemma 46.23. Let � 2 F be primitive, k even and G0 D Z.G/. Then Irr.G/ � f1Gg

is partitioned, under action of ˆ, in pn orbits of size pm � 1 and pn C 1 orbits ofsize pm�1

pnC1.

Proof. See the proof of Lemma 46.11(b).

Lemma 46.24. Let � 2 F be primitive, k even and G0 D Z.G/. Then the set of allsubgroups of Z.G/ of index p is partitioned, under action ofˆ, either in pn C1 orbitsof size pm�1

.pnC1/.p�1/or in 1

2.pn C 1/ orbits of size 2.pm�1/

.pnC1/.p�1/.

Proof. Consider a subgroup P D h.0; x/i of Z.G/ of order p. Let s 2 N be minimalsuch that �s

�..0; x// 2 P . Since o..��/Z.G// D pm�1

pnC1and P # is partitioned in p�1

t

orbits of size t under action of h's�i, st D pm�1

pnC1. Then the size of ˆ-orbit containing

P equals s. Let P1 D h.0; y/i be another subgroup of order p in Z.G/. Since Z.G/ D

f.0; b/ j b 2 Fg and � 2 F is primitive, we get .0; y/ D .0; �f x/ for some f 2 N.Next, since �s

�..0; x// 2 P , we get �s

�..0; x// D .0; .��.�//sx/ D .0; x/r D .0; rx/

for some r 2 N. Then

�s�..0; y// D .0; .��.�//sy/ D .0; .��.�//s�f x/

D .0; r�f x/ D .0; ry/ D .0; y/r :

This means that the size of the ˆ-orbit containing P1, is also equal to s. Thus, thesizes of allˆ-orbits on the set of all subgroups of order p in Z.G/ are equal. It follows


from c1.Z.G// D pm�1p�1

, that the number of ˆ-orbits on the set of such subgroups

equals pm�1s.p�1/

D t.pnC1/p�1

. Since .pn C 1; p � 1/ D 2, we get t 2 fp � 1; 12.p � 1/g.

Hence, the set of all subgroups of order p in Z.G/ is partitioned, under action of ˆ,either in pn C1 orbits of size pm�1

.pnC1/.p�1/or in 1

2.pn C1/ orbits of size 2.pm�1/

.pnC1/.p�1/.

Now let Q < G be of index p. Since Q D P1 � � Pm�1 with jPi j D p

for all i , then �s�.Q/ D Q. If for some natural s1 < s we have �s1

�.Q/ D Q,

then by Maschke’s theorem, Z.G/ D Q � P , where P is a subgroup of order p and�

s1

�.P / D P , contrary to the minimal choice of s. Now the conclusion of the lemma

follows from the previous paragraph.

Lemma 46.25. Suppose that 2 Irr1.G/, .1/ D pa. Then a � 12m. Next, a �

12m � n if the number of ˆ-orbits on the set of subgroups of index p in Z.G/ equalspn C 1 and a > 1

2m� n, if the number of ˆ-orbits on the same set equals 1

2.pn C 1/.

Proof. By Lemma 46.20, 2 Irr1.G=Z0/, whereZ0 is a subgroup of index p in Z.G/and jIrr1.G=Z0/j D .p � 1/pm�2a. It follows that a � 1

2m.

Let Z1; : : : ; Zt be the set of representatives of ˆ-orbits on the set of maximal sub-groups of Z.G/. Then cd.G=Zi / D f1; pai g, jIrr1.G=Zi /j D .p � 1/pm�2ai . Next,any two subgroups of index p in Z.G/ belonging to the same ˆ-orbit, lead to iso-morphic quotient groups. Since jIrr1.G/j D pmCn � pn, we get pmCn � pn Dpm�1t.p�1/

PtiD.p � 1/pm�2ai so tpn D

PtiD1 p

m�2ai .

Suppose that t D pn C 1. If m � 2ai > 2n for some i , then pm�2ai � p2nC1 >

pn.pn C 1/ D tpn, a contradiction. Thus, m � 2ai � 2n so ai � 12m � n for all i .

Suppose that t D 12.pn C 1/. If m � 2ai � 2n for some i , then pm�2ai � p2n >

12pn.pn C 1/ D tpn, a contradiction. Thus, ai >

12m � n for all i .

Theorem 46.26. If G0 D Z.G/ and k is even, then cd.G/ D f1; p12

m�n; p12

mg. Inthat case, jIrr

.p12

m�n/.G/j D pm�1

pnC1 p2n, jIrr

.p12

m/.G/j D pm�1

pnC1 pn.

Proof. By Lemma 46.23, the set Irr.G/ � f1Gg, under action of ˆ, is partitioned inpn orbits of size pm � 1 and pn C 1 orbits of size pm�1

pnC1. Since jLin.G/j D jG=G0j

and ˆ .G=G0/ is a Frobenius group, nonprincipal linear characters lie in the sameˆ-orbit of length pm � 1. Therefore, under ˆ, there are on Irr1.G/ exactly pn � 1

orbits of size pm � 1. By Lemma 46.21, each character from Irr1.G/ is contained inIrr1.G=Z0/, whereZ0 is a subgroup of index p in Z.G/. By Lemma 46.24, the setMof all pm�1

p�1subgroups of index p in Z.G/, under action of ˆ, is partitioned either in

pn C 1 orbits of size pm�1.pnC1/.p�1/

or in 12.pn C 1/ orbits of size 2.pm�1/

.pnC1/.p�1/.

First suppose that the set M is partitioned in pn C 1 orbits under action of ˆ.Consider 2 Irr1.G=Z0/, Z0 2 M. If the size of theˆ-orbit of equals pm �1, thenIrr1.G=Z0/ contains exactly .p � 1/.pn C 1/ characters from that orbit. If the sizeof ˆ-orbit of equals pm�1

pnC1 , then Irr1.G=Z0/ contains exactly p � 1 characters from


that orbit. Note that if, for someZ0 2 M, Irr1.G=Z0/ has no orbits of sizepm�1pnC1

, thenIrr1.G=Z0/ is partitioned in orbits of sizes .p � 1/.pn C 1/. Since jIrr1.G=Z0/j D

pm�2a, where pa 2 cd.G=Z0/, then pn C 1 divides pm�2a, which is a contradiction.It follows that for every Z0 2 M, there are in Irr1.G=Z0/ at least p � 1 characterswhose ˆ-orbits have size pm�1

pnC1. Since the number of such characters is pm � 1 and

jMj D pm�1p�1

, every group G=Z0 has exactly p � 1 characters belonging to orbits of

size pm�1pnC1

.Let cd.G=Z0/ D f1; pag and let Irr1.G=Z0/ contain exactly t .p�1/.pn C1/ char-

acters belonging to ˆ-orbits of size pm � 1. Then .p � 1/pm�2a D jIrr1.G=Z0/j D

.p � 1/ C t .p � 1/.pn C 1/. It follows that pm�2a � 1 D t .pn C 1/ so pn C 1

divides pm�2a � 1. It follows that n divides m � 2a and m�2an is even. By Lemma

46.25, 12m � n � a � 1

2m so that 0 � m�2a

n� 2. Suppose that t > 0. Then

jIrr1.G=Z0/j > p � 1 or m � 2a > 0. It follows that m�2an D 2 and so a D 1

2m � n,jIrr1.G=Z0/j D .p � 1/p2n, t D pn � 1. Since t D pn � 1, Irr1.G=Z0/ containsrepresentatives of all ˆ-orbits of size pm � 1 on Irr1.G/. Therefore, all charactersfrom aˆ-orbit of size pm �1 belong to members ofM contained in the sameˆ-orbit.Then jIrr

.p12

m�n//.G/j D pm�1

pnC1p2n. The number of remaining irreducible nonlinear

characters equals t1 D pn.pm �1/� pm�1pnC1 p2n D pm�1

pnC1 pn and the sum of squares

of their degrees is † D pm.pm � 1/ � pm�1pn�1 pm D pm�1

pn�1 pmCn. If 2 Irr1.G/,

then .1/2 � jG W Z.G/j D pm. Since † D t1pm, all other nonlinear irreducible

characters are of degree p12

m.Now suppose that the number of ˆ-orbits on M equals 1

2.pn C 1/. Recall that all

these orbits are of size 2.pm�1/.p�1/.pnC1/

. If 2 Irr1.G=Z0/ (Z0 2 M) belongs to the ˆ-

orbit of sizepm�1, then that set of characters contains also exactly .p�1/12.pnC1/ ˆ-

conjugates with . Similarly, as before, we can show that then Irr1.G=Z0/ (Z0 2 M)contains a character representing a ˆ-orbit of size pm�1

pnC1.

Suppose that, for someZ0 2 M, the set Irr1.G=Z0/ has exactly 12 .p�1/ characters

every of which is contained in a ˆ-orbit of size pm�1pnC1

. Then jIrr1.G=Z0/j D .p �

1/pm�2a D 12.p � 1/ C t .p � 1/ 1

2.pn C 1/, t � 0. It follows that 2pm�2a D

1 C t .pn C 1/, which is impossible since pn C 1 is even. Therefore, every groupG=Z0 (Z0 2 M) has at least p � 1 characters every of which belongs to a ˆ-orbit ofsize pm�1

pnC1. Since jMj D pm�1

p�1and the number of the above characters is pm � 1,

every group G=Z0 (Z0 2 M) has exactly p � 1 characters belonging to ˆ-orbits ofsize pm�1

pnC1(these characters belong to two different ˆ-orbits).

Now let G=Z0 (Z0 2 M) have a character belonging to a ˆ-orbit of size pm � 1.Then, by the above, jIrr1.G=Z0/j D .p � 1/pm�2a D .p � 1/ C t .p � 1/ pnC1

2

so 2.pm�2a � 1/ D t .pn C 1/. Since m is even then m � 2a is also even; therefore,using Lemma 46.18, we can show that either n divides 1

2.m � 2a/ , pn C 1 divides

pm�2a � 1 and t D 2.pm�2a�1/pnC1 or p D 3, n D 1 and t D 1

2 .32m�2a � 1/. However,


by hypothesis, n > 1. In the first case we get either 2n � m � 2a, a � 12m � n,

contrary to Lemma 46.25, or elsem � 2a D 0, a D 12m, t D 0, a contradiction again.

Thus, the case whereM is partitioned in 12 .p

n C1/ orbits underˆ is impossible.

Corollary 46.27. If m D 2n, then jG0j D p12

m.

Proof. Assume that G0 D Z.G/ holds. Then, by Theorem 46.26, we have cd.G/ D

f1; p12

m�n; p12

mg, which is impossible since 12m � n D 0.

Theorem 46.28. If m D 2n, then cd.G/ D f1; p12

mg.

Proof. By Lemma 46.22 and Corollary 46.27, jG0j D p12

m, jG W G0j D pmC 12

m D

pmCn D jLin.G/j so jIrr1.G/j D pm � pn D p12

m.p12

m � 1/. Every irreduciblenonlinear character of G is a character of G=Z0, where Z0 2 M and G0 6� Z0. The

number of such subgroups is pm�1p�1

� p12

m�1

p�1D p

12

m�1

p�1p

12

m, and every groupG=Z0

has at least p � 1 nonlinear irreducible characters. Comparing the number jIrr1.G/jwith the number of subgroups of index p in Z.G/ not containing G0, we see that everygroup G=Z0 has exactly p � 1 nonlinear irreducible characters. Let cd.G/ D f1; pag.Then p � 1 D jIrr1.G=Z0/j D .p � 1/pm�2a so a D 1

2m, completing the proof.

Theorem 46.17 is proven.

�47

On the number of metacyclic epimorphic imagesof finite p-groups

In this section we prove the following

Theorem 47.1 ([Ber33]). Let G be a nonmetacyclic p-group of order pm and let n <m.

(a) If p D 2 and n > 3, then the number of normal subgroups D of G such thatG=D is metacyclic of order 2n, is even.

(b) If p > 2 and n > 2, then the number of normal subgroups D of G such thatG=D is metacyclic of order pn, is a multiple of p.

Remarks. 1. Let G be a two-generator nonmetacyclic 2-group. Then all epimorphicimages ofG of order 8 are metacyclic. Since the number of normal subgroups of givenindex in a 2-group G is odd, it follows that Theorem 47.1(a) is not true for n D 3.

2. Let r � s � t > 0, r C s C t > 3 and G D hai � hbi � hci, where o.a/ D pr ,o.b/ D ps , o.c/ D pt , be an abelian p-group of rank three. Let us check Theorem47.1 for n D r C s C t � 1. Let be the number of subgroups D of G of order psuch that G=D is metacyclic. If t > 1, then �1.G/ � ˆ.G/; then D 0. Supposethat s > t D 1. If D < hai � hbi, then G=D is nonmetacyclic. If D 6� hai � hbi,then G=D is abelian of type .pr ; ps/ so metacyclic, and in that case, D p2. Nowsuppose that r > s D 1. ThenD D Ãr�1.G/ is the unique normal subgroup of orderp in G such that G=D is not metacyclic so D p2 C p. Thus, p divides in allcases.

3. Suppose that G D ha; b j a4 D b4 D c2 D 1; c D Œa; b�; Œa; c� D Œb; c� D 1i isa nonmetacyclic minimal nonabelian group of order 25. Let us check Theorem 47.1(a)for n D 4. The group G has exactly seven central subgroups of order 2:

X1 D ha2i; X2 D hb2i; X3 D ha2b2i; X4 D hci;

X5 D ha2ci; X6 D hb2ci; X7 D ha2b2ci:

Then G=Xi is metacyclic for i D 3; 4; 5; 6 and nonmetacyclic for i D 1; 2; 7.

4. Suppose that G D ha; b j a2m

D b2 D c2 D 1;m > 2; c D Œa; b�; Œa; c� D

Œb; c� D 1i is a nonmetacyclic minimal nonabelian group of order 2mC2. Set ˛ D

�47 On the number of metacyclic epimorphic images of finite p-groups 15

a2m�1

; then �1 D fhc˛i; h˛i; hcig is the set of central subgroups of order 2 in G.Then G=hci and G=hc˛i are metacyclic and G=h˛i is not metacyclic.

Let Sc.G/ D �1.Z.G// be the socle of G. Let �i denote, for all i such thatpi � jSc.G/j, the set of subgroups of order pi in Sc.G/. LetM be a set of nonidentitynormal subgroups ofG. GivenH 2 �1 [�2 [ [fSc.G/g, let ˛.H/ be the numberof members of the setM containingH . Set jSc.G/j D pt so that �t D fSc.G/g. Weclaim that the following identity holds (see �10):

˛.G/ D jMj D

tX

iD1

.�1/i�1p.i2/X

H2�i

˛.H/:(1)

Indeed, let D 2 M and jD \ Sc.G/j D pk ; then k � 1 since D > f1g. For naturalnumbers u � v, let 'u;v denote the number of subgroups of order pv in Epu . Then the

contribution of D in the right-hand side of (1) is equal toPk

iD1.�1/i�1p.

i2/'k;i , and

that number equals 1, by Hall’s identity (Theorem 5.2). Since the contribution ofD inthe left-hand side of (1) is also equal 1, identity (�) is true. In particular,

˛.G/ D jMj X

H2�1

˛.H/ .mod p/:(2)

Remark 5. Suppose that G is a noncyclic group of order pm, m > 2 and 1 < n < m.We claim that the number c.G/ of normal subgroups N of G such that G=N is cyclicof order pn, is a multiple of p. One may assume that c.G/ > 0. Let n D m�1. ThereisN � Z.G/ of order p such that G=N is cyclic. ThenG is abelian of type .pm�1; p/

so G D Z � N , where Z is cyclic of order pm�1 D pn hence c.G/ D p. Now letn < m � 1. Take H 2 �1. Suppose that G=H is cyclic; then G is abelian of type.pm�1; p/. The group G contains exactly pC1 subgroups of index pn. IfN is one ofsuch subgroups, thenG=N is cyclic if and only ifN 6� ˆ.G/ so, sinceˆ.G/ is cyclic,we get c.G/ D p. Next we assume that G is not abelian of type .pm�1; p/; thenG=His not cyclic for allH 2 �1, therefore, by induction onm, we get ˛.H/ 0 .mod p/so c.G/ 0 .mod p/, by (2).

Suppose that G is a group of order pm and pn � jG W G0j. We claim that thenthe number �.G/ of N G G such that G=N is nonabelian of order pn is a multiple ofp. One may assume that �.G/ > 0; then n > 2. The number of normal subgroupsof given index in G is 1 .mod p/ (Sylow). If N G G has index pn, then G=N isnonabelian if and only if G0 6� N . Therefore, �.G/ 0 .mod p/ (Sylow again).If a 2-group G of order 2m > 23 is not of maximal class, then the number of N GG

such that G=N is nonabelian of order 23 is even since jG W G0j � 23.

Proof of Theorem 47.1. Suppose that M is the set of normal subgroups D of G suchthat G=D is metacyclic of order pn. One may assume that M ¤ ¿. As above, ˛.H/is the number of members of the setM containing H 2 �1.


A. First we consider the most difficult case n D m � 1; then M � �1. One mayassume that G is nonabelian (Remark 2). We have to prove that p divides jMj.D

˛.G//, i.e., j�1 � Mj 1 .mod p/. By (2), we have to prove that

˛.G/ X

H2�1

˛.H/ 0 .mod p/:(3)

Let U 2 M and suppose that U — ˆ.G/. Then G D U � M , where M 2 �1 ismetacyclic. If V 2 �1 and V — M , then G D V �M so V 2 M. If W � Z.M/ is amember of the setM, thenM=W is cyclic andG is abelian of rank 3 so p divides jMj

(Remark 2). We see that jMj D j�1j � c1.Z.M// 0 .mod p/. Next we assumethat all members of the setM are contained in ˆ.G/.(i) Let p > 2. Then jG=Ã1.G/j � p3 since G is nonmetacyclic (Theorem 9.11).

Take D 2 M; then G=D is metacyclic. Assuming that jG=Ã1.G/j � p4, we con-clude that G=DÃ1.G/ is of order � p3 and exponent p so nonmetacyclic, a contra-diction. Thus, jG=Ã1.G/j D p3. Take U 2 �1 � M; then G=U is not metacyclic soj.G=U /=Ã1.G=U /j � p3 (Theorem 9.11 again) and so U � Ã1.G/. Thus, all mem-bers of the set �1 � M are contained in Ã1.G/. Conversely, if V 2 �1 is containedin Ã1.G/, then G=V is not metacyclic. Thus, j�1 � Mj equals the number of thosemembers of the set�1 that are contained in Ã1.G/. By Sylow, the last number is 1

.mod p/ so jMj j�1j � 1 D 't;1 � 1 0 .mod p/.

(ii) Suppose that p D 2. Here we have to consider the nonmetacyclic quotient groupG=Ã2.G/ (see Lemma 42.2(b)) of order � 24.

(ii1) Assume that jG=Ã2.G/j D 24. Then Ã2.G/ > f1g since m > 4. The numberof members of the set �1 contained in Ã2.G/, is odd (Sylow). Therefore, if all mem-bers of the set�1 �M are contained inÃ2.G/, then ˛.G/ D jMj is even since j�1j isodd, so we assume that there is U 2 �1 � M such that U — Ã2.G/; then NG D G=U

is nonmetacyclic. It follows from Theorem 44.2 that NG=Ã2. NG/ is nonmetacyclic soits order is � 24. Write NH D Ã2. NG/; then G=H.Š NG= NH/ is not metacyclic of order� 24 and exponent 4. It follows thatH D Ã2.G/ since jG=Ã2.G/j D 24, contrary tothe choice of U : U — Ã2.G/.

(ii2) Let jG=Ã2.G/j > 24. Take D 2 M; then, by Theorem 44.2, D — Ã2.G/.Write NG D G=D. We claim that jG=Ã2.G/j � 25. Indeed, G=DÃ2.G/, as anepimorphic image of groups NG D G=D and G=Ã2.G/, is metacyclic of exponent 4so its order is � 24, and our claim on order follows since DÃ2.G/ D D � Ã2.G/

and jDj D 2. Thus, jG=Ã2.G/j D 25, by (ii1). Since G=D is metacyclic, we havejSc.G=D/j � 22 so jSc.G/j � jDjjSc.G=D/j D 23, and we conclude that j�1j 2

f1; 3; 7g. Since j�1j is odd, to complete this case, it suffices to show that the numberj�1 � Mj is also odd. But G=D is metacyclic so D — G0 (Theorem 36.1) and henceG0 Š .G=D/0 is cyclic. Since all members of the set M are contained in ˆ.G/, weget d.G/ D 2.Let NH D Ã2. NG/ ( NG is defined in (ii1)). SinceG=DÃ2.G/ is metacyclic, its order is

� 24 soH D DÃ2.G/ D D � Ã2.G/ and j NG=Ã2. NG/j D 24 since jG=Ã2.G/j D 25.

�47 On the number of metacyclic epimorphic images of finite p-groups 17

Set N0 D fV 2 �1 j V � Ã2.G/g; then jN0j is odd (Sylow) and N0 � �1 � M.One may assume that N0 � �1 � M (otherwise, jMj D j�1j � jN0j is even). Inthat case, there is U1 2 �1 � .M [ N0/; then NG D G=U1 is not metacyclic. WriteNH D Ã2. NG/; then NG= NH is not metacyclic (Theorem 44.2) and, since U1Ã2.G/ � H

and U1 6� Ã2.G/, we get H1 D U1 � Ã2.G/ and jG=H1j D 12jG=Ã2.G/j D 1

2

25 D 24. If V < H1 is a member of the set �1, then G=V is nonmetacyclic. LetN1 D fU0 2 �1 � N0 j U0 < H1g. Then jN0 [ N1j D jN0j C jN1j, the numberof members of the set �1 contained inH1, is odd (Sylow). It follows that the numberjN1j is even since the number jN0j is odd. Since U1 2 N1, we obtain jN1j � 2 sojN0[N1j � 1C2 D 3. Assuming thatN0[N1 � �1�M, we add to the setN0[N1

the set N2 (of even cardinality > 0) of new members of the set �1 � M. Indeed, ifU2 2 �1 � .N0 [ N1 [ M/, then, writing H2 D U2Ã2.G/, we conclude, as abovewithU1 andH1, thatG=H2 is nonmetacyclic of order 24. We haveH1 \H2 D Ã2.G/

since Ã2.G/ � H1 \H2 has index 25 in G andH1;H2 are distinct of index 24 in G,soN2 is the set of members of the set�1 � .N0 [ N1/ contained inH2 but not inH1

and jN2j > 0 is even (if Vi 2 Ni , then Vi Ã2.G/ D Hi , i D 0; 1; 2). Thus, we getjN0 [ N1 [ N2j � 1C 2C 2 D 5. We claim that then the setN0 [ N1 [ N2, whichcontains exactly five members, coincides with the set�1 � M. Indeed, otherwise, wecan, acting as above, add to that sum-set at least two new members of the set�1 � M,which is impossible since j�1 � Mj � 6: the setM ¤ ¿ and j�1j D 7. Thus, in anycase, the number j�1 � Mj is odd so that the number jMj is even.

B. Now let n < m � 1. Here we consider cases p D 2 and p > 2 together. Weproceed by induction on jGj. Take H 2 �1.If G=H is metacyclic, then ˛.H/ 1 .mod p/ (Sylow). However, the number of

suchH , by part A of the proof, is a multiple of p. Therefore, the contribution of suchH in the sum on the right-hand side of formula (2), is a multiple of p. Now supposethat G=H is not metacyclic. Then, by induction, ˛.H/ 0 .mod p/. Therefore, thecontribution of such H in the sum on the right-hand side of formula (2), is a multipleof p again, Thus, congruence (3) is proven.

Supplement to Theorem 47.1. Let 1 � k � p � 1, k < n < m and a p-group G oforder pm satisfies jG=Ã1.G/j > pk . Then the number of normal subgroups D of Gsuch that j.G=D/=Ã1.G=D/j � pk and jG=Dj D pn, is a multiple of p.

2o. Here we prove the following

Theorem 47.2. If G be a nonabelian metacyclic p-group, then all representationgroups of G are also metacyclic so thatM.G/, the Schur multiplier of G, is cyclic.

Proof. Let � be a representation group of G. By definition, there is in Z.�/ \ � 0 asubgroup M Š M.G/ such that �=M Š G. Since G is nonabelian, we getM < � 0.Since �=M Š G is metacyclic, � is also metacyclic, by Theorem 36.1.


The cyclicity of Schur multipliers of metacyclic p-groups is known, their orders arealso computed in [Kar, Theorem 10.1.25].Isaacs has reported (in letter at Jan 10, 2006) that he also proved Theorem 47.3. His

proof is based on the following

Lemma 47.3 (Isaacs). Let G be a p-group and let Z � Z.G/ \ G0 be of order p. IfG=Z is metacyclic, then G is metacyclic.

Proof. Let U=Z G G=Z be with cyclic U=Z and G=U . Note that jU=Zj > p sinceG=Z is not abelian. We want to show that U is cyclic.Now let hxU i D G=U . Since U is abelian and normal in U and G=U is cyclic,

we see that G0 D ŒU; x� D hŒu; x� j u 2 U i is isomorphic to U=CU .x/ [Isa2, Lemma12.12]. But Z is contained in CU .x/ and U=Z is cyclic, and hence G0 is cyclic as anepimorphic image of U=Z. Since G0 > Z is cyclic, we have Z � ˆ.G0/ � ˆ.U / sod.U / D d.U=Z/ D 1 and U is cyclic. (In fact, Theorem 36.1 and Lemma 47.4 areequivalent.)

�48

On 2-groups with small centralizerof an involution, I

All results of this section are due to the second author. The main part of this sectioncoincides with [Jan1].In this section we give the classification of 2-groups containing an involution t such

that CG.t/ D hti � C2m , m � 1. We may assume from the start that m > 1 since, ifm D 1, G is a 2-group of maximal class, by Proposition 1.8. Note that case m D 2

was considered in [GLS, Proposition 10.27].

Exercise 1. Let G be a nonabelian 2-group and t an involution in G � Z.G/.

(a) Prove that jNG.CG.t// W CG.t/j < c1.CG.t//, where c1.G/ is the number ofsubgroups of order 2 (or, what is the same, involutions) in G.

(b) Let CG.t/ D hti�Q, whereQ is either cyclic of order 2m,m > 1, orQ Š Q2m ,m � 3. Then jNG.CG.t// W CG.t/j D 2, by (a), and

(b1) �1.Z.G// D Z.Q/.

(b2) If u is an involution inQ, then t and tu are conjugate in NG.CG.t//.

(b3) If t 2 ˆ.G/, then jˆ.G/j > 4.

Solution. (a) One may assume that CG.t/ E G. ThenG acts transitively on the conju-gacy classKt containing t . Since jG W CG.t//j > 1 is a power of 2 and c1.CG.t// > 1

is odd (Sylow), the inequality follows.

(b1, b2) Let u 2 Z.G/ be an involution. Then u is a square and t is not, so tand u are not conjugate in G. The result follows since CG.t/.> Z.G// has exactly 3involutions.

(b3) Assume that jˆ.G/j D 4. Then ˆ.G/ � Z.G/, by [BZ, Lemma 31.8], acontradiction since t 62 Z.G/.

Theorem 48.1 ([Jan1]). Let G be a nonabelian 2-group containing an involution tsuch that the centralizer CG.t/ D hti � C , where C is a cyclic group of order 2m,m � 1 (clearly, t 62 Z.G/). Then G has no elementary abelian subgroups of order 16and G is generated by at most three elements. Next, G has no t -invariant elementaryabelian subgroups of order 8.


(A1) If G has no elementary abelian subgroups of order 8, then one of the followingholds:

(a) G 2 fD2n ;SD2ng. Here we have m D 1.

(b) G Š M2n . Here we have m D n � 2.

(c) jG W CG.t/j D 2, t 2 ˆ.G/, Z.G/ is a cyclic subgroup of order � 4 notcontained in ˆ.CG.t//, G=Z.G/ is dihedral with cyclic subgroup L=Z.G/ ofindex 2, L is abelian of type .2; 2m/, m � 2, t 2 L. If x is an element ofmaximal order in G � L, then hx2i D Z.G/. (In that case CG.t/ D L.)

(d) G has a subgroup S of index � 2, where S D AL, the subgroup L is normal inG, L D hb; t j b2n�1

D t2 D 1; bt D b�1; n � 3i Š D2n . A D hai is cyclic oforder 2m, m � 2, A \ L D Z.L/, Œa; t � D 1, �1.G/ D �1.S/ D �2.A/ � L.If jG W S j D 2, then there is an element x 2 G � S so that tx D tb andCG.t/ D hti � hai. Next, G D ha; b; ti if G D S and G D ha; t; xi if S < G.

(A2) If G has an elementary abelian subgroup of order 8, then Z.G/ is of order 2,CG.t/ D hti � hai, where A D hai has order 2m (m � 2), G has a normal subgroupL such as in (A1)(d), A \ L D Z.L/ D hzi, S D AL is a normal subgroup of index� 2 in G and G is isomorphic to one of the following groups:

(e) G D ha; b; t j a2m

D b2n�1

D t2 D Œa; t � D 1; m � 3; n � 4; bt D

b�1; a2m�1

D b2n�2

D z; ba D b1C2i

; i D n �m � 2i. We have G D AL D

S and the cyclic group hai=hzi acts faithfully on L.

(f) G D ha; b; t; s j a2m

D b2n�1

D t2 D s2 D Œa; t � D 1; m � 4; n � 5; bt D

bs D b�1; a2qm�1

D b2n�2

D z; ba D b1C2i

; i D n � m C 1 � 2; t s D

tb; sa D a2m�2

b�2i�1

si. Here S D AL is a subgroup of index 2 in G and�1.S/ D �2.A/�L. AlsoG D hsiS ,M D hsiL Š D2nC1 ,NG.M/ D ha2iM

and s inverts�2.A/. The order of G is 2mCn and G D ha; t; si.

(g) Groups G D F.m; n/ (see Appendix 14). Here we have G D ha; b; t; s j

a2m

D b2n�1

D t2 D s2 D Œa; t � D Œa; b� D 1; m � 2; n � 3; bt D

bs D b�1; a2m�1

D b2n�2

D z; t s D tb; as D a�1zv; v D 0; 1; and if v D

1; then m � 3i. Here S D A �L and G D hsi S , where jG W S j D 2. We haveM D hsiL Š D2nC1 and G D ha; t; si.

(h) G D ha; b; t; s j a2m

D b2n�1

D t2 D s2 D Œa; t � D 1; m; n � 4; bt D bs D

b�1; a2m�1

D b2n�2

D z; t s D tb; ba D bz; as D a�1C2m�2

b2n�3

i. Here wehave again G D hsi S so that jG W S j D 2. Finally,M D hsi L Š D2nC1 andG D ha; t; si.

Proof. Let G be a 2-group containing a noncentral involution t such that CG.t/ D

hti � C , where C is cyclic of order 2m, m � 1. By Proposition 1.8, if m D 1, thenG 2 fD2n ; n � 3; SD2n ; n � 4g. In what follows we assume that m > 1. ThenG is not of maximal class so it contains a normal abelian subgroup U of type .2; 2/(Lemma 1.4). It follows from the structure of CG.t/ that Z.G/ is cyclic of order 2m

�48 On 2-groups with small centralizer of an involution, I 21

at most (see Exercise 1(b1)) so T D CG.U / has index 2 in G. Let u be the involutionin C ; then u 2 Z.G/ (Exercise 1(b1)). If an element x 2 NG.CG.t// � CG.t/, thentx D tu, by Exercise 1(b2).Assume that E is a t -invariant elementary abelian subgroup of G of order 8. Set

H D ht; Ei. Since H is not of maximal class, jCE .t/j > 2, by Suzuki’s theorem.Then the elementary abelian subgroup hti � CE .t/ of order � 8 is contained in (themetacyclic subgroup) CG.t/, which is a contradiction. Hence E does not exist. Itfollows that every t -invariant abelian subgroup of G is metacyclic.

(�) We examine now the case where t 2 T D CG.U /. Then ht; U i � CG.t/ sot 2 U since CG.t/ has no elementary abelian subgroups of order 8. Also we see thatT D CG.U / D hti � C has index 2 in G since U 6� Z.G/, and T D CG.t/.

(1�) Suppose that U is the only normal abelian subgroup of type .2; 2/ in G. Set�1.C / D hui D Ãm�1.T / so that U D ht; ui D �1.T / and u 2 Z.G/. Since Z.G/is cyclic, hui D �1.Z.G//. Next, Ã1.C / � Ã1.G/ � ˆ.G/.

(1.1�) Suppose, in addition, that t 62 ˆ.G/. Then there exists a maximal subgroupM of G such that t 62 M ; in that case, G D hti M , a semidirect product with kernelM . SettingM0 D T \M , we get, by the modular law, T D hti �M0. It follows fromthe structure of T thatM0 Š C2m . IfM is cyclic, then, clearly, G Š M2mC2 , and thisis the case (b) of the theorem. So assume that M is not cyclic. Since U 6� M sincet 2 U and t 62 M , the subgroup M has no G-invariant subgroups of type .2; 2/ (byassumption in (1�), U is the unique G-invariant four-subgroup) soM is of maximalclass, by Lemma 1.4. Then Z.M/ D hui, where ht; ui D U . Let hvi be the cyclicsubgroup of order 4 contained in M0 (recall that, by assumption. m � 2; obviously,v2 D u) and let y 2 M �M0; then vy D v�1 D vv2 D vu and ty D tu (see Exercise1(b2)). Hence we have .tv/y D tyvy D tuvu D tv soK D htvi is a cyclic subgroupof order 4 contained in T but not contained in M0, and .tv/2 D t2v2 D v2 D u

since, by the choice, v 2 CG.t/. We have CG.tv/ � hT; yi D G. It follows thatG D M �K, the central product,M \K D Z.M/. This group is then a group in part(d) of the theorem (with S D G and A is of order 4 centralizing L). In the case underconsideration, ˆ.G/ D Ã1.C / so C is normal in G.

(1.2�) It remains to consider the case t 2 ˆ.G/. In this case hti � Ã1.C / �

ˆ.G/ has index 4 in G so there we have equality and hence G is generated by twoelements. If m D 2, then ˆ.G/ D U , being a four-group, is contained in Z.G/,by Exercise 1(b3), which is a contradiction. Hence, we must have m � 3. Thereis an element x 2 G � T such that x2 2 ˆ.G/ � .Ã1.C / [ U/. This is due tothe fact that Z.G/ is cyclic. Indeed, there exists an element x 2 G � T such thatx2 62 C (otherwise, ˆ.G/ D Ã1.G/ D Ã1.C /, which is a contradiction). Next,if x2 2 U , then CG.x

2/ � hx; T i D G so Z.G/ is not cyclic since it containsa noncyclic subgroup hu; x2i, a contradiction. Our claim about the existence of xis justified. Then o.x2/ � 4 since x2 62 U D �1.ˆ.G//. Set C D hai so thata2m�1

D u, T D CG.t/ D hti � hai. Also, Ã1.T / D Ã1.C / D ha2i is normal


in G since T is, and ˆ.G/ D hti � ha2i, where o.a2/ D 2m�1 � 4. It followsfrom the choice of x that ˆ.G/ D ha2; x2i (however, generators x2 and a2 are notindependent). By Exercise 1(b2), we have tx D tu. If x would centralize a2, then xcentralizes ha2; x2i D ˆ.G/ and ˆ.G/ contains t so x 2 CG.t/ D T , contrary to thechoice of x. If y 2 G�T , then y2 2 Z.G/ since T is abelian and generatesG togetherwith y. Therefore, x2 2 Z.G/. Since x 62 CG.a

2/, we get a2 62 Z.G/. It follows thatall elements in .G=Z.G//�.T=Z.G// are involutions. Since Z.G/ 6� ˆ.T /, it followsthat T=Z.G/ is cyclic. In that case, G=Z.G/ is dihedral. Indeed, since d.G/ D 2 andZ.G/ 6� ˆ.G/, we conclude that G=Z.G/ is not abelian of type .2; 2/. If z is anelement of maximal order in G � T , then z2 is a generator of Z.G/ since zZ.G/ isa noncentral involution in the dihedral group G=Z.G/. As we saw, jZ.G/j � 4. Weobtain a group in part (c) of the theorem.

(��) In the rest of the proof we assume that G has a normal abelian subgroup Uof type .2; 2/ which does not contain our involution t (if G has two normal abeliansubgroups of type .2; 2/, then one of them does not contain t : otherwise t 2 Z.G/,which is not the case since G is nonabelian). Therefore, this case includes the onewhere G has at least two distinct normal four-subgroups. Set again T D CG.U / sothat G D hti T since t 62 T . By the modular law, G0 D CG.t/ D hti � A, whereA D CG.t/ \ T D CT .t/ is cyclic of order 2m, m � 2, and so Z D �1.A/ D

CU .t/ D �1.Z.G// is of order 2. We have Z.G/ � CG.U /\CG.t/ D T \CG.t/ D

CT .t/ D A. Set G1 D NG.G0/, where G0 D CG.t/; then G1 > G0 since G > G0 isnonabelian. Since t and the generator z of �1.A/ are not conjugate in view of m > 1,t has only two conjugates t and tz inG1, and we have�1.G0/ D ht; zi. It follows that2 D jG1 W CG1

.t/j D jG1 W G0j (see also Exercise 1(a)) and so G1 D G0U becauseG0U � G1 and jG0U W G0j D 2 since G0 \ U D �1.A/. Set D0 D hti U ; thenD0 Š D8 since it is nonabelian of order 8 and generated by involutions, and G1 is thecentral product G1 D A �D0 with A \D0 D hzi (indeed, A centralizes t and U soht; U i D D0; before we used the product formula). We also have U D hu; zi < D0

for some involution u. Let v be an element of order 4 in D0 and y an element oforder 4 in A; then y2 D v2 D z, and since x2 D y2v2 D y4 D 1, x D yv is aninvolution in G1 � D0. Since xt D ytvt D yv�1 D yv v2 D xz, we see thatD1 D hx; ti Š D8. Because A D Z.G1/, we also have G1 D A �D1, t 2 D1 andD1 \ U D Z.D1/ D hzi.

In the rest of the proof we consider a subgroup S of G which is maximal subject tothe following four conditions:

(i) S contains G1 D A �D1.

(ii) S D AL, where L is a normal subgroup of S and L Š D2n , n � 3,

(iii) A \ L D hzi D Z.L/, and

(iv) L \G1 D D1 Š D8.

We recall that A D hai is of order 2m, m � 2, and Z.G/ � A D Z.G1/ so Z.G/


is cyclic of order � 2m, and �1.A/ D hzi D Z.D1/ D Z.L/ D �1.Z.G//, t 2 D1,CG.t/ D hti � A. Also G1 contains U D hz; ui which is a normal abelian subgroupof G of type .2; 2/ and ut D uz. Now we act with A on the dihedral group L,where we set L D hb; t j b2n�1

D t2 D 1; bt D b�1i. Here hbi is the uniquecyclic subgroup of index 2 in L so hbi is A-admissible (recall that L is normal inS D AL). We have jhbi \ D1j D 4 (since D1 � L, by (iv)), A centralizes D1 (by(i)) and CL.a/ D hti.CL.a/ \ hbi/, by the modular law, and so CL.a/ is a dihedralsubgroup of L of order � 8 containing D1 (recall that t 2 D1). Looking at Aut.L/(see Proposition 34.8, where Aut.D2n/ is described) we see that A induces on L acyclic group of automorphisms of order at most 2n�3 and so jA W CA.L/j � 2n�3.Since S=L > f1g is cyclic, U 6� L and L is generated by involutions, we have�1.S/ D UL. It follows that �1.S/ \ A D hyi, where o.y/ D 4 (by the productformula) with y2 D z, and we have hyi D �2.A/.Let us consider at first the case where y induces on L a nonidentity automorphism

so that CA.L/ D hzi and the group hai=hzi acts faithfully on L. Then y induces anautomorphism of order 2 on L and since L > CL.A/ � D1 Š D8 we must haven � 4, Œy; t � D 1 (recall that the element y 2 A centralizes t ) and by D bz, .b2/y D

.bz/2 D b2 so that CL.y/ D ht; b2i Š D2n�1 and CL.y/ is a maximal subgroup ofL. Let v be an element of order 4 in D1; then v 2 hbi. Since U D hz; ui � AD1, wemay set u D yv. We have yb D y.D y y2 D y�1, and so utb D .uz/b D .yvz/b D

y�1vz D y�1vy2 D yv D u so that hz; u; tbi is an elementary abelian subgroupof order 8 since tb is an involution not contained in U . Obviously, C�1.S/.u/ D

CS .u/\�1.S/ D hui�htb; b2i, where htb; b2i Š D2n�1 , and since CG.t/ D hti�A

does not contain an elementary abelian subgroup of order 8, t cannot be fused in G toany involution contained in CS .u/\�1.S/ since the centralizer of any involution fromCS .u/\�1.S/ contains a subgroup isomorphic to E8. On the other hand, it is easy tocompute that any involution in S lies in L [ .CG.u/ \�1.S// since �1.S/ D UL.Naturally, t cannot be fused inG to any involution in U GG since t 62 U . It follows thatthe G-class of t contains at most 2n�2 elements so jG W CG.t/j D 2n�2 D jS ICS .t/j.Since CG.t/ D CS .t/, this forces that S D G. If m D 2, we have G D F.2; n/,stated in part (g) since ht; b2i Š D2n�1 centralizes A and the involution tb inverts A.So assume that m � 3. Then we may set (see Proposition 34.8) ba D b1C2i

, wherei � 2 because CL.a/ � D1. Since CL.a/ Š D2iC1 and CL.a

2m�1

/ Š D2iCm anda2m�1

D z 2 Z.L/, we get CL.a2m�1

/ D L. It follows that i D n � m. We haveobtained exactly the groups stated in part (e).In what follows we shall assume always, that hyi D �2.A/ centralizes L so that

�1.S/ D L ��2.A/, the central product (of order 2nC1). In this case each involutionin S is contained in L[U (see Appendix 16, where we have proved that D2n �C4 Š

Q2n � C4); it follows that U as the unique normal abelian subgroup of type .2; 2/ inS , is characteristic in S and so the subgroup L being generated by its own involutionsis characteristic in �1.S/ and so in S (see also Appendix 16). If S D G, we get somegroups stated in part (d) of the theorem.


From now on we shall assume, in addition, that S < G. Let W D NG.�1.S//;obviously, S < W . Let Kx denotes the G-class containing an element x 2 G. SinceW fusesKt \�1.S/ withKtb \�1.S/ and CG.t/;CG.tb/ < S . where both classesare of size 2n�2 and are contained in L, we get jW W S j D 2, S is normal in W soL is normal in W since L is characteristic in S , by the above. Suppose that W < G

and let g 2 NG.W / � W be such that g2 2 W . We have hKt \ W i � ht; tbi D L

(indeed, involutions t and tb are not conjugate in L so they generate L), U is normalin G and all involutions in S lie in L [ U (see Appendix 16). Therefore (replacing gwith gw, w 2 W , if necessary) one may assume that s D tg 2 W � S ; indeed, byassumption W < G and the subgroup �1.S/ is not normal in G). Since s normalizesL and CG.t/ \ .hsi L/ D ht; zi, we have hsi L Š D2nC1 : indeed, by Proposition1.8 (Suzuki’s theorem), hsi L is of maximal class and s 62 L. Now L and Lg arenormal in W (since L is characteristic in S G W ), Lg \ S D Lg \ L, LgS D W ,jLgS W S j D jLg W .Lg \ S/j D jLg W .L \ Lg/j D 2 and jLLg j D 2nC1, and soLLg D hsi L. Hence, .LLg /g D LgLg2

D LgL D LLg since g2 2 W and L isnormal inW , and so hsi L is the dihedral group of order 2nC1 which is normal inW .We have W D A.hsi L/ and this contradicts the maximality of S D AL (see (i–iv)).Hence we must have W D G in any case. If �1.S/ D �1.G/, we get the groupsstated in part (d) of the theorem.

In view of the result of the previous paragraph, in what follows we assume, inaddition, that �1.S/ ¤ �1.G/. Then for each involution s 2 G � S , s normalizes Lsince L is characteristic in �1.G/ G G, and so M D hsi L Š D2nC1 , by Suzuki’stheorem (see Proposition 1.8) since CG.t/ \ M D ht; zi is abelian of type .2; 2/.Because of the maximality of S (see (i–iv)), M is not normal in G (otherwise, Awould normalize the dihedral subgroup M containing L properly, A \ M D �1.A/

and M \ G1 D D1, against the choice of S and L); moreover, the above argumentshows also that M is not A-invariant. But L is normal in G so since t 2 L, we getA0 D CG.L/ D CA.L/ (containing �2.A/, by assumption) is also normal in G andwe have A0 D Z.S/ which is of order � 4. We look at the structure of NG D G=L

which is a group with cyclic subgroup NS D NA D .AL/=L Š A=hzi (bar convention!)of order 2m�1 and index 2 and with a subgroup NM D M=L of order 2which intersectsNA trivially. SinceM is not normal in G, we see that NG D NM NA, a semidirect product,is nonabelian. We conclude that j NAj � 4 so m � 3 since NA D jA=�1.A/j < jAj).Now jA0j � 4 and so j NA0j � 2 and NA0 is normal in NG (recall that A0 D Z.S/ ischaracteristic in S so normal in G) and NA0 � NA. Since NG= NA0 is a subgroup of theouter automorphism group of the dihedral group L, it follows that NG= NA0 is abelian(see Proposition 34.8).

Suppose at first that NG is not of maximal class; in that case, NG Š M2m , where2m D jAj (see Theorem 1.2), so that �1. NG/ D .UM/=L and UM D �1.G/ sinceU and M are generated by involutions. Since M is not normal in G, it follows thats must invert Z.�1.S// D �2.A/ and so hz; u; si Š E8. Namely, if s centralizes�2.A/, then �1.G/ D �2.A/ � M , the central product, and each involution in G


lies in M [ U and so M would be characteristic in �1.G/ so normal in G, whichcontradicts the maximal choice of S since S < AM (see (i–iv)). If we set A D hai,then NG.M/ D M ha2i which follows from the structure of NG Š M2m (see Theorem1.2). Since �2.A/ centralizes L but �2.A/=hzi acts faithfully on M , it follows thatha2i=hzi acts faithfully onM and so ha2i=�2.A/ acts faithfully on L and CG.L/ D

�2.A/ D A0. It follows that Z.G/ < A0 D �2.G/ is of order 2 and A=�2.A/ actsfaithfully on L (since the cyclic group Ã1.A/=�2.A/, as we saw, acts faithfully onL). We havem � 4 since NG Š M2m in view of the fact that NG is not of maximal class,by the assumption of this paragraph; we conclude that j NAj � 8. One may assume thatthe involution s acts on L D hb; ti as follows: bs D b�1 and ts D tb and we setb2n�2

D z, where z generates Z.G/. Replacing a with ar (r is odd), one may assumethat ba D b1C2i

, i � 2, because CL.a/ � D1 Š D8. Hence we have CL.a/ Š D2iC1

and so CL.a2m�2

/ Š D2iCm�1 Š L Š D2n , where we have taken into accountthat A=ha2m�2

i D A=�2.A/ acts faithfully on L and �2.A/ centralizes L. Thus,i Cm� 1 D n and so i D n�mC 1. Since i � 2, we must have n � mC 1. Becausem � 4, we have here n � 5. It remains to determine the commutator Œa; s�. We knowthat A D hai does not normalize M (indeed, G D AM and M is not normal in G)and so hŒ Na; Ns�i D NA0 which yields Œa; s� D a0l , where a0, a generator of A0, has order4, a2

0 D z and l 2 L. We can also write Œa; s� D .a0z/.zl/ D .a0a20/.zl/ D a�1

0 .zl/,

and so replacing l with zl , if necessary, one may assume that a0 D a2m�2

and sinverts ha0i. Hence we get Œa; s� D a�1sas D a0l and so sa D a0ls. Since, by theprevious equality, a0ls is an involution, we must have l D bj for a suitable integer j .Therefore, ta

�1sa D ta0bj s which yields ba D b1�2j . On the other hand, ba D b1C2i

and so j �2i�1 .mod 2n�2/. This yields

(1) sa D a2m�2

b�2i �1s or

(2) sa D a2m�2

b�2i �1sz.

However, replacing b with b0 D b1�2n�i�1

and t with t 0 D tb2n�i�2

, we see that allso far obtained relations in this case go into the same relations with b0 instead of b butthe relation (1) goes into relation (2). (Note that L D hb; ti D hb0; t 0i). Therefore wemay choose relation (1). We have determined in this case the structure of G uniquelyand this the group given in part (f) of the theorem.Suppose now that NG D G=L is of maximal class. Since NG= NG0 is of order 4 and

NG= NA0 is abelian, we have j NA W NA0j � 2.We consider at first the case NA D NA0 which means that A D CG.L/ and so S D

A � L (a central product) with A \ L D hzi D Z.L/. Hence A D Z.S/ and soA is normal in G. Since Ahsi=hzi is of maximal class and the case ŒA; s� � hzi isnot possible because M D hsi L is not normal in G, we have either as D a�1 oras D a�1z, which is possible only when m � 3. Our group is isomorphic to a groupF.m; n/ as stated in part (g). Also we have here hz; u; si Š E8.It remains to consider the case where j NA W NA0j D 2 so that CG.L/ D A0 is of

index 2 in A. We have A0 D ha2i D Z.S/ and so a induces an automorphism of


order 2 on L with CL.a/ � D1. This yields at once that ba D bz and so n � 4.Hence we also have ab D az. Since ŒA;L� D hzi and L=hzi is a dihedral groupof order 2n�1 � 8, it follows that CS .L=hzi/ Š A � hb0i, where hb0i is a cyclicsubgroup of order 4 in L. Hence A � hb0i is normal in G since L=hzi is G-invariant.If s normalizes A, then as D a�1z , � D 0; 1, since Ahsi=hzi is of maximal class.But from sts D tb we get acting on a: asts D atb which yields az D a. Indeed,asts D .a�1z/ts D .a�1z/s D azz D a, and this is a contradiction. Hence sdoes not normalize A. Since As � A�hb0i and .Ahb0i=hb0i/hsi is of maximal class,one may set

(3) as D a�1b0 or

(4) as D a�1C2m�2

b0.

We must have as2

D a1 D a. But in case (3) this yields .a�1b0/s D a and soaz D a, which is a contradiction. Hence we must have relation (4). Replacing a witha�1 in (4), we get .a�1/s D .a�1/�1C2m�2

.b0/�1 and other relations in this case arenot changed. Thus one may put b0 D b2m�3

in (4). This determines the group G asstated in part (h) of the theorem. Here we have also hz; u; si Š E8.An inspection of obtained groups shows that in any case G has no elementary

abelian subgroups of order 16. Indeed, S has no elementary abelian subgroups oforder 8 since S=L is cyclic and L is dihedral. It follows from jG W S j � 2 that G hasno elementary abelian subgroups of order 16, as desired. It follows from defining rela-tions of G that it is generated by 3 elements. Also in case whereG has an elementaryabelian group of order 8, we see that Z.G/ has order 2. The proof is complete.

SinceG of Theorem 48.1 has no normal subgroups isomorphic to E8, each subgroupof G is generated by 4 elements, by MacWilliams’ theorem [MacW]. (See the proofof this theorem of MacWilliams in �50.)Let G D U wrV , where U Š C2m and V D hti Š C2. Then CG.t/ D hti � C ,

where C , the diagonal of U � U t , is cyclic of order 2m, Z.G/ D C .

Exercise 2. Let L be a subgroup of maximal class in a 2-group G. Suppose that,whenever an involution s 2 G � L, then hs;Li is of maximal class. Is G of maximalclass?

Solution. Yes. If G � L has no involutions, G is of maximal class, by Theorem1.17(a). So suppose that G �L has an involution. Assume that G is a counterexampleof minimal order. Then jG W Lj > 2 and all maximal subgroups of G containing Lare of maximal class. The number of maximal subgroups of G containing L whichare of maximal class is even (see �13). Since the number of maximal subgroups of Gcontaining L is odd, we get a contradiction.

Exercise 3. Let t be an involution of a 2-group G such that CG.t/ D hti �Q, whereQ Š Q2m . Prove that (i) G has no normal elementary abelian subgroups of order 8,(ii) jNG.CG.t// W CG.t/j � 2.


Exercise 4. Let G be a nonabelian p-group, p is odd. Suppose that G has an elementt of order p such that CG.t/ D hti � C , where C is cyclic of order pm, m > 1.Prove that G has no hti-invariant subgroups of order ppC1 and exponent p. (Hint.Use Theorem 9.6.)

Problems

Problem 1. Classify the 2-groups G containing an abelian subgroup A of type .4; 2/such that CG.A/ D A.

Problem 2. Classify the 2-groups containing a cyclic subgroup Z of order 4 such thatCG.Z/ D Z � C2. (For a solution, see �77.)

Problem 3. Let a 2-groupG admit an automorphism ˛ of order 2 and set CG.˛/ D Q,whereQ is either cyclic or of maximal class. Describe the structure of G. (See �51.)

Problem 4. Classify the 2-groups G containing an involution t such that CG.t/ D

hti �M2m .

Problem 5 (Janko). Classify the 2-groups G such that CG.E/ D E for some elemen-tary abelian subgroup E of order 8. (For solution, see �51.)

Problem 6 (Blackburn). Classify the nonabelian p-groups G, p > 2, containing anelement t of order p such that CG.t/ D hti � C , where C Š Cpm , m > 1.

Problem 7. Classify the 2-groups G containing a nonabelian subgroup D of order 8such that jCG.D/j D 4.

Problem 8. Study the 2-groups G containing an extraspecial subgroup E of order> 23 such that CG.E/ D Z.E/.

Problem 9. Study the 2-groups G containing a subgroupM Š M2n such that we haveCG.M/ D Z.M/.

�49

On 2-groups with small centralizerof an involution, II

In the previous section finite 2-groups G have been classified with the property that Ghas an involution t such that CG.t/ D hti � C , where C Š C2m is a cyclic group oforder 2m, m � 1.There is one more case of the centralizer of an involution in 2-groups which is very

important. This is seen from the following “conditionless” Theorem 10.26:For a finite p-group G, one of the following holds:

(a) G has no maximal elementary abelian subgroups of order p2.

(b) j�1.G/j � p2.

(c) There exists in G an element x of order p such that CG.x/ D hxi � Q, whereQ is cyclic or generalized quaternion.

In this section we classify the finite 2-groups G which possess an involution t suchthat CG.t/ D hti�Q, whereQ Š Q2m is generalized quaternion of order 2m; m � 3.Of course, the situation becomes more complicated; however, in spite of that, theexposition remains elementary.From the start, it is clear, that such a group G cannot be of maximal class. Then, by

the known results,G must contain a normal four-subgroup U . We have two essentiallydifferent possibilities according to t 2 U or t … U .In the first case, where t 2 U , we have either CG.t/ D G or the order of G is

equal to 2mC2 and then we get four classes of 2-groups which will be given in termsof generators and relations (see Theorem 49.1).In the second case, where t 62 U (Theorem 49.2), the situation is more complicated

since the order of G is not bounded with the parameter m. In fact, in some caseswith given m � 3, we get infinitely many 2-groups with the same centralizer of ourinvolution t . The idea of the proof in the second case is to construct certain largesubgroup S of the known structure with UCG.t/ � S and then to show that S isnormal in G and that jG=S j � 4. This last “closure” argument is the main trickin the proof. The corresponding argument in the previous “cyclic” case, which wastreated in the previous subsection, was considerably simpler. As a result, we obtainthree different types of 2-groups G, since we have exactly three possibilities for thestructure of S . In two of these types, the groups G could possess elementary abelian

�49 On 2-groups with small centralizer of an involution, II 29

subgroups of order 16 and in that case the corresponding series of 2-groups will begiven in terms of generators and relations (Theorem 49.3). The first member of oneof these series is a 2-group of order 27 which is isomorphic to a Sylow 2-subgroup ofthe simple group J2 of order 604800. Finally, it turns out that the case m D 3 of anordinary quaternion groupQ Q8 is more difficult than the general casem � 4.

As a direct application of these results and the results in the previous subsection, weobtain a classification of 2-groups which have more than 3 involutions but which donot have an elementary abelian subgroup of order 8 (Theorem 49.4). Such groups havebeen considered by D. J. Rusin [Rus] by using a heavy cohomological machinery. Ourmethods are completely elementary.

Let G D QwrC , where Q Š Q2m and C D hti Š C2. Then CG.t/ D hti �D,whereD is the diagonal of the baseQ �Qt soD Š Q2m .

Let G be a 2-group possessing an involution t such that CG.t/ D hti � Q, whereQ Š Q2m . We claim that then G has no normal elementary subgroups of order 8.Assume that this is false, and let E Š E8 be normal in G. Clearly, t 62 E. SetH D hti E. Then CH .t/ contains an elementary abelian subgroup of order 8, byProposition 1.8 (Suzuki’s theorem), which is not the case. We see that the group G isone of the groups of �50. It follows from Theorem 50.3 that every subgroup of G isgenerated by four elements.

1o. The case t 2 U . In this subsection we prove the following

Theorem 49.1. Let G be a 2-group containing an involution t such that CG.t/ D

hti �Q, whereQ is a generalized quaternion group of order 2m; m � 3. We assume,in addition, that t is contained in each normal four-subgroup U of G and that G ¤

CG.t/. Then we have the following possibilities:

(A1) If t 62 ˆ.G/, then G has a maximal subgroup M so that G D M hti, whereM Š Q2mC1 orM Š SD2mC1 and CM .t/ Š Q2m .

(A2) If t 2 ˆ.G/, then one of the following assertions holds:

(a) m D 3 and G D ht; u; v;w; y j y2 D tv; w2 D v2 D u; u2 D t2 D Œt; v� D

Œt; w� D 1; Œv;w� D u; ty D tu; vy D v�1; wy D wti. We have jGj D 25,Z.G/ D hui has order 2, ˆ.G/ D hti � hvi is abelian of type .4; 2/, G0 D hu; ti

is a four-group and CG.t/ D hti�hv;wi, where hv;wi Š Q8. AlsoG D hy;wi

and G has exactly three involutions.

(b) m � 4 and G D ha; b; t; u; xi, where a2m�1

D t2 D Œt; a� D Œt; b� D Œa; x� D

1, b2 D a2m�2

D u, ab D a�1, x2 D a2m�3

, bx D bt , tx D tu. Theorder of G is 2mC2, Z.G/ D hui is of order 2, ˆ.G/ D G0 D hti � ha2i isabelian of type .2m�2; 2/ and G is not an �-group since there exist involutionsin G � ha; b; ti but G has no subgroups isomorphic to E8. Also, G=ˆ.G/ hasorder 8 and CG.t/ D hti � ha; bi, where ha; bi Š Q2m and G D ha; b; xi.


(c) m � 4 and G D ha; b; t; u; xi, where a2m�1

D t2 D Œt; a� D Œt; b� D Œa; x� D

1, b2 D a2m�2

D u, ab D a�1, x2 D a1C2m�3

, bx D bat , tx D tu. Theorder of G is 2mC2;Z.G/ D hui is of order 2, ˆ.G/ D hti � hai is abelianof type .2m�1; 2/ and G0 D hati is cyclic of order 2m�1. Also, G has exactly 3involutions , G=ˆ.G/ has order 4, CG.t/ D hti � ha; bi, where ha; bi Š Q2m

and G D hb; xi.

(d) m � 5 and G D ha; b; t; u; xi, where a2m�1

D t2 D Œt; a� D Œt; b� D x2 D 1,b2 D a2m�2

D u, ab D a�1, tx D tu, ax D a1C2m�3

t , bx D b. Theorder of G is 2mC2, Z.G/ D hui is of order 2, ˆ.G/ D G0 D hti � ha2i.Also, G is not an � -group but G has no subgroups isomorphic to E8. We haveCG.t/ D hti � ha; bi, where ha; bi Š Q2m and G D ha; b; xi, Here we haveCG.x/ Š C2 � Q2m�2 .

Proof. Suppose that G satisfies the assumptions of the theorem and let U be a normalfour-subgroup of G. By assumption, we have t 2 U . Let T D CG.U /, so thatjG W T j D 2. We have T D hti �Q, whereQ Š Q2m ; m � 3. Set�1.Q/ D hui andthen we have U D ht; ui D �1.T / and obviously Z.G/ D hui.We consider at first the easy case t … ˆ.G/. Let M be a maximal subgroup of G

such that t … M . Then G D hti M and, by the modular law, CG.t/ D hti � M0,where M0 D T \ M Š Q2m . If M is not of maximal class, then by Lemma 1.4,M contains a G-invariant four-subgroup U0. But then t … U0, which contradicts theassumption of our theorem. Thus, M is of maximal class and thereforeM Š Q2mC1

or M Š SD2mC1 (see Theorem 1.2). There is an element y 2 M � M0 such thaty2 2 hui D Z.M/. We have y�1ty D ut so that tyt D yu and this determines theaction of t on M since all such y’s together withM0 generateM . We have obtainedthe group G stated in part (A1) of the theorem.We examine now the difficult case t 2 ˆ.G/ and suppose at first that Q Š Q8 so

that hui D Z.Q/ D Z.G/ D ˆ.T / and U D ht; ui. It follows from u 2 ˆ.Q/ �

ˆ.G/ and U 6� Z.G/ that U < ˆ.G/, by Exercise 1(b3). Since ˆ.G/ < T , weget, by the modular law, ˆ.G/ D hti � hvi, where hvi is a cyclic subgroup of order4 in Q and v2 D u; then ˆ.G/ is abelian of type .4; 2/ and G is generated by twoelements. Because in case of 2-groups ˆ.G/ D Ã1.G/ and here we have ˆ.G/ > U ,it follows that there are elements y; z 2 G � T so that y2 D tv and z2 D v sincet and tu are not squares. Next, hyi 6E G (otherwise, G is metacyclic since G=hyi iscyclic in view of ˆ.G/ 6� hyi). Similarly, hzi 6E G. If z normalizes Q, then hQ; zi

is of maximal class and order 16, by Theorem 1.2, a contradiction since ˆ.G/ is notisomorphic to subgroup of a 2-group of maximal class. Thus, y does not normalizeQ so Q is not normal in G. It follows that the core hziG D hvi (let us consider therepresentation of G by permutations of left cosets of hzi). Since c2.ˆ.G// D 2,hvti is also normal in G. We have ty D tu D tz , by Exercise 1(b2). It followsfrom tv D .tv/y D tyvy D tuvy that vy D vu D v3 D v�1. Take an elementw 2 Q � hvi so thatQ D hv;wi. Since G=ˆ.G/ is abelian andQ 6E G, hvi GG, we


have wy D wvr t for some integer r . If r is odd, then Œw; y� D vr t D .vt/r . Thus,modulo hvti, elements w, y and v are pairwise permutable. Since G D hQ;yi D

hv;w; yi and v2 D w2 D u and y2 D vt are contained in hvti, we see that G=hvtiis elementary abelian, contrary to the fact that ˆ.G/ is noncyclic. Thus, r is even, andwe get wy D wt or wy D wut D w�1t . However, if wy D w�1t , then replacingy with yv, we get wyv D .w�1t /v D wt and other relations remain unchanged:tyv D tu, vyv D v�1, .yv/2 D yvyv D y2vyv D tvv�1v D tv. Hence we mayassume from the start that wy D wt . The structure of the group G of order 25 isuniquely determined and this is the group stated in part (A2)(a) of our theorem.

We turn now to the general case Q Š Q2m , m � 4. Let a be an element oforder 2m�1 in Q so that hai is the unique cyclic subgroup of index 2 in Q. HereCG.t/ D hti �Q D T D CG.U /, where U D ht; ui is the G-invariant four-subgroupin T , hui D Z.Q/ D Z.G/ and hui < hai. Now, T=U Š Q=hui Š D2m�1 sothat ha;U i=U is the unique cyclic subgroup of T=U of order 2m�2 � 4 and so L D

ha;U i D hti�hai is normal inG. If b 2 Q�hai, then b2 D u and ab D a�1. Takingany x 2 G � T , we have tx D tu and ax 2 L. If x does not normalize hai, thenhai\haxi is an x-invariant subgroup of index 2 in hai since jL W haij D 2 (here we usethe product formula). Hence, in any case, ha2i is a cyclic normal subgroup of order2m�2 � 4 of G. In particular, the cyclic subgroup hvi of order 4 contained in ha2i isnormal inG. Thus, A D hti�hvi is a normal abelian subgroup of type .4; 2/ inG. Wehave CG.A/ D L so that G=L acts as an automorphism group of order 4 on A. Sincehvi is normal in G and vb D v�1, so there is an element x in G � T with vx D v andalso tx D tu D tv2. Indeed, CG.v/ is a maximal subgroup ofG different of T so as xwe can take every element in CG.v/�T . Since tb D t , we see that b and x induce twodistinct involutory automorphisms on A and therefore G=L is an elementary abeliangroup of order 4 (recall that b 62 L and t; v 2 L). In particular, we get thatˆ.G/ � L.On the other hand, ˆ.G/ � ht; a2i. Hence we have either ˆ.G/ D L D ht; ai

or ˆ.G/ D ht; a2i.< L/. In any case, x2 2 L since exp.G=L/ D 2 and so xinduces an automorphism of order 2 on L such that tx D tu; ha2ix D ha2i and xcentralizes the subgroup hvi of order 4 which is contained in ha2i. Therefore, we haveax D aj t i ; where i D 0; 1 and an integer j is odd (otherwise, o.aj t i / < o.a/).Since x normalizes ha2i and centralizes hvi � ha2i, we have either .a2/x D a2 or.a2/x D a2u (indeed, if x does not centralize a2, then hx; a2i=Chxi.a

2/ Š M2m�1 ,by Theorem 1.2); in particular, in the case under consideration,m � 5. If .a2/x D a2,we obtain ax D at i or ax D aut i . Indeed, .a2/x D a2j so j 1 .mod 2m�2/; thenaj D a or aj D au. If .a2/x D a2u, we get ax D avt i or ax D auvt i , where we setv D a2m�3

. Indeed, then j 1C 2m�2 so aj D av or aj D auv. As we establishedabove, we must have in the case where x does not centralize a2, o.a2/ � 8 and som � 5.

Let ax D at i or ax D aut i , where m � 4 and o.a/ D 2m�1. If ax D aut i , thenreplacing a with at , we get .at/x D aut i tu D .at/t i . Note that hat; bi D Q0 Š Q2m

and also CG.t/ D hti � Q0. Hence we may assume from the start that ax D at i ,


i D 0; 1. However, if ax D at , then we compute a D ax2

D .at/x D at tu D au,which gives u D 1 and this is a contradiction. It follows that ax D a. If x2 D taj

with an integer j , then CG.x/ � ha; taj i D L, which is a contradiction since x 62

L D CG.L/. Hence we have x2 2 hai. If bx 2 Q, then hxi normalizes Q D ha; bi

and since x2 2 hai, so Qhxi would be a maximal subgroup of G which does notcontain t 2 ˆ.G/, a contradiction. It follows that bx D b0t , where b0 is an element oforder 4 contained in Q � hai. There are exactly two conjugacy classes of elements oforder 4 in Q � hai with the representatives b and ba. Thus, replacing x with xak forsome integer k, we may assume that bx D bt or bx D bat .

Let bx D bt . We compute bx2

D .bt/x D bt tu D b�1 and so x2 2 hvi, wherev is an element of order 4 in hai. Hence, we may set x2 D a2m�3

or x2 D a2m�3

u

because these elements are the only elements of order 4 in hai. If x2 D a2m�3

u, thenwe replace x with xt . Then xt 2 G � T , xt centralizes hai, bxt D .bt/t D bt andfinally .xt/2 D xtxt D x2txt D a2m�3

utut D a2m�3

. Hence, we may assume fromthe start that x2 D a2m�3

and so the structure of G is uniquely determined in thiscase. We see that here ˆ.G/ D hti � ha2i D G0 and G D ha; b; xi. Also, xa2m�4

t isan involution in G � ha; b; ti. We have obtained the group stated in part .A2/.b/.

Now let bx D bat . We compute bx2

D .bat/x D batatu D ba2u. On the other

hand, by setting v D a2m�3

, we see that ba1C2m�3

D bav D ba2u. Hence, we may setx2 D av or x2 D avu. Note that the group hai=hui produces by conjugation exactly2m�3 distinct conjugates of b. However, if x2 D avu, then similarly as before, wereplace x with xt so that we obtain .xt/2 D xtxt D x2tx t D avutut D av and allother relations remain unchanged. Hence, we may assume that x2 D av D a1C2m�3

and the group G is again uniquely determined. Here we have ˆ.G/ D hti � hai D L

and our group G D hx; bi is 2-generated. Also, we see that G0 D hati and there areno involutions in G �T . ThereforeG has exactly 3 involutions. We have obtained thegroup stated in part .A2/.c/.

Now let ax D avt i or ax D auvt i , where v D a2m�3

, m � 5. If ax D auvt i ,then replacing a with at (as above) we get .at/x D auvt i tu D .at/vt i . Hence wemay assume from the start that ax D avt i , i D 0; 1: However, if ax D av; then a D

ax2

D .av/x D .av/v D av2 D au, which gives u D 1 and this is a contradiction. Itremains only the case ax D avt:We note that x2 2 L and CL.x/ D ha4; x2i, wherehvi � ha4i since o.a/ D 2m�1 � 16. All elements in T � L are of order 4. Thereare exactly four conjugate classes of such elements in T � L and under the action ofhai each such element is conjugate to one of the following four elements: b, ba, bt ,bat , where ha; bi D Q Š Q2m . So replacing x with xak for some integer k (notethat xak also centralizes v), we may assume that we have one of the following fourpossibilities: bx D b, bx D bt , bx D ba or bx D bat . If bx D ba, then bx2

D

.ba/x D baavt D ba2vt , which is a contradiction since x2 2 L < T andQ is normalin T: Similarly, if bx D bat; then bx2

D .bat/x D batavt tu D ba2vut , which isa contradiction since Q is normal in T: If bx D bt; then we replace b with b0 D ba


and x with x0 D xa�2m�4

(which also centralizes v D a2m�3

). We compute .b0/x0

D

.ba/xa�2m�4

D .btavt/a�2m�4

D bb�1a2m�4

ba�2m�4

av D ba�2m�4

a�2m�4

av D

bv�1av D ba D b0. Hence we may assume from the start that bx D b: In this casex2 2 CL.b/ D ht; ui D U: Since CG.U / D T; we must have x2 2 hui D Z.G/: Ifx2 D u; then replacing x with xt; we get .xt/2 D xtxt D x2tx t D utut D 1 andother relations remain unchanged: axt D .avt/t D avt and bxt D bt D b, Hence wemay assume that x2 D 1 and so the structure of G is uniquely determined as given inpart (A2)(d).

2o. The case t 62 U . In this subsection we examine the difficult case, where our 2-group G has a normal four-subgroup U such that an involution t 2 G is not containedin U and CG.t/ D hti �Q withQ Š Q2m , m � 3.In order to formulate our main result, we shall define three types of 2-groups:

A.m; n/, B.m; n/ and C.m; n/, m;n 2 N .

Definition. Let S D QL be a product of two normal subgroups Q D ha; b j a2m�1

D

b4 D z2 D 1; a2m�2

D b2 D z; ab D a�1i Š Q2m and L D hc; t j c2n�1

D t2 D

z2 D 1; c2n�2

D z; ct D c�1i Š D2n , where m � 3, n � 3 andQ \ L D Z.Q/ D

Z.L/ D hzi.If ŒQ;L� D 1; then S D Q � L is the central product ofQ and L which we denote

with A.m; n/ for m � 3, n � 3.If CQ.L/ D hai and tb D t , cb D cz, then so determined group S we denote with

B.m; n/ for m � 3, n � 4.If CQ.L/ D ha2; bi Š Q2m�1 , m � 4 and ta D t , ca D cz, then so determined

group S we denote with C.m; n/ for m � 4, n � 4.Obviously, jGj D 2mCn�1.


hti �Q, whereQ is a generalized quaternion group of order 2m, m � 3. We assume,in addition, that G has a normal four-subgroup U such that t … U . Then G has anormal subgroup S containing UCG.t/ such that jG=S j � 4 and S is isomorphic toone of the groups A.m; n/, B.m; n/ or C.m; n/. If S Š B.m; n/, then we must haveS D G. If S Š C.m; n/, then jG=S j � 2 and if, in addition, jG=S j D 2, then wehave m D n. Finally, if jG=S j D 4, then we must have S Š A.m;m/ and G actstransitively on the set of involutions in S � U .

Proof. Let E4 Š U G G with t 62 U . Set T D CG.U / and then we have obviouslyt 62 T . This gives jG W T j D 2 and so G D htiT . By the modular law, G0 D

CG.t/ D hti �Q, where Q D CT .t/ Š Q2m , m � 3, and so Z D hzi D �1.Q/ D

Z.Q/ D CU .t/ D Z.G/. Set G1 D NG.G0/. Since j.UG0/ W G0j D 2, we haveUG0 � G1. It follows from �1.G0/ D ht; zi that t has exactly two conjugates tand tz in G1. Therefore, jG1 W CG1

.t/j D jG1 W G0j D 2 so G1 D UG0. SetD0 D U hti so that D0 Š D8, ŒQ;D0� D f1g withQ \D0 D hzi and U D hz; ui <


D0 so G1 D Q � D0, by the product formula. Let v be an element of order 4 inD0 and let y be an element of order 4 in hai, where Q D ha; b j a2m�1

D b4 D

z2 D 1; a2m�2

D b2 D z; ab D a�1i Š Q2m , m � 3. Since y2 D v2 D z,we get .yv/2 D y2v2 D y4 D 1 so x D yv is an involution in G1 � D0. Butxt D .yv/t D yv�1 D yvv2 D yvz D xz and therefore D1 D hx; ti Š D8 andD1 \ U D Z.D1/ D hzi. We see that CG1

.D1/ D ha; bti D Q1 Š Q2m becausexbt D .yv/bt D y�1v�1 D yv D x, .bt/2 D z and abt D a�1. Thus we haveG1 D Q1 �D1 with Q1 Š Q, Q1 \D1 D hzi, CG.t/ D hti �Q D hti �Q1 andU D hz; ui 6� D1. Denoting again Q1 with Q and bt with b, we have obtained thefollowing initial result.

(R) Supposing thatU D hz; ui is a normal four-subgroup of our 2-groupG possess-ing an involution t with t … U and CG.t/ D hti �Q, whereQ Š Q2m , m � 3, thenG1 D UCG.t/ has the following structure. We have G1 D Q �D1, whereD1 Š D8,t 2 D1,Q \D1 D hzi D Z.G/ and U 6� D1.

The next step in the proof is to “blow up” the subgroup G1 D UCG.t/ as much aspossible so that the structure of a large subgroup S containing G1 remains “about” thesame as the structure of G1.In the rest of the proof, we denote with S a subgroup of G of the maximal possible

order subject to the following conditions:

(i) S � G1 D UCG.t/ D Q �D1, whereQ Š Q2m , m � 3, D1 Š D8, t 2 D1.

(ii) S D QL, where L is normal in S , L Š D2n , n � 3,Q\L D hzi D Z.Q/ D

Z.L/, L � D1.

(iii) U D hz; ui 6� L.

In the sequel, we fix the following notation. We set Q D ha; b j a2m�1

D b4 D

z2 D 1; a2m�2

D b2 D z; ab D a�1i Š Q2m and L D hc; t j c2n�1

D t2 D z2 D

1; c2n�2

D z; ct D c�1i Š D2n , where m � 3, n � 3 and Q \ L D Z.Q/ D

Z.L/ D hzi. Then we have u D yv, where y is an element of order 4 in hai and v isan element of order 4 in hci. If a4 D 1, then we put y D a. Similarly, if c4 D 1, thenwe put v D c.At first we shall determine the structure of S . Act withQ on the dihedral group L.

Since Aut.L/=Inn.L/ is abelian of type .2n�3; 2/ (see Theorem 34.8), so Q0 D ha2i

centralizes L. Also, we know that Q centralizes hv; ti D D1 Š D8. It followsthat either Q centralizes L and then S D Q � L Š A.m; n/, m � 3, n � 3 orM D CQ.L/ is a maximal subgroup of Q, in which case n � 4. We have essentiallytwo different possibilities for M . First possibility is M D hai, tb D t , cb D cz andthen S Š B.m; n/, m � 3, n � 4. Second possibility is M D ha2; bi Š Q2m�1 ,m � 4, ta D t , ca D cz and then S Š C.m; n/, m � 4, n � 4. We see that we havein any case ŒQ;L� � hzi and soQ is a normal subgroup in S .Suppose that S Š B.m; n/. Then S has exactly six conjugacy classes of involu-

tions contained in S � U with the representatives t , tc, bv, bav, btc, batc. The


corresponding centralizers are:

CS .t/ D CG.t/ D hti �Q with Q Š Q2m ; m � 3;

CS .tc/ D htci � ha; bvi with ha; bvi Š D2m ;

CS .bv/ D hbvi � hcy; tci with hcy; tci Š SD2n ;

CS .bav/ D hbavi � hcy; tci with hcy; tci Š SD2n ;

CS .btc/ D hbtci � hby; ui with hby; ui Š D8;

CS .batc/ D hbatci � hbay; ui with hbay; ui Š D8:

It follows that t cannot be fused in NG.S/ to any of the other five conjugacy classes ofinvolutions in S � U . But CS .t/ D CG.t/ then forces that S D G.We make here the following simple observation. Since t 2 G � T , where T D

CG.U / and jG W T j D 2, t cannot be conjugate (fused) inG to any involution t 0 whichcentralizesU: Therefore, with respect to that fusion, it is enough to consider only thoseinvolutions in S which act faithfully on U .Now suppose that S Š C.m; n/. Then S has exactly four conjugacy classes of

involutions contained in S � U acting faithfully on U with the representatives t , tc,bv, and bav. The corresponding centralizers in S are:

CS .t/ D CG.t/ D hti �Q with Q Š Q2m ; m � 4;

CS .tc/ D htci � hav; bavi with hav; bavi Š SD2m ;

CS .bv/ D hbvi � hc; yti with hc; yti Š Q2n ;

CS .bav/ D hbavi � hcy; cti with hcy; cti Š SD2n :

We may assume that NG.S/ ¤ S (otherwise we are finished). Then NG.S/ canfuse the conjugacy class of t only with the conjugacy class of bv under the assump-tion that m D n and then jNG.S/ W S j D 2. Then cclS .t/, the conjugacy class ofin S containing t , equals ftc2i ; 1 � i � 2m�2g and CS .tc

2i/ D htc2ii � Q sothat .CS .tc

2i//0 D ha2i. Similarly, we have cclS .bv/ D fba2iv; 1 � i � 2m�2g,CS .ba

2iv/ D hba2ivi � hc; yti, where hc; yti Š Q2m so that .CS .ba2iv//0 D hc2i.

Hence for any x 2 NG.S/�S , we have tx D ba2jv for some j and so, by the above,ha2ix D hc2i. Assume now that NG.S/ ¤ G (otherwise we are finished). SinceCS .t/ D CG.t/ and NG.S/ already fuses the involutions in cclS .t/ with involutionsin cclS .bv/, so there is an element s 2 NG.NG.S// � NG.S/ with s2 2 NG.S/ andt 0 D t s 2 NG.S/ � S . Indeed, t s 2 cclS .tc/ or ts 2 cclS .bav/ is not possible sincea semi-dihedral group is not a subgroup of a generalized quaternion group. However,if ts 2 cclS .t/ [ cclS .bv/; then there is an n 2 NG.S/ such that t s D tn. But thentsn�1

D t , which contradicts the fact that CG.t/ � NG.S/. By the above, we have.a2/t

0

D c2k with k odd. Since y is an element of order 4 contained in ha2i and vis an element of order 4 contained in hc2i, we get that yt 0

D vzl with l D 0; 1. But


then .yvzl /t0

D yvzl which gives .uzl /t0

D uzl , recalling that yv D u. We haveproved that t 0 centralizes U D hz; ui and so CG.t

0/ contains a subgroup isomorphicto E8. This is a contradiction, since t 0 is conjugate in G to t . Hence, we must haveNG.S/ D G and so jG W S j D 2, as required.Suppose, finally, that S Š A.m; n/, m � 3, n � 3; i.e. S D Q � L is the

central product of Q and L. Then S has exactly four conjugacy classes of involutionscontained in S �U with the representatives t , tc, bv and bav and all these involutionsact faithfully on U . The corresponding centralizers are:

CS .t/ D CG.t/ D hti �Q with Q Š Q2m ; CS .tc/ D htci �Q;

CS .bv/ D hbvi � hc; yti with hc; yti Š Q2n and CS .bav/ D hbavi � hc; yti:

Note that all 2n�2 conjugates of t in S lie in the dihedral subgroup L D hc; ti and alsoall 2n�2 conjugates of tc in S lie in L. Similarly, all 2m�2 conjugates of bv in S liein the dihedral subgroup K D hbv; bavi Š D2m and also all 2m�2 conjugates of bavin S lie inK. These are all 2 2n�2 C 2 2m�2 involutions in S �U . Also, we see thatK D ha; bv j a2m�1

D .bv/2 D 1; abv D a�1i.We want to determine all dihedral subgroups of S which are generated by a pair of

distinct involutions in S � U . We recall that any two distinct involutions always gen-erate a dihedral subgroup. Any two distinct involutions in cclS .t/[ cclS .tc/ generatea dihedral subgroup which is contained in L. Similarly, any two distinct involutionsin cclS .bv/ [ cclS .bav/ generate a dihedral subgroup which is contained in K. Wenote that hci fuses (by conjugation) all involutions in cclS .t/ to t and all involutionsin cclS .tc/ to tc. Similarly, hai fuses all involutions in cclS .bv/ to bv and all invo-lutions in cclS .bav/ to bav. We have Œc; a� D 1 and a centralizes t and both tc andc centralize bv and bav. All this means that it is enough to see what is the order ofthe following four dihedral subgroups: hbv; ti; hbv; tci, hbav; ti and hbav; tci. Thecomputation shows: .bv t /2 D .bv tc/2 D .bav t /2 D .bav tc/2 D z, and so allthese four dihedral subgroups have order 8.Suppose that t is not conjugate in NG.S/ to any involution in S � L. That will

certainly happen if (for example) m ¤ n. Supposing, in addition, that S ¤ G, wesee that NG.S/ fuses t with tc and so jNG.S/ W S j D 2, since CG.t/ D CS .t/. IfNG.S/ D G, then we are finished. Therefore, we assume that NG.S/ ¤ G. Takean element s 2 G � NG.S/ such that s normalizes NG.S/ and s2 2 NG.S/. If aninvolution x 2 NG.S/ � S , then .cclL.t//x D cclL.tc/ implies that L0 D Lhxi Š

D2nC1 : If jCS .x/j D 2m; then CL.x/ D hzi and CNG.S/.x/ D hxi � CS .x/ implythat CS .x/ covers S=L and soQ normalizesL0 Š D2nC1 andQL0 D NG.S/;whichcontradicts the maximality of S . Hence there is no such involution x in NG.S/�S withjCS .x/j D 2m: Set P D cclS .t/[ cclS .tc/[ cclS .bv/[ cclS .bav/ so that hP i D S

and P � G � T . If t 0 D ts 2 NG.S/ � S , then CNG.S/.t0/ D ht 0i � CS .t

0/ and sojCS .t

0/j D 2m; a contradiction. If t 0 2 cclS .t/ [ cclS .tc/, then there is n 2 NG.S/

such that t 0 D ts D tn and so CG.t/ 6� NG.S/; a contradiction. Hence t 0 2 cclS .bv/or t 0 2 cclS .bav/. If NG.S/ fuses bv and bav, then all elements in P are conjugate in


hyiNG.S/ to t; and since hP i D S , there is x0 2 P such that x0 D .x0/s 2 NG.S/�S

and so CNG.S/.x0/ D hx0i � CS .x0/ and jCS .x0/j D 2m, a contradiction. Supposethat NG.S/ does not fuse bv and bav. Then jCNG.S/.bv/j D jCNG.S/.bav/j D 2mC2

and so n C 2 D m C 1 because t 0 2 cclS .bv/ or t 0 2 cclS .bav/. There is x0 2 P

such that x0 D .x0/s 2 NG.S/ � S and so jCNG.S/.x0/j D 2nC2 D 2mC1 whichgives jCS .x0/j D 2m, a contradiction. This gives NG.S/ D G and so jG W S j D 2, asrequired.

Now suppose that t is not conjugate in NG.S/ to tc 2 L but S ¤ G. Then NG.S/

must fuse t to an involution in S �L�U . Without loss of generality, we may assumethat t is fused in NG.S/ to bv. This gives jNG.S/ W S j D 2 and m D n becauseCS .bv/ Š C2 � Q2n . Supposing NG.S/ ¤ G, we take an element s 2 G � NG.S/

such that s normalizes NG.S/ and s2 2 NG.S/. Set again P D cclS .t/ [ cclS .tc/[

cclS .bv/[ cclS .bav/ so that hP i D S and P � G �T . There is an x0 2 P such thatt 0 D .x0/s 2 NG.S/ � S and t 0 acts faithfully on U . Suppose at first that m D n � 4.Then L D hc; ti and K D ha; bvi are the only dihedral subgroups of order 2m in Sand so for each x 2 NG.S/ � S we have Lx D K and so hcix D hai. In particular,vt 0

D yz; D 0; 1. It follows that t 0 centralizes yvz D uz and since z 2 Z.G/,so t 0 centralizes u. Hence, t 0 centralizes U D hz; ui , a contradiction. We get in thiscase NG.S/ D G and so jG W S j D 2, as required. Suppose now that m D n D 3

so that S is extraspecial of order 25 and S Š Q8 � D8. Then t (fused with bv) liesin a conjugacy class of length 4 in NG.S/. We have t t

0

2 fbv; bvzg and ht; bvi D

D Š D8. Hence Dht 0i D D0 Š D16 and so t 0 acts fixed-point-free on D � hzi. Foreach involution t0 in S � U , we have CS .t0/ D ht0i �Q0 withQ0 Š Q8. Since t

0 isa conjugate of an involution in S � U under the action os s (normalizing NG.S/), itfollows that CNG.S/.t

0/ D ht 0i � CS .t0/ contains a subgroup isomorphic to Q8 and so

CS .t0/ D Q� Š Q8 becauseQ

�\D D hzi. Since NNG.S/.D/ � hS; t 0i D NG.S/, itfollows thatQ� normalizesD0 D Dht 0i Š D16 and soD0 is normal in NG.S/ (sinceQ� covers S=D). Note that t 2 D and so we have obtained a contradiction with themaximality of S . Hence, also in this case, we have NG.S/ D G and so jG W S j D 2.

It remains to consider the case where t is fused in NG.S/ to each involution inS � U . This gives at once that m D n and jNG.S/ W S j D 4 so that NG.S/=S actsregularly on the four S -classes of involutions which are contained in S � U .

Let m D n � 4. Then S has exactly two dihedral subgroups L D hc; ti andK D ha; bvi of order 2m and L andK contain all involutions from S �U . Therefore,for each x 2 NG.S/ � S , we have either Kx D K and Lx D L or Kx D L.Supposing again that NG.S/ ¤ G, we take an element s 2 NG.NG.S// � NG.S/

with s2 2 NG.S/. Then we have t 0 D ts 2 NG.S/ � S . If Lt 0

D L and Kt 0

D K,then t 0 fuses the two classes of involutions in L� hci and also t 0 fuses the two classesof involutions in K � hai. This gives that ct 0

D c�1 and at 0

D a�1. In particular,we get ut 0

D .yv/t0

D y�1v�1 D yz vz D yv D u. Hence, the elementaryabelian subgroup hz; u; t 0i Š E8 is contained in CG.t

0/ and this is a contradiction.Now assume that Lt 0

D K. Then hcit 0

D hai and, in particular, vt 0

D yz, D 0; 1.


But then .yvz/t0

D yvz and consequently .uz/t0

D uz. Hence t 0 centralizesU D hz; ui, which is a contradiction, since t 0 is conjugate in G to t .Finally, we assume that m D n D 3 so that S Š Q8 � D8 is an extraspecial

group of order 25. We recall that NG.S/ acts transitively on the set of 8 involutionscontained in S � U . Again assuming that NG.S/ ¤ G, we take an element s 2

NG.NG.S/ � NG.S/ with s2 2 NG.S/. Then we have t 0 D ts 2 NG.S/ � S . SetM D hsiNG.S/ so that jM W NG.S/j D 2. We shall determine V D SM and wenote that S < V � NG.S/. Consider at first the possibility V < NG.S/ which givesjV W S j D 2. Then V D SS s, t 0 D t s 2 S s � S and V D Sht 0i. By the modular law,CG.t

0/ D CV .t0/ D ht 0i� OQ, where CS .t

0/ D OQ Š Q8. Then CS . OQ/ D OD Š D8 andS is the central product of OQ and OD. Since OD is t 0-invariant and t 0 acts fixed-point-freeon OD�hzi, so ODht 0i Š D16, V D . ODht 0i/� OQ, V fuses all four involutions in OD�hzi

and other V -classes of involutions in S � U have length 2. This contradicts the factthat NG.S/=S acts regularly on the four S -classes of involutions in S � U . It followsthat V D SM D NG.S/. Since SS s is normal inM , so we get V D SS s D NG.S/.Thus F D S \ S s is a normal subgroup of M of order 8. If F is nonabelian, thenS D F � CS .F /, where F1 D CS .F / is nonabelian and normal in NG.S/. SinceS Š Q8 � D8, so fF;F1g D fD8;Q8g. But this contradicts the fact that NG.S/ actstransitively an all 8 involutions from S � U . We have proved that F is abelian. SinceQ8 � D8 does not possess a subgroup isomorphic to E8, it follows that F must be oftype .4; 2/ and consequently we have z 2 F . Again, because NG.S/ acts transitivelyan all involutions from S � U , so we have U � F . Furthermore, S 0 D .S s/0 D hzi

and so if hf i is a cyclic subgroup of order 4 in F , so NG.hf i/ � hS; S si D NG.S/.Also, we have S1 D CS .f / Š C4 �Q8 and jS W S1j D 2. Hence, S1 contains exactly6 involutions which are different from z. This implies that there exist involutions inS � S1. Let t1 be an involution in S1 � F and let t2 be an involution in S � S1. Thent1 centralizes f but t2 inverts f . This is a contradiction because NG.S/ normalizeshf i and NG.S/ acts transitively on all involutions in S � U . We have proved againthat NG.S/ D G and so jG W S j D 4, as required.

3o. Groups with subgroup isomorphic to E16. As an application of Theorems 49.1and 49.2, we shall determine up to isomorphism all 2-groups with an involution t suchthat CG.t/ D hti�Q, whereQ Š Q2m ; m � 3 and which have an elementary abeliansubgroup E16 of order 16. We obtain exactly two classes of 2-groups and we presentthem in terms of generators and relations. More precisely, we shall prove the followingresult.


hti�Q, whereQ is a generalized quaternion group of order 2m,m � 3. We assume inaddition thatG has an elementary abelian subgroup of order 16. ThenG is isomorphicto one of the following groups.

(E1) G D ha; b; z; c; t; x j a2m�1

D b4 D z2 D c2m�1

D t2 D x2 D 1, m � 4,a2m�2

D b2 D c2m�2

D z, b�1ab D a�1, tct D c�1, bc D cb, bt D tb, a�1ca D


cz, at D ta, xcx D a, xtx D bc2m�3

i. Here hz, x,a2m�3

c2m�3

, ba1C2m�3

tci Š

E16 and ha; bi Š Q2m , hc; ti Š D2m , ha; b; c; ti Š C.m;m/.

(E2) G D ha; b; z; c; t; x; s j a2m�1

D b4 D z2 D c2m�1

D t2 D x2 D s2 D 1,m � 3, a2m�2

D b2 D c2m�2

D z, b�1ab D a�1, tct D c�1, bc D cb, bt D tb,ac D ca, at D ta, sx D xs, xtx D tc, xcx D c�1, xbx D ba, xax D a�1,scs D az, sts D bc2m�3

i. Here ha; bi Š Q2m , hc; ti Š D2m , ha; b; c; ti Š A.m;m/,hz, x, s, a2m�3

c2m�3

i Š E16. Finally, for m D 3, our group G is isomorphic to aSylow 2-subgroup of the simple group J2.

Proof. Suppose that a 2-group G satisfies the assumptions of Theorem 49.3 and let Ebe an elementary abelian subgroup of order 16 in G. Using theorems 49.1 and 49.2,we see that we have exactly two possibilities:

(E1) G has a normal subgroup S Š C.m;m/, m � 4 of index 2 in G, jE \ S j D 8

and so G D SE.

(E2) G has a normal subgroup S Š A.m;m/, m � 3 of index 4 in G, E \ S D U

is a normal four-subgroup of G and so G D SE.

Case .E1/. Assume that S Š C.m;m/, m � 4; is given with generators and rela-tions as in the definition. Let U D hz; yvi be the normal four-subgroup of G, wherey is an element of order 4 in hai and v D c2m�3

is an element of order 4 in hci:

In the proof of Theorem 49.2 we have found all five conjugacy classes of involutions(and their centralizers) in S � U . Their representatives are the involutions: t , tc, bv,bav and batc: It follows that t can be fused in G only to an involution in the con-jugacy class of bv:We see also that S has exactly two normal subgroups isomorphicto D2m which do not contain any conjugate of batc W L D ht; ci and K D hbv; ai.Take an involution x 2 E � S: Since x cannot fuse t with tc; it follows that x doesnot normalize the dihedral subgroup L: Hence we have Lx D K: The involution xsends the conjugacy class of t in L onto the conjugacy class of bv in K: Replacingt with tc2j (j is an integer), we may assume that tx D bv D bc2m�3

. Also wehave hcix D hai and so replacing c with ck (k is an odd integer), we may assumethat cx D a:We note that these replacements of generators did not effect the definingrelations for S Š C.m;m/: The structure of G is uniquely determined.

Case .E2/. Assume that S Š A.m;m/ is given with generators and relations asin of A.m; n/. Then we have S D Q � L; where Q Š Q2m and L Š D2m : LetU D hz; yvi be the normal four-subgroup of G, where y is an element of order 4 inhai and v D c2m�3

is an element of order 4 in hci: In the proof of Theorem 49.2 wehave found all four conjugacy classes of involutions (and their centralizers which areall isomorphic to C2 �Q2m) in S �U: Their representatives are the involutions: t , tc,bv, and bav: Since jG W S j D 4; so G fuses all involutions in S � U into a singleconjugacy class in G: This gives that E \ S D U:

Assume for a moment that m � 4: Then we have seen in the proof of Theorem49.2 that S has exactly two subgroups isomorphic to D2m : They are L D ht; ci and


K D hbv; ai and both are normal in S: If we setE D U �hx; si;where hx; si\S D 1;

then the four-subgroup hx; si acts on fL; Kg: We may assume that x normalizes Land K and Ls D K: The involution x must induce outer automorphisms on L and Kand so cx D c�1 and ax D a�1 since x fuses two conjugacy classes of non-centralinvolutions in L and also in K: Replacing s with sx (if necessary), we may assumethat s sends t to an involution in the conjugacy class of bv (instead of bav) in K:Replacing t with tc2j (where j is an integer) we may assume that t s D bv D bc2m�3

:

Also we have hcis D hai: Let tx D t 0, where t 0 D tck with k an odd integer. Thenreplacing c with ck; we see that we may assume from the start that tx D tc: Here itmight happen that .ck/2

m�3

D v�1 D vz and then the previous relation ts D bv ischanged into t s D bv�1 D bvz: In that case we replace b with b�1 D bz so thatwe may assume again ts D bv: Replacing a with al ; where l is an odd integer, wemay assume cs D az: (Here we have taken az rather than a so that the last relationlooks simpler!) Act on the relation xtx D tc with s: We get xs tsxs D tsaz and sosince Œx; s� D 1 we have xbvx D bvaz or bxv�1 D bavz and finally bx D ba:

We note that these replacements of generators did not effect the defining relations forS Š A.m;m/: The structure of G is uniquely determined.In the remaining case m D 3 we have S Š Q8 � D8:We shall determine the group

G almost in the same way as above for m � 4: The only difference is that here Shas exactly six subgroups isomorphic to D8 which do not contain the normal four-subgroup U of G (see the proof of Theorem 49.2). They are:

L D ht; tci; K D hbv; bavi; D1 D ht; bvi;

D2 D ht; bavi; D3 D htc; bvi; D4 D htc; bavi:

In S � U lie exactly eight involutions which are distributed in four S -classes: ft; tzg,ftc; tczg, fbv; bvzg, fbav; bavzg. We set againE D U�hx; si;where hx; si\S D 1:

Then the four-group hx; si acts transitively (and so regularly) on the above four S -classes. This follows from the fact that G acts transitively on the eight involutions inS �U since CG.t/ has index 8 inG: If is any involution in hx; si; then we must haveeither L� D L and K� D K or L� D K: Indeed, if L� D Di ; where i 2 f1; 2; 3; 4g;

then L\Di D ht; zi or htc; zi and L\Di is -invariant. But then the S -class ft; tzg

or the S -class ftc; tczg is -invariant which is a contradiction. Similarly, ifK� D Dj ;

where j 2 f1; 2; 3; 4g; then the S -class fbv; bvzg or the S -class fbav; bavzg is -invariant which is again a contradiction. Hence the four-group hx; si acts on fL; Kg:

We may assume that x normalizes L and K and Ls D K: Then we determine thegroup G uniquely in exactly the same way as above. The proof is complete.

4o.Groups without subgroups isomorphic to E8. As a direct application of the aboveresults, we shall prove the following

Theorem 49.4. Let G be a 2-group which is not of maximal class and such thatj�1.G/j > 4. If G has no elementary abelian subgroups of order 8, then G is


isomorphic to one of the following groups:

(a) A group G from (A1)(c) of Theorem 48.1 with involutions in G � T .

(b) A group G from (A1)(d) of Theorem 48.1.

(c) A group G from (A1) of Theorem 49.1.

(d) A group G from (A2)(b) of Theorem 49.1.

(e) A group G from (A2)(d) of Theorem 49.1.

(f) A group G with a normal subgroup S of index at most 4 in G so that S isisomorphic to the central product of Q2m and D2n , where m > 2, n > 2 and�1.G/ � S (see Theorem 49.2).

Proof. Let G be a 2-group which has more than 3 involutions and which is not ofmaximal class. Also assume that G has no subgroups isomorphic to E8. Let U be anormal four-subgroup of G and set T D CG.U /. By assumption we have�1.T / D U

and so there is an involution t 2 G � T . We have G D htiT and so CG.t/ D hti �Q,where Q D CT .t/. Since t does not centralize U , it follows that Q has only oneinvolution. Therefore, Q is either cyclic or a generalized quaternion group. Suchgroups G are then described in this and previous subsection. A simple inspectionyields the above result.

Exercise 1 (Alperin [Alp3]). Suppose that a nonabelian 2-group G has no normal ele-mentary abelian subgroups of order 8. Suppose thatG is neither cyclic nor of maximalclass. Then one of the following holds:

(a) G has a normal abelian subgroup A of type .4; 2/ with CG.A/ abelian of type.2; 2n/, n � 2. In that case,G=CG.A/ is isomorphic to a subgroup of Aut.A/ Š

D8.

(b) G has a normal abelian subgroup A of type .4; 4/ with metacyclic CG.A/. Inthat case, jG=CG.A/j � 25.

Solution. Let A be the greatest normal noncyclic abelian subgroup of G of exponent� 4. Since G is not of maximal class, either A is abelian of type .2; 4/ or .4; 4/(Lemma 1.4). By Corollary 10.2, �2.CG.A// D A. It follows from Theorem 41.1that CG.A/ is metacyclic. Note that Aut.C4 �C2/ Š D8 and jAut.C4 �C4/j D 3 25.It remains to show that, if G has not G-invariant abelian subgroup of type .4; 4/, thenCG.A/ is abelian of type .2; 2n/ for some n > 1. Indeed,�2.CG.A// D A is of order8 so our claim follows from Lemma 42.1.

Exercise 2. Let Q Š Q2m , where m � 5 and let H be the holomorph of the cyclicgroup C Š C2n , n > 2. Set G D Q �H with Q \H D Z.Q/. Show that G has anormal elementary abelian subgroup of order 8.

Solution (Janko). Let hyi be a normal (cyclic) subgroup of order 4 in Q, let hvi < C

be of order 4 and set hzi D Z.G/; then .yv/2 D y2v2 D zz D 1. Let a be an


involution in A D Aut.C / such that ha;C i Š M2nC1 (�1.A/ C has exactly threemaximal subgroups containing C ; these subgroups are isomorphic to D2nC1 , SD2nC1

and M2nC1 , respectively). Then a centralizes v 2 Z.ha;C i// and D D hz; yv; ai

is the desired normal elementary abelian subgroup of order 8 in G. Indeed, V D

hz; yvi is a normal four-subgroup of G, since hzi, hyi, hvi are normal in G ando.yv/ D 2. Next, jha; V ij D 8 and W D hz; ai D �1.ha;C i/ is characteristic in anormal subgroup ha;C i of G. It follows that D D VW is also normal in G. Since acentralizes V ,D Š E8, and we are done.

Exercise 3. LetG be a p-group containing an elementary abelian subgroup of order p3

and letE be the subgroup generated by all elementary abelian subgroups ofG of orderp3. Suppose thatG�E contains an element x of order p. Then Chx;Ei.x/ D hxi�Q,whereQ is cyclic or generalized quaternion.

Exercise 4. Classify the 2-groups G such that

(i) �2.G/ D C �M , where jC j D 2 andM is of maximal class and order > 8,

(ii) �i .G/ D C �M , where i D 1 or 2, C Š C4 andM is of maximal class.

�50

Janko’s theorem on 2-groups withoutnormal elementary abelian subgroups of order 8

The main part of this section coincides with [Jan5]. Theorem 50.1 yields a deep insightin the structure of 2-groups without normal elementary abelian subgroups of order 8.The proof is fairly difficult, however elementary. In the analogous case, for p > 2, wehave to classify the p-groups G without normal subgroup of order ppC1 and exponentp; it seems that this problem is far from a solution.

The first important result about the groups of the title is the 4-generator theoremfrom [MacW] which asserts that any subgroup of such groups can be generated byfour elements. However, this result does not say anything more about the structureof such groups. In [Kon3], the groups of the title are determined under the additionalassumption that the Frattini subgroup contains an elementary abelian subgroup of or-der 8. This determination is somewhat unfortunate, because the resulting groups aregiven in terms of generators and relations without any comments and from these it isdifficult to disclose the structure of such groups.In his long paper [Ust], Ustjuzaninov has asserted that the groups G of the title must

have a normal metacyclic subgroup N such that G=N is isomorphic to a subgroup ofthe dihedral group D8 of order 8. But this paper is completely unreadable. With anyattempt to read it, computational errors were discovered. However, it turns out thatthis result is correct after all!We shall give here a relatively short proof of a stronger result. For example, in the

case where G=N is isomorphic to D8, we shall determine completely the structure ofN by showing at first that N is either abelian or minimal nonabelian. In our proof thecomputations are reduced to a minimum. The proof is based on a method of “pushingup” normal metacyclic subgroups of G combined with a very detailed knowledgeof Aut.C4 � C4/ (Proposition 50.5). We also note that our proof of the 4-generatortheorem is character-free, i.e., it is completely elementary.

We state now our main result.

Theorem 50.1 (Janko [Jan5]). Let G be a 2-group which has no normal subgroupsisomorphic to E8. Suppose that G is neither abelian nor of maximal class. Then Ghas a normal metacyclic subgroup N such that CG.�2.N // � N and one of thefollowing holds:


(a) jG=N j � 4 and either �2.N / is abelian of type .4; 4/ or N is abelian of type.2j ; 2/, j � 2.

(b) G=N Š D8 and

(b1) N is either abelian of type .2k ; 2kC1/, k � 1 or of type .2l ; 2l /, l � 2

or(b2) N is minimal nonabelian, �2.N / is abelian of type .4; 4/ and more

precisely N D ha; b j a2m

D b2n

D 1; ab D a1C2m�1

i, where m D n

with n � 3 or m D nC 1 with n � 2.

We prove here the following easy special case of the above theorem.

Theorem 50.2. Suppose that a 2-group G has no normal elementary abelian sub-groups of order 8 and has two distinct normal four-subgroups U and V . Then D D

UV Š D8 and G is the central product G D D �C ofD and C withD\C D Z.D/and C is either cyclic or of maximal class 6Š D8. Conversely, if G D D � C , whereD Š D8, C is cyclic or of maximal class 6Š D8 and D \ C D Z.D/, then G has nonormal subgroups isomorphic to E8.

Proof. By hypothesis, ŒU; V � ¤ f1g. Hence jU\V j D 2which givesD D UV Š D8.We have C D CG.D/ D CG.U /\CG.V / so jG W C j D 4. It follows thatG D D�C

withD \C D Z.G/, where C D CG.D/. IfW is a normal four-subgroup in C , thenUW would be an elementary abelian normal subgroup of order � 8 in G. This is acontradiction and so C is either cyclic or a group of maximal class (Lemma 1.4) whichis not isomorphic to D8.Now assume thatG D D�C , whereD, C andD\C are the same as in the previous

paragraph. Clearly, Z.G/ D Z.C / is cyclic. Assume that G has a normal elementaryabelian subgroup E of order 8. Since C 6Š D8 and Z.G/ is cyclic, we get E \ C D

Z.D/. By the product formula, G D EC . Let U , V be two distinct four-subgroups inD; then they are normal in G. Since CG.U / D U � C and CG.V / D V � C , we getU 6� E and V 6� E (recall thatG D EC ). It follows that E\D D Z.D/. ThenG=Dis noncyclic since it contains a four-subgroup DE=D Š E=Z.D/; it follows that Cis of maximal class. Clearly, U is the only G-invariant four-subgroup in U � C sinceC 6Š D8 and U � C D Z � C , where Z is a subgroup of order 2 in U . However,E \ .U � C/ is a G-invariant four-subgroup. It follows that U < E, which is acontradiction. The proof is complete.

In view of Theorem 50.2, in what follows we may confine our approach to the casewhere G has exactly one normal four-subgroup.

Remark 1. Let G D D �C be a 2-group from Theorem 50.2 withG ¤ D. LetZ1 bethe cyclic subgroup of order 4 in D.Š D8/ and let Z2 be a maximal cyclic subgroupof index � 2 in C . Then N D Z1Z2 is an abelian normal subgroup of type .2j ; 2/,j � 2. and G=N is elementary abelian of order � 4. Hence this is a special case ofgroups occurring in Theorem 50.1(a).

�50 Janko’s theorem on 2-groups 45

The 4-generator theorem follows as a trivial consequence of Theorem 50.1.

Theorem 50.3 (4-generator theorem). Let G be a 2-group which has no normal ele-mentary abelian subgroups of order 8. Then every subgroup U of G is generated byfour elements.

Proof. By Theorem 50.1, G has a normal metacyclic subgroup N such that G=Nis isomorphic to a subgroup of D8. Since U=.U \ N/ is isomorphic to a subgroupof G=N , we have d.U=.U \ N// � 2. Also, U \ N is metacyclic and therefored.U \N/ � 2: Hence d.U / � 4 and we are done.

We prove three important preliminary results.

Proposition 50.4. LetG be a 2-group with a metacyclic normal subgroup N . Supposethat N has a G-invariant four-subgroup N0 which is not contained in Z.G/ but thereis no G-invariant cyclic subgroup of order 4 contained in N . If N is abelian, it is oftype .2n; 2n/ or .2nC1; 2n/, where n � 1. IfN is nonabelian, it is minimal nonabelianand moreoverN D ha; b j a2m

D b2n

D 1; ab D a1C2m�1

i, where n � 2 andm D n

or m D nC 1.

Proof. If N is abelian (of rank 2), the result is clear since N cannot contain a charac-teristic cyclic subgroup of order 4. Suppose that N 0 ¤ 1. Since N 0 is cyclic, we musthave jN 0j D 2. By Example 10.14, N is minimal nonabelian. Then a result of Redei(Exercise 1.8a) implies that N D ha; b j a2m

D b2n

D 1; ab D a1C2m�1

; m �

2; n � 1i. If n D 1, then N Š D8 or N Š M2mC1 with m � 3. In both cases Nhas a characteristic cyclic subgroup of order 4, which is a contradiction. Hence wehave n � 2. It follows that N 0 D ha2m�1

i and Z.N / D ha2; b2i. If m < n, thenZ0 D hb2n�1

i D Ãn�2.Z.N // is a G-invariant subgroup of order 2. In that case,N0 D N 0 � Z0 lies in Z.G/, contrary to our assumption. Thus m � n. However, ifm > n C 1, then Ãn�1.Z.N // is a characteristic cyclic subgroup of N of order � 4,which is a contradiction. Thus, m D n or m D nC 1.

Remark 2. Let W0 be a normal subgroup of a group W and exp.W0/ D n. LetA � Aut.W / stabilize the chainW > W0 � f1g. Then exp.A/ divides n. Indeed, takew 2 W and � 2 A. Then w� D ww0 for some w0 2 W0 so w�n

D wwn0 D w. It

follows that �n D idW so exp.A/ divides n, as claimed.

Exercise 1. Let W D ha; b j a2n

D b2n

D Œa; b� D 1; n > 1i be abelian of type.2n; 2n/ and let Y be a subgroup of index 2 inW . ThenA D f� 2 Aut.W / j Y � D Y g

is of order 24n�3. In particular, if n D 2, jAj D 25.

Exercise 2. Let W be an abelian group of type .4; 4/ and A the stabilizer of the chainW > �1.W / > f1g. Then jAj D 24.


Exercise 3. Let W be abelian of type .4; 4/. Suppose that � 2 Aut.W / stabilizes thechain W > �1.W /. Then � stabilizes the chain W > �1.W / > f1g. (Hint. UseProposition 4.9.)

Proposition 50.5. The automorphism group Aut.W / of the group W D hu; y ju4 D

y4 D Œu; y� D 1i Š C4 � C4 is of order 25 3. The subgroup A of Aut.W / of allautomorphisms fixing the subgroup Y D hu2; yi Š C2 � C4 is of order 25 and so is aSylow 2-subgroup of Aut.W /. We have

A D h�; �; � j�4 D �4 D �2 D Œ�2; �� D 1; Œ�; �� D �2�2; Œ�; �� D �2 D Œ�; ��i:

where the automorphisms �; �; � are induced with: u� D u�1y, y� D u2y; u� D uy,y� D y; u D u�1, y D y. We have A0 D Z.A/ D ˆ.A/ D h�2; �2 D �i Š E4

and so A is a special 2-group. Set W0 D �1.W / D hu2; y2i. Then the stabilizer A0

of the chain W > W0 > 1 is elementary abelian of order 24 (Exercise 2 and Remark50.2) and A0 D h�; �� D �0; �; �2i D CA.W0/. The subgroup A0 contains the“special” subset S D f�2; �; �; ; �g of five automorphisms defined by:

u� D u; y� D yu2I u� D uu2y2; y� D yu2y2I

u D uu2y2; y D yu2I u D uy2; y D yu2y2:

If X is any maximal subgroup of A0, then X \ S is nonempty. Each 2 S hasthe property that does not invert any element of order 4 in W . In addition, the“superspecial” automorphisms and � have also the property that they do not fix(centralize) any element of order 4 in W . We have CA.Y / D h�; �2i Š E4 is normalin A and A=h�; �2i Š D8. In fact A is a splitting extension of h�; �2i by h��; �i Š

D8. Finally, if U is any subgroup of A covering A=h�; �2i (i.e. U h�; �2i D A), thenU does not normalize any of the six cyclic subgroups of order 4 in W .

Proof. The number of elements x, x0 of order 4 in W such that hx; x0i D W (by theproduct formula, fx; x0g is a basis ofW ) is 128 and so jAut.W /j D 25 3. By Exercise1, jAj D 25 and therefore A is a Sylow 2-subgroup of Aut.W /. The stabilizer A0 ofthe chain W > W0 > 1 is elementary abelian (Remark 50.2) of order 24 (Exercise2) and so jA W A0j D 2. Any automorphism from A � A0 acts faithfully on W=W0

(and so also on W0; see Exercise 3) and so we see that A0 D CA.W0/ (recall thatW0 D �1.W /). Defining the automorphisms �, � , �, � D �2, �0 D �� as above,we verify the defining relations for A. Since �, �0, �, �2 all lie in A0 and generate asubgroup of order 24, we get A0 D h�; �0; �; �2i. For any x 2 A � A0, x2 2 Z.A/which gives ˆ.A/ � Z.A/: We have h�2; �2i � ˆ.A/ � Z.A/ and so the four-subgroup h�2; �2i is normal in G. On the other hand, the relations for A show thatA=h�2; �2i is elementary abelian and h�2; �2i � A0. Hence h�2; �2i D ˆ.A/ D A0.Also, h�2; �2i � Z.A/ < A0 and we check that no element in A0 � h�2; �2i lies inZ.A/. Hence Z.A/ D A0 and so A is special.


Since CA.Y / � A0, so for each 2 CA.Y / we have u� D uw0 with w0 2 W0.It follows that jCA.Y /j � 4. On the other hand, h�; �2i � CA.Y / and so CA.Y / D

h�; �2i Š E4. Hence h�; �2i is normal in A and since A=h�; �2i acts faithfullyon Y Š C4 � C2, it follows that A=h�; �2i Š Aut.C4 � C2/ Š D8. Since �� is aninvolution and �� D ��1, it follows that h��; �i Š D8. Because h��; �i\h�; �2i D

1, A is a splitting extension of the four-group h�; �2i by h��; �i.LetU be any subgroup ofA coveringA=h�; �2i. Then we have U=.U \h�; �2i/ Š

D8 and so there exists an element ˛ of order 4 in U . Since ˛ 2 A � A0 (recall thatexp.A0/ D 2), ˛ acts faithfully on W=W0. But ˛ fixes Y D W0hyi (since Y is A-invariant) and so ˛ sends two cyclic subgroups of order 4 in W0hui onto two cyclicsubgroups of order 4 in W0huyi since W0hui is not A-invariant. It remains to beshown that U does not fix the cyclic subgroup hyi contained in Y . Suppose that Ufixes hyi so that jU W CU .y/j � 2. Sine c1.Y / D 3, we get jU W CU .u

2/j � 2 andso CU .Y / D CU .y/ \ CU .u

2/ has index � 4 in U . This is a contradiction, sincejU W .U \ h�; �2i/j D 8 (recall that CA.Y / D h�; �2i). We have proved that U doesnot fix any cyclic subgroup of order 4 in W .Let X be any maximal subgroup of A0. Suppose that S \ X is empty, where

S D fx1; : : : ; x5g is the “special” subset of automorphisms in A0 with x1 D �2,x2 D � , x3 D �, x4 D and x5 D �. Then we verify that the 10 products xixj .1 �

i < j � 5/ give 10 pairwise distinct elements in A0. But all these products lie inX . This is a contradiction since jX j D 23. Other statements about “special” and“extraspecial” elements in A0 are self-explanatory.

Proposition 50.6. Let a 2-group G have no normal elementary abelian subgroups oforder 8. Suppose that G is neither abelian nor of maximal class. The one of thefollowing holds:

(a) G has a normal abelian subgroup A of type .4; 2/ with CG.A/ abelian of type.2n; 2/, n � 2. In that case,G=CG.A/ is isomorphic to a subgroup of Aut.A/ Š

D8. If G=CG.A/ Š D8, then CG.A/ D A.

(b) G has a normal abelian subgroup W of type .4; 4/ with metacyclic CG.W /. Inthat case �2.CG.W // D W and G=CG.W / is isomorphic to a subgroup ofAut.W /, i.e., jG=CG.W /j � 25 (see Proposition 50.5).

Proof. Let A be a greatest normal noncyclic abelian subgroup of G of exponent � 4.Since G is not of maximal class, A is abelian of type .4; 2/ or .4; 4/. By Corollary10.2, �2.CG.A// D A. Then Theorem 41.1 implies that CG.A/ is metacyclic.Suppose that A Š C4 � C2. Then j�2.CG.A//j D 8 and Lemma 42.1 gives that

CG.A/ is abelian of type .2n; 2/, n � 2. Suppose in addition that G=CG.A/ Š D8.If CG.A/ > A, then CG.A/ contains a characteristic cyclic subgroup Z of order 4.We have Z < A and Z is normal in G. But G=CG.A/ Š Aut.A/ and Aut.A/ Š D8

contains an automorphism ˛ which permutes two cyclic subgroups of order 4 in A(Remark 50.3), a contradiction and so we must have CG.A/ D A.


Proof of Theorem 50.1. Let a 2-groupG have no normal elementary abelian subgroupsof order 8. Assume that G is neither abelian nor of maximal class. The starting pointis Proposition 50.6. If we are in case (a) of Proposition 50.6, then G has a normalabelian subgroup A of type .4; 2/ with N D CG.A/ abelian of type .2j ; 2/, j � 2. Inthat case G=N is isomorphic to a subgroup of D8 and if G=N Š D8, then N D A

and we are done.We assume that we are in case (b) of Proposition 50.6. Then G has a normal abelian

subgroup W D hu; y ju4 D y4 D Œu; y� D 1i Š C4 � C4 with metacyclic C D

CG.W / and �2.C / D W . Set W0 D �1.W / D hu2; y2i so that W0 is a normalfour-subgroup of G. Let Y be a normal subgroup of G such that W0 < Y < W .Then Y is abelian of type .4; 2/ and we may choose the generators u and y of W sothat Y D hu2; yi. Also set D D CG.Y / so that G=D is isomorphic to a subgroupof Aut.Y / Š D8. It follows from hy2i D Ã1.Y / that y2 2 Z.G/. Now, G=C isisomorphic to a subgroup of A D h�; �; �i which is the Sylow 2-subgroup of Aut.W /fixing Y . In what follows we use freely Proposition 50.5 about the structure of A andthe action on W together with the notation introduced there. It follows that D=C iselementary abelian of order � 4 since D=C induces on W a subgroup of h�; �2i Š

E4. We note that A=h�; �2i Š D8 andW0 � Z.G/ if and only if G=C induces onWa subgroup of A0 in which case G=C is elementary abelian of order � 24. If C D D,then G=C is isomorphic to a subgroup of D8. Suppose in addition that G=C Š D8.Then W0 6� Z.G/ and G=C induces on W a subgroup U Š D8 of automorphismswhich covers A=h�; �2i. By Proposition 50.5, there is no cyclicG-invariant subgroupof order 4 contained in C . Using Proposition 50.4, we see that the structure of thenormal metacyclic subgroup C is completely determined, as stated in Theorem 50.1.From now on, letD=C ¤ f1g. Suppose at first thatD=C contains a subgroup F=C

of order 2 such that F=C induces the automorphism �2 on W , where u�2

D uy2,y�2

D y and .uy/�2

D uyy2 . Since �2 does not invert any element of order 4 in W ,it follows that �2 does not invert any element in C �W0. Because �2 2 Z.A/, so F isnormal in G. We claim that �2.F / D W and so, by Theorem 41.1, F is metacyclic.If not, then �2.F / ¤ W and so �2.F / 6� C . There exists an element s 2 F � C

so that o.s/ � 4 which gives s2 2 W0. It is easy to see that there exist involutions in.W hsi/ � W . Indeed, we have ys D y, us D uy2. If s2 D 1, then we are finished.If s2 D y2, then .sy/2 D s2y2 D 1. If s2 D u2, then .syu/2 D s2.yu/syu D

u2 yuy2 yu D 1. If s2 D y2u2, then .su/2 D s2usu D y2u2 uy2 u D 1.Hence we may assume that s 2 F � C is an involution. For any c 2 C , sc is aninvolution if and only if .sc/2 D 1 or cs D c�1, i.e., s inverts c. But s inverts inC only the elements in W0. (Actually, s centralizes W0.) It follows that there areexactly 4 involutions in F � C and they all lie in the elementary abelian subgroupE D hsi � W0. Thus E is characteristic in F and so E is normal in G, which is acontradiction. We have proved that �2.F / D W and so F GG is metacyclic.Assume now that jD=C j D 2. If D=C induces the automorphism �2 on W , then

by the above, D D F is a normal metacyclic subgroup of G with �2.D/ D W .


If jG=Dj � 4, then we are done. It remains to consider the case where G=D Š

D8. In this case G=C induces on W a subgroup U of automorphisms which coversA=h�; �2i. In fact, we have in addition U \ h�; �2i D h�2i. Again by Proposition50.5, there is no cyclic G-invariant subgroup of order 4 contained in D. In this case,W0 6� Z.G/. By Proposition 50.4, the structure of the normal metacyclic subgroup Dis uniquely determined, as stated in Theorem 50.1.It remains to consider here (where jD=C j D 2) the case that D=C induces on

W the automorphism � or ��2. Note that neither � nor ��2 is central in A and soNA.h�i/ D A0 and NA.h��

2i/ D A0. Hence G=C induces on W a subgroup of A0

which does not contain �2. If jG=C j � 4, we are done. Hence we may assume thatG=C induces on W a maximal subgroup X of A0 which does not contain �2. In thatcaseG=C Š E8. By Proposition 50.5, X contains a “special” element 2 f�; �; �; g,where does not invert any element of order 4 in W .LetH=C be a subgroup of order 2 inG=C Š E8 so thatH=C induces the automor-

phism D � onW , where y� D yu2 and u� D u. If�2.H/ > W , then there is an el-ement s 2 H�C with s2 2 W0. By replacing s with sw for a suitablew 2 W , we mayassume that s is an involution. Indeed, if s2 D u2, then us D u and ys D yu2 implythat .su/2 D s2u2 D 1. If s2 D y2, then .syu/2 D s2.yu/syu D y2 yu2u yu D 1.If s2 D y2u2, then .sy/2 D s2.y/sy D y2u2 yu2 y D 1. Then sc (with c 2 C ) isan involution if and only if s inverts c and so there are exactly 4 involutions inH � C

and therefore they all lie in E D hsi �W0 Š E8. Hence E is characteristic in H andso E is normal in G, which is a contradiction. Thus �2.H/ D W and so H must bemetacyclic. Since G=H Š E4, we are done.Let K=C be a subgroup of order 2 in G=C Š E8 so that K=C induces the auto-

morphism D � on W , where y� D yu2y2 and u� D uu2y2. If �2.K/ > W , thenthere is an element s 2 K � C with s2 2 W0. By replacing s with sw for a suitablew 2 W , we may assume that s is an involution. If s2 D y2, then .su/2 D 1. Ifs2 D u2, then .sy/2 D 1. If s2 D y2u2, then .suy/2 D 1. Then sc (with c 2 C ) isan involution if and only if s inverts c and so there are exactly 4 involutions in K � C

and therefore they all lie in E D hsi �W0 Š E8. Hence E is characteristic in K andso E is normal in G, which is a contradiction. Thus �2.K/ D W and so K must bemetacyclic. Since G=K Š E4, we are done.Let L=C be a subgroup of order 2 in G=C Š E8 so that L=C induces a “super-

special” automorphism 2 f�; g on W . Then has the additional property that itdoes not fix (centralize) any element of order 4 in W . Suppose that �2.L/ D W sothat L is metacyclic. In that case an element l 2 L � C commutes with an element oforder 4 in W , which contradicts the above property of . Hence there is s 2 L � C

so that s2 2 W0. Again we may assume that s is an involution. If D �, thenus D uy2, ys D yu2y2. In this case, if s2 D u2, then .sy/2 D 1. If s2 D y2u2,then .su/2 D 1 and if s2 D y2, then .suy/2 D 1. If D , then us D uu2y2,ys D yu2. In this case, if s2 D u2y2, then .sy/2 D 1. If s2 D u2, then .suy/2 D 1

and if s2 D y2, then .su/2 D 1. It follows that L � C contains exactly 4 involutions


and so E D hsi � W0 Š E8 (being characteristic in L ) is normal in G, which is acontradiction. This finishes completely the case jD=C j D 2.We make here the following observation. If G=C Š E16 and G=C induces on

W the group A0 of automorphisms, then we get a contradiction. Indeed, we considera subgroup L=C of order 2 in G=C so that L=C induces the automorphism � onW . Then exactly the same proof as above shows that L has an elementary abeliansubgroup E of order 8 which contains all involutions in L and so E would be normalin G, which is a contradiction.It remains to consider the case whereD=C is a four-group inducing on W the four-

group h�; �2i. If C < F < D and F=C induces �2 on W , then we have alreadyproved that�2.F / D W and F is a normal metacyclic subgroup of G. If jG=Dj � 2,then jG=F j � 4 and we are finished. Hence we may assume that jG=Dj � 4 and soG induces on W either the whole group A (in which case G=D Š D8) or one of themaximal subgroups of A containing h�; �2i:

A. h�; �2; �; �0 D ��i D A0,

B. h�; �2; �; �i D h�i � h�; �i Š C2 � D8 with � D ��1,

C. h�; �2; �; �i, where �2 D � and Œ�; �� D �2.

Case A is impossible, by the previous paragraph.Suppose that we are in case B. We note that h�i is normal in h�; �2; �; �i. Let

K0 be a subgroup of G such that F < K0 < G and K0=C induces the cyclicgroup h�i Š C4 on W . Hence K0=C Š C4 and K0 is normal in G. Suppose that�2.K0/ > W . Since �2.F / D W , it follows that there is an element x 2 K0 � F sothat o.x/ � 4 and hxiC D K0. But then o.x2/ � 2 and x2 2 F and so x2 2 W0 � C .This is a contradiction sinceK0=C Š C4. Hence�2.K0/ D W and soK0 is a normalmetacyclic subgroup of G with jG=K0j D 4 and so we are finished in this case.Finally, we assume that we are in case C or G=D Š D8 (in which case G induces

the full group A onW ). In any case, there is an element k 2 G such that k induces theautomorphism � onW and so uk D u�1y and yk D yu2. Hence k does not normalizeany cyclic subgroup of order 4 in W . Also, W0 6� Z.G/. We know that F is a G-invariant nonabelian metacyclic subgroup with �2.F / D W . By Proposition 50.4, Fis minimal nonabelian and we may set: F D ha; b j a2m

D b2n

D 1; ab D a1C2m�1

i,where n � 2 and m D n or m D n C 1. If n � 3, then W � ha2; b2i D Z.F /.This is a contradiction since F=C induces the automorphism �2 on W . It followsthat n D 2. But we have F > C � W and so m D n C 1 D 3. In particular,W D C and so W is self-centralizing in G. Since CW .a/ D Y D hy; u2i, we havea2 2 Y � W0. Replacing y with some other element of order 4 in Y (if necessary),we may assume that a2 D y. The structure of F is uniquely determined. We haveF D ha; u j a8 D u4 D 1; au D aa4i andW D hu; yi, where y D a2 and CG.W / D

C D W . Set y2 D z so that hzi D Z.G/ because CW .k/ D hy2i. The element k2

induces the automorphism � D �2 on W and so k2 inverts W . Set S D F hki so thatS=F Š C4 and k4 2 C D W . Since h�2; �i is normal in A, so S is normal in G.


Since k4 2 CW .k/ D hzi, we have k4 D 1 or k4 D z. Because �2.F / D W , itfollows that all elements in F � W have the order 8. We shall determine the actionof k on F . We have ak D auiyj , where i , j are some integers and so uiyj is anarbitrary element in W . We get

.a2/k D yk D yu2 D .ak/2 D .auiyj /2 D auiyj auiyj

D a2.uiyj /auiyj D y.uz/iyjuiyj D yu2iy2j zi :

Thus u2 D u2iy2j zi and so .u2/1�i D ziCj . Hence 1�i 0 .mod 2/ and iCj 0

.mod 2/. This gives that we must have i j 1 .mod 2/. Let x be an element inG inducing the automorphism � on W . Then x normalizes S , x2 2 W , yx D y

and ux D u�1. Since .�2/ D �2 and � D ��2, we may set ax D aumyn andkx D kaupyq , wherem;n; p; q are some integers. We get

.a2/x D yx D y D .ax/2 D .aumyn/2 D aumyn aumyn

D a2.umyn/aumyn D y.uy2/mynumyn D yu2my2mC2n:

Thus u2my2mC2n D 1 and so m n 0 .mod 2/. Then the relation ak D

auiyj with i j 1 .mod 2/ should hold under the action of x, i.e. .ax/kx

D

ax.ux/i .yx/j . Since m and n are even and au D ay2 and so aup

D ay2p , this gives:

.aumyn/kaupyq

D aumynu�iyj D aum�iynCj D .auiyj .u�1y/mynu2n/aupyq

D .a.uy2/iyju�mymyn/upyq

D ay2puiy2iyju�mymyn

D aui�my2pC2iCj CmCn:

Hence um�i D ui�m and so u2i�2m D 1. Since m 0 .mod 2/, we get u2i D 1

and so i 0 .mod 2/, which contradicts the above statement.

Exercise 4. LetW be a homocyclic p-group of exponent 2n and rank d and let L be asubgroup of index 2 in W . Find the order of AutL.G/ D f� 2 Aut.W / j L� D Lg.

Exercise 5. Suppose that a 2-group G has a subgroup isomorphic to E8. If G hasno subgroups isomorphic to C2 � D8, it has an odd number of subgroups isomorphicto E8.

�51

2-groups with self centralizing subgroupisomorphic to E8

In ��48, 49 the 2-groups G have been described which have the property that theypossess an involution t such that CG.t/ D hti � C , where C is either a non-trivialcyclic group or a generalized quaternion group. It is natural to ask what happens if Cis isomorphic to one of the other two groups of maximal class, i.e., C is isomorphic toD2n (dihedral group) or to SD2n (semi-dihedral group) (Question 49.4). It is easy tosee that in that case the 2-group G has a self-centralizing elementary abelian subgroupE of order 8 but the problem of classifying 2-groups G which possess such a subgroupE is far more general.

In this section we classify 2-groups G which possess a self-centralizing elemen-tary abelian subgroup E of order 8. If E is normal in G, then G=E is isomorphicto a subgroup of GL.3; 2/ and therefore G=E is isomorphic to a subgroup of thedihedral group D8 of order 8. We determine here the unique 2-group G such thatE D ˆ.G/ (Theorem 51.3). There are exactly five 2-groups G, where E is normal inG and E � ˆ.G/. We shall present these groups in terms of generators and relations(Theorems 51.3 and 51.4). All these groups exist because we find some faithfultransitive permutation representation of degree 8 or 16 for these groups. We assumein the sequel that E is not normal in G. Then we have two essentially different possi-bilities according to E � ˆ.G/ or E 6� ˆ.G/, where ˆ.G/ is the Frattini subgroupof G.

Suppose that E � ˆ.G/. Then our first non-trivial result is that G has no normalelementary abelian subgroups of order 8 (Theorem 51.5). We are now in a positionto use the main theorem in �50 which allows us to describe the structure of G veryaccurately (Theorem 51.6).

Suppose that E 6� ˆ.G/. In that case, we see easily that for each involution t 2

E�ˆ.G/, we have CG.t/ D hti�M0, whereM0 is isomorphic to one of the followinggroups: E4, D2n .n � 3/, or SD2m .m � 4/. Then we try to pin down the structureofG by using the structure of CG.t/. Assume in addition thatG has normal elementaryabelian subgroups of order � 8. In that special case we see that M0 is isomorphicto E4 or D8 (Theorem 51.7). An exceptional case is treated in Theorem 51.8. Thegroup G could possess also a normal elementary abelian subgroup B of order 16 suchthat G=B Š D8 (Theorem 51.9). In all other cases, the structure of G is described

�51 2-groups with self centralizing subgroup isomorphic to E8 53

in Theorem 51.10. Finally, the example produces a group of order 28 (in terms ofgenerators and relations) which satisfies the assumptions of Theorem 51.10.It remains to consider the caseE 6� ˆ.G/ and G has no normal elementary abelian

subgroups of order 8. In this case Theorem 50.1 does not give a very precise infor-mation about the structure of G. Therefore, we use here the fusion arguments forinvolutions which have been introduced in ��48,49. In this way we show that G has inmost cases a large normal subgroup S with jG=S j � 4 and for the structure of S wehave exactly seven possibilities (Theorems 51.14 and 51.15). The exceptional casesare treated in Theorems 51.11–51.13.We use here the standard notation introduced in ��48. 49. In particular, for x 2 G

we denote with cclG.x/ the conjugacy class of x in G.For the sake of completeness, we give here some known results which will be used

often in this book.

Proposition 51.1. Let be an involutory automorphism acting on an elementary abe-lian 2-group A. Then we have jCA./j � jA W CA./j.

Proof. Consider the Jordan normal form of the linear transformation induced by onA (considered as a vector space over GF.2/).Then the result follows.

Proposition 51.2. Let be an involutory automorphism acting on an abelian 2-groupB so that CB./ D W0 � �1.B/. Then inverts Ã1.B/ and B=W0.

Proof. If x 2 B; then .xx� /� D x�x�2

D x�x D xx� . Hence, xx� D w0 with w0 2

W0 and so x� D x�1w0. This gives .x2/� D .x� /2 D .x�1w0/2 D x�2w0

2 D x�2

and our result follows.

1o. The case E � ˆ.G/. In this subsection we suppose that a 2-group G containsa self-centralizing elementary abelian subgroup E of order 8 which is contained inˆ.G/. Obviously, Z.G/ � E and also Z.ˆ.G// � E. If jZ.ˆ.G//j D 2 then, byProposition 1.13, we get that ˆ.G/ is cyclic, which is not the case. It follows that4 � jZ.ˆ.G//j � 8.We consider at first the possibility jZ.ˆ.G//j D 8 which gives that Z.ˆ.G// D

E and since CG.E/ D E, it follows that E D ˆ.G/ is a normal self-centralizingelementary abelian subgroup of order 8 ofG. On the other hand,G=E is an elementaryabelian subgroup of D8 and so G=E is a four-group. Thus jGj D 25. For eachx 2 G � E, we must have jCE .x/j D 4 (since x acts faithfully on E and hx;Ei isnot of maximal class according to Theorem 1.7, Propositions 1.8 and 51.1), CG.x/ D

hxiCE .x/ with x2 2 CE .x/. Otherwise, CG.x/ would cover G=E, which is notpossible since E D ˆ.G/. It follows that for each x 2 G � E, jcclG.x/j D 4 andso jG0j � 4. If G0 D E, then by the Taussky’s theorem (Proposition 1.6), G is ofmaximal class, which is not the case. Thus jG0j D 4. Suppose that jZ.G/j D 4.Then for each x 2 G � E we have CG.x/ D hxiZ.G/ which gives x2 2 Z.G/.But then E D ˆ.G/ D Ã1.G/ � Z.G/, a contradiction. Hence Z.G/ D hzi is of


order 2 and Z.G/ < G0. Since Z.G/ < G0 < E, we have ŒG;G0� D Z.G/, so Gis a 2-group of class 3. The group G=G0 is abelian of type .4; 2/ and so there is anelement x 2 G �E so that x2 D v 2 E � G0. Since x centralizes z and v D x2, andE D hv;G0i, it follows that x does not centralize G0 (otherwise x would centralizeE). Let y 2 G � .Ehxi/ which centralizes G0 so that y2 2 G0 and G D hx; yi. Notethat jG W CG.G

0/j D 2. If Œx; y� 2 hzi, then G=hzi is abelian, a contradiction. Hencewe may put Œx; y� D u, where G0 D hz; ui. From Œx; y� D u we get xy D xu and sovy D .x2/y D .xy/2 D .xu/2 D xuxu D x2.x�1ux/u D v.uz/u D vz. Hence weget zy D z, uy D u, vy D vz, xy D xu, x2 D v, zx D z, ux D uz, vx D v. Therelation Œx; y� D u gives also yx D yu D uy since y centralizesu 2 G0. From this weget .y2/x D .yx/2 D .uy/2 D y2. Hence x centralizes y2 2 G0. But x acts faithfullyon G0 and so we must have y2 2 hzi. Suppose now that y2 D z. Then we replace ywith y0 D yv. We get .y0/2 D .yv/2 D yvyv D y2vyv D z.vz/v D 1 and otherrelations remain unchanged zy0

D z, uy0

D u, vy0

D vz, xy0

D xyv D .xu/v D xu,and so we may assume from the start that y2 D 1. The structure of G is uniquelydetermined. It is easy to see that this group G is isomorphic to a subgroup of A8. Itis enough to set x D .2; 7; 6; 8/.4; 5/, y D .1; 2/.3; 6/.4; 7/.5; 8/, and we computefurther permutations x2 D v D .2; 6/.7; 8/, Œv; y� D z D .1; 3/.2; 6/.4; 5/.7; 8/,Œx; y� D u D .1; 4/.2; 7/.3; 5/.6; 8/. Then we simply check that all above relations aresatisfied for these permutations. Since z is represented with a non-trivial permutation,so the induced permutation representation of degree 8 is faithful. Note that all abovepermutations are even. Hence our group G exists and is isomorphic to a subgroup ofA8. We have proved the following result.

Theorem 51.3. Suppose that a 2-group G contains a self-centralizing elementaryabelian subgroup E of order 8 and E D ˆ.G/. Then G is uniquely determined andwe have G D hx; y; z; u; v jx4 D y2 D z2 D u2 D v2 D Œz; u� D Œz; v� D Œz; x� D

Œz; y� D Œu; v� D Œu; y� D Œv; x� D 1; x2 D v; vy D vz; xy D xu; ux D uzi. HereE D hz; u; vi is a self-centralizing normal elementary abelian subgroup of order 8 ofG, E D ˆ.G/, G D hx; yi is of order 25, G0 D hz; ui, and Z.G/ D ŒG;G0� D hzi isof order 2. The group G exists and is isomorphic to a subgroup of A8.

It remains to consider the second possibility jZ.ˆ.G//j D 4. We set Z.ˆ.G// D

W0 so that W0 is a G-invariant four-group contained in E. Therefore the subgroupT D CG.W0/ is of index � 2 in G. Since CG.E/ D E, so for each t 2 E � W0

we have CT .t/ D E. Assume that T D G, in which case W0 D Z.G/. Let A be aG-invariant subgroup so that W0 < A � ˆ.G/ and jA W W0j D 2. Then A is abelianof order 8 and G stabilizes the chain A > W0 > 1 so that G=CG.A/ is elementaryabelian (of order � 4/. In particular, ˆ.G/ � CG.A/ and so A � Z.ˆ.G//, which isa contradiction. It follows that jG W T j D 2 and so Z.G/ D hzi is of order 2 and wesetW0 D hz; ui. For each x 2 G � T , ux D uz. Since jˆ.G/j � 24, so jGj � 26.Assume now in addition that E is normal in G. Since G=E is isomorphic to a

subgroup of D8 and jGj � 26, so jGj D 26 and G=E Š D8. The subgroup T


stabilizes the chain E > W0 > 1 and jG W T j D 2, so T=E is a 4-group. We haveZ.T / D W0. For each x 2 T � E, x2 2 W0 since x acts non-trivially on E. Thusˆ.T / D Ã1.T / � W0 and so T 0 � W0. On the other hand, if t 2 E � W0, thenCT .t/ D E implies that jcclT .t/j D 4 and so cclT .t/ D E � W0 and jT 0j � 4.It follows that Z.T / D T 0 D ˆ.T / D W0 and therefore T is a special 2-group oforder 25. Let x; y 2 T � E be such that Ex and Ey are two distinct elements inT=E Š E4. Then M D W0hx; yi is a maximal subgroup of T not containing E.If M is nonabelian, then Œx; y� D v is an involution in W0. Let t 2 E � W0. Weknow that CT .t/ D E and so Œx; t �, Œy; t �, and Œxy; t � D Œx; t �Œy; t � are three distinctinvolutions in W0. Note that T is of class 2. If, for example, Œx; t � D Œy; t �, thenŒxy; t � D Œx; t �Œy; t � D 1, a contradiction. Therefore, we may assume that Œx; t � D v.It follows Œx; ty� D Œx; t �Œx; y� D v2 D 1. Hence, replacing y with ty 2 Ey, wemay assume from the start that Œx; y� D 1 and so M D W0hx; yi is abelian. Weclaim that M is the unique abelian maximal subgroup of T . Indeed, if M0 ¤ M isanother abelian maximal subgroup of T , thenM \M0 � Z.T / and jM \M0j D 23,a contradiction. Since T D CG.W0/ and Z.ˆ.G// D W0, so T is a characteristicsubgroup of G. HenceM is also a characteristic subgroup of G. Since G=E Š D8

and D8 has five involutions, there is an element s 2 G � T such that s2 2 E. Notethat s normalizes M and so if s2 2 W0 � M , then M hsi is a maximal subgroup ofG which does not contain E, a contradiction. Thus, we must have s2 D t 2 E �W0.Since G=E Š D8, s acts non-trivially on M=W0 Š T=E Š E4. Hence, there isan element x 2 M � W0 such that y D xs has the property M D W0hx; yi. Also,x�1y D x�1xs D Œx; s� 2 ˆ.G/ and so x2.x�1y/ D xy 2 ˆ.G/. Hence ˆ.G/ D

Ehxyi and G D hs; xi. Now, ys D xs2

D xt D xu, where u 2 W0 � hzi. Indeed, ifxt D txt D xz, then we act on this relation with s. We get t sxs ts D xszs which givestyt D yz and then .xy/t D xzyz D xy. But this contradicts the fact that CT .t/ D E

and so we must have ys D xu, with u 2 W0 � hzi. Also note that us D uz since sacts non-trivially on W0. It remains to determine x2 and y2. There are exactly threepossibilities for the structure of the abelian groupM .(a) �1.M/ D M0 is of order 23. Since M0 is normal in G, so x and y D xs are

elements of order 4 contained in M � M0. On the other hand, jÃ1.M/j D 2 and soÃ1.M/ D hzi D Z.G/. This gives x2 D y2 D z. If we set

s D .2; 5; 8; 12/.4; 6; 10; 11/.7; 9/.13; 15; 14; 16/;

x D .1; 2; 3; 4/.5; 13; 6; 14/.7; 8; 9; 10/.11; 15; 12; 16/;

then we see (in the same way as in the proof of Theorem 51.3) that G exists in thiscase and is isomorphic to a subgroup of A16.(b) �1.M/ D M is of order 24. Here we have x2 D y2 D 1. If we set s D

.1; 4; 3; 5/.2; 7; 6; 8/, x D .4; 7/, then we see easily that G exists in this case as asubgroup of S8.(c) �1.M/ D W0 is of order 22 and so M Š C4 � C4. Here we have Ã1.M/ D

W0 D hx2; y2i and hx; xsi D M so that we have two possibilities for x2 and y2.


(c1) If x2 D u, then y2 D uz and so ys D xu D x�1. This gives yt D ys2

D

.x�1/s D y�1 and xt D xs2

D ys D x�1 and so t inverts M . If we set s D

.1; 5/.2; 8; 4; 7/.3; 6/, x D .5; 7; 6; 8/, then we see easily that G exists in this case asa subgroup of S8.

(c2) If x2 D uz, then y2 D u and then ys D xu D .xuz/z D x�1z. This givesyt D ys2

D .x�1z/s D y�1z and xt D xs2

D ys D x�1z and so t does not invertsM . If we set

s D .2; 5; 12; 10/.3; 6/.4; 7; 14; 15/.9; 13; 16; 11/;

x D .1; 2; 3; 4/.5; 9; 10; 11/.6; 12; 8; 14/.7; 13; 15; 16/;

then we see as before that G exists in this case as a subgroup of A16.

We have proved the following result.

Theorem 51.4. Suppose that a 2-group G contains a self-centralizing elementary abe-lian subgroup E of order 8, E < ˆ.G/, and E is normal in G. Then we have G D

hx; y; u; z; t; si with the following initial relations

u2 D z2 D t2 D s4 D Œx; y� D Œx; u� D Œx; z� D Œy; u� D Œy; z� D Œu; z�

D Œu; t � D Œz; t � D Œz; s� D 1; s2 D t; us D uz; xs D y; ys D xu:

HereE D ht; z; ui is a self-centralizing elementary abelian normal subgroup of order8 in G, G=E Š D8, Z.G/ D hzi is of order 2, G is of order 26, ˆ.G/ D Ehxyi

and G D hs; xi. Also, Z.ˆ.G// D W0 D hz; ui is a four-group, M D hx; y; z; ui

is a characteristic abelian subgroup of G of order 24 and exponent at most 4 andM is the unique abelian maximal subgroup of T D CG.W0/, jG W T j D 2 andT=E Š E4. There are exactly four possibilities for the structure of G (with twoadditional relations).

(a) x2 D y2 D z, M is abelian of type .4; 2; 2/, and the group G is in this caseisomorphic to a subgroup of A16.

(b) x2 D y2 D 1,M is abelian of type .2; 2; 2; 2/, and the group G is in this caseisomorphic to a subgroup of S8.

(c1) x2 D u, y2 D uz,M Š C4 �C4, t invertsM , and the group G is in this caseisomorphic to a subgroup of S8.

(c2) x2 D uz, y2 D u,M Š C4 � C4, t does not invert onM , and the group G isin this case isomorphic to a subgroup of A16.

In what follows, we assume that E is not normal in G. We recall that W0 D

Z.ˆ.G// is a four-group and jG W T j D 2, where T D CG.W0/. We set W0 D hz; ui

and we know that Z.G/ D hzi is of order 2. Finally, for each involution t 2 E �W0,CT .t/ D E and jGj � 26 since jˆ.G/j � 24. Our aim is to prove that G does notpossess in this case a normal elementary abelian subgroup of order 8.


Suppose that A1 is a normal elementary abelian subgroup of order 16 in T . Lett 2 E �W0. No involution in E �W0 could be contained in A1 and, by Proposition51.1, jCA1

.t/j � 22. Hence, jCA1.t/j D 4 and CA1

.t/ D W0 so that E \ A1 D W0.Set C D EA1 and so jC j D 25 with jC W A1j D 2. Take any a 2 A1. Then ta is aninvolution if and only if .ta/2 D 1 or equivalently at D a�1 D a. It follows that allinvolutions in C � A1 lie in E �W0. Since hE �W0i D E, we see that E is normalin NG.C /. If C D T , then E would be normal in G, a contradiction. Let D=C be asubgroup of order 2 in NT .C /=C . ThenE is normal inD and note that all involutionsin E�W0 lie in a single conjugacy class in C since CG.t/ D E and jC W Ej D 4. Butthen CD.t/ 6� C , a contradiction. We have proved that T has no normal elementaryabelian subgroups of order 16.Let A2 be a normal elementary abelian subgroup of order 16 in G. By the previous

paragraph, A2 6� T and so B2 D A2 \ T is a normal elementary abelian subgroupof order 8 in G. If W0 6� B2, then W0B2 is a normal elementary abelian subgroup oforder � 16 contained in T , a contradiction. Hence we must have W0 � B2. But thenCG.W0/ � hT;A2i D G, which is a contradiction. We have proved that G has nonormal elementary abelian subgroups of order 16.Suppose that A0 is a normal elementary abelian subgroup of order 8 in G but

A0 6� T . Then QA0 D A0 \ T is a G-invariant 4-subgroup. If W0 D QA0, thenCG.W0/ � hA0; T i D G, a contradiction. Hence W0 ¤ QA0 and so QA0W0 is a G-invariant elementary abelian subgroup of order 8 contained in T . We have provedthat if G has a normal subgroup isomorphic to E8, then G has also such one which iscontained in T .Now assume that G has a normal subgroup A Š E8. By the previous paragraph,

we may assume that A � T . Since E is not normal in G, so we get E ¤ A. IfW0 6� A, then W0A would be a G-invariant elementary abelian subgroup of order� 16, a contradiction. Hence W0 � A and so E \ A D W0. It is well known thatthere is a self-centralizing normal abelian subgroup B of T such that A � B and Bis normal in G. We have �1.B/ D A and E \ B D E \ A D W0. Since G=T actsfaithfully on W0 (and so on B), B is also self-centralizing in G. For any x 2 B , tx isan involution if and only if t inverts x.Suppose thatB D A. In this special case,A is a self-centralizing normal elementary

abelian subgroup of order 8 of G and soG=A Š D8 since jGj � 26. Set V0 D EA andnote that V0 � A contains exactly four involutions and they are contained in E �W0.Hence E is normal in NG.V0/ and so V0=A does not lie in Z.G=A/ since E is notnormal in G. Let R=A be the cyclic subgroup of index 2 in G=A so that we haveR \ V0 D A and R is a maximal subgroup of G. Hence E 6� R and this is acontradiction since E � ˆ.G/. We have proved that B ¤ A and so jBj � 24.We shall determine now the structure of B . If an involution e 2 A � W0 would

be a square in B , then t would centralize e, since t inverts on Ã1.B/ (Proposition51.2). This is a contradiction. Suppose that all three involutions in W0 are squares inB . Then �2.B/ D hb1i � hb2i � heiq, where o.b1/ D o.b2/ D 4, e is an involution


in A �W0, and W0 D hb21; b

22i. We act with t 2 E �W0 on �2.B/. By Proposition

51.2, bt1 D b�1

1 w with w 2 W0. This gives bt1 D b1.b

21w/, where b

21w D w1 2 W0.

Similarly, bt2 D b2w2 with w2 2 W0 and so .b1b2/

t D .b1b2/.w1w2/. Here w1; w2,and w3 D w1w2 must be three pairwise distinct involutions in W0 since C�2.B/.t/ D

hb21; b

22i D W0. But then et D ewj with j 2 f1; 2; 3g and so t centralizes one of

the elements b1e, b2e, or b1b2e of order 4, which is a contradiction. We have provedthat at most one involution in W0 is a square in B . Hence, we have obtained thefollowing result. The subgroup B .> A/ is abelian of type .2m; 2; 2/, m � 2, and soB D hbi � hwi � hei, where o.b/ D 2m � 4, b2m�1

D z, hzi D Z.G/ D Ãm�1.B/,W0 D hz;wi, e 2 A �W0.

Set V D EB D htiB , where t is an involution in E � W0 D E � B . Our aimis to show that V D T . Therefore, we assume V ¤ T and set QV D NT .V / so thatj QV =V j � 2. We note that jBj D 2mC2 .m � 2/ and so V �B is a normal subset of QV

and jV �Bj D 2mC2 D jtBj. If j QV =V j � 4, then j QV W C QV.t/j D j QV W Ej � 2mC2 and

so j QV =V j D 4 and all elements in V � B are involutions since they are all conjugatein QV to t . This is not the case since te is not an involution in view of Œt; e� ¤ 1.Hence we must have j QV =V j D 2. Assume for a moment that QV D NG.V /. ByTheorem 1.7 and Proposition 1.8, G=B is of maximal class and V=B is a non-centralsubgroup of order 2 in G=B . In fact G=B must be dihedral of order � 8 or semi-dihedral. Let R�=B be the cyclic subgroup of index 2 in G=B . Then jG W R�j D 2

and R� \ V D B . But then E 6� R� and so E 6� ˆ.G/, a contradiction. Hence wemust have NG.V / > QV . We set V � D NG.V / so that jV � W QV j D 2 and V �T D G.If CV �.t/ D E, then we see (since jV � W Ej D 2mC2 ) that all 2mC2 elements inthe normal subset V � B in V � are conjugate to t and so they are all involutions. Butthis is not the case, since te (as above) is not an involution. Thus CV �.t/ D QE isof order 24, j QE W Ej D 2 and QE \ T D E. Hence all elements y 2 QE � E lie inG � T , y2 2 E and wy D wz, where W0 D hz;wi. Suppose at first that there isy 2 QE � E with y2 2 W0. In this case W0hyi Š D8 and since there are involutionsin .W0hyi/ � W0, we may assume that y2 D 1. We act with the involution y onA. Since CW0

.y/ D hzi and jCA.y/j � 4 (Proposition 51.1), we see that there isan element e 2 A � W0 so that Œy; e� D 1. We have 1 ¤ Œt; e� 2 W0 and sinceŒt; e�y D Œty; ey � D Œt; e�, so Œt; e� D z which gives et D ez. Set v D b2m�2

so thathvi is the cyclic subgroup of order 4 in hbi. Since et D ez, we must have vt D vw foran element w 2 W0 � hzi. Otherwise, t would centralize ve, which is not possible. Inparticular, t does not invert v, and so v 62 Ã1.B/. Recall that, by Proposition 51.2, tinverts Ã1.B/. But this forces that o.b/ D 4; hvi D hbi; jBj D 24, and E is normalin V . The last statement follows from the fact that there are only four involutionsin V � B (since t inverts in B only the involutions in W0) and they lie in E � W0.Suppose now that for each y 2 QE � E, y2 2 E � W0 and so we may assume inthis case that y2 D t 2 E � W0. The cyclic group hyi of order 4 acts faithfully onA and hyi acts fixed-point-free on the set A � W0. If ey D ez with e 2 A � W0,then ey2

D .ez/y D ezz D e and so et D e, a contradiction. Hence, ey D ew


with w 2 W0 � hzi. This gives ey2

D et D .ew/y D ewwz D ez. It followsthat vt D vw1 with some w1 2 W0 � hzi, where again hvi is the cyclic subgroup oforder 4 in hbi. Otherwise, t would centralize ve. But this means that t does not inverthvi. As before, this forces that o.b/ D 4; hbi D hvi, jBj D 24, and E is normal inV . Again, the last statement follows from the fact that there are only four involutionsin V � B (since t inverts in B only the involutions in W0) and these lie in E � W0.In both cases for the structure of QE D hyiE, we have jBj D 24, j QV j D 26, and QV

normalizes the subset V � B containing exactly four involutions. On the other hand,j QV W C QV

.t/j D j QV W Ej D 23 which means that V � B contains eight conjugates of t ,which is the final contradiction in this paragraph. We have proved that we must haveV=T and so T D htiB , where t 2 E �W0.

Set again v D b2m�2

so that o.v/ D 4 and v2 D z. Since t 2 ˆ.G/ and ˆ.G/ D

Ã1.G/, there is an element x 2 G � T such that x2 D t 0 2 T � B . Because t 0 D l t

with l 2 B , we have for all y 2 B , yt 0

D ylt D yt and .t 0/2 2 B . Hence xinduces an automorphism of order 4 on �2.B/ D Ahvi since x2 induces the sameinvolutory automorphism on �2.B/ as the automorphism induced by t . Since t actsfixed-point-free on A�W0, so x induces a 4-cycle on A�W0 and so ex D ew, wherew 2 W0 �hzi. Indeed, if ex D ez, then et D et 0

D ex2

D .ez/x D ezz D e, which isa contradiction. From ex D ew follows ex2

D et 0

D et D .ew/x D .ew/.wz/ D ez.But then we must have vt D vw0 with w0 2 W0 � hzi. Namely, if vt D vz D v�1,then .ve/t D .vz/.ez/ D ve, a contradiction. It follows from vt D vw0 that t doesnot invert v. By Proposition 51.2, t inverts Ã1.B/. Hence v is not a square in B andso �2.B/ D B is of order 24. Since t inverts in B only the elements in W0, so thereare exactly four involutions in tB and they lie in E �W0. It follows that Ex D E andso E is normal in G. This is the final contradiction.We have proved the following major result.

Theorem 51.5. Suppose that a 2-group G contains a self-centralizing elementary abe-lian subgroup E of order 8, E � ˆ.G/, and E is not normal in G. Then G has nonormal elementary abelian subgroups of order 8.

In the rest of this subsection, we study the structure of a 2-group G satisfying theassumptions of Theorem 51.5.We recall that W0 D Z.ˆ.G// is a normal four-subgroup contained in E and jG W

T j D 2, where T D CG.W0/. We set W0 D hz; ui and we know that Z.G/ D hzi isof order 2. Finally, for each involution t 2 E �W0, CT .t/ D E and jGj � 26. SinceG has no normal elementary abelian subgroups of order 8, we may apply Theorems50.1 and 50.2.Suppose that G has two distinct normal 4-subgroups. Then Theorem 50.2 implies

that G D D � C , where D Š D8, D \ C D Z.D/ and C is either cyclic or ofmaximal class but not isomorphic to D8. We get ˆ.G/ � C and so E � C , which isnot possible. We have proved that W0 is the unique normal four-subgroup of G.


Since G is neither abelian nor of maximal class, we may apply Proposition 50.1.It follows that G has a normal metacyclic subgroup N such that CG.�2.N // � N ,G=N is isomorphic to a subgroup of D8 and W D �2.N / is abelian of type .4; 2/or .4; 4/. In any case, W0 D �1.N / is the unique normal four-subgroup of G andW0 6� Z.G/. It follows that E \ N D W0. Since E � ˆ.G/, G=N is not elementaryabelian. Hence G=N is either cyclic of order 4 or G=N Š D8.Assume that G=N Š C4. If N is abelian of type .2j ; 2/, j � 2, then G=N

acts faithfully on �2.N / Š C4 � C2. If, in addition, N > �2.N /, then thereis a characteristic cyclic subgroup Z of order 4 contained in �2.N / so that Z isnormal in G. But then acting with G=N on Z, we see that t 2 E � W0 central-izes Z � T D CG.W0/, which is a contradiction. Thus, N D �2.N / and sojGj D 25, which is again a contradiction. It follows that �2.N / D W is abelianof type .4; 4/. We want to show that N does not possess a G-invariant cyclic sub-group Z of order 4. Suppose that there is such Z so that Z � W � T . ButjG W CG.Z/j � 2 and so E centralizes Z since E � ˆ.G/, a contradiction. Itfollows that we may use Proposition 50.4. Hence in case G=N Š C4, N is eitherabelian of type .2n; 2n/ or .2nC1; 2n/ with n � 2 or N is minimal nonabelian andmore precisely N D ha; b j a2m

D b2n

D 1; ab D a1C2m�1

i, where m D n withn � 3 or m D nC 1 with n � 2.Assume that G=N Š D8. Then the structure of N is determined by Theorem 50.1.

We shall show here that the minimal case with N Š C4 � C2 cannot occur. Indeed,suppose that N is abelian of type .4; 2/. Let L=N D Z.G=N/ so that jL=N j D 2 andE � L since E � ˆ.G/. Set W0 D �1.N / so that W0 D Z.ˆ.G//, W0 < E andsince E < ˆ.G/, we get ˆ.G/ D L. By the structure of G=N Š D8 Š Aut.N /,we know that t 2 E � W0 inverts N . Thus, all elements in L � N are involutions.Set N D hy;w jy4 D w2 D Œy;w� D 1; y2 D zi so that W0 D hz;wi, E D

ht; z;wi with hzi D Z.G/. Let k be any element in G such that hkiN=N Š C4. SetK D hkiN . It follows that k2 2 L � N and since k2 is an involution, so k4 D 1.By the structure of Aut.N / Š D8, we have yk D yw, wk D wz. Then we see thatÃ1.K/ D ˆ.K/ D hk2; w; zi. Hence the elementary abelian subgroup hk2; w; zi oforder 8 is normal in G. But then the other elementary abelian subgroup hk2y;w; zi

of order 8 in K must be also normal in G. Note that K has exactly two elementaryabelian subgroups of order 8 since there are exactly eight involutions in L � N . Inparticular, E is normal in G, a contradiction. This proves that the case G=N Š D8

with N Š C4 � C2 cannot occur.We have proved the following final result of this subsection.

Theorem 51.6. Let G be a 2-group which has a self-centralizing elementary abeliansubgroup E of order 8. Assume that E � ˆ.G/ and E is not normal in G. Then thegroup G has the following properties.

(a) G has no normal elementary abelian subgroups of order 8.

(b) G has the unique normal four-subgroup W0 and W0 < E.


(c) G has a normal metacyclic subgroup N such that �2.N / D W is abelian oftype .4; 4/, CG.W / � N , �1.W / D W0, and G=N is either cyclic of order 4 orG=N Š D8.

(d) N is either abelian of type .2k ; 2kC1/ or .2k ; 2k/, k � 2, or N is minimalnonabelian and more precisely N D ha; b j a2m

D b2n

D 1; ab D a1C2m�1

i,where m D n with n � 3 or m D nC 1 with n � 2.

2o. The case E 6� ˆ.G/ and G has a normal elementary abelian subgroup oforder � 8. We suppose throughout this subsection that our 2-group G contains aself-centralizing non-normal elementary abelian subgroup E of order 8 such that E 6�

ˆ.G/. (If E is normal in G, then G=E is isomorphic to a subgroup of D8.) Also, weassume that G has a normal elementary abelian subgroup of order � 8.Let t 2 E � ˆ.G/ and letM be a maximal subgroup of G such that t 62 M . Then

CG.t/ D hti � CM .t/. But M0 D CM .t/ contains the four-subgroup E0 D E \M

and CM0.E0/ D E0. If E0 ¤ M0, then Theorem 1.7 and Proposition 1.8 imply that

M0 Š D2n orM0 Š SD2n . Let A be a normal elementary abelian subgroup of order8 in G. If t 2 A, then A is a normal subgroup of CG.t/. If t 62 A, then Proposition51.1 implies that CA.t/ D A0 is of order � 4 and so hti � A0 is a normal elementaryabelian subgroup of order � 8 of CG.t/. In any case, M0 has a normal 4-subgroupand so E0 ¤ M0 implies thatM0 Š D8.We have proved the following result.

Theorem 51.7. Let G be a 2-group which has a self-centralizing elementary abeliansubgroup E of order 8 which is not contained in ˆ.G/. Assume that G has a normalelementary abelian subgroup of order � 8. Then for each t 2 E � ˆ.G/, we haveeither CG.t/ D E or CG.t/ D hti �M0, whereM0 Š D8.

Suppose that there is an involution t 2 E � ˆ.G/ such that CG.t/ contains anelementary abelian G-invariant subgroup A of order 8. Since E is not normal in G,so E ¤ A and therefore C D CG.t/ D hE;Ai D hti � D, where D Š D8. Thesubgroup C has exactly two elementary abelian subgroups of order 8 and so they areA and E. Also, t 2 A \ E. Since CC .A/ D A, so A is self-centralizing in G andtherefore G=A is isomorphic to a subgroup of D8. Obviously, E is normal in NG.C /.But E is not normal in G and so NG.C / ¤ G. This forces that jG W C j � 4 andtherefore G=A Š D8.We have proved the following result which describes an exceptional case.

Theorem 51.8. Let G be a 2-group which has a self-centralizing non-normal elemen-tary abelian subgroup E of order 8 which is not contained in ˆ.G/. Suppose thatthere is an involution t 2 E � ˆ.G/ such that CG.t/ contains an elementary abelianG-invariant subgroup A of order 8. Then A is self-centralizing in G and G=A Š D8.

We assume in the rest of this subsection that there is no involution t 2 E � ˆ.G/

such that CG.t/ contains an elementary abelian G-invariant subgroup of order 8.


Suppose that G has a normal elementary abelian subgroup B of order � 16. Lett be an involution in E � ˆ.G/. By the structure of CG.t/, we have jCB.t/j � 4.Then Proposition 51.1 forces that jBj D 16 and CB.t/ D W0 is of order 4. Let Cbe a maximal normal abelian subgroup of G containing B . Then B D �1.C / and,by the structure of CG.t/, we have CC .t/ D CB.t/ D W0. Also, G=C acts faithfullyon C . Set W0 D hz1; z2i and let W1 D hu1; u2i be a complement of W0 in B . Setz3 D z1z2 and u3 D u1u2. We claim that we may choose the generators of W0 andW1 in such a way that ut

i D uizi for i D 1; 2; 3. Indeed, suppose that there exist twoelements u ¤ u0 2 W1 so that ut D uz and .u0/t D u0z, where 1 ¤ z 2 W0. Thenuu0 ¤ 1 and .uu0/t D .uz/.u0z/ D uu0, which contradicts the fact that CC .t/ D W0.Hence ut D uz and .u0/t D u0z0, where z; z0 are distinct involutions in W0. Thisgives .uu0/t D .uu0/.zz0/, where uu0 ¤ 1 and zz0 ¤ 1. It remains to set u1 D u,u2 D u0, u3 D uu0, z1 D z, z2 D z0, z3 D zz0, and we see that the above claim isproved. Suppose that C ¤ B . Let v be an element of order 4 contained in C � B .Then v2 2 B and, by Proposition 51.2, .v2/t D v�2 D v2 and so v2 2 W0. ByProposition 51.2, vt D v�1w0 with w0 2 W0 and so vt D v.v2w0/ D vzj , wherej 2 f1; 2; 3g. But then .vuj /

t D vzjuj zj D vuj , which is a contradiction. We haveproved that C D B is a self-centralizing normal elementary abelian subgroup of order16 of G. Set E1 D hti � W0 and L D htiB . Then all elements in L � .E1 [ B/

have order 4 and four involutions in E1 � W0 form a single conjugacy class in Lsince jL W CL.t/j D 4. It follows L0 D Z.L/ D ˆ.L/ D W0. If L D G, thenCG.t/ D E1 D E is normal in G. This is a contradiction and so L ¤ G. SetK D NG.L/. ThenK normalizes theL-classE1�W0. Hence CK.t/must coverK=L.Since N D CK.t/ D CG.t/ Š C2 � D8, so jK=Lj D 2 and N \ L D E1. We haveNG.E1/ � hN;Li D K and so K ¤ G by our last assumption. Indeed, if K D G,then N D CG.t/ would contain the G-invariant subgroup E1. Now, Theorem 1.7 andProposition 1.8 imply that G=B is of maximal class (acting faithfully on B). Also,L=B is a non-central subgroup of order 2 in G=B and so G=B is not a generalizedquaternion group. On the other hand, GL.4; 2/ does not possess elements of order 8and so jG=Bj D 8 and consequently G=B Š D8. We have Z.G/ � CG.t/ D N andso Z.G/ � Z.N / D ht; z0i, where hz0i D N 0 < W0. Hence Z.G/ � Z.L/ D W0 andso Z.G/ D ht; z0i \W0 D hz0i is of order 2.We have proved the following result.

Theorem 51.9. Let G be a 2-group which has a self-centralizing non-normal elemen-tary abelian subgroup E of order 8 which is not contained in ˆ.G/. Assume thatthere is no involution t 2 E � ˆ.G/ such that CG.t/ contains an elementary abelianG-invariant subgroup of order 8. Suppose that G has a normal elementary abeliansubgroup B of order � 16. Then jBj D 16, B is self-centralizing in G, G=B Š D8,and Z.G/ is of order 2.

We suppose also in the rest of this subsection that G has no normal elementaryabelian subgroups of order � 16.


LetA be a normal elementary abelian subgroup of order 8 inG. LetB be a maximalnormal abelian subgroup of G containing A. Then G=B acts faithfully on B and wehave �1.B/ D A. Let t be any involution in E � ˆ.G/. By our assumptions, CG.t/

does not contain a G-invariant elementary abelian subgroup of order 8. Hence t 62 B .By Theorem 51.7, we have either CG.t/ D E or CG.t/ D hti �D, where D Š D8.By Proposition 51.1, jCA.t/j � 4 and so CB.t/ D CA.t/ D W0 is a four-group.Assume that B D A. Then G=A is isomorphic to a subgroup of D8. Set htiA D L

and hti �W0 D E1. All elements in L� .E1 [A/ are of order 4 and so E1 is normalin NG.L/ since hE1 � W0i D E1. If L is normal in G, then E1 is normal in G.But then CG.t/ contains a G-invariant elementary abelian subgroup of order 8, whichcontradicts our assumptions. It follows that L=A is a non-central subgroup in G=Aand so G=A Š D8. In what follows we shall assume that B ¤ A.By Proposition 51.2, no element in A � W0 is a square in B . Assume that all

three involutions in W0 are squares in B . Then �2.B/ D hb1i � hb2i � hei, whereo.b1/ D o.b2/ D 4, e 2 A � W0, and W0 D hb2

1 ; b22i. By Proposition 51.2, bt

1 D

.b1/�1w0 with w0 2 W0 and so bt

1 D b1.b21w0/, where 1 ¤ w1 D b2

1w0 2 W0.Similarly, bt

2 D b2w2 with 1 ¤ w2 2 W0 and so .b1b2/t D .b1b2/.w1w2/. Since

C�2.B/.t/ D W0, so w1; w2; w3 D w1w2 must be pairwise distinct involutions inW0. But then et D ewj with j 2 f1; 2; 3g and so t centralizes one of the elementsb1e; b2e, or b1b2e (all of order 4), which is a contradiction. It follows that exactly oneinvolution in W0 is a square in B . We have proved that B D hbi � hwi � hei, whereo.b/ D 2m, m � 2, b2m�1

D z, hzi � Z.G/, W0 D hz;wi, and e 2 A �W0.Set V D htiB and suppose that CG.t/ D hti � D, where D Š D8. Then W0 is

normal in CG.t/, Z.CG.t// D ht; zi and .CG.t//0 D hzi. Hence CG.t/ contains an

involution u 2 G�V so thatwu D wz and soW0hui D D Š D8. Acting with u onA,we see that jCA.u/j � 4 (Proposition 51.1). Since CW0

.u/ D hzi, so u centralizes aninvolution in A �W0. We may assume eu D e and so CA.u/ D hz; ei. If Œt; e� D w0

with w0 2 W0 � hzi, then Œt; e�u D Œtu; eu� D Œt; e� D w0 D wu0 D w0z and so

z D 1, a contradiction. Hence, we must have Œt; e� D z. If m � 3, then t invertsthe element v D b2m�2

of order 4 (Proposition 51.2) and so vt D v�1 D vz. Butthen .ve/t D .vz/.ez/ D ve, which is a contradiction since CB.t/ D W0. Hence wemust have m D 2, o.b/ D 4, b2 D z, bt D bw with w 2 W0 � hzi, and et D ez. Itfollows that t inverts inB exactly four elements, namely the elements inW0 (which areactually centralized by t ). Hence there are exactly four involutions in V � B and theyall lie in E1 D hti �W0. These four involutions in E1 �W0 lie in a single conjugacyclass in V since jV W CV .t/j D 4. Set QV D V CG.t/ D V hui so that NG.V / D QV

(since CG.t/must cover NG.V /=V ) andE1 is normal in QV because hE1 �W0i D E1.If QV D G, then CG.t/ would contain a G-invariant elementary abelian subgroupE1 of order 8, contrary to our assumptions. Hence QV ¤ G and then Theorem 1.7and Proposition 1.8 imply that G=B is of maximal class and QV =B Š ht; ui Š E4.Replacing u with ut (if necessary), we may assume that huiB=B D Z.G=B/. Weclaim that CG.A/ D B . Set CG.A/ D QB so that QB � B and QB is normal in G.


If QB ¤ B , then QB � huiB . This is a contradiction since wu D wz and so u actsfaithfully on A. Hence we have CG.A/ D B and so G=B Š D8 since QV ¤ G.It remains to consider the case where CG.t/ D E D hti�W0 for each t 2 E�ˆ.G/.

We set again V D htiB . Since jBj D 2mC2,m � 2, so also jV �Bj D jBt j D 2mC2.Set QV D NG.V /. If j QV =V j � 4, then j QV W C QV

.t/j D j QV W Ej � 2mC2 and so all 2mC2

elements in V �B are involutions. This is a contradiction since te is not an involutionin view of Œt; e� ¤ 1. Therefore we must have j QV =V j � 2. If QV D G, then jG=Bj D 2

or 4.We assume that G ¤ QV and so in this case j QV =V j D 2. Since B is abelian, so the

set B0 of elements of B which are inverted by t is a subgroup of B . The number ofinvolutions in V � B is equal to jB0j. Since t does not invert e 2 A �W0, so jB0j �

2mC1. On the other hand, jccl QV.t/j D j QV W Ej D 2mC1 and so jB0j D 2mC1, V � B

has exactly 2mC1 involutions and they lie in a single QV -class. By Proposition 51.2,Ã1.B/ D hb2i � B0 and so B1 D Ã1.B/W0 D hb2; wi � B0, where w 2 W0 � hzi.SinceB=B1 Š E4 and hB1; ei 6� B0, soB0 D hB1; bi orB0 D hB1; bei. Replacing bwith be (if necessary), we may assume that t inverts b and so B0 D hB1; bi D hb;wi.Since t inverts v D b2m�2

with v2 D z, so vt D v�1 D vz. Then et D ez wouldimply .ve/t D .vz/.ez/ D ve, which is a contradiction since CB.t/ D W0 D hz;wi.Therefore et D ew with w 2 W0 � hzi.Since NG.V /=B is of order 4 and G ¤ NG.V / D QV , Theorem 1.7 and Proposition

1.8 imply thatG=B is of maximal class. The subgroup V=B (of order 2) is non-centralin G=B and so G=B is dihedral or semi-dihedral. In particular, QV =B is a 4-group. Letx 2 G � QV be such that x normalizes QV and x2 2 QV . Set D� D h QV ; xi so that (bythe structure of G=B)D�=B Š D8 and therefore we may assume that x2 2 B . SinceV=B is noncentral inD�=B , u D tx 2 QV � V .We claim that W0 D hz;wi is normal in G. Obviously Z.G/ � CB.t/ D W0. If

Z.G/ D W0, then our claim is clear. Suppose that Z.G/ < W0 so that Z.G/ D hzi is oforder 2. LetU be a normal 4-subgroup contained inA and assume that U ¤ W0. Thenwe have U > hzi D Z.G/ and so U \W0 D hzi. But then for an element u 2 U �hzi,we have ut D uz since CA.t/ D W0. This gives .vu/t D .vz/.uz/ D vu, where vis the cyclic subgroup of order 4 in hbi. This is a contradiction and so W0 D hz;wi isnormal in G.Since CA.t/ D W0 and x normalizes A and W0, it follows CA.u/ D W0. Hence

CG.u/ D hui � W0 and Z. QV / D W0 since QV D Bht; ui. Note that ht; bi Š D2mC1

and te D tw. There are exactly two V -classes of involutions contained in V � B

with the representatives t and tb containing 2m elements each. The V -class of t is theset ftwib2j g and the V -class of tb is the set ftbwib2j g, where i; j are any integers.Now u fuses these two V -classes and so tu D tbwrb2s for suitable integers r; s.This gives .tu/2 D tutu D t tu D bwrb2s D b2sC1wr and so o.tu/ D 2mC1 andht; ui Š D2mC2 . Since u inverts htui, so u inverts .tu/2 and w. But h.tu/2; wi D B0

and so u inverts B0. It follows that bu D b�1 and wu D w. Thus tu centralizesB0 D hb;wi. Since tu acts faithfully on B and B D B0A, tu must act faithfully


on A. Note that htuiB=B D Z.D�=B/ D Z.G=B/. If QB D CG.A/ > B , then tu 2 QB

since QB is normal in G and G=B is of maximal class. This is a contradiction and soCG.A/ D B ,D� D G, and G=B Š D8.We have proved the following final result of this subsection.

Theorem 51.10. Let G be a 2-group which has a self-centralizing non-normal ele-mentary abelian subgroup E of order 8 which is not contained in ˆ.G/. Assume thatthere is no involution t 2 E �ˆ.G/, such that CG.t/ contains an elementary abelianG-invariant subgroup of order 8. Suppose that G has a normal elementary abeliansubgroup of order 8 but none of order 16. Then G has a maximal normal abeliansubgroup B of type .2m; 2; 2/, m � 1 such that G=B is isomorphic to a non-trivialsubgroup of D8.

Example.We give here an example of a 2-group G of order 28 satisfying the as-sumptions of Theorem 51.10. We set G D hb; z;w; e; t; u; x j b8 D z2 D w2 D

e2 D t2 D u2 D x2 D Œb;w� D Œb; e� D Œw; e� D Œt; w� D Œu;w� D Œe; x� D

1; b4 D z; u D tx; bt D b�1; bu D b�1; et D ew; eu D ewz; .tu/2 D b; bx D

b�1; wx D wzi. Here B D hb;w; ei is abelian of type .8; 2; 2/ and G=B Š D8. Acoset enumeration computer program assures that such a group G exists!

3o. The case E 6� ˆ.G/ and G has no normal elementary abelian subgroups oforder � 8. We suppose throughout this subsection that a 2-group G contains a self-centralizing elementary abelian subgroupE of order 8which is not contained inˆ.G/.We assume in addition that G has no normal elementary abelian subgroups of order 8.Let t 2 E � ˆ.G/ and M be a maximal subgroup of G such that t 62 M . Then

(by modular law) CG.t/ D hti �M0, whereM0 D CM .t/ contains the four-subgroupE0 D E\M and CM0

.E0/ D M0. Hence we have either CG.t/ D E or (by Theorem1.7 and Proposition 1.8)M0 Š D2n , n � 3 orM0 Š SD2m , m � 4.Suppose that G has (at least) two distinct normal 4-subgroups. Then Theorem 50.2

implies that G is the central product G D D � C of D Š D8 and C with D \

C D Z.D/ and C is either cyclic or of maximal class different from D8. However,if C is cyclic or generalized quaternion, then G does not possess a self-centralizingelementary abelian subgroup of order 8. Hence C Š D2n or C Š SD2n , n � 4.We have proved the following result.

Theorem 51.11. Let G be a 2-group which has a self centralizing elementary abeliansubgroup E of order 8 with E 6� ˆ.G/. Assume in addition that G has no normalelementary abelian subgroups of order 8. Then for each involution t 2 E �ˆ.G/, wehave CG.t/ D hti �M0, whereM0 is isomorphic to one of the following groups: E4,D2n .n � 3/, or SD2m .m � 4/. If G has (at least) two distinct normal 4-subgroups,then G D D �C is the central product ofD and C , whereD Š D8,D \C D Z.D/,and C Š D2n or C Š SD2n with n � 4.

In view of Theorem 51.11, we assume in the rest of this subsection that G has theunique normal four-subgroup U (see Lemma 1.4).


Suppose that CG.t/ D E for an involution t 2 E � ˆ.G/. This is an exceptionalcase. Since G is neither abelian nor of maximal class, we may apply Theorem 50.1.Let N be a metacyclic normal subgroup of G such that G=N is isomorphic to asubgroup of D8, W D �2.N / is abelian of type .4; 4/ or .4; 2/, and CG.W / � N .SetW0 D �1.W / so that W0 is the unique normal four-subgroup of G. Suppose thatjE \ N j � 2. In that case jE \ N j D 2 and E \ N D E \W0. Since CG.t/ D E,t 2 E � N induces on W an involutory automorphism, where W � D CW .t/ is oforder 2. We may apply Proposition 51.2 which shows that t inverts Ã1.W / and alsoon W=W �. If W Š C4 � C4, then t inverts Ã1.W / D W0 contrary to the fact thatjCW .t/j D 2. Suppose that W D ha; s j a4 D s2 D Œa; s� D 1; a2 D zi Š C4 � C2.In this caseW � D CW .t/ D hzi D Ã1.W /. We must have st D sz and at D a�1z�

with � D 0; 1. If � D 1, then at D a�1z D a, a contradiction. If � D 0, thenat D a�1 D az and so .as/t D .az/.sz/ D as, a contradiction. We have proved thatin this special case jE \N j D 4 and so E > W0, which gives the following result.

Theorem 51.12. Let G be a 2-group which has a self-centralizing elementary abeliansubgroup E of order 8 with E 6� ˆ.G/. Assume that G has no normal elementaryabelian subgroups of order 8. Suppose in addition that G has the unique normal 4-subgroup W0 and that for an involution t 2 E � ˆ.G/, CG.t/ D E. Then we haveE > W0 and Theorem 50.1 gives further information about the structure of G.

In the rest of this subsection we assume also that there is an involution t 2 E�ˆ.G/

such that G ¤ CG.t/ D hti �D withD Š D2m , m � 3 or D Š SD2n , n � 4.

Remark. If G is a 2-group possessing an involution t such that CG.t/ D hti �D withD Š D2m or D Š SD2m , where m � 4, then G has no normal elementary abeliansubgroup A of order 8. Indeed, if A is such a subgroup, then Proposition 51.1 impliesthat jCA.t/j � 4 and so CG.t/ would contain a normal elementary abelian subgroupof order 8, which is not the case.

Let U be the unique normal 4-subgroup of G. Set T D CG.U / so that jG W

T j � 2. Suppose at first that t 2 U . Since G ¤ CG.t/, we have jG W T j D 2

and T D CG.t/ D hti � D with D \ U D hui D Z.D/ D Z.G/ and D Š D2m ,m � 3 or D Š SD2n , n � 4. LetM be a maximal subgroup of G such that t 62 M .Set M0 D T \ M so that CG.t/ D hti � M0 and M0 Š D. If M were not ofmaximal class, then Lemma 1.4 implies thatM contains aG-invariant 4-subgroup U0.Since U0 ¤ U , we have obtained a contradiction. Hence M is of maximal class andthereforeM Š D2mC1 or M Š SD2mC1 and M0 Š D Š D2m , m � 3. There is anelement y 2 M �M0 such that y2 2 hui D Z.M/ D Z.M0/. We get ty D tu so thatyt D yu and this determines the action of t onM .We have proved the following result.

Theorem 51.13. Let G be a 2-group which has a self-centralizing elementary abeliansubgroup E of order 8 with E 6� ˆ.G/. Assume that G has no normal elementary


abelian subgroups of order 8. Suppose in addition that G has the unique normal 4-subgroup U and there is an involution t 2 E � ˆ.G/ such that t 2 U and G ¤

CG.t/ D hti � D with D Š D2m , m � 3 or D Š SD2n , n � 4. Then G has amaximal subgroup M such that G D htiM , where M Š D2mC1 or M Š SD2mC1

and CM .t/ Š D2m , m � 3.

In what follows we assume also that t 62 U , where U is the unique normal four-subgroup of G.Suppose for a moment that t 2 T � U , where T D CG.U /. Then CG.t/ has the

normal elementary abelian subgroup E1 D hti � U of order 8. This forces CG.t/ D

hti �D, where D Š D8. Since E1 6� Z.CG.t//, it follows CG.t/ 6� T and CT .t/ D

E1. In particular, jG W T j D 2 and there is an involution t 0 2 CG.t/ � T . Sethzi D CU .t

0/ so that z 2 Z.G/. It follows that E2 D ht 0; t; zi is another elementaryabelian subgroup of order 8 contained in CG.t/ and so CG.E2/ D E2. Also, t 0 is aninvolution in E2 � ˆ.G/. If CG.t

0/ D E2, then applying Theorem 51.12 we get acontradiction since E2 does not contain U . It follows that G ¤ CG.t

0/ D ht 0i �M0

withM0 Š D2m , m � 3 orM0 Š SD2n , n � 4. Replacing E with E2 and t with t 0,we see that we may assume from the start that t 2 G � T .It remains to consider only the case t 2 G � T , where CG.t/ D hti � M0 with

M0 Š D2m , m � 3 or M0 Š SD2n , n � 4. Here T D CG.U /, jG W T j D 2,where U is the unique normal four-subgroup of G. It follows that G D htiT andso, by the modular law, G0 D CG.t/ D hti � D, where D D CT .t/ Š M0. Wehave hzi D Z.D/ D CU .t/ D Z.G/ is of order 2. Set G1 D NG.G0/. Sincej.UG0/ W G0j D 2, we have UG0 � G1. But Z.G0/ D ht; zi, so t has exactly twoconjugates t and tz in G1. It follows G1 D UG0. Set D0 D U hti and so D0 Š D8

and G1 D D �D0 with D \D0 D hzi and U D hz; ui < D0. We fix in the rest ofthis subsection the notation for the structure of D. We set

D D ha; b j a2m�1

D b2 D z2 D 1; a2m�2

D z; ab D a�1z�; � D 0; 1i;

where m � 3, and if � D 1, then m � 4. Let v be an element of order 4 in D0

and let y be an element of order 4 in hai. Since y2 D v2 D z, so x D yv is aninvolution in G1 � .D0 [ D/. We compute xt D .yv/t D yv�1 D yvz D xz,and so D1 D hx; ti Š D8 and D1 \ U D Z.D1/ D Z.G/ D hzi. We see thatCG1

.D1/ D ha; bti D D� Š D because xbt D .yv/bt D y�1v�1 D yv D x,.bt/2 D 1 and abt D a�1z� . Thus G1 D D� �D1 with D� Š D, D� \D1 D hzi,CG.t/ D hti �D D hti �D� and U D hz; ui 6� D1. Denoting againD� withD andbt with b, we have obtained the following initial configuration.(R) The subgroup G1 D UCG.t/ is the central product G1 D D � D1, where

D1 Š D8, t 2 D1, D \ D1 D hzi D Z.G/, CG.t/ D hti � D with D Š D2m orD Š SD2m , m � 3, and U 6� D1, U 6� D.

In the rest of this subsection we denote with S a subgroup of G of the maximalpossible order subject to the following three conditions.


(i) S � G1 D UCG.t/ D D � D1, where D Š D2m or D Š SD2m , m � 3,D1 Š D8, t 2 D1.

(ii) S D DL, where L is normal in S , L Š D2n , n � 3,D \L D hzi D Z.D/ D

Z.L/ and L � D1.

(iii) U D hz; ui 6� L and U 6� D.

We set for the rest of this subsection:

L D hc; t j c2n�1

D t2 D z2 D 1; c2n�2

D z; ct D c�1; n � 3i Š D2n :

We may set u D yv, where y is an element of order 4 in hai and v is an element oforder 4 in hci. If a4 D 1, then we put a D y and if c4 D 1, then we set c D v.It is now easy to determine all possibilities for the structure of S . Act withD on the

dihedral subgroup L. Since Aut.L/=Inn.L/ is abelian of type .2n�3; 2/, it follows thatD0 D ha2i centralizes L. Also we know thatD centralizes hv; ti D D1 Š D8, whereD1 � L and so by the structure of Aut.L/ eitherD centralizesL (and so S D D �L)orM D CD.L/ is a maximal subgroup of D in which case n � 4. In that caseD=Minduces such an involutory automorphism onL so that CL.D=M/ D hc2; ti Š D2n�1 .In any case, ŒD;L� � hzi and so D is also normal in S . If � D 1 (i.e. D is semi-dihedral), then we have three possibilities for the maximal subgroupM ofD. If � D 0

(i.e. D is dihedral), then two distinct maximal subgroups of D are isomorphic and sowe have in this case only two possibilities for the maximal subgroup M of D. Thisgives (together with central products) exactly seven possibilities Si , i D 1; 2; : : : ; 7 forthe structure of S and they are given explicitly (in terms of generators and relations)in the following definition.

Definition 1. Let S D DL be a product of two normal subgroups

D D ha; b j a2m�1

D b2 D z2 D 1; a2m�2

D z; ab D a�1z�; � D 0; 1; m � 3;

and if � D 1; then m � 4i

and

L D hc; t j c2n�1

D t2 D z2 D 1; c2n�2

D z; ct D c�1; n � 3i Š D2n ;

whereD\L D Z.D/ D Z.L/ D hzi and Œa; t � D Œb; t � D 1 so that CS .t/ D hti�D.Here if � D 1, thenD Š SD2m , m � 4, and if � D 0, thenD Š D2m , m � 3.

(1) If � D 1 and Œa; c� D Œb; c� D 1, then S D D � L D S1, m � 4, n � 3.

(2) If � D 1, CD.L/ D hai, and cb D cz, then S D S2, m � 4, n � 4.

(3) If � D 1, CD.L/ D ha2; bi Š D2m�1 , and ca D cz, then S D S3, m � 4,n � 4.

(4) If � D 1, CD.L/ D ha2; bai Š Q2m�1 , then S D S4, m � 4, n � 4.

(5) If � D 0, CD.L/ D hai, and cb D cz, then S D S5, m � 3, n � 4.


(6) If � D 0, CD.L/ D ha2; bi Š E4 or D2m�1 , and ca D cz, then S D S6,m � 3, n � 4.

(7) If � D 0, and CD.L/ D D, then S D D � L D S7, m � 3, n � 3.

We make here the following simple observation. Since t 2 G � T , where T D

CG.U / and jG W T j D 2, t cannot be conjugate (fused) inG to any involution t 0 whichcentralizes U . Therefore, with respect to that fusion, it is enough to consider onlythose involutions in S which act faithfully on U . We want to show in the sequel thatS must be a normal subgroup of G and jG=S j � 4. In some cases for the structure ofS this is difficult.

4o. The case S Š S1. We have exactly four conjugacy classes of involutions con-tained in S � U which act faithfully on U with the representatives t , tc, b, and bav.The corresponding centralizers in S are CS .t/ D CG.t/ D hti �D withD Š SD2m ,m � 4; CS .tc/ D htci�D withD Š SD2m m � 4; CS .b/ D hbi�L withL Š D2n ,n � 3; CS .bav/ D hbavi � hc; tyi with hc; tyi Š Q2n , n � 3.We may assume NG.S/ ¤ S (otherwise we are finished). Then NG.S/ can fuse

cclS .t/ (the conjugacy class of t in S ) only with cclS .tc/ and so jNG.S/ W S j D 2.Obviously, L�hci is a normal subset of involutions in NG.S/ andL D hL�hciiwhichimplies that L is normal in NG.S/. Suppose that N—G.S/ ¤ G (otherwise we arefinished). Note that each x in NG.S/�S sends (by conjugation) cclS .t/ onto cclS .tc/.By the above, NG.S/ cannot fuse cclS .b/ and cclS .bav/ to any other conjugate classof involutions in S . This gives that CNG.S/.b/ covers NG.S/=S and also CNG.S/.bav/

covers NG.S/=S . In particular jCNG.S/.b/j D jCNG.S/.bav/j D 2nC2. The subgroupK D hb; bavi is dihedral of order 2m and S D KL. Since bothK andL are generatedby its non-central involutions, we have hcclS .t/, cclS .tc/, cclS .b/, cclS .bav/i D S .Let x be any involution in NG.S/�S . Since .cclL.t//x D cclL.tc/, so x induces an

outer involutory automorphism on L which inverts hci. Indeed, if tx D tck .k odd/,then x inverts t .tck/ D ck and so x inverts c. Hence CL.x/ D hzi and so, by Theorem1.7 and Proposition 1.8, Lhxi D L0 is of maximal class. Since L0 is generated by itsinvolutions, it follows L0 Š D2nC1 . We have NG.S/ D Shxi and so CNG.S/.x/ D

hxi � CS .x/. On the other hand CL.x/ D hzi and so if jCS .x/j D 2m, then CS .x/

covers S=L which implies that L0 Š D2nC1 is normal in NG.S/ D DL0 > S . Thiscontradicts the maximality of S . (See the construction of S .) It follows that there isno involution x 2 NG.S/� S with the property jCS .x/j D 2m.Let y0 be an element in G � NG.S/ such that y0 normalizes NG.S/ and .y0/2 2

NG.S/. If x D ty0

2 NG.S/ � S , then CNG.S/.x/ D hxi � CS .x/ Š hti � D andso jCS .x/j D 2m, which is a contradiction. If ty

0

2 cclS .t/ [ cclS .tc/, then thereis an n 2 NG.S/ such that ty

0

D tn. But then ty0n�1

D t and so CG.t/ 6� NG.S/,a contradiction. Hence we must have ty

0

2 cclS .b/ or ty0

2 cclS .bav/. We knowthat jCNG.S/.b/j D jCNG.S/.bav/j D 2nC2 and on the other hand jCNG.S/.t

y0

/j D

jCNG.S/.t/j D 2mC1. Hence m D n C 1. It follows that for each x 2 .cclS .t/ [

cclS .tc/ [ cclS .b/ [ cclS .bav// D P , we have jCNG.S/.x/j D 2nC2 D 2mC1.


Finally, since Sy0

¤ S , there is an x 2 P such that x0 D xy0

2 NG.S/� S . But thenCNG.S/.x

0/ D hx0i � CS .x0/ and CS .x

0/j D 2m, which is the final contradiction. Itfollows that NG.S/ D G and so in case S Š S1, the subgroup S is normal in G andjG=S j � 2.

5o. The case S Š S2. We have exactly four conjugacy classes of involutionscontained in S � U which act faithfully on U with the representatives t , tc, b, andbav. The corresponding centralizers in S are CS .t/ D CG.t/ D hti � D withD Š SD2m , m � 4; CS .tc/ D htci � ha; bavi with ha; bavi Š SD2m , m � 4;CS .b/ D hbi � hyc; ti with hyc; ti Š SD2n , n � 4; CS .bav/ D hbavi � hcy; tci

with hcy; tci Š SD2n , n � 4.

We assume NG.S/ ¤ S (otherwise we are finished). Set P D cclS .t/[ cclS .tc/[

cclS .b/ [ cclS .bav/. Obviously, hP i D S and P � G � T , where T D CG.U / andjG W T j D 2. Suppose also that t is conjugate in NG.S/ to tc but t is not conjugatein NG.S/ to any involution in S � L. Then jNG.S/ W S j D 2 and since NG.S/

normalizes the set L� hci and hL� hcii D L, so L is normal in NG.S/. Let x be anyinvolution in NG.S/ � S . Since .cclL.t//x D cclL.tc/, we get (as in case S Š S1)L0 D Lhxi Š D2nC1 . We have CNG.S/.x/ D hxi � CS .x/ and CL.x/ D hzi. IfjCS .x/j D 2m, then CS .x/ covers S=L and so L0 is normal in DL0 > S . Thiscontradicts the maximality of S . Hence there is no involution x 2 NG.S/ � S withjCS .x/j D 2m. Suppose in addition that NG.S/ ¤ G (otherwise we are finished). Lety0 be an element in G � NG.S/ such that y0 normalizes NG.S/ and .y0/2 2 NG.S/.If x D ty

0

2 NG.S/ � S , then jCNG.S/.x/j D 2mC1 and so jCS .x/j D 2m, acontradiction. If ty

0

2 cclS .t/ [ cclS .tc/, then CG.t/ 6� NG.S/ (as in case S Š S1),which is a contradiction. Hence we must have ty

0

2 cclS .b/ or ty0

2 cclS .bav/.Since y0 does not normalize S , there is x0 2 P such that x0 D .x0/y

0

2 NG.S/ � S .If NG.S/ fuses b with bav, then all involutions in P are fused in hy0iNG.S/ to t andso jCNG.S/.x0/j D 2mC1 and jCS .x0/j D 2m, a contradiction. If NG.S/ does notfuse b with bav, then jCNG.S/.b/j D jCNG.S/.bav/j D 2nC2. Since jCNG.S/.t

y0

/j D

2mC1, so nC 2 D mC 1 and therefore jCNG.S/.x0/j D 2nC2 D 2mC1 and so againjCS .x0/j D 2m, a contradiction. Hence we must have NG.S/ D G and so we aredone in this case.

Suppose now that t is not conjugate to tc in NG.S/. Then t must be conjugatein NG.S/ to b or to bav. This implies jNG.S/ W S j D 2 and m D n � 4. Eachelement x 2 NG.S/ � S sends .CS .t//

0 D ha2i onto .CS .tx//0 D hc2i. Suppose

that NG.S/ ¤ G and let y0 be an element in G � NG.S/ such that y0 normalizesNG.S/ and .y0/2 2 NG.S/. Since y0 does not normalize S , there is an x0 2 P suchthat x0 D .x0/y

0

2 NG.S/ � S , x0 2 G � T , and so x0 does not centralize U . Onthe other hand, x0 sends hyi (contained in ha2i) onto hvi (contained in hc2i). Wehave yx0 D vz� with � D 0; 1. But then .yvz�/x0 D yvz� and so x0 centralizesU D hz; u D yvi. This is a contradiction and so we have NG.S/ D G also inthis case.


It remains to consider the case that t is fused in NG.S/ to all three involutions tc,b, and bav. In that case m D n � 4 and jNG.S/ W S j D 4. Note that L andK D hb; bavi are two dihedral normal subgroups both of order 2m of S containingcclS .t/, cclS .tc/, cclS .b/, and cclS .bav/ with NG.S/=S acting regularly on thosefour S -classes. Hence for each x 2 NG.S/� S we have either Lx D L andKx D K

or Lx D K. Suppose now that NG.S/ ¤ G and let y0 be an element in G � NG.S/

such that y0 normalizes NG.S/ and .y0/2 2 NG.S/. Then t 0 D ty0

2 NG.S/ � S .Indeed, if t 0 2 S , then t 0 2 P and so there is an n 2 NG.S/ such that t 0 D ty

0

D tn.But then ty

0n�1

D t and so CG.t/ 6� NG.S/, a contradiction. If Lt 0

D L andK t 0

D K, then .cclL.t//t0

D cclL.tc/ and .cclK.b//t0

D cclK.bav/. This gives (asin case S Š S1) Lht 0i Š D2mC1 and Kht 0i Š D2mC1 . In particular, t 0 inverts hci

and on havi. Hence t 0 inverts v 2 hc2m�3

i and on y 2 h.av/2m�3

i D ha2m�3

i. Fromyt 0

D y�1 and vt 0

D v�1, we get ut 0

D .yv/t0

D y�1v�1 D yv D u. Hencet 0 centralizes U D hz; ui and so t 0 2 T . But t 2 G � T and jG W T j D 2, whichis a contradiction. If Lt 0

D K, then hcit 0

D havi and so vt 0

D yz, D 0; 1.This gives .yvz/t

0

D yvz and .uz/t0

D uz and so t 0 centralizes U , which is acontradiction. It follows that we must have NG.S/ D G also in this case. We haveproved that in case S Š S2, the subgroup S is normal in G, jG W S j � 4, and ifjG=S j D 4, then m D n � 4.

6o. Cases S Š S3 and S Š S4. We have exactly four conjugacy classes of involu-tions contained in S � U which act faithfully on U with the representatives t , tc, b,and bav. The corresponding centralizers in S in case S Š S3 are CS .t/ D CG.t/ D

hti � D with D Š SD2m , m � 4; CS .tc/ D htci � hav; bi with hav; bi Š D2m ,m � 4; CS .b/ D hbi � L with L Š D2n , n � 4; CS .bav/ D hbavi � hcy; tci withhcy; tci Š SD2n , n � 4.

The centralizers in S in case S Š S4 are CS .t/ D CG.t/ D hti � D with D Š

SD2m , m � 4; CS .tc/ D htci � hav; bai with hav; bai Š Q2m , m � 4; CS .b/ D

hbi � hcy; ti with hcy; ti Š SD2n , n � 4; CS .bav/ D hbavi � hc; tyi with hc; tyi Š

Q2n , n � 4.

Suppose that S ¤ NG.S/. Then NG.S/ can fuse t only with an involution incclS .bav/ when S Š S3 and only with an involution in cclS .b/ when S Š S4. Thisgivesm D n and jNG.S/=S j D 2. Set P D cclS .t/[cclS .tc/[cclS .b/[cclS .bav/.Obviously, hP i D S and P � G � T , where T D CG.U / and jG W T j D 2. Assumethat NG.S/ ¤ G and let y0 be an element in G � NG.S/ such that y0 normalizesNG.S/ and .y0/2 2 NG.S/. Since y0 does not normalize S , there is x0 2 P suchthat x0 D .x0/y

0

2 NG.S/ � S , x0 2 G � T and so x0 does not centralize U . Since.cclS .t//x0 D cclS .tx0/, where tx0 2 cclS .bav/ in case S Š S3 and tx0 2 cclS .b/in case S Š S4, so x0 sends .CS .t//

0 D ha2i onto .CS .tx0//0 D hc2i. Indeed,

.CS .bav//0 D hc2i in case S Š S3 and .CS .b//

0 D hc2i in case S Š S4 and in bothcases hc2i is normal in S . We get ha2ix0 D hc2i and so, in particular, yx0 D vz� ,� D 0; 1 and so .yvz�/x0 D yvz� or .uz�/x0 D uz� and so x0 centralizes U D


hz; u D yvi. This is a contradiction and so we must have NG.S/ D G. We haveobtained in both cases S Š S3 and S Š S4 that jG W S j � 2 and if jG W S j D 2, thenm D n � 4. We have proved the following result.

Theorem 51.14. Let G be a 2-group which has a self-centralizing elementary abeliansubgroupE of order 8 withE 6� ˆ.G/. We assume thatG has the unique normal four-subgroup U and set T D CG.U /. Suppose that there is an involution t 2 E � ˆ.G/

such that CG.t/ D hti � D with D Š SD2m , m � 4. Finally, assume t 62 U andin that case we have proved that we may assume t 2 G � T . Then G has a (large)normal subgroup S containing UCG.t/ which is isomorphic to one of the groups Si

for i D 1; 2; 3; 4 (see Definition 1) so that jG=S j � 4. More precisely,

(a) if S Š S1, then jG=S j � 2;

(b) if S Š S2, then jG=S j � 4 and if jG=S j D 4, then m D n � 4;

(c) if S Š S3, then jG=S j � 2 and if jG=S j D 2, then m D n � 4;

(d) if S Š S4, then jG=S j � 2 and if jG=S j D 2, then m D n � 4.

We have to analyze three remaining cases S5, S6, and S7 for the structure of S .

7o. The case S Š S5. We have exactly four conjugacy classes of involutions con-tained in S � U which act faithfully on U with the representatives t , tc, b, and ba.The corresponding centralizers in S are CS .t/ D CG.t/ D hti �D with D Š D2m ,m � 3; CS .tc/ D htci � ha; bvi with hv; bvi Š Q2m , m � 3; CS .b/ D hbi � hyc; ti

with hyc; ti Š SD2n , n � 4; CS .ba/ D hbai � hyc; ti with hyc; ti Š SD2n , n � 4.We see that t cannot be fused in NG.S/ to any of the involutions tc, b, or ba and

this implies S D G. By our assumption, G has the unique normal 4-subgroup U andsom � 4.

8o. The case S Š S6. We have exactly four conjugacy classes of involutions con-tained in S � U which act faithfully on U with the representatives t , tc, b, and ba.The corresponding centralizers in S are CS .t/ D CG.t/ D hti �D with D Š D2m ,m � 3; CS .tc/ D htci � hav; bi with hav; bi Š SD2m , m � 4 and hav; bi Š D8 ifm D 3 (in which case tc centralizes U ), CS .b/ D hbi � L with L Š D2n , n � 4;CS .ba/ D hbai � hcy; ti with hcy; ti Š SD2n , n � 4.Suppose that NG.S/ ¤ S . Since NG.S/ can fuse t only with an involution in

cclS .b/, we have m D n � 4 and jNG.S/=S j D 2. Set P D cclS .t/ [ cclS .tc/ [

cclS .b/ [ cclS .ba/. Obviously, hP i D S and P � G � T , where T D CG.U / andjG W T j D 2. Assume that NG.S/ ¤ G and let y0 be an element in G � NG.S/ suchthat y0 normalizes NG.S/ and .y0/2 2 NG.S/. Since y0 does not normalize S , thereis an x0 2 P such that x0 D .x0/y

0

2 NG.S/ � S , x0 2 G � T and so x0 does notcentralize U . Then x0 sends (by conjugation) t onto an S -conjugate of b and so x0

sends .CS .t//0 D ha2i onto .CS .b//

0 D hc2i. Indeed, each S -conjugate b0 of b hasthe property CS .b

0/ D hb0i � L because L is normal in S . In particular, yx0 D vz� ,� D 0; 1 and so yvz� D uz� is centralized by x0. Hence x0 centralizes U D hz; ui,


which is a contradiction. Hence we must have NG.S/ D G which implies that in caseS Š S6 we get jG=S j � 2 and if jG=S j D 2, then m D n � 4.

It remains to consider the case S Š S7 which is rather difficult since there areseveral possibilities for the fusion of the involution t with other involutions in S � U

which act faithfully on U .

9o. The case S Š S7. We have exactly four conjugacy classes of involutions con-tained in S � U which act faithfully on U with the representatives t , tc, b, and ba.The corresponding centralizers in S are CS .t/ D CG.t/ D hti �D with D Š D2m ,m � 3; CS .tc/ D htci �D withD Š D2m ,m � 3; CS .b/ D hbi �L with L Š D2n ,n � 3; CS .ba/ D hbai � L with L Š D2n , n � 3.

Suppose that NG.S/ ¤ S (otherwise we are finished). SetP D cclS .t/[cclS .tc/[cclS .b/ [ cclS .ba/. Obviously, hP i D S and P � G � T , where T D CG.U / andjG W T j D 2. Suppose also that t is not fused in NG.S/ to any involution in S � L.Then NG.S/ fuses t with tc and so jNG.S/ W S j D 2 and L is normal in NG.S/ sinceL � hci is a normal subset in NG.S/. Let x be any involution in NG.S/ � S . Since.cclL.t//x D cclL.tc/, we get L0 D Lhxi Š D2nC1 (as in case S Š S1). We haveCNG.S/.x/ D hxi � CS .x/ and CL.x/ D hzi D Z.L/ D Z.L0/. If jCS .x/j D 2m,then CS .x/ covers S=L and so L0 is normal in DL0 > L. This contradicts themaximality of S . Hence there is no involution x 2 NG.S/�S such that jCS .x/j D 2m.Suppose in addition that NG.S/ ¤ G. Let y0 be an element in G � NG.S/ suchthat y0 normalizes NG.S/ and .y0/2 2 NG.S/. If x D ty

0

2 NG.S/ � S , thenjCNG.S/.x/j D 2mC1 and so jCS .x/j D 2m, a contradiction. If ty

0

2 cclS .t/ [

cclS .tc/, then (as before) CG.t/ 6� NG.S/, a contradiction. Hence ty0

2 cclS .b/ orty

0

2 cclS .ba/. Since y0 62 NG.S/, there is an x0 2 P with x0 D .x0/y0

2 NG.S/�S .If NG.S/ fuses b and ba, then all involutions in P are fused in hy0iNG.S/ to t and sojCNG.S/.x0/j D 2mC1 and jCS .x0/j D 2m, a contradiction. If NG.S/ does not fuse band ba, then jCNG.S/.b/j D jCNG.S/.ba/j D 2nC2 and since jCNG.S/.t

y0

/j D 2mC1,so we get nC 2 D mC 1 and therefore jCNG.S/.x0/j D 2nC2 D 2mC1 and so againjCS .x0/j D 2m, a contradiction. Hence we must have NG.S/ D G in this case.

Suppose that t is not conjugate to tc in NG.S/. Then t must be conjugate in NG.S/

to b or ba. This implies jNG.S/ W S j D 2 and m D n � 3.

We make here the following simple observation. All involutions in S � .D [ L/

centralize U and so lie in T . All noncentral involutions in D and L lie in G � T .Since S is the central product S D D �L ofD and L, it follows thatD and L are theonly dihedral normal subgroups of order 2m of S all of whose noncentral involutionsbelong toG�T . Therefore for each element x 2 NG.S/�S , we have Lx D D sincetx 2 cclS .b/ or tx 2 cclS .ba/. In particular, hcix D hai.

Suppose again that NG.S/ ¤ G. Let y0 be an element in G � NG.S/ such that y0

normalizes NG.S/ and .y0/2 2 NG.S/. Since y0 62 NG.S/, there is an x0 2 P suchthat x0 D .x0/y

0

2 NG.S/ � S and x0 2 G � T and so x0 does not centralize U .Note that hP i D S and P � G � T . On the other hand (by the above), hcix0 D hai


and so vx0 D yz, D 0; 1. But then .yvz/x0 D yvz, where yv D u. Hence x0

centralizes U D hz; ui, a contradiction. Thus in this case we have also NG.S/ D G.It remains to analyze the case that t is fused in NG.S/ to all three involutions tc, b,

and ba. In that case m D n � 3 and jNG.S/ W S j D 4. By the above observation, foreach x 2 NG.S/ � S , we have either Lx D L and Dx D D or Lx D D. Supposenow that NG.S/ ¤ G. Let y0 be an element in G � NG.S/ such that y0 normalizesNG.S/ and .y0/2 2 NG.S/. Then t 0 D ty

0

2 NG.S/ � S . Otherwise, ty0

2 P andso there is an n 2 NG.S/ with ty

0

D tn and so ty0n�1

D t , which is a contradictionsince y0n�1 62 NG.S/. If Lt 0

D L and Dt 0

D D, then .cclL.t//t0

D cclL.tc/and .cclD.b//t

0

D cclD.ba/ imply (as before) that Lht 0i Š Dht 0i Š D2mC1 . Inparticular, t 0 inverts v and y which gives ut 0

D .yv/t0

D y�1v�1 D yv D u and so t 0

centralizesU . This is not possible since t 0 (together with t ) lies inG�T . If Lt 0

D D,then hcit 0

D hai and so in particular vt 0

D yz� , � D 0; 1. But then t 0 centralizesyv D u and so t 0 centralizes U , which is again a contradiction. Hence we must havealso in this case NG.S/ D G.We have proved the following final result.

Theorem 51.15. Let G be a 2-group which has a self-centralizing elementary abeliansubgroup E of order 8 with E 6� ˆ.G/. We assume that G has the unique normal4-subgroup U and set T D CG.U /. Suppose that there is an involution t 2 E�ˆ.G/

such that CG.t/ D hti � D with D Š D2m , m � 3. Also, in case D Š D8 weassume in addition that G does not possess a normal elementary abelian subgroup oforder 8. Finally, assume t 62 U and in that case we have proved that we may assumet 2 G � T . Then G has a (large) normal subgroup S containing UCG.t/ which isisomorphic to one of the groups Si for i D 5; 6; 7 (see Definition 1) so that jG=S j � 4.More precisely,

(a) If S Š S5, then S D G and m � 4, n � 4;

(b) if S Š S6, then jG=S j � 2 and if jG=S j D 2, then m D n � 4;

(c) if S Š S7, then jG=S j � 4 and if jG=S j D 4, then m D n � 3.

�52

2-groups with�2-subgroup of small order

The main results of this section are due to Janko [Jan6].In Lemma 42.1 finite 2-groups G have been determined with j�2.G/j � 8. In this

section we do the next step and determine the finite 2-groups G with jGj > 16 andj�2.G/j D 16 (D �2-groups). All nonmetacyclic �2-groups will be given in termsof generators and relations. For more general result, see �55.If jGj D 16, then �2.G/ < G if and only if G 62 fC16;C8 � C2;M16g. Therefore,

we consider in this section only groups of order > 16.If j�2.G/j D 16, then the numbers c1.G/ and c2.G/ are small. This observation

allows us to generalize the main result of this section (see �55).In subsection 1o we study �2-groups G which have no normal subgroups isomor-

phic to E8. Using the main theorem from �50, we obtain three classes of such 2-groupsand one exceptional group of order 25 (Theorem 52.1).In subsection 2o we study nonabelian �2-groups G with normal subgroup isomor-

phic to E8. It is easy to see that then G=E is either cyclic or generalized quaternion.If G=E is cyclic, then we get three classes of groups (Theorem 52.2). If G=E isgeneralized quaternion and E 6� ˆ.G/, then we get one class of groups (Theorem52.4). If G=E is generalized quaternion and E � ˆ.G/, we get two classes of groups(Theorem 52.5).In subsection 3o we investigate 2-groups G with jGj > 16 which have exactly one

subgroup of order 16 and exponent 4. It turns out that a 2-group G has this property ifand only if jGj > 16 and j�2.G/j D 16 (Theorem 52.6).It is interesting to note that, for a�2-group G, there are exactly five possibilities for

the structure of �2.G/: Q8 � C4, Q8 � C2, D8 � C2, C4 � C2 � C2, C4 � C4 (notethat Q8 �C4 Š D8 �C4). In fact, if�2.G/ Š Q8 �C2 or D8 �C2 and jGj > 16, thenwe get exactly one group of order 25 in each case (see Theorem 52.1(d) and Theorem52.2(a) for n D 2). In other three cases for the structure of �2.G/ we get infinitelymany groups.In subsection 4o we consider a similar problem. In [Ber25, �48] the 2-groups G

have been considered which have exactly one abelian subgroup of type .4; 2/. It wasshown that either j�2.G/j D 8 (and then G is isomorphic to one of the groups inLemma 42.1) or G has a self-centralizing elementary abelian subgroup of order 8 (see�51). We improve this result by determining completely the groups of the secondpossibility (Theorem 52.7). Finally, Theorem 52.8 shows that our result also slightly


improves the classification of 2-groups with exactly two cyclic subgroups of order 4given in Theorem 43.4.

Lemma A ([Ber25, �48]). LetG be a metacyclic 2-group of order> 24. If j�2.G/j D

24, then �2.G/ Š C4 � C4.

Proof. Let jGj D 2m > 24. Assume that �2.G/ D H is nonabelian. We want toshow that this assumption leads to a contradiction. If H is not minimal nonabelian,then Proposition 10.19 implies that G is of maximal class. But then �2.G/ D G, acontradiction. Suppose that H has a cyclic subgroup of index 2. Since in that caseH is minimal nonabelian, H Š M24 (Theorem 1.2). But then �2.G/ D �2.H/

is abelian of type .4; 2/ which contradicts the fact that H D �2.G/. Exercise 1.8Aimplies H D ha; b j a4 D b4 D 1; ab D a�1i. It follows that Z.H/ D ha2; b2i D

�1.H/ D �1.G/, H 0 D ha2i, and Z D hb2i is a characteristic subgroup of order 2in H so Z G G. Indeed, b2 is a square in H , a2b2 is not a square in H (check!), andZ.H/ � H 0 D fb2; a2b2g. Thus Z is central in G. Since H=Z Š D8, Proposition10.19 implies that G=Z is of maximal class. All involutions in G=Z lie in H=Z andso G=Z Š SD16 and jGj D 25. But there are elements of order 4 in .G=Z/� .H=Z/

whose square lies in Z.G=Z/ D ha2; b2i=Z. Hence there is an element x 2 G � H

such that x2 2 ha2; b2i and so o.x/ � 4, which is the final contradiction. Thus,H Š C4 � C4.

1o. G has no normal elementary abelian subgroups of order 8. In this subsectionwe prove the following result.

Theorem 52.1. Let G be a 2-group of order > 24 satisfying j�2.G/j D 24. Suppose,in addition, that G has no normal elementary abelian subgroups of order 8. Then wehave one of the following four possibilities:

(a) G D D � C (the central product) with D Š D8, D \ C D Z.D/, and C iscyclic of order � 8. Here we have �2.G/ Š Q8 � C4 Š D8 � C4. (For suchgroups, see Appendix 16; it is proved there, that also G Š Q8 � C .)

(b) G is metacyclic of order � 25 (in that case, by Lemma A, �2.G/ Š C4 � C4).

(c) G D QS , where Q Š Q8, Q is normal in G, S is cyclic of order � 16,Q \ S D Z.Q/, and if C is the subgroup of index 2 in S , then CG.Q/ D C .Setting Q D ha; bi and S D hsi, we have as D a�1 and bs D ba. Here wehave �2.G/ Š Q8 � C4.

(d) The exceptional group G of order 25 has a maximal subgroup M D hui � Q,where u is an involution, Q Š Q8, CG.u/ D M and G is isomorphic to thegroup A2(a) from Theorem 49.1. We have �2.G/ D M Š Q8 � C2.

Proof. We may assume that G is nonabelian. If G were of maximal class, then�2.G/ D G, a contradiction since jGj > 24. Since G is neither abelian nor ofmaximal class, we may apply main theorem from �50 about 2-groups without nor-

�52 2-groups with �2-subgroup of small order 77

mal elementary abelian subgroup of order 8. Then G has a normal metacyclic sub-group N such that CG.�2.N // � N , G=N is isomorphic to a subgroup of D8, andW D �2.N / is either abelian of type .4; 4/ or N is abelian of type .2j ; 2/, j � 2.However, if W Š C4 � C4, then W D �2.G/ since j�2.W /j D 24 D j�2.G/j, andTheorem 41.1 and Remark 41.2 imply that G is metacyclic. This gives the case (b) ofour theorem, by Lemma A. In what follows we assume that G is not metacyclic.We assume in the sequel that N is abelian of type .2j ; 2/, j � 2 and so W D

�2.N / is abelian of type .4; 2/. By hypothesis, j�2.G/ W W j D 2.Suppose that G has at least two distinct normal four-subgroups. Then it follows

from Theorem 50.2 that G D D � C , where D is dihedral of order 8 and C is eithercyclic or of maximal class andD \ C D Z.D/. In view of j�2.G/j D 16, C must becyclic of order � 8 since jGj > 24, and this gives the case (a) of our theorem.Next we assume that G has the unique normal four-subgroup W0 D �1.W / D

�1.N /. Also we have CG.W / D N (N is abelian of type .2j ; 2/, j � 2).Set QW D �2.G/ so that QW \ N D W , j QW W W j D 2, and QW is nonabelian (the

last assertion follows from CG.W / D N � QW ). If QW is metacyclic, then Theorem41.1 and Remark 41.2 imply that G is metacyclic. But then Lemma A gives a con-tradiction. Thus QW is nonabelian non-metacyclic. In particular, exp. QW / D 4 (seeTheorem 1.2). Proposition 1.6 gives j QW = QW 0j D 8 and so j QW 0j D 2. By Lemma 1.1,j QW j D 2j QW 0jjZ. QW /j and so jZ. QW /j D 4 and Z. QW / < W .First assume that Z. QW / D hvi Š C4 is cyclic and set z D v2, W0 D hz; ui D

�1.W /. Since f1g < QW 0 < hvi (by the previous paragraph, j QW j D 2), we haveQW 0 D hzi. For any x 2 QW � W , we have x2 2 Z. QW / D hvi (indeed, QW =Z. QW / iselementary abelian). But QW has no elements of order 8 and so x2 2 hzi and ux D uz

since C QW.u/ D W . It follows that the nonabelian subgroup W0hxi Š D8 (it is

nonabelian since W0 6� Z. QW /) and so, since D8 has five involutions, we may assumethat x is an involution. Then QW is the central product of hu; xi Š D8 and hvi D

Z. QW / Š C4 with hvi \ hu; xi D hzi and .ux/2 D z. We setQ D hux; vxi Š Q8 andsee that QW D Q � hvi is also the central product of Q8 and C4 (see also Appendix 16,subsection 1o). All six elements in QW � .Q [ hvi/ are involutions and thereforeQ isthe unique quaternion subgroup of QW D �2.G/ and soQ is characteristic in QW hencenormal inG (see also Appendix 16). There are exactly three four-subgroups in QW andone of them is W0 D hz; ui and this one is normal in G. Since G has exactly onenormal four-subgroup, it follows that the other two four-subgroups in QW are conjugateto each other in G. Set C D CG.Q/. Obviously C is normal in G and C \ QW D hvi.But hvi is the unique subgroup of order 4 in C (indeed, all subgroups of G of order4 are contained in QW ) so C is cyclic, by Proposition 1.3, and QWC D Q � C withQ \ C D hzi. Set Q D ha; bi, where we choose the generator a so that u D av.Since W0 D �1.W / D hz; ui and hvi D Z. QW / are both normal in G, it followsthat hai .� W / is normal in G (indeed, W has exactly three subgroups of order 4).Every four-subgroup L is normal in QW . Indeed, L is characteristic in LZ. QW /, whichis normal in QW ; it follows that all four-subgroups are normal in Q � C . The four-


subgroups hbv; zi and hbav; zi must be conjugate inG since they are not normal in Gand are contained in QW .� Q � C/ ( QW has exactly three four-subgroups and exactlyone of them is normal in G). Therefore Q � C < G so jG W .Q � C/j D 2, by theproduct formula, and G=.Q �C/ induces an outer automorphism ˛ onQ normalizinghai and sending hbi onto hbai. A Sylow 2-subgroup of Aut.Q/ is isomorphic to D8. Itfollows that G=C Š D8 and so there is an element s 2 G � .Q �C/ such that s2 2 C

and bs D ba. If as D a, then .ba/s D ba2 D b�1 and so ba D .bs/�1 D .ba/�1,which is a contradiction. It follows that as D a�1. It remains to determine s2. SethsiC D S and we see that C is a cyclic subgroup of index 2 in S . Since o.s/ � 8

(indeed, all elements of G of order � 4 are contained in QW � Q � C ), we haveo.s2/ � 4 and therefore jZ.S/j � 4. The same argument shows that all elements inS � C have order � 8. It follows from Theorem 1.2 that S is cyclic. We may sets2 D c, where hci D C . We compute

.sb/2 D sbsb D s2bsb D cbab D cabzb D cab2z D cazz D ca:

If jC j D 4, then C D hvi and so ca is an involution. But then o.sb/ D 4, by thedisplayed formula, and this is a contradiction since sb 62 QW . Thus jC j � 8 and soo.s/ � 16. We have obtained the groups stated in part (c) of our theorem.Now suppose that Z. QW / Š E4, the four-group, so that Z. QW / D W0 D �1.W / D

�1.N / D hz; ui, where z D v2 and W D hui � hvi Š C2 � C4.If QW is minimal nonabelian, then the fact that QW is nonmetacyclic implies with the

help of Exercise 1.8A that QW D ha; b j Œa; b� D c; a4 D b2 D c2 D Œa; c� D

Œb; c� D 1i. We see that Z. QW / D ˆ. QW / D ha2; ci Š E4 and QW has exactly threemaximal subgroups ha2; c; bi Š E8, ha; ci Š C4 � C2, hba; ci Š C4 � C2 since.ba/2 D aŒa; b�a D aca D a2c. Hence ha2; c; bi is normal in G, contrary to theassumption of this subsection.We have proved that QW is not minimal nonabelian. LetD be a minimal nonabelian

subgroup of QW (of order 8). If u 2 Z. QW / � D, then QW D hui � D and ˆ. QW / DQW 0 D ˆ.D/ D Ã1.D/ D hzi since z D v2.Suppose first that D Š D8. Let t be an involution in QW � W (t exists since

c1. QW / D 7 > 5 D c1.D/ > 3 D c1.W /). Since Œv; t � D z D v2 (hvi is normal inQW and it is not contained in Z. QW / Š E4), t inverts hvi and so we may setD D hv; ti,where QW D hui � D. Since t inverts �2.N / D W (indeed, t centralizes �1.N /

and both cyclic subgroups of W of order 4 are t -invariant but not centralized by t ),it follows that CN .t/ D W0 D hz; ui. Assume that N � W contains an element xof order 8 (i.e., j > 2; by the above, N is abelian of type .2j ; 2/). Then Theorem51.2 implies xt D x�1w0 with w0 2 W0. But then .tx/2 D .txt/x D xtx D

x�1w0x D w0 and therefore o.tx/ � 4 with tx 62 QW , a contradiction. Thus, x doesnot exist so N D W is abelian of type .4; 2/. Since t inverts N D W , all eightelements in QW � N D QW �W are involutions: indeed, QW D hti W . In particular,no involution in QW � N is a square in G (otherwise, G has an element of order 4 notcontained in QW ) and so G=N Š E4 (indeed, G=N is isomorphic to a subgroup of


Aut.W / Š D8 and jG W N j > 2) and jGj D 25. All elements in G � QW must be oforder 8 and so for any y 2 G � QW , y2 2 N � W0 D W � W0 since G=N Š E4.Replacing huvi with hvi and v with v�1, if necessary, we may assume that y2 D v

since c2.W / D 2. Since CG.N / D N.D hy2; ui/, we get uy D uz, and so yu D yz.Thus hy; ui Š M24 and hy; ui > N . The cyclic group hvi D Z.hy; ui/ of order 4 isnormal in G and CG.v/ D hyiN Š M24 . Since ty is also an element of order 8, wehave .ty/2 2 N �W0. If .ty/2 2 hvi, then CG.v/ � hN;y; tyi D G, a contradiction.Thus .ty/2 2 huvi. If .ty/2 D .uv/�1 D .uz/v, then replacing u with uz, we mayassume from the start that .ty/2 D uv. We have uv D tyty D ty2ty D tvty . Itfollows that ty D v�1tuv D tvuv D tuz, and so y normalizes the elementary abeliansubgroup E D hz; u; ti of order 8. Since E is normal in QW , E is also normal inG D h QW ;yi, contrary to the assumption of this subsection.Finally, letD Š Q8 so that QW D hui�D, Z.D/ D hzi, z D v2, andD\N D hvi.

Take r 2 D � hvi. We have r2 D z with z 2 Z.G/ and so r induces an involutoryautomorphism on the abelian group N of type .2j ; 2/, j � 2. Since r inverts W D

hu; vi D �2.N /, it follows that CN .r/ D W0 D hu; zi. Then Theorem 51.2 impliesthat r inverts N=W0. Suppose that W < N (i.e., j > 2) and let x 2 N �W . We getxr D x�1w0 with w0 2 W0 and so .rx/2 D rxrx D r2xrx D zx�1w0x D zw0 2

W0. Since rx 62 QW D �2.G/ and o.rx/ � 4, we get a contradiction. We have provedthat N D W D CG.N / is abelian of type .4; 2/. We have jGj � 26.Set S D CG.W0/ so that S � QW and S=N stabilizes the chain N > W0 > f1g

since jN=W0j D 2. Hence S=N is elementary abelian of order � 4. Suppose thatS > QW ; then S=N is a four-group. Let QS=N be a subgroup of order 2 in S=N distinctfrom QW =N . Thus QS \ QW D N . Take an element y 2 QS � N so that o.y/ D 8 andtherefore y2 2 N �W0. But then y centralizes hy2;W0i D N , a contradiction. HenceS D QW D CG.W0/, jG W QW j D 2 and jGj D 25. In particular, CG.u/ D hui�D withD Š Q8. Note that the involution u is contained in the unique normal four-subgroupW0 of G and CG.u/ ¤ G. Then our group G is isomorphic to the group A2(a) of theTheorem 49.1. The proof is complete.

2o. G has a normal E8. Throughout this subsection we assume that G is a non-abelian 2-group of order > 24 satisfying j�2.G/j D 24. We assume in addition thatG has a normal elementary abelian subgroup E of order 8. In that case, G=E hasexactly one subgroup �2.G/=E of order 2 so G=E is either cyclic of order � 4 orgeneralized quaternion (see Proposition 1.3). In the case where G=E is cyclic, G isdetermined in Theorem 52.2. In the case where G=E is generalized quaternion andE 6� ˆ.G/, G is determined in Theorem 52.4. Finally, if G=E is generalized quater-nion and E � ˆ.G/, then G is determined in Theorem 52.5.

Theorem 52.2. LetG be a nonabelian 2-group of order> 24 satisfying j�2.G/j D 24.Suppose, in addition, that G has a normal subgroup E Š E8 such that G=E is cyclicof order 2n, n � 2. Then one of the following holds:


(a) G D ha; e1 j a2nC1

D e21 D 1; n � 2; a2n

D z; Œa; e1� D e2; e22 D Œe1; e2� D

1; Œa; e2� D zi is a group of order 2nC3, Z.G/ D ha4i, G0 D hz; e2i, ˆ.G/ D

ha2; e2i, E D hz; e1; e2i is a normal elementary abelian subgroup of order 8,and �2.G/ D Eha2n�1

i is abelian of type .4; 2; 2/ for n � 3, and �2.G/ Š

C2 � D8 for n D 2.

(b) G Š M2nC2 � C2 with n � 2, and �2.G/ is abelian of type .4; 2; 2/.

(c) G D ha; e1 j a2nC1

D e21 D 1; n � 2; Œa; e1� D e2; e

22 D Œe1; e2� D Œa; e2� D

1i being minimal nonabelian, jGj D 2nC3, ˆ.G/ D ha2; e2i, G0 D he2i, and�2.G/ D Eha2n�1

i is abelian of type .4; 2; 2/, where E D ha2n

; e1; e2i is anormal elementary abelian subgroup of order 8.

Proof. By assumption, G D Ehai for some a 2 G � E; let jG=Ej D 2n > 2.ObviouslyG does not split overE (otherwise, if G D T E, then�2.G/ D �2.T /E

is of order 25 > 24) and so o.a/ D 2nC1 and hai \ E D hzi is of order 2, wherez D a2n

. Note that a Sylow 2-subgroup of Aut.E/ is isomorphic to D8, so a inducesan automorphism of order � 4 on E hence a4 2 Z.G/. It is easy to check, that ifCE .a/ is a four-subgroup, then a induces an involutory automorphism on E. Indeed,then a stabilizes the chain f1g < CE .a/ < E with factors of exponent 2.Suppose that a induces an automorphism of order 4 on E. In that case, by what has

been said in the previous paragraph, CE .a/ D hzi so that Z.G/ D ha4i � hzi andCG.E/ D Z.G/E. There are involutions e1; e2 2 E such that hz; e1; e2i D E andea

1 D e1e2, ea2 D e2z. Then CE .a

2/ D hz; e2i D G0, G D ha; e1i, ˆ.G/ D ha2; e2i,

�2.G/ D Eha2n�1

i. We have determined the groups stated in part (a).Suppose that a induces an involutory automorphism on E. We have CE .a/ D

E1 D hz; e2i Š E4 (see Exercise 19, below), Z.G/ D ha2; e2i and G0 D Œa;E� is asubgroup of order 2 in E1 (by Lemma 1.1, since G has an abelian subgroup ha2; Ei

of index 2 and Z.G/ is of index 4 in G). If G0 D hzi, then hai is normal in G. Lete1 2 E�E1. Then ha; e1i Š M2nC2 andG D he2i�ha; e1i which is the group statedin part (b) of our theorem. If G0 D he2i, where e2 2 E1 � hzi and e1 2 E �E1, thenŒa; e1� D e2 and this group is stated in part (c) of our theorem.

In the rest of this subsection we assume that G=E Š Q2n ; n � 3. Then CG.E/ >

E, since a Sylow 2-subgroup of Aut.E/ is isomorphic to D8. Therefore QW D �2.G/

centralizes E, where QW =E D Z.G=E/ (it is important that j�2.G/j D 24), QW isabelian of type .4; 2; 2/ and hzi D Ã1. QW / is of order 2. All elements in QW � E areof order 4 and so no involution in E � hzi is a square in G. Indeed, if x2 2 E � hzi,then o.x/ D 4 and x 62 QW since z is the unique nontrivial square on QW , and this is acontradiction. We first prove the following useful result.

Lemma 52.3. Let, as above,G=E Š Q2n , whereG is an�2-group. We have CG.E/�QW D �2.G/, CG.E/=E is cyclic of order� 2, andG=CG.E/ is isomorphic to C2;E4

or D8.


Proof. Suppose that CG.E/ contains a subgroup H > E such that H=E Š Q8.Then Z.H=E/ D QW =E . Let A=E and B=E be two distinct cyclic subgroups oforder 4 in H=E . Then A and B are abelian maximal subgroups of H . It follows thatA \ B D QW � Z.H/ and therefore QW D Z.H/ is of order 24. Using Lemma 1.1,we get 26 D jH j D 2jZ.H/jjH 0j, which gives jH 0j D 2. Since �1.H/ D E, weget H 0 � E and H=E is abelian, a contradiction. It follows that CG.E/=E must bea normal cyclic subgroup of G=E and so G=CG.E/ is isomorphic to a nonidentityepimorphic image of Q2n of order � 8 such that G=CG.E/ 6Š Q8 (the last conditionis superfluous if n > 3). We conclude that G=CG.E/ 2 fC2;E4;D8g.

Theorem 52.4. Let G be a 2-group of order > 24 satisfying j�2.G/j D 24. Suppose,in addition, that G has a normal elementary abelian subgroup E of order 8 such thatG=E Š Q2n , n � 3, and E 6� ˆ.G/. Then we have

G D ha; b; e j a2n

D b8 D e2 D 1; n � 3; a2n�1

D b4 D z; ab D a�1;

Œe; b� D 1; Œe; a� D z; D 0; 1i:

Here G is a group of order 2nC3, G0 D ha2i, ˆ.G/ D ha2; b2i is abelian of type.2n�1; 2/. The subgroup E D he; a2n�2

b2; zi is a normal elementary abelian sub-group of order 8 in G and G=E Š Q2n . If D 0, then Z.G/ D he; b2i is abelian oftype .4; 2/. If D 1, then Z.G/ D hb2i Š C4. Finally,�2.G/ D Ehb2i is abelian oftype .4; 2; 2/.

Proof. Let QW =E D Z.G=E/; then QW D �2.G/ since j�2.G/j D 16 and the sub-group QW has order 16 and exponent � 4. It follows from Aut.E/ D GL.3; 2/ that aSylow 2-subgroup of Aut.E/ is dihedral of order 8 so CG.E/ > E. We conclude thatQW is abelian of type .4; 2; 2/. Since E 6� ˆ.G/, there is a maximal subgroupM of Gsuch thatE 6� M ; thenG D EM . We haveE0 D E\M Š E4 (obviously, E0 is nor-mal in G since E andM are), QW0 D QW \M is abelian of type .4; 2/, QW0 D �2.M/,and M=E0 Š Q2n , n � 3. If ˆ. QW0/ D hzi, then z 2 E0. By Lemma 42.1, M isisomorphic to a group (c) in that proposition. HenceM is metacyclic with a cyclic nor-mal subgroup hai of order 2n and the cyclic quotient groupM=hai of order 4 so thatthere is an element b of order 8 inM � .E0hai/ with b2 2 QW0 �E0, b4 D z D a2n�1

,hbi \ hai D hzi, QW0 D hb2iha2n�2

i � ˆ.M/. We have used the following facts:�1. QW / D E, Ã1. QW / D hzi, hai \ hbi is of order 2 (by the product formula),ˆ.G/ D Ã1.G/. We have (see Lemma 42.1(c))ˆ.M/ D ha2; b2i � QW0,M 0 D ha2i,ab D a�1. Also, Z.M/ D hb2i is cyclic of order 4 and so E0hai D hb2; ai is abelianof index 2 in M , and CM .E0/ D E0hai (again by Lemma 42.1(c)). The element binduces an involutory automorphism on E (because b2 2 Z.G/). The element b doesnot centralize E0 since b 62 ha;E0i D CM .E0/. For every u 2 E0 � f1; zg, wehave hb; ui Š M16 (we have jhb;E0ij D 16 so E0 normalizes hbi). We conclude thatZ.hb;E0i/ D hzi. The subgroup QW D hb2; Ei is abelian so b induces an involutoryautomorphism on E. Therefore, by Exercise 19, Z D CE .b/ is of order 4. By the


above, Z ¤ E0 so there is an element e 2 E � E0 such that b centralizes e. Setu D a2n�2

b2 2 E0 � hzi and we have ub D uz. We act with the cyclic group hai onE. By the above, ha;E0i is abelian. It follows from jE W E0j D 2 that hai stabilizesthe chain E > E0 > f1g and so a2 centralizes E and ea D ey with y 2 E0. Thisgives ae D ay. We may set ab D a�1. Then we act on b�1ab D a�1 with e. Itfollows b�1ayb D .ay/�1 and so a�1yb D a�1y which gives yb D y and y 2 hzi.If y D 1, then G D hei � M . We set ae D az, where D 0; 1. The group G iscompletely determined.

Theorem 52.5. Let G be a 2-group of order > 24 satisfying j�2.G/j D 24. Suppose,in addition, that G has a normal elementary abelian subgroup E of order 8 such thatG=E Š Q2n , n � 3, and E � ˆ.G/. Then one of the following holds:

G D ha; b j a2n

D b8 D 1; a2n�1

D b4 D z; b2 D a2n�2

e;(a)

ab D a�1ue�; � D 0; 1;

u2 D e2 D Œe; u� D Œe; z� D Œu; z� D 1; ea D ez;

ua D u; eb D ez; ub D uz; n � 3; and if � D 1; then n � 4i:

Here G is a group of order 2nC3, G0 D hue�i � ha2i is abelian of type .2n�1; 2/,ˆ.G/ D Eha2i, where E D hz; e; ui is a normal elementary abelian subgroup oforder 8 in G with G=E Š Q2n . Finally, Z.G/ D hzi is of order 2, and �2.G/ D

Eha2n�2

i is abelian of type .4; 2; 2/.

G D ha; b j a2n

D b8 D 1; n � 3; a2n�1

D b4 D z;(b)

b2 D a2n�2

u; ab D a�1e;

u2 D e2 D Œe; u� D Œe; z� D Œu; z� D Œa; e� D Œa; u� D Œb; e� D 1;

ub D uzi:

HereG is a group of order 2nC3, G0 D ha2ei is cyclic of order 2n�1,ˆ.G/ D Eha2i,where E D hz; e; ui is a normal elementary abelian subgroup of order 8 in G withG=E Š Q2n and CG.E/ D Ehai. Finally, Z.G/ D he; b2i, and �2.G/ D Ehb2i isabelian of type .4; 2; 2/.

Proof. We have in this case ˆ.G/ � QW D �2.G/, QW =E D Z.G=E/ and QW isabelian of type .4; 2; 2/ since CG.E/ > E (a Sylow 2-subgroup of Aut.E/ is iso-morphic to D8) and j QW j D 24. Hence hzi D Ã1. QW / is a central subgroup of order2 contained in E. By Lemma 52.3, since every normal cyclic subgroup of G=E iscontained in a cyclic subgroup of index 2 in G=E, there is a subgroup M of index2 in G such that M=E is cyclic and CG.E/ � M . (If n > 3, then M is uniquesince G=E Š Q2n has only one cyclic subgroup of index 2. However, if n D 3


and CG.E/ D QW , then we have three possibilities for M since G=E Š Q8 has ex-actly three cyclic subgroups of index 2.) Let a be an element in M � E such thatha;Ei D M . Then o.a/ D 2n since jM=Ej D 2n�1 and M does not split over E(indeed, all involutions of G lie in E). Hence a0 D a2n�2

2 QW � E, a20 D z 2 E.

By the structure of G=E Š Q2n , all elements in .G=E/ � .M=E/ have order 4, sowe have for each x 2 G �M , x2 2 QW � E, x4 D z, x induces an involutory auto-morphism on E since CG.E/ � M and x 62 M , and so CE .x/ Š E4 (see Exercise19). CE .x/ > hzi, and C QW

.x/ D hx2iCE .x/ is abelian of type .4; 2/. It followsthat all elements in G � M have the same order 8. We have M 0 < E (recall thatM=E is cyclic) so jM 0j � 4, and M 0 D Œhai; E�. For each aie 2 M , where e 2 E

and i is an integer, we get .aie/2 D aieaie D a2i .a�ieai /e D a2i Œai ; e� and soˆ.M/ D Ã1.M/ D ha2iM 0. We have exactly the following four cases for the actionof the cyclic subgroup hai of order 2n on E:

(1) a induces an automorphism of order 4 on E.

(2) a induces an automorphism of order 2 on E and ŒE; a� D hui, where u 2

E � hzi.

(3) a induces an automorphism of order 2 on E and ŒE; a� D hzi.

(4) a centralizesE (in that case, obviously,M is abelian).

Case (1). Since a induces an automorphism of order 4 on E, we get n � 4 (recallthat a2n�2

centralizesE) and CE .a/ D hzi (see, in the first paragraph, the proof of theequality CG.x/ Š E4). In this caseM 0 is a four-subgroup contained in E,M 0 > hzi

(if hzi 6� M 0, then hai \M 0 D f1g and a centralizes some element in .M 0/#, whichis not the case) and for each Qe 2 E � M 0, Qea D Qe Qu with some Qu 2 M 0 � hzi andQua D Quz. We see that ha;M 0i Š M2nC1 so CE .a

2/ D M 0 (this is true again since a2

induces an involutory automorphism on E). Since jG W CG.M0/j D 2, it follows that

CG.M0/ D Eha2; b0i, where b0 2 G �M and Eha2; b0i is a maximal subgroup of G.

Set b D b0a so that Qub D Quz for each Qu 2 M 0 � hzi and therefore CE .b/ D he; zi

for some e 2 E �M 0 (see the proof of Theorem 52.4), ea D eu with u 2 M 0 � hzi,ua D uz, and ub D uz. Since G=E Š Q2n , b�1ab D a�1y with y 2 E. We havetwo subcases according to y 2 E �M 0 or y 2 M 0.

Subcase (1a). Here y D e0 2 E �M 0 and so b�1ab D a�1e0. We compute

ab2

D .ab/b D .a�1e0/b D .a�1e0/�1.e0/b D e0a.e0/b D a.e0/a.e0/b D au:

Indeed, if e0 D ez . D 0; 1/, then .e0/a D e0u, .e0/b D e0, and if e0 D euz . D

0; 1/, then .e0/a D eauaz D euuzz D e0uz, .e0/b D e0z. On the other hand,b2 D a0x with x 2 E. But M 0hai D hu; ai is isomorphic to M2nC1 and so ab2

D

au D aa0x D ax implies that x 2 E �M 0. We may set x D eu0 with u0 2 M 0 and sob2 D a0eu

0. We compute

b�1a2b D .a�1e0/2 D a�1e0a�1e0 D .a�1e0a/a�2e0

D .e0/a.e0/a2

a�2 D u0za�2;


since .e0/a D e0u0 .u0 2 M 0 � hzi/ and .e0/a2

D e0z. Therefore .a4/b D ..a2/b/2 D

.u0za�2/2 D a�4 because CE .a

2/ D M 0. Since a0 D a2n�2

2 ha4i (recall thatn � 4), we get ab

0 D a�10 . Obviously, b centralizes b2 D a0eu

0 .u0 2 M 0/ andso a0eu

0 D .a0eu0/b D a�1

0 e.u0/b D a0ze.u0/b which gives .u0/b D u0z and

u0 2 M 0 � hzi. Therefore u0 D uz� .� D 0; 1/ and b2 D a0euz� . On the other hand,

au D ab2

D aa0euz�

D aeu D a.a�1eua/eu D a.eu/.uz/eu D auz, which is acontradiction.

Subcase (1b). Here b�1ab D a�1y with y 2 M 0. This gives .a2/b D .a�1y/2 D

a�1ya�1y D .a�1ya/a�2y D yaya2

a�2 D yzya�2 D za�2 .� D 0; 1/, andso .a4/b D .za�2/2 D a�4. Since n � 4, we get a0 2 ha4i so ab

0 D a�10 . If

b2 2 M 0ha0i, then S D M 0ha; bi is a maximal subgroup of G not containing Esince S \ E D M 0, a contradiction (recall, that, by hypothesis, E � ˆ.G/). Thusb2 D a0e0 with e0 2 E � M 0. Since b centralizes b2, we get a0e0 D .a0e0/

b D

a�10 eb

0 D a0zeb0 and so e

b0 D e0z. Hence e0 D euz� .� D 0; 1/ and b2 D a0euz

� .From b�1ab D a�1y with y 2 M 0, we get b�2ab2 D .a�1y/b D .ab/�1yb D

yayb D a.a�1ya/yb D a.yz�/.yz�/ D a, where � D 0 if y 2 hzi and � D 1 ify 2 M 0 � hzi. On the other hand, we compute

a D ab2

D aa0euz�

D aeu D euaeu D a.a�1eua/eu D a.eu/.uz/eu D auz;

which is a contradiction.

Case (2). Here, by assumption, a induces an automorphism of order 2 on E andM 0 D ŒE; a� D hui, where u 2 E � hzi. It follows hu; zi � Z.G/. For eachQe 2 E � hu; zi, Qea D Qeu. Since ˆ.M/ D hui � ha2i and ˆ.G/ � E, there existsb 2 G � M such that b2 D a0e, where e 2 E � hu; zi and a0 D a2n�2

. We haveab D a�1y with y 2 E (see the last two sentences of the first paragraph of Case(1)). This gives ab2

D .a�1y/b D .ab/�1yb D yayb D ayayb D aa0e D ae D

eae D a.a�1ea/e D aeue D au. Hence yayb D u. If y 2 hu; zi � Z.G/, thenu D yayb D y2 D 1, a contradiction. Therefore y D es with s 2 hu; zi. But thenu D yayb D .es/a.es/b D .eus/.ebs/ D ueeb and so eb D e. Hence b centralizesE D he; u; zi, a contradiction.

Case (3). Here a induces an automorphism of order 2 on E and M 0 D ŒE; a� D

hzi. It follows ˆ.M/ D ha2i � ha0i and so ha0i is normal in G since the cyclicsubgroup ˆ.M/ is G-invariant. Since Ã1.G/ D ˆ.G/ � QW D Eha0i, it followsha0; x

2 jx 2 G � M i D QW . Since hx2i is normal in G for each x 2 G � M andx4 D z (indeed, M=M 0 D M=hzi is an abelian maximal subgroup of G=M 0), weget QW =hzi � Z.G=hzi/. Set U D CE .a/ Š E4 and obviously Z.M/ D U ha2i.Also, U D Z.M/ \ E is normal in G. Suppose that U � Z.G/. Then the four-group G=Eha2i acts as the full stability group of the chain E > U > f1g sinceG=Eha2i acts faithfully on E. (Note that CG.E/ � M .) This is a contradiction sinceE=hzi < QW =hzi � Z.G=hzi/. Hence for each x 2 G � M and each Qu 2 U � hzi,we have Qux D Quz. There exists b 2 G � M such that b2 D a0e with e 2 E � U .


Also, ab D a�1y with y 2 E. Then we compute ab2

D .a�1y/b D yayb D

a.a�1ya/yb D ayayb D aa0e D eae D a.a�1ea/e D aeze D az. This gives

(1) yayb D z.

On the other hand, b centralizes b2 and so a0e D .a0e/b D ab

0eb which gives

(2) a0e D ab0e

b .

We compute further b�1a2b D .a�1y/.a�1y/ D .a�1ya/.a�2ya2/a�2 D yaya�2,and so we get

(3) .a2/b D yaya�2. There are two subcases according to y 2 U or y 2 E � U .

Subcase (3a). Suppose that y 2 U D CE .a/. Then (1) implies yb D yz and soy D u 2 U�hzi and b�1ab D a�1u. From (3) we get .a2/b D yaya�2 D yya�2 D

a�2 and so ab0 D a�1

0 since a0 is a power of a2. From (2) follows eb D a�b0 a0e D

a20e D ez and we have determined the group stated in part (a) of our theorem for� D 0.

Subcase (3b). Suppose that y 2 E � U , where ab D a�1y. From (1) we getyzyb D z and so yb D y. The relation (3) implies .a2/b D a�2z. We compute, foreach x 2 E and each integer i , .b.a2/ix/2 D b2z� and .ba.a2/ix/2 D b2yz� , where�; � D 0; 1. Hence the square of each element in G �M lies in hb2i D ha0ei and inha0eyi D hb2yi. Since ˆ.G/ � E, we must have y D eu with some u 2 U � hzi.From yb D y follows eu D ebuz (since ub D uz) and so eb D ez. Then (2) impliesa0e D ab

0ez and so ab0 D a0z D a�1

0 . From .a2/b D a�2z follows that ha2i > ha0i

which gives n � 4. We get the group stated in part (a) with � D 1 and n � 4.

Case (4). Here a centralizes E and so M is abelian of type .2n; 2; 2/, n � 3; andha0i is normal in G, where a0 D a2n�2

. Since ˆ.G/ � QW D Eha0i, there existsb 2 G �M such that b2 D a0u, where u is an involution in E � hzi (recall that allelements in G � M have the same order 8). Indeed, if such b does not exist, we getb2 D a0 or a3

0 for all b 2 G � M . Then ˆ.G/ D Ã1.G/ � E, a contradiction.Since CG.b

2/ � hM;bi D G, we get b2 2 Z.G/, ha0; b2i D ha0i � hui is abelian

of type .4; 2/ and ˆ.G/ � ha2; ui. Again ˆ.G/ � QW and ha2; b2i � QW impliesthat there exists c 2 G � M such that c2 2 QW � ha0; b

2i and (as before) c2 2

Z.G/. If a0 2 Z.G/, then ha0; b2; c2i D QW lies in Z.G/ contrary to Lemma 52.3.

Hence a0 62 Z.G/ and therefore ab0 D a�1

0 and ub D uz (recall that u D a�10 b2 62

Z.G/). The subgroup Z.G/ D hb2; c2i is abelian of type .4; 2/. Lemma 1.1 impliesjGj D 2nC3 D 2jZ.G/jjG0j and so jG0j D 2n�1. Since j.G=E/0j D 2n�2 andˆ.G/ > G0 > hzi, G0 is cyclic of order 2n�1, by the modular law. Set ab D a�1e

with e 2 E. If e 2 ha0; ui D ha0; b2i, then ha; bi, a metacyclic maximal subgroup

of G, does not contain E, a contradiction. Thus e 2 E � hz; ui and we compute.ba/2 D baba D b2.b�1ab/a D b2.a�1e/a D b2e, Since both b2 and b2e lie inZ.G/, it follows e 2 Z.G/ and therefore Z.G/ D he; b2i. Finally, G0 D ha2ei iscyclic of order 2n�1. We get the group from part (b).


3o. 2-groups with exactly one subgroup of order 24 and exponent 4.

Theorem 52.6. The following two statements for a 2-groupG of order> 24 are equiv-alent: (a) j�2.G/j D 24 and (b) G has exactly one subgroup of order 24 and expo-nent 4.

Proof. Suppose that (a) holds. The results of Subsections 1 and 2 imply that�2.G/ isisomorphic to one of the following groups Q8 �C4, Q8 �C2, D8 �C2, C4 �C2 �C2,C4 � C4. In particular, exp.�2.G// D 4 and so .b/ holds.Suppose now that (b) holds. LetH be the unique subgroup of order 24 and exponent

4 inG, where jGj > 24. We want to show thatH D �2.G/. Suppose that this is false.Then there exist elements of order� 4 inG�H , whereH is a characteristic subgroupof G. In particular, there is an element a 2 G �H such that o.a/ � 4 and a2 2 H .Set QH D H hai so that j QH j D 25. SinceH is neither cyclic nor a 2-group of maximalclass, there exists a G-invariant four-subgroup W0 contained inH , by Lemma 1.4.Let x be any element in QH � H with o.x/ � 4. Then hx;W0i is a subgroup

of order � 24 in QH and so there exists a maximal subgroup M of QH containinghx;W0i. Since exp.M \ H/ � 4 and M D hM \ H;xi, we have �2.M/ D M .But jM j D 24 and M ¤ H , so M is either elementary abelian or exp.M/ D 8.Suppose that exp.M/ D 8. Then M is a nonabelian group of order 24 with a cyclicsubgroup of index 2. SinceM has the normal four-subgroup W0, it follows thatM isnot of maximal class. Thus M Š M24 . But �2.M24/ is abelian of type .4; 2/ whichcontradicts the above fact that�2.M/ D M .We have proved thatM Š E24 . In particular, x must be an involution and exp. QH/ D

4 because for each y 2 QH , y2 2 M . Hence all elements in QH � H are of order� 4 and so (by the above) all these elements must be involutions. In that case, everyinvolution x 2 QH � H inverts H . Let k be any element of order 4 in H . Sincekx D k�1, hk; xi D D Š D8. Let QM be a maximal subgroup of QH containing D.Then j QM j D 24, exp. QM/ D 4, QM ¤ H , and this is a final contradiction.

4o. Generalization of Theorem 43.4. Here we improve Theorem 43.4 and someresults of [Ber25, �48]. In Theorem 52.7 we classify the 2-groups with exactly onabelian subgroups of type .4; 2/; the proof is independent of the proof of Theorem43.4. The following remark shows that there are few 2-groups without abelian sub-groups of type .4; 2/.

Remarks. 1. Let us classify the 2-groups without abelian subgroup of type .4; 2/.Of course, if G is cyclic, elementary abelian or of maximal class, it has no abeliansubgroups of type .4; 2/. So in what follows suppose that G is of exponent > 2 and isneither cyclic nor of maximal class. Let A be a cyclic subgroup of order 4 in G. ThenCG.A/ � A has no involutions. It follows that CG.A/ is cyclic (Proposition 1.3). BySuzuki’s theorem (Proposition 1.8), jCG.A/j > 4. By Lemma 1.4, G has a normalabelian subgroup R of type .2; 2/. Since Z.G/ < CG.A/, we get R \ CG.A/ is of


order 2. It follows that H D RCG.A/ Š M2n for some n > 3. Let B be a subgroupof order 4 in Z.H/. In that case, RB is abelian of type .4; 2/, which is not the case.Thus, our G is elementary abelian, cyclic or of maximal class.

2. Let G be a p-group without abelian subgroup of type .p2; p/, p > 2, exp.G/ >p. Then, as in Remark 1, we obtain exp.G/ D p2 and CG.x/ D hxi so G is ofmaximal class (Proposition 1.8). In particular, jGj � p2p�1 (Theorem 9.6). We claimthat jGj � ppC1. Assume that this is false. Let G1 be a fundamental subgroup of G.Then exp.G1/ > p and Z.G1/ is noncyclic. In that case, obviously, G1 contains anabelian subgroup of type .p2; p/, a contradiction. Thus, jGj � ppC1. It is easy tocheck that a Sylow p-subgroup of the symmetric group of degree p2 has no abeliansubgroups of type .p2; p/.

Theorem 52.7. Let G be a 2-group containing exactly one abelian subgroup of type.4; 2/. Then one of the following holds:

(a) j�2.G/j D 8 and G is isomorphic to one of the metacyclic groups (a), (b) or (c)in Lemma 42.1.

(b) G Š C2 � D2nC1 , n � 2.

(c) G D hb; t j b2nC1

D t2 D 1; bt D b�1C2n�1

u; u2 D Œu; t � D 1; bu D

b1C2n

; n � 2i. Here jGj D 2nC3, Z.G/ D hb2n

i is of order 2, ˆ.G/ D

hb2; ui < hb; ui Š M2nC2 , E D hb2n

; u; ti Š E8 is self-centralizing in G,�2.G/ D hui � hb2; ti Š C2 � D2nC1 , G0 D hb2n

; ui Š E4 in case n D 2,and G0 D hb2ui Š C2n for n � 3. Finally, the group G for n D 2 (of order25) is isomorphic to the group (a) in Theorem 52.2 for n D 2 (since �2.G/ Š

C2 � D8).

Proof. Let G be a 2-group with exactly one abelian subgroup A of type .4; 2/; then Ais characteristic inG. SetC D CG.A/. ThenC is normal inG andG=C is isomorphicto a subgroup of Aut.A/ Š D8. We claim that �2.C / D A. Indeed, let y 2 C � A

such that o.y/ � 4 and y2 2 A. Then Ahyi is an abelian subgroup of order 24 andexponent 4. But then Ahyi contains an abelian subgroup of type .4; 2/ distinct fromA, a contradiction. Thus �2.C / D A, as claimed. Lemma 42.1 implies that C mustbe abelian of type .2n; 2/, n � 2. If �2.G/ D �2.C / D A, then G is metacyclic andG is isomorphic to one of the groups in Lemma 42.1.We assume from now on that �2.G/ > A. Set U D �1.A/ and hzi D ˆ.A/ so

that z 2 Z.G/ and U is a normal four-subgroup of G. Let a 2 G � C be such thato.a/ � 4 and a2 2 C . Obviously, a2 2 U since U D �1.C /. Let D D haiC ; thenjD W C j D 2. Since D is a nonabelian group (of order � 24) containing a normalfour-subgroup U , it follows that D is not of maximal class. If o.a/ D 4, then a doesnot centralize U (otherwise U hai would be abelian of type .4; 2/ distinct from A) andsoU hai Š D8. But then there exists an involution in .U hai/�U . In any case the cosetCa .� Ua/ contains involutions and let t be one of them. If t centralizes an elements of order 4 in C , then hs; ti is an abelian subgroup of type .4; 2/ distinct from A, a


contradiction. Hence CC .t/ is of exponent 2 and (since D is not of maximal class)CC .t/ D U (Proposition 1.8) and E D CD.t/ D U � hti Š E8. It is easy to checkthat, for each x 2 D�C D C t , CC .x/ D U . Hence x2 2 U and hU; xi is elementaryabelian (of order 8). Thus, all elements inD�C are involutions and therefore t invertsC . Let B be a cyclic subgroup of index 2 in C and an involution u 2 C � B . ThenhB; ti Š D2nC1 and D Š C2 � D2nC1 , n � 2, where C2 D hui. Obviously, we maychoose a 2 G � C so thatD D ha;C i GG.Suppose that G=C has a cyclic subgroup K=C of order 4. Then K > D since

D=C , in the case under consideration, is contained in ˆ.G=C/, and let k 2 K besuch that K D hk;C i. We have k2 2 D � C and so k2 is an involution. It followso.k/ D 4 and so hk; zi is an abelian subgroup of type .4; 2/ distinct from A sincehki \ C D f1g, and this is a contradiction. We have proved that G=C is elementaryabelian. If jG=C j D 2, then G D D and we are done.It remains to study the possibility G=C Š E4. By the above, if y 2 G � D

and o.y/ � 4, then y2 2 C and y is an involution inverting C (indeed, as in thesecond paragraph of the proof, all elements in hy;C i � C are involutions). But thenyt 62 C and yt centralizes C , a contradiction (C D CG.A/ is self centralizing in G).Hence, for each element y 2 G �D, o.y/ � 8 and y2 2 C . This gives�2.G/ D D,CG.t/ D E D hti��1.A/ and soE is a self-centralizing elementary abelian subgroupof order 8 in G.For each y 2 G � D, we have y2 2 C and o.y2/ � 4 and so y centralizes a

cyclic subgroup of order 4 in A. Since y does not centralize A, it follows that y doesnot centralize U D �1.A/ since y 62 C D CG.A/. Hence D D CG.U / and soZ.G/ D hzi D ˆ.A/ is of order 2. Since �2.hyiC/ D A and hyiC is nonabelian,Lemma 42.1 implies that hyiC is isomorphic to a group (a) or (c) of that lemma.Assume that there is b 2 G �D such that hbiC is isomorphic to a group of Lemma

42.1(c). In particular, jhbiC j � 25, C is the unique abelian maximal subgroup ofhbiC (since Z.hbiC/ Š C4), o.b/ D 8, Z.hbiC/ D hb2i < A, hb4i D hzi D ˆ.A/,C contains a cyclic subgroup hai of index 2 such that hai is normal in hbiC , o.a/ �

23, A \ hai Š C4, hbi \ hai D hzi, and ab D a�1. Also, we see that for eachy 2 .hbiC/ � C , CC .y/ D hb2i Š C4, and y is of order 8. Since t inverts C ,we get abt D a and so jCC .bt/j � 23. This implies that hbtiC Š M2nC2 becauseCC .bt/ � Z.hbtiC/.Replacing b with bt (if necessary), we may assume from the start that there is an

element b 2 G � D such that hbiC Š M2nC2 .n � 2/, which is a group in part (a)of Lemma 42.1. We have o.b/ D 2nC1, hbi is a cyclic subgroup of index 2 in hbiC ,z D b2n

, U D hz; ui D �1.hbiC/, and ub D uz. This gives also bu D bz D b1C2n

,Z.hbiC/ D hb2i Š C2n , A D hb2n�1

D v; ui, C D hb2i � hui. We see thatCC .bt/ D hvui Š C4 and so Z.hbtiC/ D hvui. It follows that hbtiC Š M24 forn D 2 and hbtiC is isomorphic to a group (c) of Lemma 42.1 for n > 2. In any case,.bt/2 2 hvui � hzi and so .bt/2 D vu or .bt/2 D v�1u D vuz. Replacing u with uz(if necessary), we may assume that .bt/2 D vu, where v D b2n�1

. From this crucial


relation follows bt D b�1C2n�1

u. The structure of G D hb; ti is determined and ourtheorem is proved.

Theorem 52.8. The following two statements for a 2-group G are equivalent: (a)c2.G/ D 4; (b) G has exactly one abelian subgroup of type .4; 2/.

Proof. Suppose that G is a 2-group having exactly two cyclic subgroups U;V of order4. Then jG W NG.U /j � 2 and so V � NG.U /. Similarly, U � NG.V /. We getŒU; V � � U \ V . Since A D hU;V i has exactly two cyclic subgroups of order 4, Amust be abelian of type .4; 2/. But A is generated by its two cyclic subgroups of order4 and so (b) holds.Let (b) hold. Then the groupG is completely determined by Theorem 52.7. Looking

at�2.G/, we see that G has exactly two cyclic subgroups of order 4.

Exercise 1. Let G be a 2-group, exp.G/ � 2n with n > 1. Let H < G contains allcyclic subgroups of G of order 2n. If x 2 H with o.x/ D 2n, then �n�1.CG.x// �

H .

Exercise 2. Let G be a 2-group with �2.G/j � 25. Prove that dl.G/, the derivedlength of G, is at most 5.

Solution. If G has no normal elementary abelian subgroups of order 8, then thedl.G/ � 4, by �50. Now let E be a normal elementary abelian subgroup of G oforder 8. Then j�1.G=E/j � 4 so dl.G=E/ � 4, and we are done.

Exercise 3. Let G be a 2-group and n > 3. Denote by Mn the set of noncyclicsubgroups of order 2n in G which are abelian of type .2n�1; 2/ or M2n . Study thestructure of G in the case jMnj D 1.

Exercise 4. Study the 2-groups without abelian subgroups (i) of type .4; 4/, (ii) of type.4; 2; 2/.

Exercise 5. Suppose that a noncyclic p-group G has no abelian subgroups of type.p2; p/. If exp.G/ > p, then either G is a 2-group of maximal class or p > 2 and Gis of maximal class and order � ppC1.

Exercise 6. Let G be a noncyclic p-group of order > p4 and exponent > p, p > 2.Prove that if G has no subgroups of order p4 and exponent � p2, then G has a cyclicsubgroup of index p.

5o. 2-groups G with j�3.G/j D 25. The ultimate goal is to classify finite 2-groups G of order > 26 with j�3.G/j D 26 (see #525 in Research problems andthemes I). However, if �3.G/ 6� ˆ.G/, there is a maximal subgroup M of G suchthat j�3.M/j � 25 and without the exact knowledge of the structure ofM there is nochance to determine G in that case.All results of this subsection are due to Z. Bozikov [Boz].


In this subsection we determine up to isomorphism all 2-groups G with j�3.G/j �

25 and G > �3.G/. The cases j�3.G/j D 23 and j�3.G/j D 24 are easy (Propo-sition 52.9). If j�3.G/j D 25, we show first that exp.�3.G// D 23 and j�2.G/j D

24 (Proposition 52.10). This allows us to use the classification of 2-groups G withj�2.G/j D 24 and jGj > 24 given in first two subsections. Indeed, in Theorem 52.11we consider nonmetacyclic 2-groups G and get our result almost immediately. In The-orem 52.12 we consider metacyclic 2-groups G with the above property. This case isdifficult since in �50 there is no single information for metacyclic groups but we haveto determine the groups G up to isomorphism also in this case.In conclusion, we classify the 2-groups G containing exactly one subgroup of order

25 and exponent � 8 (see #437 in Research problems and themes I). We show thatsuch groups satisfy j�3.G/j D 25. In fact, we get a more general result (Theorem52.14).

5.1o. For the convenience we prove two preliminary results (Propositions 52.9 and52.10).

Proposition 52.9. Let G be a 2-group with G > �3.G/. If j�3.G/j � 23, thenj�3.G/j D 23 and G is cyclic. If j�3.G/j D 24, then �3.G/ is abelian of type .8; 2/and G is either abelian of type .2m; 2/ or G Š M2mC1 , m � 4.

Proof. Set �3.G/ D H . Assume that G is noncyclic. If G is of maximal class,then H � �2.G/ D G, which is a contradiction. Therefore, G has a normal abeliansubgroup R of type .2; 2/. By hypothesis, H=R is cyclic of order 4 so G=R cyclic. Itfollows that R D �1.G/ so G contains a cyclic subgroup of index 2. Now the resultfollows from Theorem 1.2.

Proposition 52.10. Let G be a 2-group with G > �3.G/. If j�3.G/j D 25, thenexp.�3.G// D 23 and j�2.G/j D 24.

Proof. Set�3.G/ D H so that jH j D 25 and G > H . IfH is cyclic, then�3.H/ <

H , a contradiction. If H is of maximal class so G is since c1.G/ D c1.H/ (seeTheorem 1.17); then H � �2.G/ D G, a contradiction. If exp.H/ D 24, then(Theorem 1.2)H is either abelian of type .16; 2/ orH Š M32. In any case,�3.H/ <

H which is again a contradiction.We have proved that exp.H/ D 23 (since exp.H/ � 22 is obviously impossible).

Since H is neither cyclic nor of maximal class, there is a G-invariant four-subgroupR contained in H . We assume (by way of contradiction) that H D �2.G/. Then�2.G=R/ D H=R is of order 23 and jG=Rj > 23. Lemma 42.1 implies that H=Ris abelian of type .4; 2/. In particular, H 0 � R and H 0 is elementary abelian. If theclass cl.H/ of H is � 2, then for each x; y 2 H with o.x/ � 4, o.y/ � 4, we get.xy/4 D x4y4Œy; x�6 D 1, which implies that exp.H/ � 4, a contradiction. Thus(sinceH is not of maximal class) cl.H/ D 3 and soH 0 D R and R 6� Z.H/. We setC D CH .R/ and D D CG.R/. Since jH W C j D 2, D covers G=H and G D HD.


If exp.C / � 4, then taking x 2 D � C such that x2 2 C , we get o.x/ � 8, acontradiction. It follows exp.C / D 8. Let a 2 C with o.a/ D 8. Then C D Rhai,jR \ haij D 2, and so C is abelian of type .8; 2/. Since �3.D/ D C , our Proposition52.9 implies that D is either abelian of type .2m; 2/ or D Š M2mC1 , m � 4. Inany case, ˆ.D/ is cyclic of order 2m�1 and so ˆ.D/ contains a characteristic cyclicsubgroup Z of order 8. It follows that Z � C and Z is normal in G. But jH=Zj D 4

and soH 0 � Z. This contradicts the fact thatH 0 D R and jR \Zj D 2.We have proved that �2.G/ < H . On the other hand, R < �2.G/ and so

j�2.G/j D 8 or 16. Set K D �2.G/ and assume jKj D 8. Then Lemma 42.1shows that either j�3.G/j D 24 (in cases (a) or (b)) or�3.G/ D G (in case (c)). Thisis a contradiction and so jKj D 16 and we are done.

5.2o. Now let G be a nonmetacyclic 2-group of order > 25 with j�3.G/j D 25.We set H D �3.G/ and apply Proposition 52.10. We see that exp.H/ D 23 and�2.G/ D K is of order 24 so that jH W Kj D 2. By Theorem 41.1, K is nonmeta-cyclic. We are in a position to use the first part of this section. Since jGj > 25, we geteither K Š Q8 � C4 or K Š C4 � C2 � C2. All elements inH �K are of order 8.Assume first that K D Q �Z, whereQ Š Q8, Z Š C4, andQ \ Z D Z.Q/. All

elements inK � .Q[Z/ are involutions and soQ is the unique quaternion subgroupin K. Thus Q is normal in G and so also C D CG.Q/ is normal in G. SincejG=.QC/j � 2 and jGj > 25, we have jC j � 23 and C \K D Z. Also,�2.C / D Z

and so Lemma 42.1 implies that C is cyclic and H D Q � �3.C /. Suppose thatjG=.QC/j D 2. Since G=C Š D8, there is an element c 2 G � .QC/ such thatc2 2 C . Set P D C hci. Again, �2.P / D Z and so P D hci is cyclic. But hci

induces on Q an “outer” involutory automorphism and so we may set Q D ha; bi sothat ac D a�1 and bc D ba.Assume nowK Š C4 �C2 �C2 so thatE D �1.K/ is a normal elementary abelian

subgroup of order 8 inG. We haveÃ1.K/ D hzi is a central subgroup of order 2 inG.Obviously,K=E is the unique subgroup of order 2 inG=E and soG=E is either cyclic(of order � 8) or generalized quaternion. Suppose that G=E is generalized quaternionand letL=E be a cyclic subgroup of index 2 inG=E. For each x 2 G�L, x2 2 K�E

and so o.x/ D 8 which forces�3.G/ D G, a contradiction.We have proved that G=E is cyclic. Let a 2 G � K be such that hai covers G=E.

Then hai is cyclic of order 2m,m � 4, G D Ehai,K\ hai Š C4, and E \ hai D hzi.We may set E D he; u; zi so that we have the following possibilities for the action ofhai on E:

(i) ea D eu, ua D uz (here a induces an automorphism of order 4 on E);

(ii) ea D e, ua D uz (here G D hei � ha; ui Š C2 �M2mC1 ;

(iii) ea D eu, ua D u (here G is minimal nonabelian);

(iv) ŒE; a� D 1 (here G is abelian of type .2m; 2; 2/).


We have proved the following result.

Theorem 52.11 ([Boz]). Let G be a non-metacyclic 2-group of order > 25 withj�3.G/j D 25. Then one of the following holds.

(a) G D QP , where Q D ha; bi is a normal quaternion subgroup of G, P D hci

is cyclic of order 2m, m � 4, Q \ P D Z.Q/ and c either centralizes Q orac D a�1 and bc D ba.

(b) G D EP , where E D he; u; zi is a normal elementary abelian subgroup oforder 8, P D hai is cyclic of order 2m, m � 4, and E \ P D hzi with z D

a2m�1

. For the action of hai on E we have one of the following possibilities:

(i) ea D eu, ua D uz, and here a induces an automorphism of order 4 onE;

(ii) ea D e, ua D uz, and here G Š C2 �M2mC1 ;

(iii) ea D eu, ua D u, and here G is minimal nonabelian;

(iv) ea D e, ua D u, and here G is abelian of type .2m; 2; 2/.

5.3o. Now we assume thatG is a metacyclic 2-group of order> 25 with j�3.G/j D

25. We setH D �3.G/. By Proposition 52.10, exp.H/ D 23 and �2.G/ D K is oforder 24. The first part of this section implies that K Š C4 � C4 and all elements inH � K are of order 8. It follows that R D �1.K/ is a normal 4-subgroup. Supposethat CH .R/ D K. Since CG.R/ covers G=H , there is an element x 2 CG.R/ � H

with x2 2 K. But then o.x/ � 8, a contradiction. We have proved that R � Z.H/.Let Y=K be a subgroup of order 2 in G=K. Since Y � �3.G/, we get Y D H .

Hence H=K is the unique subgroup of order 2 in G=K. This implies that G=K iseither cyclic (of order � 4) or generalized quaternion.We study first the case, whereG=K is cyclic. Let a 2 G�K be such that hai covers

G=K. Then hai\H Š C23 , hai\K Š C22 , and hai\R D hzi � Z.G/. Thus a is oforder 2m, m � 4, and we set s D a2m�3

and v D a2m�2

so that s2 D v, v2 D z, ando.s/ D 8. Set C D CG.K/. Since G=C is cyclic and acts faithfully on K, we havejG=C j � 4. For the structure of Aut.C4 � C4/ see Proposition 50.5. Also, G0 � K

and G0 is cyclic which implies that jG0j � 4.If G0 D f1g, then G is abelian of type .2m; 4/, m � 4.Suppose jG0j D 2. Then Exercise 10.13 implies that G is minimal nonabelian.

Since jGj > 8, Exercise 1.8a together with the fact that j�3.G/j D 25 gives G D

ha; b j a2m

D b4 D 1; m � 4; ab D a1C2m�1

or ba D b�1i.Suppose jG0j D 4 and jG W C j D 2. Then a induces an involutory automorphism

on K, H D hK; si � C , and so H is abelian of type .8; 4/. If a centralizes K=R,then G0 � R. But G0 is cyclic and so jG0j � 2, a contradiction. Hence a acts non-trivially on K=R and so a acts also non-trivially on R D Ã1.K/. There is an elementw 2 K � R such that w2 D u 2 R � hzi, wa D w0, .w0/2 D uz, K D hwi � hw0i,and ua D uz. Since a induces an involutory automorphism on K, .w0/a D wa2

D w


and therefore CK.a/ D hww0i D hvi with .ww0/2 D z. From waw�1 D Œa;w�1� D

w0w�1 D w0wu D ww0u follows that G0 D hww0ui Š C4, .ww0u/2 D z, andG0 6� Z.G/. From the above, ww0 D vz� .� D 0; 1/ and set l D swu�C1. Thenl2 D s2w2 D vu D ww0uz� 2 G0, .vu/2 D z, and so hli Š C8 is normal in G withhli \ hai D hzi. We compute

la D .swu�C1/a D sw0u�C1z�C1 D s�1s2w0u�C1z�C1

D s�1vw0u�C1z�C1 D s�1.ww0z�/w0u�C1z�C1

D s�1w�1w2.w0/2u�C1z D s�1w�1uuzu�C1z

D s�1w�1u�C1 D l�1:

We have obtained the following metacyclic groupG D ha; l j a2m

D l8 D 1, a2m�1

D

l4, la D l�1, m � 4i, whereH D hl; si D hl; a2m�3

i is abelian of type .8; 4/.It remains to study here the case jG0j D 4 and jG W C j D 4. Hence a induces on

K an automorphism of order 4 and so, in particular, a acts nontrivially on R. Thisimplies that CK.a/ D hvi, where hv2i D hzi D CR.a/. We may set K D hwi � hw0i,where wa D w0, w2 D u 2 R � hzi, .w0/2 D uz, ua D uz, and .w0/a D ws0 withs0 2 R. We have s0 ¤ 1 (otherwise a induces on K an involutory automorphism).Since .ww0/2 D w2.w0/2 D uuz D z, it follows that ww0 2 Rhvi. We compute.ww0/a D w0ws0 D .ww0/s0, which implies hww0i ¤ hvi and so hww0ui D hvi.Here we have used the fact that Rhvi contains exactly two cyclic subgroups of order4 and they are hvi and hvui. This gives (since a centralizes v) ww0u D .ww0u/a D

w0ws0uz, and consequently s0 D z. Hence .w0/a D wz and so the action of hai onKis uniquely determined. We see that Œa;w�1� D waw�1 D w0w�1 D w0wu D ww0u

and so G0 D hww0ui. But hww0ui D hvi � hai and so hai is normal in G. Also, hwi

induces on hai an automorphism of order 4 with hwi \ hai D 1 and jhai W Chai.w/j D

4. This determines uniquely the structure of our “splitting” metacyclic group G D

ha;w j a2m

D w4 D 1; m � 4; aw D a1C2m�2

i, where H D hw; a2m�3

i. Form D 4, H is nonabelian and for m > 4, H is abelian of type .8; 4/. Here G0 D

ha2m�2

i � Z.G/ D ha4i.We have to study the difficult case, where G=K is generalized quaternion. Since a

generalized quaternion group is not a subgroup of Aut.K/ (see Proposition 50.5), wehave CG.K/ � H and consequently H is abelian of type .8; 4/. Note that H=K D

Z.G=K/. We set L=K D .G=K/0 D ˆ.G=K/ so that L=K is cyclic containingH=K .We have G=L Š E4 and since G is metacyclic (and so d.G/ D 2), we get L D ˆ.G/

and G0 is cyclic. On the other hand, G0 covers L=K and all elements in H � K areof order 8. Hence G0 \ H Š C8, G0 \ K Š C4, and jL W G0j D 4. Again, sinceG is metacyclic, there is a cyclic normal subgroup S of G such that G0 < S andjS W G0j D 2. If S � L, then jS \ Kj D 8. But a maximal subgroup S \ K ofK is not cyclic since exp.K/ D 4. This contradiction shows that S \ L D G0. SetT D KS D LS so that T=K is a cyclic subgroup of index 2 in G=K. For each


x 2 G � T , we have x2 2 H �K and so all elements in G � T are of order 16. Now,ˆ.T / � L and ˆ.T / � hG0; Ã1.K/ D Ri since G0 D Ã1.S/. But jL W .G0R/j D 2

and T is noncyclic and so ˆ.T / D G0R. Then F D ˆ.T / \ H is abelian of type.8; 2/ and F1 D ˆ.T / \K is abelian of type .4; 2/.Set S D hai, where o.a/ D 2m, m � 4. Also, set s D a2m�3

, v D a2m�2

, andz D a2m�1

so that z 2 Z.G/, hsi D S \ H , hvi D S \ K, and hzi D S \ R.Obviously, hsi and hvi are normal subgroups in G. Since ˆ.G/ D L, there is anelement b 2 G � T (of order 16) such that b2 2 H �K and b2 62 F . Then b2 D ws,wherew 2 K�F1 andK D hwi� hvi. We setw2 D u and so R D �1.K/ D hu; zi.We compute b4 D w2s2 D uv and b8 D z. Hence G D haihbi, where hai is acyclic normal subgroup of order 2m .m � 4/ of G, hbi is cyclic of order 16, andhai \ hbi D hzi (of order 2).It remains to determine the action of hbi on hai. Now, hbi induces a cyclic auto-

morphism group on hai so that b inverts hai=hvi since G=K is generalized quater-nion. Thus ab D a�1v0 with v0 2 hvi and so .a4/b D .ab/4 D .a�1v0/

4 D

a�4. In particular, b inverts hvi. On the other hand, b centralizes b4 D uv andso uv D .uv/b D ubvb D ubv�1, which gives ub D uz. We compute ab2

D

.a�1v0/b D .a�1v0/

�1v�10 D av�2

0 D az� , � D 0; 1. Since b2 D sw, this gives

asw D aw D az� and au D aw2

D .az�/w D az�z� D a and so Œa; u� D 1. Re-placing a with au we get o.au/ D o.a/ D 2m and .au/b D a�1v0uz D .au/�1v0z.Hence replacing a with au (if necessary), we may assume that v0 D v D a2m�2

orv0 D 1. We have obtained exactly two possibilities for our group G D ha; b j a2m

D

b16 D 1; m � 4; a2m�1

D b8; ab D a�1C��2m�2

; � D 0; 1i. The following resultwas proved.

Theorem 52.12 ([Boz]). LetG be a metacyclic 2-group of order> 25 with j�3.G/j D

25. Then one of the following holds.

(a) G is abelian of type .2m; 4/, m � 4.

(b) G D ha; b j a2m

D b4 D 1; m � 4; ab D a1C2m�1

or ba D b�1i is minimalnonabelian.

(c) G D ha; l j a2m

D l8 D 1; m � 4; a2m�1

D l4; la D l�1i, where �3.G/ D

hl; a2m�3

i is abelian of type .8; 4/, G0 D hl2i Š C4, Z.G/ D ha2i Š C2m�1 ,and G0 6� Z.G/.

(d) G D ha;w j a2m

D w4 D 1; m � 4; aw D a1C2m�2

i; where �3.G/ D

hw; a2m�3

i, G0 D ha2m�2

i Š C4, Z.G/ D ha4i, and so G0 � Z.G/. Also, form D 4, �3.G/ is nonabelian and for m > 4, �3.G/ is abelian of type .8; 4/.

(e) G D ha; b j a2m

D b16 D 1; m � 4; a2m�1

D b8; ab D a�1C��2m�2

; � D

0; 1i, where �3.G/ D hb2; a2m�3

i is abelian of type .8; 4/ and G=�2.G/ Š

Q2m�1 . If � D 0, then jG W CG.�2.G//j D 2 and if � D 1, then jG W

CG.�2.G//j D 4.


5.4o. We prove here the following result.

Theorem 52.13 ([Boz]). Let G be a 2-group containing exactly one subgroup of order25 and exponent � 8. Then we have j�3.G/j D 25.

In fact, with the similar proof, we get the following more general result.

Theorem 52.14 ([Boz]). Let G be a 2-group containing exactly one subgroup of order2nC2 and exponent � 2n, where n � 2 is a fixed integer. Then we have j�n.G/j D

2nC2.

Proof. Let H be the unique subgroup of order 2nC2 and exponent � 2n .n � 2/.ThenH is a characteristic subgroup of G andH � �n.G/. Suppose that the theoremis false. Then there exists an element a 2 G � H of order � 2n with a2 2 H . SetQH D H hai so that j QH j D 2nC3. Since H is neither cyclic nor of maximal class,H has a G-invariant four-subgroup R. We have jRhaij � 2nC2 and so there is amaximal subgroup M of QH such that M � Rhai. We have exp.M \ H/ � 2n

and since M D .M \ H/hai, we get �n.M/ D M , jM j D 2nC2, M ¤ H , andexp.M/ � 2nC1.The uniqueness of H forces exp.M/ D 2nC1 soM has a cyclic subgroup of index

2. But jM j � 24 (since n � 2) and M has the normal 4-subgroup R. This impliesthatM is not of maximal class and so M is either abelian of type .2nC1; 2/ orM Š

M2nC2 . In both cases �n.M/ < M and this contradicts our result in the previousparagraph. The theorem is proved.

�53

2-groups G with c2.G/ D 4

For a finite 2-group G and a fixed integer n � 1 we denote with cn.G/ the number ofcyclic subgroups of order 2n. The starting point is the following result: if a 2-groupG is neither cyclic nor of maximal class and n > 1, then cn.G/ is even. Thus, forany 2-group G we have c2.G/ D 1 if and only if G is either cyclic or dihedral andc2.G/ D 3 if and only if G 2 fQ8;SD16g.In this section we shall determine (up to isomorphism) all 2-groups G of order

> 24 with c2.G/ D 4. If, in addition, j�2.G/j D 24, then such groups G havebeen determined in �52. In the sequel we assume that j�2.G/j > 24. If moreoverG has a quaternion subgroup, then we get an infinite class of 2-groups and they havethe properties �2.G/ D G and jZ.G/j D 2 (Theorem 53.6). Therefore we assumealso in what follows that G has no quaternion subgroups. Then we show first thatA D hU1; U2; U3; U4i Š C4 � C2 � C2, where fU1; U2; U3; U4g is the set of fourcyclic subgroups of order 4 in G (Theorem 53.7). It is interesting to note here thateach subgroup Ui turns out to be normal in G. If in addition �2.G/ D G, thenG D Bhti, where B is abelian of type .2m; 2; 2/, m � 2, and t is an involutioninverting B and here Z.G/ Š E8 (Theorem 53.8). If jG W �2.G/j � 4, then jG W

�2.G/j D 4 and G is a uniquely determined group of order 27 with jZ.G/j D 2

(Theorem 53.9). Finally, if jG W �2.G/j D 2, then we get an infinite class of 2-groupsG with the properties jGj � 27, G0 6� �1.A/, and Z.G/ Š E4 (Theorem 53.10(a))and we get an exceptional group G of order 26 with Z.G/ Š E4 and G0 � �1.A/

(Theorem 53.10(b)). All groups will be given in terms of generators and relations.The above exceptional groups of orders 26 and 27 have peculiar structure but theyexist as subgroups of the alternating group A16.

Exercise 1 (see Theorem 6.10). Let A be an abelian 2-group of type .2n; 2; : : : ; 2/

and of order 2nCd�1, n > 1, d > 1. Prove that jAut.A/j D 2nCd�2.2d � 2/.2d �

22/ : : : .2d � 2d�1/.

We need the following information about the structure of Aut.C4 � C2 � C2/.

Proposition 53.1. Let S 2 Syl2.Aut.A//, where A Š C4 � C2 � C2. Let F be thestability group of the chain A > �1.A/ > f1g. Then F Š E8 is normal in Aut.A/and S D D F , where F \D D f1g and D Š D8. If X is any subgroup of S suchthat X \ F D hti is of order 2, then t is either a square in X or t 62 ˆ.X/.

�53 2-groups G with c2.G/ D 4 97

Proof. By Exercise 1 with n D 2 and d D 3, we get jAut.A/j D 8 6 4 D 26 3

and so jS j D 26. Let A D hv; u; ei, where o.v/ D 4, o.u/ D 2 D o.e/. The stabilitygroup F of the chain A > �1.A/ > f1g is normal in Aut.A/, F Š E8, and so F � S .Consider the automorphisms ˛ and ˇ of A given by v˛ D v, u˛ D uz, e˛ D eu;vˇ D v, uˇ D uz, eˇ D e, where z D v2. Then o.˛/ D 4, o.ˇ/ D 2, ˛ˇ D ˛�1,and so D D h˛; ˇi Š D8 with D \ F D f1g. Since jFDj D 26 D jS j, we may setS D D F .Let X � S be such that X \ F D hti is of order 2 and assume that t is not

a square in X . Since F is normal in S , hti � Z.X/ and X=hti is isomorphic toa subgroup of S=F Š D8. If X=hti is elementary abelian, then ˆ.X/ � hti andtherefore ˆ.X/ D f1g since t is not a square in X ; in that case t 62 ˆ.X/. If X=htiis cyclic of order 4, then, since X is not cyclic, we get t 62 ˆ.X/.D Ã1.X//. Finally,suppose that X=hti Š D8. Let U=hti be the cyclic subgroup of index 2 in X=hti.Then U is noncyclic so t 62 ˆ.U / D Ã1.U /. For each x 2 X � U , x2 2 hti

(indeed, hx; ti=hti is of order 2), and so (by our assumption) x2 D 1. It follows thatˆ.X/ D Ã1.U /. Hence in any case t 62 ˆ.X/.

Theorems 52.7 and 52.8 imply the following

Proposition 53.2. LetK be a 2-group possessing exactly two cyclic subgroups of order4 and assume that neither of them is a characteristic subgroup of K. Then one of thefollowing holds: (a) K Š C4 � C2; (b) K Š D8 � C2; (c) K D hb; t j b8 D t2 D

1; bt D bu; u2 D Œu; t � D 1; bu D bz; z D b4i and this is a group of order 25 with�2.K/ D hb2; ti � hui Š D8 � C2.

The following two propositions describe the results from �52 (see Theorems 52.4and 52.5 for the Proposition 53.3 and Theorem 52.2 for Proposition 53.4) adopted foran application in the proofs of Theorems 53.9 and 53.10. Since the notation introducedin these propositions will be used in the proofs (which are quite involved), the explicitstatement of these results is unavoidable.

Proposition 53.3. LetH be a 2-group of order > 24 and suppose that A D �2.H/ Š

C4 � C2 � C2. We set E D �1.A/ Š E8, Ã1.A/ D hzi, B D CH .A/. ThenCH .E/=E is cyclic and so B is abelian and H=E is either cyclic or generalizedquaternion. Suppose that H=E Š Q2n , n � 3, and set E D hz; u; ei, H D hE; a; bi,where hE; ai=E is a cyclic subgroup of index 2 in H=E , o.a/ D 2n, a2n�1

D z,a2n�2

D v is of order 4, b2 2 A � E, o.b/ D 8, and A D hE; vi. Then we have oneof the following possibilities:

(a) If E 6� ˆ.H/, then b2 D uv, Œa; u� D 1, Œa; e� D z, D 0; 1, Œb; e� D 1,ub D uz, and ab D a�1. If D 0, then B D hE; ai, and if D 1, thenB D hE; a2i.

(b) If E � ˆ.H/ and ŒE; a� ¤ f1g, then b2 D ev, Œa; u� D 1, Œa; e� D z, Œb; e� D z,Œb; u� D z, ab D a�1ue�, � D 0; 1, and if � D 1, then n � 4. Here B D

hE; a2i.


(c) If E � ˆ.H/ and ŒE; a� D f1g, then b2 D uv, Œb; e� D 1, ub D uz, andab D a�1e. Here B D hE; ai.

Proposition 53.4. Let H be a nonabelian 2-group of order > 24 and suppose thatA D �2.H/ Š C4 � C2 � C2. We set E D �1.A/ Š E8, Ã1.A/ D hzi, and B D

CH .A/. Suppose that H=E Š C2n�1 , n � 3, and set E D hz; u; ei, H D hE; ai,where o.a/ D 2n, a2n�1

D z, a2n�2

D v is of order 4, and A D hE; vi. Then we haveone of the following possibilities:

(a) If jŒE; a�j D 4, then Œa; e� D u, Œa; u� D z, n � 4, and B D hE; a4i.

(b) If jŒE; a�j D 2 and ŒE; a� D hzi, then Œa; e� D 1, Œa; u� D z, and B D hE; a2i.

(c) If jŒE; a�j D 2 and ŒE; a� ¤ hzi, then Œa; e� D u, Œa; u� D 1, and B D hE; a2i.

Let G be a 2-group of order > 24 with c2.G/ D 4. Then j�2.G/j � 24. Ifj�2.G/j D 24, then from results of �52 follows that �2.G/ is isomorphic to Q8 � C4

or C4 � C2 � C2 and the following result follows at once.

Theorem 53.5. Let G be a 2-group of order > 24 with c2.G/ D 4 and j�2.G/j D 24.Then one of the following holds:

(a) �2.G/ Š Q8 � C4 and G is isomorphic to a group (a) or (c) of Theorem 52.1.

(b) �2.G/ Š C4�C2�C2 andG is isomorphic to a group described in Propositions53.3 and 53.4 or G Š C2m � C2 � C2, m � 2.

In the rest of this section we assume that j�2.G/j > 24. First we prove the follow-

ing easy special result.

Theorem 53.6. Let G be a 2-group of order > 24 with c2.G/ D 4 and j�2.G/j > 24.

If G has a subgroup Q Š Q8, thenQ GG, C D CG.Q/ is cyclic of order 2n, n � 2,G D .Q�C/hti, where t is an involution such thatQhti Š SD24 and htiC Š D2nC1 .We have jZ.G/j D 2, jGj D 2nC3 and �2.G/ D G.

Proof. If Q were not normal in G, there is g 2 G such that Q ¤ Qg . But thenjQ \ Qg j � 4 and so Q [ Qg contains at least five cyclic subgroups of order 4, acontradiction. HenceQ is normal in G. If CG.Q/ � Q, then jGj � 24, a contradic-tion. Therefore C D CG.Q/ > Z.Q/ and C is normal in G. If t0 is an involutionin C � Z.Q/, then c2.Q � ht0i/ D 6, a contradiction. Since C cannot be general-ized quaternion (otherwise c2.G/ � 6), C is cyclic of order 2n, n � 2. We see thatc2.Q � C/ D 4 and �2.Q � C/ D Q � �2.C / is of order 24. But j�2.G/j > 24

and so there is an involution t 2 G � .QC/, G D .QC/hti and t induces an “outer”automorphism on Q. It follows Qhti Š SD24 . Since there are no elements of order4 in G � .QC/, CQC .t/ is elementary abelian. Hence CC .t/ D Z.Q/ and so hti C

is of maximal class. The only possibility is that hti C is dihedral of order 2nC1. Thestructure of G is completely determined.


Remark. Here we will give another proof of Theorem 53.6, retaining the same no-tation. Since G is not of maximal class, C D CG.Q/ 6� Q, by Proposition 10.17.If z is an involution in C � Z.Q/, then c2.Q � hzi/ D 6 > 4 D c2.G/, a con-tradiction. If c2.C / > 1, then, obviously, c2.QC/ � 5 > 4 D c2.G/. Thus,c1.C / D c2.C / D 1 so, by Theorem 1.17(b), C is cyclic, C \ Q D Z.Q/. Wehave c2.Q ��2.C // D 4 D c2.G/ soQ ��2.C / is normal in G. SinceQ is charac-teristic inQ��2.C / (Appendix 16), it is normal inG. Since jAut.Q/j2 D 8, we havejG W .Q � C/j � 2. Since j�2.G/j > 16 D jQ ��2.C /j, there exists an involutiont in G � .Q � C/, and we get G D hti .QC/. Using the condition c2.G/ D 4, weget hti Q Š SD16 and hti C Š D2n for some n 2 N. The group G is completelydetermined.

Next we assume also that G has no subgroups isomorphic to Q8. The subgroupgenerated by four cyclic subgroups of order 4 will be determined in our next basicresult.


Suppose that G has no subgroups isomorphic to Q8. Then A D hU1; U2; U3; U4i isabelian of type .4; 2; 2/, where U1; U2; U3; U4 < G are cyclic of order 4.

Proof. We first show the following easy fact. If a cyclic subgroup V of order 4 nor-malizes another cyclic subgroup U of order 4, then jU \V j D 2 and hU;V i is abelianof type .4; 2/. Indeed, if U \ V D f1g, then j.UV /0j � 2 and so UV is metacyclicof order 24 and class � 2. This implies that exp.UV / D 4 and �1.UV / is abelian oftype .2; 2/. If N is a maximal subgroup of UV containing U , then all eight elementsin .UV /�N are of order 4 and so c2.UV / > 4, a contradiction. Hence jU \ V j D 2

and jUV j D 8. But UV cannot be nonabelian, i.e., quaternion or dihedral, and so UVis abelian of type .4; 2/.If each Ui is normal in G, then (by the above) A D hU1; U2; U3; U4i is abelian

of type .4; 2; 2/, and we are done. Therefore we may assume that U1 is not normalin G. Set K D NG.U1/ and we have 2 � jG W Kj � 4. By Theorem 1.17(b),c2.K/ is even. Let M be a subgroup of G such that K < M , jM W Kj D 2 andjG W M j � 2. For each m 2 M � K, Um

1 D U2 ¤ U1 and NG.U2/ D K. By theabove, A0 D hU1; U2i Š C4 � C2 and A0 is normal inM . If there is a further cyclicsubgroup U3 of order 4 contained in K, then U3 6� A0, U3 normalizes U1 and U2 andso, by the above, U3 centralizesU1 and U2 and jA0 \U3j D 2. HenceA0U3 is abelianof type .4; 2; 2/ and we are done.We assume in the sequel that K has exactly two cyclic subgroups U1 and U2 of

order 4. Since U1 and U2 are conjugate in M , it follows that neither U1 nor U2 is acharacteristic subgroup of K. By Proposition 53.2, we have exactly three possibilitiesfor the structure of K.Assume (by way of contradiction) that K > A0 and set L D �2.K/. By Lemma

42.1, jLj > 23 (if jLj D 23, then Z.L/ is a characteristic cyclic subgroup of order 4 in


K, which is not the case). Therefore, by Proposition 53.2, we have L Š D8 � C2 andL is normal inM . If K > L, then jK W Lj D 2 (Proposition 53.2(c)) and all elementsin K � L are of order 8. In any case, by Proposition 53.2, CK.A0/ D A0 and K=A0

is elementary abelian of order 2 or 4.Suppose that K > L. Then M=A0 Š Aut.A0/ Š D8 since jM=A0j D 8. Since

L=A0 D ˆ.M=A0/, there is an element k 2 M � K such that k2 2 L � A0 (ifk2 2 A0 for all k 2 M � K, then A0 � ˆ.M/, a contradiction sinceM=A0 Š D8).All elements in L � A0 are involutions and so k is an element of order 4. In fact, all16 elements in .hkiA0/� L are of order 4. But then c2.hkiA0/ � 8, a contradiction.We have K D L Š D8 � C2; in that case, jM j D 25. Assume thatM=A0 Š C4.

Let x 2 M � K. If o.x/ D 8, then hxi \ K is cyclic of order 4, say U1. ThenNG.U1/ � hx;Ki > K, which is a contradiction. Thus, M � K has no elementsof order 8. Suppose that o.x/ D 2. Then hx;A0i D K since M=A0 is cyclic; inthat case x 2 K, again a contradiction. Thus, in the case under consideration, all16 elements in M � K are of order 4, a contradiction. Thus M=A0 Š E4. For eachm 2 M�K,Um

1 D U2 and som2 2 �1.A0/ andm is an element of order 2 or 4. Sincec2.A0/ D 2, it follows thatM � K contains at most four elements of order 4. Hencethere exists an involution s 2 M �K since jM �Kj D 16 (in fact, there are at least 12involutions inM �K). Let t be an involution in K � A0. We set A0 D ha; u j a4 D

u2 D Œa; u� D 1; a2 D zi and so �1.A0/ D hu; zi and at D a�1, ut D u, as D au,us D u, because U s

1 D U2 and s induces an involutory automorphism on A0. Sinces inverts (actually centralizes) exactly the elements in�1.A0/, it follows that A0s hasexactly four involutions. Hence the other four elements in A0s are of order 4. Thismeans that we have found all four elements of order 4 in M � K. Hence the cosetA0.st/ consists only of involutions. But then the involution st must invert A0. On theother hand, we have ast D .au/t D a�1u, which is a contradiction.We have proved that K D A0. Since jGj > 24, it follows jG W M j D 2 and

jGj D 25. In this case jG W Kj D 4 and so fU1; U2; U3; U4g forms a single conjugacyclass in G. This implies that NG.A0/ D M (otherwise, A0 is normal in G so U1 andU3 are not conjugate in G) andM �A0 contains exactly four elements of order 4, andthe other four elements in M � A0 are involutions. Since j�2.G/j > 24, there areinvolutions in G �M .Let m be an involution inM �A0. We set again A0 D ha; u j a4 D u2 D Œa; u� D

1; a2 D zi. Since M is not of maximal class (see Proposition 1.8), Um1 D U2, and

m induces an involutory automorphism on A0, we may set am D au ( D ˙1) andum D u. Thus CA0

.m/ D �1.A0/ D hu; zi and so E D hu; z;mi D �1.M/ (indeed,jhu; z;mi � jhu; zij D 4 is the number of involutions in M � K D M � A0) is anormal elementary abelian subgroup of order 8 in G.Let � be an involution in G � M . If � does not centralize E, then E� would

contain some elements of order 4, a contradiction since M \ E� � M \ M� D ¿.Thus hE; �i D E � h�i is an elementary abelian maximal subgroup of G. But thenexp.G/ D 4 and so all elements in G � M must be involutions. It follows that �


invertsM and soM is abelian. In this case, as it is easy to check since jM j D 24,Mis abelian of type .4; 2; 2/, and the theorem is proved.

In the rest of this section we set A D hU1; U2; U3; U4i Š C4 � C2 � C2, wherefU1; U2; U3; U4g is the set of four cyclic subgroups of order 4 inG,E D �1.A/ Š E8,hzi D Ã1.A/ � Z.G/, B D CG.A/. Let v 2 A � E so that v2 D z. If there is aninvolution s 2 B�A, then sv is an element of order 4 in B�A, a contradiction. HenceA D �2.B/ and Proposition 53.3 implies that B=E is cyclic and so B is abelian oftype .2m; 2; 2/, m � 2. It follows that CG.B/ D B . If c 2 B with o.c/ D 2m, thenc2m�1

D hzi, B D Ehci, E \ hci D hzi, A D Ehc2m�2

i. Since j�2.G/j > 24, there

exists an involution t 2 G � B . If there is an element x 2 Bt with o.x/ > 2, theno.x/ � 8 since B \Bx D ¿, and so o.x2/ � 4 and x2 2 B . Thus CB.t/ D CB.x/ isnot elementary abelian and so Bt would contain elements of order 4, a contradiction(namely, if y is an element of order 4 in CB.t/, then yt 2 Bt is of order 4). Hence allelements in Bt are involutions and consequently t inverts B . If there is an involutiont 0 2 G � Bhti, then t 0 also inverts B and so t t 0 2 G � B would centralize B , acontradiction since CG.B/ D B . Thus D D Bhti D �2.G/. If �2.G/ D G, thestructure of G is determined and we have proved the following

Theorem 53.8. Let G be a 2-group of order > 24 with c2.G/ D 4 and j�2.G/j >

24. Suppose that G has no subgroups isomorphic to Q8. Then D D �2.G/ D

Bhti, where B is abelian of type .2m; 2; 2/, m � 2, and t is an involution invertingB . We have Z.D/ D E D �1.B/, CG.t/ D hti � E D F , and so F is a self-centralizing elementary abelian subgroup of order 16. If�2.G/ D G, the structure ofG is completely determined.

In what follows we assume, in addition, that G > �2.G/. We show next thatCG.E/ D D D �2.G/. Indeed, suppose that CG.E/ > D. Since CG.E/ stabi-lizes the chain A > E > f1g in view of jA W Ej D 2, it follows that CG.E/=B

.D CG.E/=CG.A// is elementary abelian and 4 � jCG.E/=Bj � 8. Let F0=B be asubgroup of order 2 in CG.E/=B such that D \ F0 D B . Then �2.F0/ D A (oth-erwise, c2.F0/ > 4 D c2.G/) and F0 is nonabelian of order > 24 since F0 6� B D

CG.A/). Proposition 53.3 implies, however, that F0=E is cyclic and so F0 is abelian,a contradiction.We have proved that CG.E/ D D .D �2.G/). Hence X D G=B is a subgroup of

Aut.A/ (see Proposition 53.1) and if F1.Š E8/ is the stability group of A > E > f1g

(note that F1 < Aut.A/), then X \ F1 D D=B is of order 2. Suppose that theinvolution in D=B is a square in X . Then there exists x 2 G such that x2 D t 0 2

D � B . But t 0 is an involution (indeed, all elements in D � B , by the above, areinvolutions) and so x is an element of order 4 in G �B , a contradiction since A � B .Proposition 53.1 implies that there exists a maximal subgroup H of G containing Bsuch that H \D D B (indeed, by the above proposition, D=B 6� ˆ.G=B/), and soG=B D .D=B/ � .H=B/ since D is normal in G. Note that �2.H/ D A (indeed,H\�2.G/ D H\D D B and�2.B/ D A) andH > B is nonabelian of order> 24,


where B D CG.A/ D CH .A/. In this situation we shall apply Propositions 53.3 and53.4 which describe all possibilities for the structure of H . Also we shall use freelythe notation introduced in these propositions. In particular, 2 � jH=Bj � 4 and soG=B.D .D=B/ � .H=B// is abelian of order � 4. We also setD D Bhti, where t isan involution inverting B .It is possible at this stage to eliminate the possibility (c) of Proposition 53.3 and the

possibilities (a) and (c) of Proposition 53.4 (thus, we intend to consider these threepossibilities using the notation introduced in corresponding parts).Suppose that H is isomorphic to a group (c) of Proposition 53.3. Consider the

subgroup K D Bhtbi, where .tb/2 2 B and �2.K/ D A. We compute atb D

.a�1/b D .ab/�1 D .a�1e/�1 D ae and so tb centralizes B=E and jB=Ej � 4.Hence K=E cannot be generalized quaternion and so K=E is cyclic. It follows thathtbi covers K=E which gives .tb/2 D ai s, where s 2 E and i is an odd integer. Onthe other hand, tb centralizes .tb/2 and so ais D .ai s/tb D .a�i s/b D .a�1e/�i sb D

aiesb . Thus sb D se and Œb; s� D e. This is a contradiction since Œb;E� D hzi

according to Proposition 53.3(c).Suppose that H is isomorphic to a group (a) of Proposition 53.4; in that case,

H=B Š C4 sinceH D hE; ai and B D hE; a4i. Consider the subgroup K D Bhtai.Since G=B Š C2 � C4, it follows K=B Š C4 and �2.K/ D A. We compute.a4/ta D .a4/�1 and so if jB=Ej � 4, K=E is nonabelian and soK=E is generalizedquaternion. But K=E has the cyclic factor group .K=E/=.B=E/ Š K=B of order 4,a contradiction. Hence we must have B D A, n D 4, jGj D 27, and K=E is cyclic oforder 8. Since htai covers K=E, we get .ta/4 D sv with s 2 E. But ta centralizes.ta/4 and so we get (recall that v D a4) sv D .sv/ta D .sv�1/a D sav�1 D savz

which implies sa D sz. This gives s D uz� with � D 0; 1, and .ta/4 D uz�v.On the other hand, .ta/2 2 .Bha2i/ � B , and so .ta/2 D a2s0, where s0 2 A.Indeed, G=B is abelian and so .ta/2 D t2a2s0 with s0 2 B D A. We computeuz�v D .ta/4 D .a2s0/2 D a2s0a2s0 D a4.a�2s0a2/s0 D v.s0/a

2

s0, which gives.s0/a

2

D .s0/�1uz� and Œa2; .s0/�1� D u.z�.s0/�2/ with z�.s0/�2 2 hzi. This is acontradiction since, according to Proposition 53.4(a), Œa2; A� D hzi.Suppose thatH is isomorphic to a group (c) of Proposition 53.4. HereB D hE; a2i

and we consider the subgroup K D Bhtai. We see .a2/ta D .a2/�1 and so tainverts B=E. Note that �2.K/ D A and so in case jB=Ej � 4, K=E is generalizedquaternion and consequently .ta/2 2 A �E (since A=E D Z.K=E/). If jB=Ej < 4,then B D A and again .ta/2 2 A � E (since o.ta/ � 8). It follows .ta/2 D sv withs 2 E and v D a2n�2

. Since ta centralizes .ta/2, we get sv D .sv/ta D .sv�1/a D

sav�1 D savz. This gives sa D sz and Œa; s� D z. But Proposition 53.4(c) impliesŒa;E� D hui ¤ hzi, a contradiction.We are now ready to prove the following deep result.


Suppose that G has no subgroups isomorphic to Q8 and jG W �2.G/j � 4. Then we


have jG W �2.G/j D 4 and G is a uniquely determined group of order 27:

G D ha; b; t j a8 D b8 D t2 D 1; a2 D v; a4 D z; b2 D ev; ab D a�1u;

e2 D u2 D Œe; v� D Œu; v� D Œe; u� D Œa; u� D Œt; e� D Œt; u� D 1;

ea D ez; eb D ez; ub D uz; vt D v�1; at D eva�1;

bt D euvb�1i:

Here E D hz; u; ei is a normal elementary abelian subgroup of order 8. Four cyclicsubgroups of order 4 in G generate the abelian subgroup A D hE; vi of type .4; 2; 2/which is self-centralizing inG and each cyclic subgroup of order 4 is normal inG. Wehave D D �2.G/ D Ahti, where the involution t inverts A and CG.t/ D E � hti Š

E16 is self-centralizing in G. We have ˆ.G/ D A, all elements in G �D are of order8, and Z.G/ D hzi is of order 2. For one of the four maximal subgroupsM ofG whichdo not contain D we have M=E Š Q8 and E < ˆ.M/ (so this M is generated bytwo elements) but for the other three such maximal subgroupsM we haveM=E Š Q8

and E 6� ˆ.M/ (so these M are not generated by two elements). For each of thethree maximal subgroups N of G which contain D we have N=E Š D8. The groupG exists as a subgroup of the alternating group A16 and 16 is the smallest degree forany faithful permutation representation of G.

Proof. SinceG=B D .D=B/�.H=B/ we getG=�2.G/ D G=D Š H=B , so we haveto examine here only the cases, where H is isomorphic to a group (b) of Proposition53.3 or to a group (a) with D 1 of Proposition 53.3. In these cases we have jH=Bj D

4 and so we have already proved that jG W �2.G/j D 4.Suppose first that H is isomorphic to a group (b) of Proposition 53.3. Note that

in this case H=E Š Q2n , n � 3, E � ˆ.H/, and if n D 3, then � D 0, i.e.,ab D a�1ue� D a�1u. We have hereG=B Š E8. Consider the subgroup L D Bhtai,where .ta/2 2 B and �2.L/ � D \ L � A so �2.L/ D A (recall that Ahti D D D

�2.G/ and involution t invertsA). We compute .a2/ta D .a�2/a D .a2/�1, and so tainverts B=E (recall that B D hE; a2i). If n � 4, then jB=Ej � 4, L=E is generalizedquaternion, and consequently .ta/2 2 A�E (since o.ta/ � 8). If n D 3, then B D A

and o.ta/ � 8 implies again .ta/2 2 A � E. Thus .ta/2 D sv, where s 2 E andv D a2n�2

. From this relation we get sv D tata D ata so at D sva�1. Since tacentralizes .ta/2, we have sv D .sv/ta D .sv�1/a D sav�1 D savv2 D savz andsa D sz, which gives s D eu0 with u0 2 hu; zi since E D he; u; zi.On the other hand, set B1 D Bhai D Ehai and consider the subgroup L1 D

B1htbi; then �2.L1/ D A since L ¤ D D �2.G/ and A � �2.L/. We compute,taking into account the above obtained equalities, atb D .sva�1/b D sbvb.ab/�1 D

.sbvbe�u/a D a0a, where � D 0; 1 and a0 D sbvbe�u 2 A�E. If n � 4, then B1=E

is cyclic of order � 8 and so L1=E Š M2n . This is a contradiction since L1=E mustbe either cyclic or generalized quaternion. It follows (see Proposition 53.3(b)) n D 3,B D A, � D 0, a2 D v, b2 D ve, ab D a�1u, and jGj D 27. We get vb D .a2/b D


.ab/2 D .a�1u/2 D a�2 D v�1, and so vb D v�1. Since .tb/2 2 B D A ando.tb/ � 8 (indeed, tb 2 G �D D G ��2.G/), we have .tb/2 D s0v, where s0 2 E.Since tb centralizes .tb/2, we get s0v D .s0v/

tb D .s0v�1/b D sb

0v, and thereforesb0 D s0, which implies s0 2 heu; zi. We get s0v D tbtb D btb and bt D s0vb

�1.The above elements s; s0 2 E are connected with a relation which we obtain in the

following way. We act with t (by conjugation) on the relation b�1ab D a�1u and get(taking into account the above obtained equalities)

.s0vb�1/�1.sva�1/.s0vb

�1/ D .sva�1/�1u;

bs0sa�1s0vb

�1 D av�1su;

s0sa�1s0v D b�1.av�1su/b D a�1uvsbuz D a�1vsbz;

.as0sa�1/s0v D vsbz;

and since sa�1

0 D sa0 , s

a�1

D sz, we get sa0szs0v D vsbz, and this implies the desired

relation

sa0 s

b D s0s:(1)

Note that s D eu0 with u0 2 hu; zi and s0 2 heu; zi.If s0 D euz˛ .˛ D 0; 1/, then (1) gives sb D sz and so s D ezˇ .ˇ D 0; 1/. If

s0 D z˛ .˛ D 0; 1/, then (1) gives sb D s and so s D euzˇ .ˇ D 0; 1/. We consideragain the relations at D sva�1, bt D s0vb

�1, where s; s0 are the above elements inE. Replacing t with tue, we get atue D .sva�1/ue D sva�1z D .sz/va�1, btue D

.s0vb�1/ue D s0vb

�1. Replacing t with tu, we get atu D .sva�1/u D sva�1,btu D .s0vb

�1/u D .s0z/vb�1. The above facts show that in our expressions for s

and s0 we may choose ˛ D ˇ D 0. Hence we get either s D e and s0 D eu or s D eu

and s0 D 1.We have obtained exactly two groups G1 and G2 (of order 27) which differ only in

the last two relations:

(a) G1 with relations at D eva�1, ; bt D euvb�1;

(b) G2 with relations at D euva�1, bt D vb�1.

For both groups G1 and G2 we have obtained the following common relations:

(c) a8 D b8 D t2 D 1, a2 D v, a4 D z, b2 D ev, ab D a�1u, e2 D u2 D Œe; v� D

Œu; v� D Œe; u� D Œa; u� D Œt; e� D Œt; u� D 1, ea D ez, eb D ez, ub D uz,vt D v�1.

It is possible to show that the groups G1 and G2 are isomorphic. In the group G1

we first obtain by computation (using the relations for G1)

.ab/t D zv.ab/�1:(2)

Then in G1 we replace the elements a; v; b; e; t with a0 D a�1, v0 D v�1, b0 D ab,e0 D eu, t 0 D te, and see that these new elements satisfy the same common relations


(c). But from the relations (a) and (2), we get .a0/t0

D e0uv0.a0/�1, .b0/t0

D v0.b0/�1,which are the last two relations (b) for the group G2. Hence the groups G1 and G2 areisomorphic and we have obtained the unique group G D G1 of order 27 as stated inour theorem.We have ˆ.G/ D A. In order to investigate the structure of seven maximal sub-

groupsM of G, we first see a2 D v, b2 D ev, .ab/2 D euzv, .ta/2 D ev, .tb/2 D

euv, and .tab/2 D zv. If M D hA; a; bi D H , then ˆ.M/ D A Š C4 � C2 � C2

and so E � ˆ.M/. If M D hA; a; tbi, then ˆ.M/ D hv; eui Š C4 � C2. IfM D hA; b; tai, then ˆ.M/ D hv; ei Š C4 � C2. If M D hA; ta; tbi, thenˆ.M/ D hev; ui Š C4 � C2. Hence in the last three cases we have E 6� ˆ.M/.For these four maximal subgroups M (which do not contain D D Ahti) we haveM=E Š Q8. For the other three maximal subgroups N (which containD) we see thatN=E Š D8.We get a faithful permutation representation of degree 16 ofG in the following way.

We set:

a D .1; 2; 3; 4; 5; 6; 7; 8/.9; 16; 12; 13; 11; 14; 10; 15/;

b D .1; 9; 3; 10; 5; 11; 7; 12/.2; 13; 8; 14; 6; 15; 4; 16/;

t D .3; 7/.4; 8/.10; 12/.14; 16/:

We see that the above permutations a; b; t are even and check that they satisfy thedefining relations for G. The obtained representation is faithful since

z D a4 D .1; 5/.2; 6/.3; 7/.4; 8/.9; 11/.14; 16/.10; 12/.13; 15/ ¤ 1:

It remains to show that ı.G/, the minimal degree of a faithful permutation repre-sentation ofG (minimal representation), equals 16 (in the previous paragraph we haveshowed that ı.G/ � 16). Assume that this is false. Since Z.G/ is cyclic, every min-imal representation of G, is transitive so ı.G/ is a power of 2. Then G is isomorphicto a 2-subgroup of Sı.G/ so ı.G/ D 8. Moreover, G Š †3 2 Syl2.S23/. However,c1.G/ D c1.�2.G// D jD � Aj C c1.A/ D 23 (note that c1.A/ D 7). On the otherhand, c1.†3/ D jD8j C Œc1.D8/ C 1�2 � 1 D 8 C 62 � 1 D 43 > 23, and this is acontradiction. Thus, ı.G/ D 16.In the second half of the proof we investigate the remaining possibility, where H

is isomorphic to a group (a) with D 1 of Proposition 53.3. Note that in this caseH=E Š Q2n , n � 3, and E 6� ˆ.H/. In exactly the same way as in the first part ofthe proof we get n D 3, B D A, jGj D 27, a2 D v, b2 D vu, and ab D a�1. Alsowe get at D sva�1 where s 2 E, sa D sz, s D eu0 with u0 2 hu; zi, bt D s0vb

�1,where s0 2 E, sb

0 D s0, and s0 2 he; zi.We conjugate the relation b�1ab D a�1 with the element t and obtain .bt/�1atbt D

.at /�1 which gives (as in the first part of the proof) the following connection betweens and s0:

sa0 s

b D s0sz:(3)


If s0 D ez˛ .˛ D 0; 1/, then (3) gives sb D s and s D ezˇ .ˇ D 0; 1/. If s0 D z˛

.˛ D 0; 1/, then (3) gives sb D sz and so s D euzˇ .ˇ D 0; 1/. Replacing t with te,we get ate D .sva�1/e D .sz/va�1, bte D .s0vb

�1/e D s0vb�1. Replacing t with

tu, we get atu D .sva�1/u D sva�1, btu D .s0vb�1/u D .s0z/vb

�1. This showsthat in our expressions for s and s0 we may choose ˛ D ˇ D 0 and so we have eithers D e and s0 D e or s D eu and s0 D 1.We have obtained again exactly two new groups G1 and G2 (of order 27) which

differ only in the last two relations:

(a’) G1 with relations at D eva�1, bt D evb�1;

(b’) G2 with relations at D euva�1, bt D vb�1.

It is possible to show that the new groups G1 and G2 are isomorphic. In G1 wereplace the elements e, b, t with e0 D eu, b0 D ab, t 0 D tu, respectively, and seethat the common relations for G1 and G2 remain valid. But from the relations (a’) weget .a/t

0

D e0uv.a/�1, .b0/t0

D v.b0/�1, which are the last two relations (b’) for thegroup G2. Hence the groups G1 and G2 are isomorphic and we have to study furtheronly the new group G D G1.We consider the maximal subgroup H� D hA; b; tai and see that ˆ.H�/ D

huv; ev; vi D A. We have �2.H�/ D A, H�=E Š Q8, and E � ˆ.H�/. But

this is exactly the starting point for the first part of the proof with the subgroup H�

instead ofH . It follows that our new group G D G1 must be isomorphic to the uniquegroup of order 27 stated in our theorem.

Remark. Let us show that every subgroup T � G of order � 24 contains z (here G isthe group of Theorem 53.9). Since Z.G/ D hzi, it suffices to show that TG D f1g. As-sume that this is false. Then jG W T j D 8 and G is isomorphic to a Sylow 2-subgroupof the symmetric group S8. As the proof of Theorem 53.9 shows, G and a Sylow 2-subgroup of S8 have distinct numbers of involutions, and this is a contradiction. Thus,z 2 T .

In our next result the remaining 2-groups G with c2.G/ D 4 will be determined.

Theorem 53.10. LetG be a 2-group of order> 24 with c2.G/ D 4 and j�2.G/j > 24.

Suppose that G has no subgroups isomorphic to Q8 and jG W �2.G/j D 2. Then wehave the following two possibilities:

(a) G D hb; e; ti with

b8 D e2 D t2 D 1; .tb/2 D a; a2n

D 1; n � 3;

a2n�2

D v; a2n�1

D z; b2 D uv; u2 D 1

Œb; e� D Œa; e� D Œa; u� D Œu; e� D Œt; e� D Œt; u� D 1;

ub D uz; ab D a�1; at D a�1:


Here jGj D 2nC4, n � 3, G D hei � ha; b; ti, and E D hz; u; ei is a normal ele-mentary abelian subgroup of order 8 in G. Four cyclic subgroups of order 4 generatethe abelian subgroup A D hE; vi of type .4; 2; 2/ and each cyclic subgroup of or-der 4 is normal in G. We have B D CG.A/ D hE; ai is abelian of type .2n; 2; 2/,�2.G/ D Bhti is of order 2nC3, where the involution t inverts B , and CG.t/ D

E � hti Š E16 is self-centralizing in G. We have ˆ.G/ D hui � hai Š C2 � C2n ,G0 6� E, Z.G/ D he; zi Š E4. Finally, .Eha; bi/=E Š Q2n and .Ehtbi/=E Š C2n .

(b) G D ha; t; ei with

a8 D e2 D t2 D 1; a2 D v; a4 D z; .ta/2 D uv;

u2 D Œa; e� D Œu; e� D Œt; e� D Œt; u� D 1; ua D uz:

Here jGj D 26 and E D hz; u; ei is a normal elementary abelian subgroup of order8 in G. Four cyclic subgroups of order 4 generate the abelian subgroup A D hE; vi

of type .4; 2; 2/ and each cyclic subgroup of order 4 is normal in G. We have A D

CG.A/, �2.G/ D Ahti is a maximal subgroup of G, where the involution t inverts A.We have ˆ.G/ D hu; vi Š C4 � C2, G0 D hz; ui � E, Z.G/ D he; zi Š E4. Thisexceptional group exists as a subgroup of A16. The subgroup hei is a direct factorof G. The minimal degree of a faithful permutation representation of G equals 10(obviously, this representation is intransitive: every permutation representation of a2-group G such that its degree does not equal to a power of 2, is intransitive).

Proof. Here H is isomorphic to a group (a) with D 0 of Proposition 53.3 or toa group (b) of Proposition 53.4 since these are the only remaining possibilities withjH=Bj D 2, where B D CG.A/ and G=�2.G/ Š H=B .Suppose that H is isomorphic to a group (a) with D 0 of Proposition 53.3. Here

CG.A/ D B D hE; ai is abelian of type .2n; 2; 2/, n � 3, and H=E Š Q2n . Wehave atb D .a�1/b D a and so looking at K D Bhtbi with �2.K/ D A, weget that K=E must be cyclic. Thus htbi covers K=E and so .tb/2 D ai s, wheres 2 E and i is an odd integer. Replacing a with ai we may assume from the start.tb/2 D as which gives bt D asb�1. Since tb centralizes .tb/2 we obtain as D

.as/tb D .a�1s/b D asb , which gives sb D s and so s 2 he; zi. We computebtu D .asb�1/u D asb�1z D a.sz/b�1, and so replacing t with the involution tu(if necessary), we may assume that s D 1 or s D e. Suppose that s D e and so.tb/2 D ae. But then replacing a with a0 D ae, all other relations remain unchangedand the last relation is transformed into .tb/2 D a0. Hence we may assume from thestart that .tb/2 D a and our groupG is uniquely determined as stated in part (a) of ourtheorem.Suppose, finally, that H is isomorphic to a group (b) of Proposition 53.4. Here

CG.A/ D B D hE; a2i is abelian of type .2n�1; 2; 2/, n � 3, and H=E Š C2n�1 .We set K D Bhtai and see that .a2/ta D .a2/�1 and therefore ta inverts B=E. IfjB=Ej � 4, then �2.K/ D A implies that K=E is generalized quaternion. But thenreplacingH with K, we know that such groups have been determined before.


It follows that we may assume that jB=Ej D 2 and so B D A, K=E Š C4, andjGj D 26. Hence o.ta/ D 8 and .ta/2 D sv, where s 2 E and at D sva�1. Sinceta centralizes .ta/2, we get sv D .sv/ta D .sv�1/a D sav�1 D savz, which givessa D sz and so s D ue0 with e0 2 he; zi. We note that atu D .sva�1/u D .sz/va�1

and so replacing t with tu (if necessary), we may set s D u or s D ue. However,if s D ue, then .ta/2 D uev. In this case we replace u with u0 D eu and seethat all other relations remain valid (with u0 instead of u) and only the last relation istransformed in .ta/2 D u0. Hence we may assume from the start that .ta/2 D u. Ourgroup G is uniquely determined as stated in part (b) of our theorem.We get a faithful representation of degree 16 of G in the following way. We set

a D .1; 2; 3; 4; 5; 6; 7; 8/.9; 10; 11; 12; 13; 14; 15; 16/;

e D .1; 9/.2; 10/.3; 11/.4; 12/.5; 13/.6; 14/.7; 15/.8; 16/;

t D .2; 6/.3; 7/.10; 14/.11; 15/:

We see that the permutations a; e; t are even and they satisfy the defining relations forG. The obtained representation is faithful since the central permutations z D a4, e,and ez are nontrivial.Since e 62 ˆ.G/, we get G D hei �M for some maximal subgroup M of G. The

subgroup Z.M/ D hzi is cyclic so a faithful permutation representation of minimaldegree ı.M/ is transitive, i.e., ı.M/ is a power of 2. We claim that ı.M/ D 8.Clearly, ı.M/ � 8. Set U D hai and V D hu; e; ti. Since U \ V D f1g, we haveı.M/ � jM W U j C jM W V j D 4 C 4 D 8. It follows that ı.M/ D 8. We getı.G/ D ı.M � hei/ � ı.M/C ı.hai/ D 8C 2 D 10. Since, by [Ber26, page 740],

ı.G/ � ı.hai � hei/ D ı.hai/C ı.hei/ D o.a/C o.e/ D 8C 2 D 10;

we get ı.G/ D 10. The proof is complete.

Problem. Classify the 2-groups G such that c2.G/ 6 0 .mod 4/. In particular, clas-sify the 2-groups G with c2.G/ D 6.

�54

2-groups G with cn.G/ D 4, n > 2

In this section we study a 2-groupG with cn.G/D 4, n > 2 [Jan8]. If fU1; U2; U3; U4g

is the set of four cyclic subgroups of order 2n, then we describe first the structure ofthe subgroup X D hU1; U2; U3; U4i (Theorems 54.1 and 54.2). It is interesting tonote that always jX j D 2nC2. If G has a normal elementary abelian subgroup of order8, the structure of G is described in Theorem 54.6. If G has no normal elementaryabelian subgroups of order 8, the structure of G is described in Theorem 54.7. Theproofs of these theorems are quite involved and therefore we prepare the stage withPropositions 54.3, 54.4, and Theorem 54.5 which are also of independent interest.If cn.G/ D 2, then the 2-groups are known (Corollary 43.6). The next interesting

and important case is cn.G/ D 4 (see Research Problems #188 and #425). However,this problem is essentially more difficult.Throughout this section we assume that G is a 2-group with cn.G/ D 4, n > 2. We

consider first the case that G has a normal subgroup isomorphic to E8.

Theorem 54.1. Let G be a 2-group with cn.G/ D 4, n > 2, and suppose that G hasa normal subgroup E Š E8. Let fU1; U2; U3; U4g be the set of four cyclic subgroupsof order 2n in G. Then X D hU1; U2; U3; U4i D EU1, where E \ U1 > f1g and sojX j D 2nC2. Moreover,�2.X/ Š C4 �C2 �C2 or�2.X/ Š D8 �C2 in which casen D 3.

Proof. Assume that for each cyclic subgroup U of order 2n, U \E D 1. Let U D hai

be one of them. Then .EU /=E Š C2n and all 2nC2 elements in .EU /� .Eha2i/ areof order 2n. Indeed, if x 2 .EU / � .Eha2i/ and o.x/ D 2nC1, then o.x2/ D 2n andhx2i \ E ¤ f1g, a contradiction. But then cn.G/ � 2nC2

'.2n/D 8 which contradicts our

assumption (here '.�/ is Euler’s totient function; '.2n/ is the number of generators ofa cyclic group of order 2n).We have proved that there is a cyclic subgroup hbi of order 2n with jhbi \ Ej D 2

so that .Ehbi/=E Š C2n�1 . Let H D Ehbi, z D b2n�1

, and v D b2n�2

; thenhbi \E D hzi.Assume that CH .E/ D E. Then n D 3, H=E Š C4 acts faithfully on E, and

we may set E D he; u; zi, eb D eu, ub D uz. The structure of H is uniquelydetermined, H D hb; ei. We have uv D .ub/b D .uz/b D uzz D u, ev D .eu/b D

eu uz D ez; ve D eve D veve D veze D vz D vv2 D v�1. It follows thatEhvi D he; vi � hui Š D8 � C2. Let us check that �2.H/ D Ehvi, ˆ.H/ D


hui � hvi Š C4 � C2. We have .be/2 D b2ebe D veue D vu 2 ˆ.G/. Sincev D b2 2 ˆ.G/ we get u D v�1 vu 2 ˆ.G/. It follows from d.H/ D 2 that ˆ.G/has order 23 so is abelian, and the second isomorphism in the displayed formula isproved. It remains to prove the first equality in the displayed formula. Assume that itis false. Then there exists an element y of order 4 in H � .Ehvi/. We may assumethat y 2 fbe; bu; beu; bez; buz; beuzg. However, it is easy to check that all these sixelements have order 8. Thus, �2.H/ D Ehvi Š D8 � C2. and we see that all 16elements in H � .Ehvi/ are of order 8. Thus c3.H/ D c3.G/ D 4 and therefore, bythe product formula, X D EU1, where U1 D hbi.Suppose now that CH .E/ > E . Then A D Ehvi is abelian of type .4; 2; 2/,

all elements in A � E are of order 4, and Ã1.A/ D hzi. For each element y 2

H � .Ehb2i/, we have y2n�2

2 A � E and so o.y/ D 2n. Hence all 2nC1 elementsin H � .Ehb2i/ are of order 2n which gives cn.H/ D 2nC1

'.2n/D 4 D cn.G/. We get

again X D EU1 and �2.X/ D Ehvi Š C4 � C2 � C2.

We assume now that G is a nonabelian 2-group which has no normal subgroupsisomorphic to E8. Since G is not of maximal class, we may apply the main theoremin �50 to conclude that G has a normal subgroup W which is either abelian of type.4; 4/ with CG.W / being metacyclic orW is abelian of type .4; 2/ with CG.W / beingabelian of type .2j ; 2/, j � 2.

Theorem 54.2. Let G be a nonabelian 2-group with cn.G/ D 4, n > 2, and supposethat G has no normal elementary abelian subgroups of order 8. Let fU1; U2; U3; U4g

be the set of four cyclic subgroups of order 2n. Then the subgroup X D hU1; U2; U3;

U4i is of order 2nC2 and for the structure of X we have the following possibilities:

(a) n > 2, X D WU1 withW \U1 Š C4, whereW is an abelian normal subgroupof type .4; 4/. We have �2.X/ D W and X is metacyclic.

(b) n > 2, X D QU1, where Q Š Q8 is a normal quaternion subgroup of X ,Q \ U1 D Z.Q/, U1 D hbi Š C2n , and b either centralizesQ or b induces onQ an involutory outer automorphism in which case n > 3.

(c) n D 3, jX j D 25, and X has a self-centralizing (in X) abelian normal subgroupA of type .4; 2/. Furthermore, �2.X/ Š Q8 � C2 or �2.X/ Š D8 � C2 andso (according to �52) the group X is uniquely determined in each of the twopossibilities for �2.X/.

Proof. By Proposition 50.6, G has a normal abelian subgroup W of exponent 4 suchthat either W is of type .4; 4/ or of type .4; 2/ and such that �2.CG.W // D W ; inthe first case CG.W / is metacyclic and in the second case CG.W / is abelian and hasa cyclic subgroup of index 2.

(i) We examine first the possibility that G has a normal subgroup W Š C4 � C4,where CG.W / is metacyclic. Set W0 D �1.W /, H D WU1, where U1 D hai

is a cyclic subgroup of order 2n, n > 2. The structure of Aut.W / is described in

�54 2-groups G with cn.G/ D 4, n > 2 111

Proposition 50.5. In particular, exp.Aut.W /2/ D 4, where Aut.W /2 2 Syl2.Aut.W //

(moreover, Aut.W /2 is special) and if ˛ ¤ ˇ are two involutions in Aut.W /2 whichdo not stabilize the chainW > W0 > 1, then h˛; ˇi is a four-group.We want to show thatW \U1 Š C4. So suppose that jW \U1j � 2. IfW \U1 D 1,

then CH .W / contains the involution z D a2n�1

since jU1j > 4 D exp.Aut.W //2 andW0 � hzi Š E8, contrary to the fact that CH .W / is metacyclic. Hence we haveW \ U1 D hzi is of order 2. Set v D a2n�2

so that v2 D z. Since �2.CG.W // D

W , v does not centralize W . Hence in this case U1=hzi acts faithfully on W , whereU1 D hvi. This gives o.a/ D 8, n D 3, a2 D v, H=W Š C4 and jH j D 26. Weshall determine the structure of H D WU1, where U1 D hai. Since a induces anautomorphism of order 4 on W , a cannot stabilize the chainW > W0 > f1g (Remark50.2). Set W D hxi � hyi, x2 D z, y2 D u; then W0 D hzi � hui. Assume thata stabilizes the chain W0 > f1g. Then xa is an element containing z so xa 2 xW0.Similarly, ya 2 yW0. We conclude that a stabilizes the chain W > W0 > f1g, whichis not the case. Thus, we have ua D uz. If w 2 W is such that w2 D u, then settingwa D y, we get y2 D .wa/2 D .w2/a D .ua/ D uz and W D hwi � hyi with.wy/2 D z. Since W0hai Š M24 , we get c3.W0hai/ D 2. It remains to determine theelement ya D ws, where s 2 W0. We compute

wv D wa2

D ya D ws; yv D ya2

D .ws/a D wasa D ysa;

and so s ¤ 1 since a2 D v induces an involutory automorphism on W . Comput-ing again .a.wy//2 D awyawy D a2.wy/awy D vywswy D vzs, we see thato.awy/ D 8 since o.vzs/ D 4 noting that W0hvi is abelian of type .4; 2/. Since.wy/a D .wy/s, we conclude thatW0hwyihai contains exactly four cyclic subgroupsof order 8. Therefore all other elements in aW must be of orders 2 or 4. We compute

.aw/2 D awaw D a2waw D vyw;

.vyw/2 D v2.yw/v.yw/ D zysaws.yw/ D sas:

It follows that sas D 1, and so sa D s (otherwise aw would be of order 8). Sinces 2 Z.ha;W0i/# D fzg (recall, that ha;W0i Š M24), we get s D z, and so ya D wz

and .wy/a D wyz D .wy/�1 since z D .wy/2. The structure of H is uniquelydetermined and X D hU1; U2; U3; U4i D W0hwyihai is a normal subgroup of G.The group X is an extension of the normal abelian subgroup W0hwyi of type .4; 2/by C4 and .W0hwyi/ \ hai D hzi. We have ua D uz and .wy/a D .wy/�1 and sov D a2 centralizesW0hwyi. Therefore A D hu;wy; vi is abelian of type .4; 2; 2/ and�1.X/ D �1.A/ D hu; z;wyvi Š E8 is normal in G. This is a contradiction.We have proved that for each cyclic subgroup U of order 2n we haveW \U Š C4.

Set again H D WU1 so that we have H=W Š C2n�2 . Set U1 D hai, y D a2n�3

,y2 D v, and y4 D z.Assume that CH .W / > W . ThenW hyi is abelian of type .8; 4/ and so all elements

in .W hyi/ � W are of order 8; in particular, all elements in the coset Wy are of


order 8. For each x 2 H � .W ha2i/, we have x2n�3

2 Wy since all elements in.H=W / � .W ha2i=W / have order 2n�2, and so o.x2n�3

/ D 8 and o.x/ D 2n. Weget cn.H/ � jH j�jW haij

'.2n/D 2nC1

2n�1 D 4 D cn.G/ and so X D H D WU1 as stated inpart (a) of our theorem. We have in this case�2.X/ D W .Until the end of part (i), we may assume that CH .W / D W . Since exp.Aut.W /2/ D

4, where Aut.W /2 is a Sylow 2-subgroup of Aut.W /, jH=W j � 4, and so n D 3 or4. Set W0 D �1.W / and assume that the involutory automorphism of W induced byy stabilizes the chain W > W0 > f1g. (This will certainly happen if n D 4 becausein that case y D a2.) Let w 2 W be such that W D hwi � hvi. We have wy D ws,where s 2 W0 and we compute

.y.wivj //2 D ywivjywivj D y2.wivj /ywivj D v.ws/ivjwivj

D vsiw2iv2j ;

where i , j are any integers. Since o.vsiw2iv2j / D 4, all elements in yW are of order8. If n D 4, then for each x 2 H � .hyiW /, x2 2 yW and so c4.H/ D 32 W 8 D 4.If n D 3, then H D hyiW and so c3.H/ D 4. Hence we get again X D H D WU1

and �2.X/ D W , as stated in part (a).It remains to consider the case CH .W / D W , where the involutory automorphism

of W induced by y does not stabilize the chain W > W0 > f1g. By the above,n D 3, H D hyiW , and if we set W0 D hu; zi, then uy D uz and z 2 Z.G/.If w 2 W is such that w2 D u, then setting w0 D wy , we get .w0/2 D uz andW D hwi�hw0i. Note that CW .y/ D hvi and so .ww0/y D ww0 impliesww0 D v˙1.But replacing w with w�1 (if necessary), we may assume from the start that ww0 D v

and so w0 D w�1v.Since W0hyi Š M24 , we get c3.W0hyi/ D 2 and we claim that c3.H/ D 2. We

compute (for any s 2 W0)

.yws/2 D y2.ws/yws D vw0syws D v2sys D zsys;

.yw0s/2 D y2.w0s/yw0s D vwsyw0s D v2sys D zsys:

Since o.zsys/ � 2, the above claim is proved. Hence there exists an element y0 oforder 8 in G � H such that v0 D .y0/2 2 W and y0 induces on W an involutoryautomorphism which does not stabilize the chain W > W0 > f1g. Otherwise, weget (as above) c3.hy

0iW / D 4, which is a contradiction (since c3.H/ D 2). Sincez 2 Z.G/, we get uy0

D uz and therefore .y0/4 D z and v0 D .y0/2 2 W0hvi.Hence both y and y0 normalize the abelian subgroup A D W0hvi of type .4; 2/. SinceW0hy0i Š M24 , we get c3.W0hy0i/ D 2.It is easy to see that CG.W / D W . Indeed, if W < S � CG.W / and jS W

W j D 2, then S is abelian metacyclic and so S is of type .8; 4/ and c3.S/ D 4, acontradiction. Hence y and y0 induce on W two distinct involutory automorphismswhich do not stabilize the chain W > W0 > f1g. By our remark about Aut.W /at the beginning of the proof, yy0 induces an involutory automorphism on W which


obviously stabilizes the chainW > W0 > f1g. Since the cosetW.yy0/ cannot containelements of order 8, we get .yy0/2 2 W0 � A. The subgroup Ahy; y0i is of order 25

and since c3.Ahy; y0i/ D 4, we get X D Ahy; y0i.Suppose that B D Ahyy0i is abelian. Then exp.B/ D 4 and all 16 elements in

X � B are of order 8. Since X D hU1; U2; U3; U4i is normal in G, B is normal inG, and so (by assumption) �1.B/ D �1.A/ D W0 Š E4 and B is of type .4; 4/. (If�1.B/ Š E8, then G would possess a normal elementary abelian subgroup of order8.) But uy D uz and so y does not stabilize the chain B > W0 > f1g. By the above,c2.Bhyi/ D 2 which is a contradiction since Bhyi D X D hU1; U2; U3; U4i. HenceCX .A/ D A and we have obtained a group stated in part (c) with X=A Š E4.

(ii) It remains to examine the second possibility, where G has a normal abeliansubgroup W1 of type .4; 2/ such that N D CG.W1/ is abelian of type .2j ; 2/, j � 2,andG=N is isomorphic to a subgroup of D8. If j > 2, thenG=N is elementary abelianof order � 4. Indeed, if j > 2, then N contains a characteristic subgroup Z Š C4.Since Z < W1 and Z is normal in G, it is easy to see that the nontrivial elements inG=N induce on W1 only involutory automorphisms. We setW0 D �1.W1/ Š E4 andhzi D Ã1.W1/ � Z.G/.Suppose that G=N is not elementary abelian. Then j D 2, W1 D N D CG.W1/,

and G=N Š C4 or G=N Š D8. In that case n � 4. Suppose that U � G andU Š C24 . Then U=.U \ W1/ Š C4 acts faithfully on W1. But U centralizes thecyclic subgroup U \ W1 Š C4, a contradiction. It follows n D 3. If L > W1 issuch that L=W1 Š C4, then (by the structure of Aut.C4 �C2/)�1.L=W1/ D L0=W1

invertsW1. In particular, L0 does not contain a cyclic subgroup of order 8.Suppose first thatL contains a cyclic subgroup U1 D hai of order 8. ThenU1 covers

L=W1 and U1 \W1 D hzi is of order 2, where z D a4 and L0 D W1hvi with v D a2

invertingW1. For eachw 2 W1 we compute .vw/2 D v2wvw D v2w�1w D v2 D z,and so all elements in L0 � W1 D vW1 are of order 4. For each x 2 L � L0,x2 2 L0 �W1 and so all 16 elements in L�L0 are of order 8. Hence c3.L/ D 4 andsoX D L is a group of order 25 as stated in part (c) of our theorem withX=W1 Š C4.We may assume that exp.L/ < 8. In that case G=W1 Š D8. Let U1 � G and

U1 D ha1i Š C8. Then U1 \ W1 D hvi Š C4, where v D a21. We may set

W1 D hv; u j v4 D u2 D Œv; u� D 1; v2 D zi, and then va1 D v, ua1 D uz

(since a1 does not centralize W1). Obviously, M D W1U1 D ha1; ui Š M24 andso c3.M/ D 2. Hence there is another cyclic subgroup U2 D ha2i in G which isnot contained in M and U2 \ W1 Š C4. Then a2 must induce on W1 an involutoryautomorphism which is distinct from that one induced by a1. The only possibility is.vu/a2 D vu, ua2 D uz, and so va2 D vz D v�1, with U2 \ W1 D hvui. Again,P D W1U2 D ha2; ui Š M24 and so c3.P / D 2. Since a1a2 inverts W1, ha1; a2i

induces a four-group of automorphisms on W1 and therefore .MPW1/=W1 Š E4,whereMPW1 D ha1; a2iW1. Note that c3.ha1a2iW1/ D 0 (by a remark above). Weget X D MPW1 and so we have obtained a group of order 25 stated in part (c) of ourtheorem with X=W1 Š E4.


It remains to consider the case where G=N is elementary abelian (of order � 4)with N abelian of type .2j ; 2/, j � 2. Assume first j D 2. In that case we have againN D CG.W1/ D W1 Š C4 � C2. Let U1 � G with U1 Š C8. Then U1 \W1 Š C4

and (as before) W1U1 Š M24 which gives c2.W1U1/ D 2. Hence there is U2 � G

with U2 Š C8 and U2 6� W1U1. Then again W1U2 Š M24 and c2.W1U2/ D 2. Wehave G=W1 Š E4 and we have obtained again a group X D G of order 25 as stated inpart (c).

We assume now j � 3 and set N D hu; a j u2 D a2j

D Œu; a� D 1i. Then putv D a2j �2

and z D v2. We see that W1 D hu; vi, z 2 Z.G/, hvi is a characteristicsubgroup of N , and so hvi is normal in G. Obviously, n � j C 1. Assume first thatn D j C 1. Let x 2 G � N be such that o.x/ D 2n D 2j C1. Then x centralizesx2 2 N and o.x2/ D 2j . In particular, x centralizes hvi and so ux D uz (since x doesnot centralizeW1). We get hxiN D hx; ui Š M2nC1 and so c2.hxiN/ D 2. It followsthat G=N Š E4 and there is another element y 2 G � .hxiN/ with o.y/ D 2n.We get again that y centralizes hvi and uy D uz. But then xy centralizes W1 andxy 2 G �N , a contradiction.We have proved that n � j . Then cn.N / D 2. Take a0 2 hai � N with o.a0/ D

2n. Then A D hui � ha0i is an abelian normal subgroup of type .2n; 2/ and cn.A/ D

cn.N / D 2. There is b0 2 G � N with o.b0/ D 2n. Since o.b20/ D 2n�1, it follows

that b20 2 A, b0 centralizes a cyclic subgroup of order 4 inW1 and so ub0 D uz. Also,

hu; b0i Š M2nC1 and so cn.hu; b0i/ D 2. We get X D Ahb0i.We see that hb2

0i (of order 2n�1) is contained in Z.X/. Set Z D Z.X/ and assumefirst that Z > hb2

0i. Then Z is a cyclic subgroup of order 2n contained in A andhb0iZ is abelian of order 2nC1. Since hb0iZ is not cyclic (because it contains twodistinct cyclic subgroups Z � A and hb0i 6� A of order 2n), there is an involutiont 0 2 .hb0iZ/ � A. Since t 0 acts faithfully on W0, it follows that W0ht 0i D D� Š D8

and X D D� � Z. There is a subgroup Q� Š Q8 contained in X (Appendix 16) sothat X D Q� � Z withQ� \ Z D Z.Q�/ D hzi and Z Š C2n . We have obtained agroup X of order 2nC2 as stated in part (b) of our theorem.Assume now that Z.X/ D hb2

0i is cyclic of order 2n�1. Suppose at the momentthat �2.X/ D W1. In that case we can apply the classification result of (Lemma42.1) in which all 2-groups X with jX j > 23 and j�2.X/j � 23 are determined.Since jX W Z.X/j D 23, X is not isomorphic to M2nC2 . Hence X is isomorphic tothe metacyclic group G� from part (c) of the above lemma. Since Z.G�/ Š C4, itfollows that n D 3 and jX j D jG�j D 25. But in that case c3.G

�/ D 6, whichis a contradiction. We have proved that �2.X/ > W1. Hence there is an elementx0 2 X � A with o.x0/ � 4. Since o.x2

0/ � 2, it follows that x20 2 W0 and D D

hW0; x0i Š D8 because ux0 D uz. Thus X0 D D � hb20i is the central product of

D Š D8 and Z.X/ D hb20i, whereD \ hb2

0i D Z.D/ D hzi. Since jX0j D 2nC1, X0

is a maximal subgroup of X and obviously exp.X0/ D 2n�1. All 2nC1 elements inX � X0 must be of order 2n (since cn.X/ D 4). There exists (as above) a quaternionsubgroupQ � X0 so that X0 D Q � hb2

0i. SinceQ is a characteristic subgroup in X0


(because it is a unique quaternion subgroup ofX0),Q is normal inX andX D Qhb0i.From Z.X/ D hb2

0i follows that CX .Q/ D hb20i and so b0 induces on Q D hx; yi

an outer involutory automorphism with xb0 D y and .xy/b0 D .xy/�1. We compute.b0x/

2 D b20x

b0x D b20.yx/. If n D 3, then b2

0.yx/ is an involution. In that caseo.b0x/ D 4, which is a contradiction (since all elements in X � X0 must be of order23). Hence we must have n > 3. In that case o.b0x/ D 2n, as required. We haveobtained a group stated in part (b) of our theorem.It remains to determine �2.X/, where X is a group (of order 25) given in part (c)

of our theorem with A being a self-centralizing (in X) abelian normal subgroup oftype .4; 2/. We know that Aut.A/ Š D8 and Z.Aut.A// inverts A. Since jX=Aj D 4

and X=A acts faithfully on A, there is a subgroup B .> A/ of order 24 such that B=Ainverts A. Then B � A cannot contain elements of order 8 and so exp.B/ D 4. Sincec3.X/ D 4, all 16 elements in X � B must be of order 8 and so B D �2.X/. But Bis nonabelian and so using the results in �52, we see that B is isomorphic to one of thefollowing groups: Q8 � C2, D8 � C2, and Q8 � C4. It is easy to see that the last casecannot occur. Indeed, if B Š Q8 �C4, then, by �52, X Š Q8 �C8. But in that case Ais not self-centralizing. Using again the results in �52, we see that in each of the firsttwo cases the group X is uniquely determined.

Proposition 54.3 (see �18). Let G be a 2-group possessing exactly one Ls-subgroupH of order 2sCm with a fixed integer m � 2. Then G is either an Ls-group or aUs-group with respect to the kernel R D �1.H/.

Proof. Set R D �1.H/ so that R Š E2s , s � 2, H=R is cyclic of order 2m, m � 2,and R is normal in G. Set S=R D �1.H=R/. Then we have S=R � ˆ.H=R/ �

ˆ.G=R/ and�1.S/ D R.We use induction on jGj. If H D G, we are done. In the sequel let H < G. Let

H � M < G, whereM is maximal in G. By induction,M is either an Ls-group or aUs-group with respect to the kernel R D �1.H/.Note that the Frattini subgroup of an arbitrary p-group is not of maximal class.

Since ˆ.G=R/ � M=R, ˆ.G=R/ is cyclic (of order � 2 ) or abelian of type .2; 2/,where the last case is possible only if M=R Š D8. However, if ˆ.G=R/ is abelianof type .2; 2/, then a result of Nekrasov (see Proposition 4.9) implies ˆ.G=R/ �

Z.G=R/, and then M=R is abelian, a contradiction. Thus ˆ.G=R/ is cyclic and soS=R D �1.ˆ.G=R//.Let F=R < G=R be cyclic of order 2m. Now, �1.F=R/ � ˆ.F=R/ � ˆ.G=R/

and so �1.F=R/ D S=R which gives �1.F / D �1.S/ D R. It follows that F is anLs-group of order 2sCm. By assumption, F D H and so G=R has exactly one cyclicsubgroup of order 2m. By Exercise 1.104, G=R is either cyclic or of maximal class.If G=R is cyclic, �1.G/ D �1.S/ D R and G is an Ls-group. Suppose that

G=R is of maximal class and let T=R be a cyclic subgroup of index 2 in G=R. SinceS=R � ˆ.G=R/ < T=R, it follows S=R D �1.T=R/ and �1.T / D �1.S/ D R.This means that G is a Us-group with the kernel R and we are done.


Proposition 54.4. Let G be a 2-group with cn.G/ D 2, n > 2. Then G is an L2-groupor a U2-group.

Proof. SinceG is neither cyclic nor of maximal class, it has a normal abelian subgroupR of type .2; 2/. Let U1 be a cyclic subgroup of order 2n. Set H D U1R and assumethat R \ U1 D f1g. Let K=R be the subgroup of index 2 in H=R and let Z=R be thesubgroup of order 2 inK=R. Then Z Š E8 and for each x 2 H �K, x2n�1

2 Z �R

which gives o.x/ D 2n. But then cn.H/ � 2nC1 W '.2n/ D 2nC1 W 2n�1 D 4, acontradiction.Hence jU1\Rj D 2 and soH is an L2-group of order 2nC1, i.e.,H is either abelian

of type .2n; 2/ orH Š M2nC1 . In any caseH is generated by its two cyclic subgroupsof order 2n and so H is the unique L2-subgroup of G of order 2nC1. By Proposition54.3, G is either an L2-group or a U2-group with the kernel R.

Theorem 54.5. LetX be a 2-group of order> 24 with�2.X/ Š D8�C2 or�2.X/ Š

Q8 � C2. Then jX j D 25 and for each of the two possibilities for �2.X/, X isuniquely determined and is given explicitly in Theorems 52.2(a) and 52.1(d). In bothcases ˆ.X/ is abelian of type .4; 2/, X 0 D �1.ˆ.X// Š E4, Z.X/ D Ã1.ˆ.X//

is of order 2, and c3.X/ D 4. Also, each maximal subgroup of X is nonabelian andtherefore A D ˆ.X/ is self-centralizing in X .Suppose that G is a 2-group with G > X and c3.G/ D 4. Then we have the

following possibilities:

(a) CG.A/ D A, G=A Š D8, and G has a normal elementary abelian group oforder 8.

(b) CG.A/ D C Š C4 � C4, C is self-centralizing in G, and G=C Š E4 or D8.

Proof. The first part of this proposition is a direct consequence of the results in�52. Suppose that X has an abelian maximal subgroup. Then the formula jX j D

2jZ.X/jjX 0j (see Lemma 1.1) gives a contradiction (since jX j D 25).We have to determine the structure of G, where G > X and c3.G/ D 4 D c3.X/.

This implies that X is normal in G. If CG.A/ D A, then G > X and Aut.A/ Š D8

imply G=A Š D8. We have �2.X/ > A, j�2.X/ W Aj D 2, and �2.X/=A D

Z.G=A/. SinceX=A Š E4, there is an element x 2 G�X such that x2 2 �2.X/�A.But o.x2/ � 4 and x is not of order 8. Hence x2 D t is an involution which gives�2.X/ Š D8 � C2 and Z.�2.X// D �1.A/ D X 0. Thus E D hti � �1.A/ isan elementary abelian normal subgroup (of order 8) of X . Since x centralizes t andnormalizes �1.A/ D X 0, it follows that E GG, as required.We have to study the case C D CG.A/ > A so that we get C \ X D A, C is

normal in G, .XC/=C Š X=A Š E4, and G=C Š E4 or D8. It remains to determinethe structure of C . Since C has no elements of order 8, we get exp.C / D 4. Let sbe an element of order 8 in X . Then s2 2 A D ˆ.X/, D D Ahsi is of order 24,and �1.A/ D �1.D/ is a normal 4-subgroup of D. Since every maximal subgroupof X is nonabelian, it follows that D Š M24 and c3.D/ D 2. Set K D C hsi D CD.


BecauseK \ X D D, we have c3.K/ D 2 andD is the unique L2-subgroup of K oforder 24. By Proposition 54.3, K (of order � 25) is an L2-group or a U2-group withthe kernel R D �1.A/ D �1.D/. If K is an L2-group, then K Š M2m , m > 4, acontradiction. Indeed,M2m is minimal nonabelian contrary to the fact thatK containsthe nonabelian proper subgroup D.We have proved that K is a U2-group with the kernel R D �1.A/, where K=R is

of maximal class. Let T=R be a cyclic subgroup of index 2 in K=R. Then jT=Rj � 4

and �1.T / D R. Let T1=R be the cyclic subgroup of order 4 in T=R. Then T1 is anL2-subgroup of K of order 24 and so T1 D D. If T1 ¤ T , then CT .R/ � T1 and soT1 D D would be abelian, a contradiction. Hence T1 D T D D and so jKj D 25.Since jK W C j D 2, C is of order 24. But A � Z.C / and so C is abelian.Assume that E0 D �1.C / > �1.A/ so that E0 is a normal elementary abelian

subgroup of order 8 inG. Consider the subgroup X0 D E0hsi, whereE0 \hsi D hs4i

is of order 2. We see (as in the proof of Theorem 54.1) that c3.X0/ D 4. This impliesX0 � X , a contradiction. We have proved that C is of rank 2 and so C Š C4 � C4

(since exp.C / D 4). Our proposition is proved.

Next we shall classify 2-groups G appearing in Theorems 54.1 and 54.2.

Theorem 54.6. Let G be a 2-group with cn.G/ D 4, n > 2, and suppose that G hasa normal elementary abelian subgroup E of order 8. Let fU1; U2; U3; U4g be the setof four cyclic subgroups of order 2n. Then X D hU1; U2; U3; U4i D EU1, whereE \ U1 ¤ 1 and we have the following possibilities.

(a) �2.X/ Š C4 �C2 �C2 andG=E is either cyclic (of order� 2n�1) orG=E is ofmaximal class (and order � 2n) and for n D 3 the last group must be dihedral.

(b) n D 3, �2.X/ Š D8 � C2, and one of the following holds:

(b1) G D X , which is a uniquely determined group of order 25;

(b2) G has a self-centralizing abelian normal subgroup A of type .4; 2/ suchthat G=A Š D8;

(b3) G has a self-centralizing abelian normal subgroup W of type .4; 4/ suchthat G=W Š E4 or D8.

Proof. Theorem 54.1 implies that X D EU1 with E \ U1 ¤ 1 and �2.X/ Š C4 �

C2 � C2 or�2.X/ Š D8 � C2, where in the second case n D 3.Suppose that�2.X/ Š C4�C2�C2. Then�1.X/ D E Š E23 andX=E Š C2n�1

with jX=Ej � 4. Hence X is an L3-group of order 2nC2. Since X is generated byits cyclic subgroups of order 2n, it follows that X is the unique L3-subgroup of order2nC2 inG. By Proposition 54.3, G is either an L3-group or a U3-group with respect tothe kernel R D �1.X/ D E. In particular, G=E is either cyclic (of order � 2n�1) orG=E is of maximal class (and order� 2n). If in the second case n D 3, thenG=E mustbe dihedral. Indeed, if n D 3 andG=E is of maximal class but not dihedral, then there


is an element x 2 G � X such that x2 2 �2.X/ � E, where �2.X/=E D Z.G=E/.But then o.x/ D 8, a contradiction.Assume now that �2.X/ Š D8 � C2 in which case n D 3. Our result follows at

once from Theorem 54.5.

Theorem 54.7. Let G be a 2-group with cn.G/ D 4, n > 2, and suppose that G doesnot have a normal elementary abelian subgroup of order 8. Let fU1; U2; U3; U4g bethe set of four cyclic subgroups of order 2n. Then the subgroup X D hU1; U2; U3; U4i

is of order 2nC2 and we have the following possibilities.

(a) X D WU1 with W \ U1 Š C4, where W is an abelian normal subgroup oftype .4; 4/, �2.X/ D W , X is metacyclic, CG.W / is metacyclic and one of thefollowing holds:

(a1) n > 3, and G=W is either cyclic (of order � 2n�2) or G=W is ofmaximal class (and order � 2n�1) and for n D 4 the last group must bedihedral;

(a2) n D 3, G has a normal metacyclic subgroup N such that CG.W / �

N , X � N , and G=N is isomorphic to a subgroup of D8. Moreover,CG.W /=W is cyclic, N=W is either cyclic or generalized quaternion,and in the second case G=N is elementary abelian of order � 4.

(b) n > 2, X D QU1, where Q Š Q8 is a normal quaternion subgroup of G,G D QP withQ \ P D Z.Q/, U1 � P , CG.Q/ � P , jP W CG.Q/j � 2, andP is either cyclic or of maximal class.

(c) n D 3, jX j D 25, �2.X/ Š Q8 � C2 or �2.X/ Š D8 � C2 and if G > X ,then G has a self-centralizing normal abelian subgroup W of type .4; 4/ withG=W Š E4 or D8.

Proof. We may assume that G is nonabelian.

(i) We consider first the groups G from Theorem 54.2(a). In this case X D WU1

with U1 Š C2n ,W \U1 Š C4, whereW is an abelian normal subgroup of type .4; 4/and �2.X/ D W . Also, X and CG.W / are metacyclic.Assume that n > 3. Then X=W is cyclic of order 2n�2 � 4. Set S=W D

�1.X=W / so that S=W � ˆ.X=W / � ˆ.G=W /. Since S=W stabilizes the chainW > �1.W / > f1g, it follows (as in the proof of Theorem 54.2) that all elements inS �W are of order 8. Indeed, set U1 \ S D hd i, where d is an element of order 8.For each w 2 W , wd D wl with l 2 �1.W /. Then .dw/2 D dwdw D d2wdw D

d2wlw D d2w2l , where w2l 2 �1.W / and so o.dw/ D 8. Also, X is generated byits elements of order 2n.We want to show that G=W is either cyclic or of maximal class and we use an in-

duction on jGj. If X D G, we are done. In the sequel let X < G. Let X � M < G,where M is maximal in G. By induction, M=W is either cyclic or of maximalclass. We have ˆ.G=W / � M=W and since ˆ.G=W / cannot be of maximal class


(Burnside), ˆ.G=W / is either cyclic (of order � 2) or abelian of type .2; 2/ (thelast case being possible only if M=W Š D8). However, if ˆ.G=W / Š E4, thenˆ.G=W / � Z.G=W / (see Proposition 4.9). But then M=W is abelian, a contradic-tion. Thus ˆ.G=W / is cyclic and so S=W D �1.ˆ.G=W //.Let F=W be a cyclic subgroup of order 2n�2 in G=W . Now, �1.F=W / �

ˆ.F=W / � ˆ.G=W / and so �1.F=W / D S=W . But all elements in S �W are oforder 8 and so cn.F / D 4 and F is generated by its elements of order 2n. Indeed, letF0=W D ˆ.F=W / and x 2 F � F0. Then x2n�3

2 S �W and so o.x/ D 2n. Thisgives F D X and so X=W is the unique cyclic subgroup of order 2n�2 in G=W . Asin the proof of Proposition 54.3, G=W is either cyclic (of order � 2n�2) or G=W isof maximal class (and order � 2n�1) and for n D 4 the last group (obviously) must bedihedral.It remains to consider the case n D 3. In this case jX j D 25 and all 16 elements in

X � W are of order 8. Suppose first that X is nonabelian. Then CG.W / \ X D W

and CG.W / is metacyclic. If CG.W / > W , then CG.W / (being metacyclic) haselements of order 8 (which are not contained in X), a contradiction. We have provedthat CG.W / D W . By Theorem 50.1, the group G has a normal metacyclic subgroupN such that W � N , �2.N / D W , and G=N is isomorphic to a subgroup of D8

and we assume that N is maximal subject to these conditions. In that case N > W .Indeed, if N D W , then G=W is isomorphic to a subgroup of D8. But X is a normalmetacyclic subgroup of G, W � X , �2.X/ D W , and G=X (being a factor-groupof D8) is also isomorphic to a subgroup of D8. This contradicts the maximality ofN and so N > W . In that case there are elements of order 8 in N and so X � N .Suppose that Y=W is a subgroup of order 2 inN=W . Since�2.Y / D W , all elementsin Y � W are of order 8. Hence Y D X and so N=W has exactly one subgroup oforder 2. By the structure of Aut.W / (Proposition 50.5), we see that Aut.W / has noquaternion subgroups. Thus N=W is cyclic (of order � 4). We have obtained allproperties stated in part (a2) of our theorem.We consider now the remaining case, where X D WU1 is abelian of order 25 (and

of type .8; 4/). According to Theorem 50.1, G has a normal metacyclic subgroup Nsuch that �2.N / D W , C D CG.W / � N , and G=N is isomorphic to a subgroup ofD8. Obviously, X � C . Suppose that Y=W is a subgroup of order 2 in N=W . Since�2.Y / D W , all elements in Y �W are of order 8. Hence Y D X and so N=W hasexactly one subgroup of order 2. It follows that N=W is either cyclic or generalizedquaternion.We claim that C=W is cyclic. Suppose false. Then C=W is generalized quaternion.

Let C0=W Š Q8 be a quaternion subgroup in C=W , whereX < C0 (becauseX=W isthe unique subgroup of order 2 inN=W ). LetZ1=W andZ2=W be two distinct cyclicsubgroups of order 4 in C0=W . Then Z1 and Z2 are both abelian, hZ1; Z2i D C0,and Z1 \ Z2 D X . This implies that X � Z.C0/ and so X D Z.C0/. Since C0 ismetacyclic, C 0

0 is cyclic and C00 covers X=W . All elements in X �W are of order 8,

so jC 00j D 8. But then the formula (see Lemma 1.1) jC0j D 27 D 2jZ.C0/jjC

00j gives


a contradiction. We have proved that CG.W /=W is cyclic. Thus CG.W / D CG.X/

is abelian.We have to consider the case, where N=W is generalized quaternion. We set

K=W D .N=W /0 D ˆ.N=W / so that K=W is cyclic and Z.N=W / D X=W .We have N=K Š E4 and since N is metacyclic (and therefore d.N / D 2), we getK D ˆ.N/ and N 0 is cyclic. On the other hand, N 0 covers K=W and all elementsin X �W are of order 8. Hence N 0 \ X Š C8, N 0 \ W Š C4, and jK W N 0j D 4.Again, since N is metacyclic, there is a cyclic normal subgroup S of N such thatN 0 < S and jS W N 0j D 2. If S � K, then jS \ W j D 8. But a maximal sub-group S \ W of W is not cyclic since S \ W � ˆ.W / D �1.W / Š E4. Thiscontradiction shows that S \K D N 0. Set T D WS D KS so that T=W is a cyclicsubgroup of index 2 in N=W . For each x 2 N � T , we have x2 2 X � W andso all elements in N � T are of order 16. Also, T=S Š W=.S \ W / Š C4. Now,ˆ.T / � K and ˆ.T / � hN 0;�1.W /i since N 0 D Ã1.S/ and �1.W / D ˆ.W /.Since jK W .N 0�1.W //j D 2 and T is not cyclic, we get ˆ.T / D N 0�1.W /,F D ˆ.T / \ X is abelian of type .8; 2/, and F1 D ˆ.T / \ W is abelian of type.4; 2/.Set S D hai, where o.a/ D 2m, m � 4. Also set s D a2m�3

, v D a2m�2

, andz D a2m�1

, so that z 2 Z.G/, hsi D S \ X , hvi D S \W , and hzi D S \�1.W /.Obviously, hsi and hvi are normal subgroups of G. Since ˆ.N/ D K, there is anelement b 2 N � T (of order 16) such that b2 2 X �W and b2 62 F . Then b2 D ws,where w 2 W � F1 and W D hwi � hvi. We set w2 D u and so �1.W / D hu; zi.We compute b4 D w2s2 D uv and b8 D z. Hence N D haihbi, where hai isa cyclic normal subgroup of order 2m .m � 4/ of N and hbi is cyclic of order 16with hai \ hbi D hzi (of order 2). It remains to determine the action of hbi on hai.Now, hbi induces a cyclic automorphism group on hai so that b inverts hai=hvi (sinceN=W is generalized quaternion). Thus ab D a�1v0 with v0 2 hvi and so .a4/b D

.ab/4 D .a�1v0/4 D a�4. In particular, b inverts hvi. On the other hand, b centralizes

b4 D uv. We compute uv D .uv/b D ubvb D ubv�1, and so ub D uz. The lastrelation is crucial and shows that b does not stabilize the chain W > �1.W / > f1g.If A (of order 25) is a Sylow 2-subgroup of Aut.W / and B is the full stabilizer

group of the chain W > �1.W / > f1g, then B is elementary abelian of order 24

and jA=Bj D 2 (see Proposition 50.5). We note that CG.W / � N and CG.W /=W

is a cyclic normal subgroup of N=W and so CG.W / � T . Hence, if L is the setof all y 2 G which stabilize the chain W > �1.W / > f1g, then the subgroup Lcovers G=N (since b 62 L) and so G=N (being isomorphic to a subgroup of D8) iselementary abelian of order � 4. We have obtained all properties stated in part (a2) ofour theorem.

(ii) We consider now the groups G from Theorem 54.2(b). In this case n > 2,X D QU1, whereQ Š Q8 is a normal quaternion subgroup of X , Q \ U1 D Z.Q/,U1 D hbi Š C2n , and b either centralizes Q or b induces on Q an involutory outerautomorphism in which case n > 3.


Assume first that b centralizes Q. In that case X D Q � hbi with Q \ hbi D

Z.Q/ D hzi. Set v D b2n�2

so that �2.X/ D Q � hvi. Since Q is the uniquequaternion subgroup in �2.X/, it follows that Q is characteristic in X and so Q is anormal subgroup in G. Set C D CG.Q/ so that X \ C D U1 D hbi, C is normal inG, and jG W .Q � C/j � 2. Since U1 is the unique cyclic subgroup of order 2n in C ,it follows that C is either cyclic or of maximal class. If G D Q � C , we are done.Suppose that jG W .Q � C/j D 2. Since G=C Š D8 (a Sylow 2-subgroup of

Aut.Q8//, there is an element d 2 G � .QC/ with d2 2 C . Set P D C hd i so thatG D QP and P \X D U1. Since U1 is the unique cyclic subgroup of order 2n in P ,we get again that P is either cyclic or of maximal class.It remains to consider the case, where b induces on Q an outer involutory auto-

morphism (and in that case n > 3). Obviously, Q is the unique quaternion subgroupof Qhb2i D Q � hb2i, where Q \ hb2i D hzi. We have exp.Qhb2i/ D 2n�1 andall elements in X � .Qhb2i/ are of order 2n. Hence Q is also the unique quaternionsubgroup of X and so Q is normal in G. Again set C D CG.Q/ so that C is normalinG and X \C D hb2i. Also, jG W .QC/j D 2 and so G D XC . Set P D C hbi. Wehave jP W C j D 2, P \ X D hbi D U1, and G D QP withQ \ P D hzi. Since U1

is the unique cyclic subgroup of order 2n in P , it follows that P is either cyclic or ofmaximal class and we are done.

(iii) Finally, we consider the groups G from Theorem 54.2(c). In this case n D 3,jX j D 25, and �2.X/ Š Q8 � C2 or �2.X/ Š D8 � C2. Since G has no normalsubgroups isomorphic to E8, Proposition 54.5 gives at once the result stated in part (c)of our theorem.

�55

2-groups G with smallsubgroup hx 2 G j o.x/ D 2ni

In �52, 2-groups G with j�2.G/j D 16 are classified. In this section we consideressentially more difficult problem to classify the 2-groups in which all elements oforder 4 generate the subgroup of order 16.In what follows we suppose thatG is a 2-group of order > 24 all of whose elements

of order 4 generate a subgroup H of order 24. In that case, obviously, H is not ofmaximal class so it has no cyclic subgroups of index 2 (see Theorem 1.2). It is obviousthat, in the case under consideration, c2.G/ � 4. Indeed, if the above assertion isfalse, then c2.G/ D 2, by Theorem 1.17(b). In that case, by �52, the elements of order4 generate the abelian subgroup of type .4; 2/, contrary to our hypothesis (this alsofollows from �43). If c2.G/ D 4, then H Š Q8 � C4 or H Š C4 � C2 � C2 and thegroup G was completely determined in �52. If c2.G/ D 5, then G is of maximal class(Theorem 1.17(b)) and so H Š Q16 and (since jGj > 24) G Š SD32. It remains toexamine the case c2.G/ D 6 since there is no 2-group G with c2.G/ D 7 (Theorem1.17(b)). The 2-groups G with j�2.G/j D 24 have been determined in �52, and so wemay also assume j�2.G/j > 2

4.

Exercise 1. Let G be a 2-group, H D hx 2 G j o.x/ D 4i. If exp.Z.H// D 2, thenCG.H/ D Z.H/.

Solution. Assume that CG.H/ > Z.H/. Let x 2 CG.H/ � Z.H/ be such thatx2 2 Z.H/. Then o.x/ � 4 so x is an involution since x 62 H . If a 2 H is of order 4,then o.ax/ D 4 and ax 62 H , a contradiction.

Let G be a 2-group and let ��2.G/ D hx 2 G j o.x/ D 22i. Suppose that

j��2.G/j D 24 and c2.G/ > 4; then c2.G/ � 6 (Theorem 1.17(b)) so G contains

at least twelve elements of order 4. Then c2.G/ D 6 since c1.G/ � 3.

Theorem 55.1 ([Jan8]). Let G be a 2-group of order > 24 all of whose elements oforder 4 generate a subgroup H of order 24, i.e., H D ��

2.G/. Assume, in addition,that c2.G/ D 6 and j�2.G/j > 2

4. Then we have the following possibilities:

(a) H Š Q8 � C2 and G Š SD16 � C2.

(b) H Š ha; b j a4 D b4 D 1; ab D a�1i is metacyclic and G D hb; t j b4 D

t2 D 1; bt D ab; a4 D 1; ab D a�1; at D a�1i. Here jGj D 25, H D

ha; bi, ˆ.G/ D ha; b2i Š C4 � C2, �2.G/ D G, Z.G/ D ha2; b2i Š E4.

�55 2-groups G with small subgroup hx 2 G j o.x/ D 2ni 123

(c) H Š C4�C4 andG has a metacyclic maximal subgroupM such that�2.M/ D

H , G D M hti, where t is an involution with CM .t/ D �1.M/ Š E4.

Proof. SinceH is neither cyclic nor of maximal class,H has a normal four-subgroupH0 (Lemma 1.4) and so all twelve elements inH �H0 are of order 4. IfH is abelian,then H0 D �1.H/ implies that H D C4 � C4. Suppose that H is nonabelian. ThenH=H0 is elementary abelian and so ˆ.H/ � H0.Suppose, in addition, that H is not minimal nonabelian and let Q be a nonabelian

subgroup of order 8. Since H has only three involutions, we have Q Š Q8. IfCH .Q/ � Q, thenH is of maximal class (Proposition 10.17), a contradiction. HenceH D Q � Z, where jZj D 4 andQ \ Z D Z.Q/. If Z is cyclic, then c2.H/ D 4, acontradiction. Hence Z Š E4 andH Š Q8 � C2.Now suppose that H is minimal nonabelian. Then d.H/ D 2 so ˆ.H/ D H0 and

jH 0j D 2. In that case,H=H 0 is abelian of type .4; 2/. Let Z=H 0 be a direct factor ofH=H 0 of order 2. SinceH=Z is cyclic of order 4, we get Z ¤ ˆ.H/ D �1.H/ soZis cyclic. LetZ D hai andH=Z D hbZi; then hbi \ hai D f1g. Since Aut.Z/ Š C2,we getH D ha; b j a4 D b4 D 1; ab D a�1i.We have to determine the structure of G for each of the above three possibilities for

the structure ofH .

(i) Assume that H D Q � hui, where Q Š Q8 and u is an involution. Sethzi D Z.Q/ so that Z.H/ D hu; zi. There are exactly three maximal subgroupsof H containing Z.H/ and they are abelian of type .4; 2/. One of them, say hu; ai

(a 2 Q � Z.Q/), is normal in G (consider the action of G by conjugation on the setof above three maximal subgroups of H ). If b 2 Q � hai, then the other two suchabelian subgroups of type .4; 2/ are hu; bi D hui � hbi and hu; abi D hui � habi.Since�2.G/ > H , there is an involution t 2 G �H sinceH contains all elements

of order 4 in G. Let hu; xi, where x 2 Q�Z.Q/, be any t -invariant abelian subgroupof type .4; 2/ in H . If t acts nontrivially on Z.H/, then ht;Z.H/i Š D8 has a cyclicsubgroup of order 4 that is not contained in H , a contradiction. Thus t centralizesZ.H/. Since CH .t/ must be elementary abelian (otherwise, .hti � CH .t// � CH .t/

has an element of order 4), we get CH .t/ D Z.H/ and so t acts fixed-point-free onhu; xi � Z.H/. If xt D xs with s 2 Z.H/ � hzi, then .tx/2 D .txt/x D xtx D

xsx D sx2 D sz 2 Z.H/#, and so o.tx/ D 4, a contradiction since tx 62 H . Hencext D x�1 and so t inverts the subgroup hu; xi. In particular, t inverts hu; ai. If tnormalizes hu; bi, then t also normalizes hu; abi. But in that case t inverts hu; bi andhu; abi and so t invertsH D hu; ai [ hu; bi [ hu; abi. This implies thatH is abelian(Burnside) which is not the case. Hence we must have hu; bit D hu; abi.Let M D NG.hu; bi/ so that H � M , jG W M j D 2 and G D M hti (see the

last sentence of the previous paragraph). By the previous paragraph, the set M � H

has no involutions so �2.M/ D H since exp.H/ D 4 and H contains all elementsof order 4 from G. Assume that M > H so that we can apply the results of �52(see Theorem 52.1(d)). It follows thatM is a uniquely determined group of order 25


which is given explicitly in Theorem 49.1, (A2)(a)). There is an element y 2 M �H

of order 8 (indeed, exp.M/ D 8 and H D �2.M/) such that y2 D ua, uy D uz,ay D a�1, by D bu. Hence y acts nontrivially on Z.H/ and y normalizes all threeabelian subgroups of type .4; 2/ inH .We have G=H Š E4 sinceM=H and H hti=H are two distinct subgroups of order

2 in G=H . Hence .yt/2 2 H and we consider the subgroup N D H hyti of order 25.Since yt acts nontrivially on Z.H/ (t centralizes Z.H/), it follows that there are noinvolutions in N �H (indeed, if w is an involution in N �H , then hw;Z.H/i Š D8

contains an element of order 4 which is not contained in H , a contradiction). ThusH D �2.N / and so, by the above, N is uniquely determined and therefore N Š M .In particular, N normalizes all three abelian subgroups of type .4; 2/ in H . SinceM ¤ N , we get that G D hM;N i normalizes all three abelian subgroups of type.4; 2/ inH (see the last sentence of the previous paragraph), a contradiction.We have proved that H D M and so G D H hti. Since hu; bit D hu; abi, we

have bt D abs0 with s0 2 Z.H/. Hence t normalizes Q� D hb; bt i Š Q8 andH D hui �Q�. ReplacingQ withQ�, we may assume from the start that t normal-izesQ and so t induces on Q an outer automorphism (indeed, hbi is not t -invariant).We get Qhti D D Š SD24 (if D 6Š SD24 , then cl.D/ D 2; in that case, hbi G D

so t -invariant, which is not the case) and G D hui �D. We have obtained the possi-bility (a).

(ii) Now suppose that H D ha; b j a4 D b4 D 1; ab D a�1i is metacyclic oforder 16. Here ˆ.H/ D Z.H/ D ha2; b2i D �1.H/ Š E4, H 0 D ha2i, all elementsin H � Z.H/ are of order 4 and H=H 0 is abelian of type .4; 2/. It follows that H isminimal nonabelian. Therefore all three maximal subgroups Wi , i D 1; 2; 3, of H areabelian of type .4; 2/: W1 D hZ.H/; ai with Ã1.W1/ D ha2i; W2 D hZ.H/; bi withÃ1.W2/ D hb2i; W3 D hZ.H/; abi with Ã1.W3/ D hb2i since .ab/2 D ab2ab D

ab2a�1 D b2. It follows that a2b2 is not a square in H.D W1 [ W2 [ W3/ andW1 is normal in G. (Note, however, that all involutions in a homocyclic 2-group ofcomposite exponent are squares; see, for example, Theorem 6.1.) Hence ha2i D H 0,hb2i, and ha2b2i are characteristic subgroups in H and therefore they are normal inG. This gives Z.H/ � Z.G/. In fact we have Z.H/ D Z.G/, by Exercise 1.Since �2.G/ > H , there is an involution t 2 G � H . By Exercise 1, CH .t/ D

Z.H/ D Z.G/. If t normalizes a maximal subgroup Wi of H , then t inverts Wi , byExercise 1. We know that t normalizesW1 and so t invertsW1 and therefore at D a�1:

If t normalizes W2, then t also normalizes W3 and consequently t inverts H (see theprevious paragraph). But then H is abelian, which is not the case. We have provedthatW t

2 D W3.Let K D NG.W2/ .D NG.W3// so that H � K, jG W Kj D 2, and G D Khti

(see the last sentence of the previous paragraph). But then the set K � H has noinvolutions (otherwise, that involution inverts H so H is abelian, which is not thecase) and so �2.K/ D H . Since H is metacyclic, it follows from Theorem 41.1and Remark 41.2 that K is also metacyclic. If K > H , then [Ber25, Remark R48.2]


implies that H Š C4 � C4, which is not the case. Hence we have K D H and soG D H hti is of order 25.Since W t

2 D W3, we have bt 2 fab; .ab/�1; .ab/a2; .ab/�1a2g. Note thatba D a2b and so

.ab/a D a.a2b/ D .ab/a2; ..ab/�1/a D ..ab/a/�1 D ..ab/a2/�1 D .ab/�1a2:

Hence, replacing t with the involution ta (if necessary), we see that we may assumethat bt 2 fab; .ab/�1 D .ab/b2 D ab�1g. If bt D ab�1, then we replace awith a0 D ab2. We get bt D ab�1 D ab3 D .ab2/b D a0b, .a0/b D .a0/�1,.a0/t D .a0/�1, and see that one may assume from the start (writing again a instead ofa0) that bt D ab. The structure of G is uniquely determined as stated in part (b).

(iii) Finally, assume thatH D hai� hbi Š C4 �C4. Set C D CG.H/. By Exercise1, �2.C / D H and therefore C is metacyclic. Since, by assumption, �2.G/ > H ,there is an involution t 2 G � C . Suppose that the coset C t contains an element ywhich is not an involution. Then o.y/ � 8 and o.y2/ � 4. In particular, yo.y/=4 D d

is an element of order 4 in H . We have y D ct with c 2 C . But then d D dy D

d ct D d t , contrary to Exercise 1. Hence, all element in C t are involutions. Thisimplies that t inverts C and C is abelian (Burnside). Since�2.C / D H , we concludethat C is of rank 2.Suppose that t 0 is an involution inG� .C hti/. Then, by the above, t 0 inverts C . But

then t t 0 2 G � C and t t 0 centralizes C , a contradiction. Thus C hti D �2.G/. SetD D �2.G/. We have CG.t/ D hti�H0 D CD.t/, whereH0 D �1.H/ D ha2; b2i.Indeed, if CG.t/ 6� �2.G/, then there is an element y0 2 CG.t/ � CD.t/ such that.y0/2 2 CD.t/ D �1.H/ so o.y0/ � 4 and y0 2 H , which is a contradiction.The last result gives an upper bound for the order of G=C . The conjugacy class of

t in G is locked in D � C and so there are at most jD � C j D jC j conjugates of t inG. This gives 1

8jGj D jG W CG.t/j � jC j and so jG=C j � 8. Also note thatD=C is a

normal subgroup of order 2 in G=C .Suppose that the involution in D=C is a square in G=C . Then there is an element

k 2 G � D such that k2 2 D � C . But all elements in D � C are involutions andso k is an element of order 4, a contradiction. It follows that G=C is abelian andD=C 6� ˆ.G=C/. There is a maximal subgroup M of G such that D \ M D C .Since �2.M/ D H , it follows that M is metacyclic. We have G D M hti, whereCM .t/ D �1.M/ D �1.H/ Š E4. We have obtained a group stated in part (c) of ourtheorem which is now completely proved.

Let n > 1 and let G be a p-group. Denote ��n.G/ D hx 2 G j o.x/ D pni. We

have ��n.G/ D f1g if exp.G/ < pn.

Proposition 55.2 (Berkovich). Let G be an irregular p-group. Suppose that we havej��

2.G/j D ppC1 < pm D jGj. Then the following assertions hold:

(a) If p D 2, then G Š SD24 and H Š Q8 (see Theorems 43.4 and 52.8). In thesequel we assume that p > 2.


(b) If �2.G/ D ��2.G/, then G is not of maximal class and ��

2.G/ is regular (seeLemma 42.1). Moreover, G is a group of Lemma 42.1.

(c) If G is of maximal class and p > 3, then ��2.G/ D G1, where G1 is the

fundamental subgroup of G. In this case every irregular maximal subgroup ofG contains exactly p subgroups of order pp and exponent p, and G containsexactly p2 such subgroups. Next,m D p C 2.

(d) If p D 3 and G is of maximal class, then ��2.G/ D �2.G1/ is metacyclic, all

nonmetacyclic subgroups of order 33 are nonabelian and their number equals3m�3. In the sequel we assume that G is not of maximal class.

(e) If �2.G/ is regular, then ��2.G/ D �2.G/ (see (b)). In the sequel we assume

that ��2.G/ < �2.G/.

(f) Suppose that p > 3. Let t be an element of order p in G � ��2.G/ and set

D D ht;��2.G/i. Then d.D/ D 3, exactly p2 maximal subgroups of D are of

maximal class; other maximal subgroups of D are of exponent p.

(g) All pC1maximal subgroups of��2.G/ are normal inG. ˆ.�

�2.G// is contained

in aG-invariant subgroup L of order pp and exponent p such that L 6� ��2.G/.

Proof. (a) Let p D 2. Then c2.G/ D c2.��2.G// 2 f2; 3g. If c2.G/ D 3, then G is of

maximal class by Theorem 1.17(b). In that case, G Š SD24 . If c2.G/ D 2, then suchG are described in Theorems 43.4 and 52.8. In what follows we assume that p > 2.

(b) The p-groups with j�2.G/j D ppC1 are classified in Lemma 42.1. In all cases,�2.G/ is regular.

(c) Let G be of maximal class and p > 3. In this case, j�2.G1/j � ppC1 sincej�1.G1/j � pp�1 and G=�1.G1/ has no cyclic subgroups of index p (see Theorems9.5 and 9.6). It follows that��

2.G/ D �2.G1/ and j�2.G1/j D ppC1 < p2p�2 sincep > 3. In that case, G1 D �2.G1/ D ��

2.G/. IfM is an irregular maximal subgroupof G, then ��

2.M/ D M \ ��2.G/ < M and ��

2.M/ is the unique absolutelyregular maximal subgroup ofM . We conclude thatM has exactly p distinct maximalsubgroups of exponent p. Since ˆ.G/, the intersection of any two distinct maximalsubgroups of G, is absolutely regular, all remaining assertions of (c) are true.

(d) Now let G be a 3-group of maximal class and order � 35. Since, �2.G1/ D

��2.G/ is of order 3

4, we get ��2.G/ D �2.G1/. The remaining assertions in (d) are

trivial since the number of irregular subgroups of order 34 inG equals 3m�4 (Exercise9.27) and every such subgroup contains exactly three subgroups of order 33 and expo-nent 3 (see the proof of (c)). All subgroups of order 33 and exponent 3 are nonabeliansince m > 4 (this is an easy consequence of Theorems 9.5 and 9.6).

(e) follows from Theorem 7.2. In what follows we assume that ��2.G/ < �2.G/.

(f, g) By hypothesis, G ��2.G/ has an element t of order p. SetD D ht;��

2.G/i.Assume thatD is of maximal class. LetD1 be the fundamental subgroup ofD (see �9,Theorems 9.5 and 9.6). Since p > 3 and D1 is absolutely regular of width p � 1 and


order � ppC1, we get��2.D1/ D �2.D1/ D D. It follows that c2.G/ D c2.�

�2.G//

is not divisible by pp�1 so G is of maximal class (Theorem 13.2(b)), contrary tothe assumption. Thus, D is not of maximal class. By Theorem 12.12, D containsexactly p2 subgroupsM1 D ��

2.G/; : : : ;Mp2 of maximal class and index p. We havejMi j D ppC1 for all i . All other pC 1 maximal subgroups ofD are not generated bytwo elements so they are regular (Theorem 12.12). LetM be one of regular maximalsubgroups of D. Supposing that exp.M/ D p2, we get ��

2.M/ D M 6� ��2.G/, a

contradiction. Thus, all regular maximal subgroups of D have exponent p. We seethat N D M \ ��

2.G/ is a maximal subgroup of exponent p in ��2.G/. If M1 is

another maximal subgroup of exponent p inD, thenN1 D M1 \��2.G/ is a maximal

subgroup of exponent p in ��2.G/. By Theorem 12.12(c), M \ M1 D �.D/ and

�.D/ 6� ��2.G/. It follows that N ¤ N1. We see that the numbers of subgroups of

order pp and exponent p is not divisible by p. By hypothesis, ��2.G/ has at least

two absolutely regular subgroups of index p. It follows that the number of absolutelyregular subgroups of index p in ��

2.G/ is not divisible by p. Therefore, all maximalsubgroups of ��

2.G/ are normal in G. Since the number of normal subgroups oforder pp and exponent p in G is 1 .mod p/ (Theorem 13.5), the last assertionin (h) follows.

Exercise 2. Study the irregular p-groups G D ��n.G/ of order p

pCn, generated byelements of order pn, n > 3, p > 2. Is it true that exp.G/ D pn?

Exercise 3. Classify the metacyclic 2-groups G such that ��2.G/ D G.

Exercise 4. Classify the nonmetacyclic 2-groups G with metacyclic��2.G/.

Theorem 55.3 ([Jan8, Theorem 4.2]). Let G be a 2-group with j��n.G/j D 2nC2,

n > 2. Then cn.G/ D 4 or 6. If cn.G/ D 4, then such groups G are classified in �54.If cn.G/ D 6, then n D 3 and the structure of H D ��

3.G/ is uniquely determined.We have H D ha; b j a8 D b8 D 1; ab D a�1; a4 D b4 D zi, where jH j D 25 andE D �1.H/ D hz; a2b2 D ui Š E4. If G > H , then jGj D 26 and we have for thestructure of G one of the following possibilities:

(a) G is the “splitting” metacyclic group G D hw; c jw16 D c4 D 1; wc D w11i,whereH D hwc�1; w2i and Z.G/ D hw8i is of order 2.

(b) G D hH; ci with c2 D z� .� D 0; 1/, .bc/2 D a, and ac D a�1. HereCG.E/ D hui � ha; ci, where ha; ci Š D16 or Q16 for � D 0 or 1, respectively.Also, Z.G/ D hzi is of order 2.

(c) G is a non-metacyclic U2-group with the kernel E Š E4 and G=E Š SD16.More precisely, G D hd; s jd16 D s2 D 1; d4 D v; d8 D z; d s D d�1vu;

u2 D 1; ud D du; us D uzi, where E D hu; zi and H D hsd; d2i withZ.H/ D Z.G/ D hvui Š C4. (This is the group (c) with n D 4 of Theorem67.3.)

All above groups are determined up to isomorphism and they exist.


Proof. Let G be a 2-group with j��n.G/j D 2nC2, n > 2. This implies that G is

neither cyclic nor of maximal class. By Theorem 1.17(b), cn.G/ is even. If cn.G/ D 2,then Proposition 54.4 gives that j��

n.G/j D 2nC1, a contradiction. If cn.G/ D 4, thensuch groups G have been classified in �54. Hence, we may assume that cn.G/ � 6.However, if cn.G/ � 8, then the number of elements of order 2n is � 8'.2n/ D

8 2n�1 D 2nC2, a contradiction.

We assume in the sequel that cn.G/ D 6 and set ��n.G/ D H . We shall determine

the structure of H . The number of elements of order 2n in H equals l D 6'.2n/ D

6 2n�1 D 3 2n D 2nC1 C 2n. Let E be a normal four-subgroup of H and let Z bea cyclic subgroup of order 2n�1 in H . It is easy to see that H0 D ZE is of exponent2n�1. Indeed,H 0

0 < E and soH0 is of class� 2with jH 00j � 2. Also,H0 is generated

with elements of orders � 2n�1 and 2n�1 � 4. If x; y 2 H0 and o.x/ � 2n�1,o.y/ � 2n�1, then .xy/2

n�1

D x2n�1

y2n�1

D 1. This shows that exp.H0/ D 2n�1.Since jH j � l D 2nC2 � 2nC1 � 2n D 2n, it follows that jH0j � 2n and so jH0j D 2n

and Z \ E ¤ f1g. All elements in H � H0 are of order 2n. This implies that�2.H/ D �2.H0/ is of order � 8 and so we may use Lemma 42.1. It follows that Hmust be isomorphic to a group (c) of that proposition. In particular, H is a U2-groupwith the kernel E, H=E is generalized quaternion, �2.H/ is abelian of type .4; 2/,and �2.H/=E D Z.H=E/. If n > 3, then cn.G/ D cn.H/ D 2, a contradiction.Hence n D 3, H=E Š Q8, �2.H/ D H0, and all 24 elements in H � H0 are oforder 8. More precisely, H D ha; b j a8 D b8 D 1; ab D a�1; a4 D b4 D zi, andso jH j D 25, Z.H/ D hb2i Š C4, H 0 D ha2i Š C4, H0 D �2.H/ D ha2; b2i,E D �1.H/ D hz; a2b2 D ui Š E4, and CH .E/ D CH .H0/ D H1 D hb2; ai isabelian of type .8; 2/. The following subgroups E, hzi D Ã1.H0/, H0, ha2i, hb2i,and H1 are characteristic in H and so they are normal in G. We assume G > H andwe intend to determine the structure of G up to isomorphism.

Set C D CG.E/ so that C is a maximal subgroup of G, C > H1, G D CH D

C hbi, C \ H D H1, and c3.C / D 2. By Proposition 54.4, C is an L2-group or aU2-group with kernelE. Note that U2-groups contain a unique normal four-subgroup.If C is a U2-group, thenH0=E is a normal subgroup of order 2 in C=E and thereforeH0=E D Z.C=E/. If C=E in this case is not dihedral, then there is an elementy 2 C � H1 such that y2 2 H0 � E. But then o.y/ D 8, which is a contradiction(since ��

3.G/ D H ). Here we have also used the fact that H1=E is a normal cyclicsubgroup of order 4 in C=E and so H1=E is contained in a cyclic subgroup of index2 in C=E. We have proved that C is either an L2-group (in which case C is abelianof type .2i ; 2/, i � 4) or C is a U2-group with C=E dihedral. In any case, C=Ehas the unique maximal cyclic subgroup T=E of index � 2. Then T is normal in G,T is abelian of type .2j ; 2/, j � 3, and T � H1 (since H1=E is the unique cyclicnormal subgroup of order 4 in C=E). Since b2 2 H1 � T , it follows that in casejG=T j D 4, C=T and hbiT=T are two distinct subgroups of order 2 in G=T andso G=T Š E4. In any case, G=T is elementary abelian of order � 4. We haveCH1

.b/ D hb2i Š C4 and so CT .b/ D hb2i. Indeed, if b centralizes an element


y 2 T �H1, then o.y/ � 16 and so b centralizes an element of order 8 and this onemust lie in H1, a contradiction.Set K D hbiT and assume K > H or equivalently T > H1. We shall determine

the structure of K. Suppose that there is an element k of order 16 in K � T . Thenk2 2 T and o.k2/ D 8. But CT .b/ D CT .k/ and so b centralizes an elementof order 8 in T , which contradicts our last result in the previous paragraph. Hencec4.K/ D c4.T / D 2. Proposition 54.4 implies that K is either an L2-group or aU2-group with the kernel E. But H is a U2-group with H=E Š Q8 and so K is aU2-group. We have H0=E D Z.K=E/ and K=E is of maximal class containing theproper subgroup H=E isomorphic to Q8. We know that T=E is the cyclic subgroupof index 2 in K=E. If x 2 K � T is such that x2 2 H0 � E, then o.x/ D 8 andso x 2 H . Hence, if s 2 K � .H [ T /, then s2 2 E and CT .s/ D hb2i impliess2 2 E \ hb2i D hzi. This forces K=E Š SD16 and jT=Ej D 8. Note that SD16

is the unique 2-group of maximal class and order > 8 containing Q8 as its maximalgeneralized quaternion subgroup. Since s 62 C , it follows us D uz, where E D hz,u D a2b2i. Thus Ehsi Š D8 and so (since D8 has five involutions) we may assumethat s is an involution. We may set T D hui�hd i, where o.d/ D 16 and d8 D z. Alsoset d4 D v so that o.v/ D 4 and v2 D z. Note that E � ˆ.H/ � ˆ.K/ and so theinvolution s does not normalize hd i (otherwise hd; si would be a maximal subgroupof K not containing E). On the other hand, K=E Š SD16 and so d s D d�1ve,where e 2 E � hzi. Replacing b with b�1 leads to the replacement of u D a2b2

with a2b�2 D a2b2z D uz and so we may assume from the start that e D u andd s D d�1vu. It is easy to see that K is not metacyclic. We have Ã1.T / D hd2i andso s normalizes hd2i. Indeed, .d2/s D .d s/2 D .d�1vu/2 D d�2v2u2 D d�2z,and so hs; d2i Š SD16 and Ehs; d2i is a non-metacyclic maximal subgroup of K.We locate H as the subgroup of K generated by elements of order 8 in K. We setb0 D sd and compute b2

0 D .sds/d D d�1vud D vu. Hence b0 and d2 areelements of order 8, hd2i is normal in K, hb0i \ hd2i D hzi, and so H D hsd , d2i

with Z.H/ D Z.K/ D hvui Š C4.Suppose in addition that C > T so that C is a U2-group with C=E Š D16,

G D KC , and G=T Š E4. Let c 2 C � T . Then c2 2 E, c inverts T=E ,c centralizes E (since C D CG.E/), and d c D d�1f with f 2 E. We com-pute .d2/c D .d�1f /2 D d�2, and so vc D v�1 because v D d4 2 hd2i.Hence d sc D .d�1vu/c D df v�1u D dvuzf and so .T hsci/=E Š M24 . Onthe other hand, .T hsci/ \ H D H1 and therefore c3.T hsci/ D c3.H1/ D 2. ByProposition 54.4, T hsci is either an L2-group or a U2-group with the kernel E. But.T hsci/=E Š M24 , a contradiction.We have proved that in this caseG D K and so it remains to prove the existence of

K. We set:

d D .1; 2; 3; 4; 5; 6; 7; 8; 9; 10; 11; 12; 13; 14; 15; 16/

.17; 21; 24; 25; 28; 20; 23; 26; 27; 29; 31; 32; 30; 18; 19; 22/;


and

s D .2; 17/.3; 7/.4; 23/.5; 13/.6; 30/.8; 24/.10; 27/

.11; 15/.12; 19/.14; 28/.16; 31/.18; 20/.22; 32/.25; 26/:

Then we see that the even permutations d and s (of degree 32) satisfy the definingrelations for G D K D hd; si:

d16 D s2 D 1; d4 D v; d8 D z; d s D d�1vu;

u2 D 1; ud D du; us D uz:

Since the permutation z D d8 is non-trivial, z 2 Z.K/, and Z.K/ Š C4, it followsthat the induced permutation representation of G is faithful and so G is realized as asubgroup of the alternating group A32. Thus, G is the group from (c).It remains to analyze the caseK D H or equivalently T D H1. Since G > H , we

have C > T and so C=E Š D8. We know that H=E Š Q8, G=T D G=H1 Š E4,and jGj D 26. Let c 2 C � T . Since C=E Š D8 and T=E is the unique cyclicsubgroup of order 4 in C=E, it follows c2 2 E. Both elements b and c invert T=E andtherefore bc centralizes T=E . Set N D T hbci, where .bc/2 2 T . Then N \H D T

and so c3.N / D c3.T / D 2. Proposition 54.4 implies that N is either an L2-group ora U2-group. But bc centralizes T=E and therefore N=E is abelian. This shows thatN is an L2-group and so N=E Š C23 with �1.N / D E. Since bc 62 C , ubc D uz

and so N Š M25 and .bc/2 is an element of order 8 in T D H1. Hence .bc/2 D ai

or .bc/2 D aiu, where i is an odd integer. Set a0 D ai and we see that (replacing awith a0) the relations forH remain unchanged:

b8 D .a0/8 D 1; b4 D .a0/4 D z; and .a0/b D .a0/�1;

except possibly .a0/2b2 D uz, where u D a2b2. But in that case we replace b withb0 D b�1 and obtain the old relations for H : .b0/8 D .a0/8 D 1, .b0/4 D .a0/4 D z,.a0/b

0

D .a0/�1 and .a0/2.b0/2 D .a0/2b�2 D .a0/2b2z D u. Hence we may writeagain a and b instead of a0 and b0 and so we may assume from the start that: .bc/2 D a

or .bc/2 D au, where u D a2b2, and the relations forH D ha; bi remain unchanged.Assume first .bc/2 D au. Then bc centralizes au and so we get au D .au/bc D

.a�1uz/c D .a�1/cuz, which gives ac D a�1z and .a2/c D a�2. Note that c 2

C D CG.E/, where E D hu; zi and so c centralizes u D a2b2. We get u D uc D

.a2b2/c D a�2.b2/c D a2b2, and so .b2/c D zb2 D b�2. From .bc/2 D au follows(since c2 2 E and c4 D 1)

bcbc D au; cbc D b�1au; c2.c�1bc/ D b�1au; bc D c2b�1au;

and so noting that b2 2 Z.H/ we get

.b2/c D c2b�1auc2b�1au D b�2; c2.b�1auc2b/au D 1;

c2a�1uz.c2/bau D 1; c2z.c2/b D 1; .c2/b D c2z;


and so c2 2 E � hzi and therefore we may set c2 D uz� .� D 0; 1/ and bc D

uz�b�1au. Conjugating the last relation with c�1 we get

b D uz�.bc�1

/�1a�1zu; bc�1

D uz�C1a�1b�1u;

bc�1

D uz�C1b�1au; bc�1

D bcz:

We compute

uzb�1au D uz.b�1aub/b�1 D uza�1uzb�1 D ua�1ub�1

D ua�1.a2b2/b�1 D .ua/b D .au/b D .bc/2b;

and noting that .bc/8 D .au/4 D a4 D z we get

.bc/c D bcc D uz�b�1auc D z�C1.uzb�1au/c D z�C1.bc/2bc

D .bc/8.�C1/.bc/3 D .bc/11C8�:

Since hbci\hci D f1g (which follows from .bc/8 D .au/4 D a4 D z and c2 D uz�),we see that G D hbc; ci is a “splitting” metacyclic group with the cyclic normalsubgroup hbci of order 16 and a complement hci Š C4. Since bc�1

D bcz, we get

.bc/c�1

D bc�1

c D bczc D .bc/cz D .bc/11C8�.bc/8 D .bc/3C8�:

If � D 0, then .bc/c D .bc/11 and .bc/c�1

D .bc/3. If � D 1, then .bc/c D .bc/3

and .bc/c�1

D .bc/11. Hence, replacing c with c�1 (if necessary), we may assumefrom the start that .bc/c D .bc/11. We set bc D w and so we have obtained theunique metacyclic group G D hw; c jw16 D c4 D 1; wc D w11i, which was statedin part (a) of our theorem. This group obviously exists because w ! w11 induces anautomorphism of order 4 of the cyclic group hwi of order 16.Assume now .bc/2 D a. Then bc centralizes a and so we get a D abc D .a�1/c

which implies ac D a�1 and .a2/c D a�2. Note that c 2 C D CG.E/, whereE D hu; zi and so c centralizes u D a2b2. We get

u D uc D .a2b2/c D a�2.b2/c D a2b2 and so .b2/c D zb2 D b�2:

From .bc/2 D a follows (since c2 2 E and c4 D 1)

bcbc D a; cbc D b�1a; c2.c�1bc/ D b�1a; bc D c2b�1a;

and so noting that b2 2 Z.H/ we get

.b2/c D c2b�1ac2b�1a D b�2; c2.b�1ac2b/a D 1;

c2a�1.c2/ba D 1; .c2/b D c2;


and so c2 2 hzi. We set c2 D z� .� D 0; 1/ and have C D CG.E/ D hui � ha; ci.If � D 0, then ha; ci Š D16. If � D 1, then ha; ci Š Q16. Also, Z.G/ D hzi sinceZ.H/ D hb2i and .b2/c D b�2.If � D 0, then we set (identify):

b D .1; 9; 7; 10; 5; 11; 3; 12/.2; 13; 8; 14; 6; 15; 4; 16/;

c D .2; 8/.3; 7/.4; 6/.9; 13/.10; 16/.11; 15/.12; 14/:

Since b4 D z is a nontrivial permutation, we have realized the group G D hb; ci as asubgroup of S16.If � D 1, then we set

b D .1; 9; 7; 10; 5; 11; 3; 12/.2; 13; 8; 17; 6; 18; 4; 19/

.14; 15; 23; 24; 28; 22; 31; 27/.16; 30; 25; 32; 29; 20; 21; 26/;

c D .1; 16; 5; 29/.2; 22; 6; 15/.3; 25; 7; 21/.4; 27; 8; 24/

.9; 14; 11; 28/.10; 31; 12; 23/.13; 20; 18; 30/.17; 32; 19; 26/:

Since b4 D z is a nontrivial permutation, we have obtained our group G D hb; ci as asubgroup of A32.We have obtained two groups stated in part (b) of our theorem which is now com-

pletely proved.

In conclusion we prove the following

Theorem 55.4. Let G be a p-group such that �2.G/ is extraspecial. Then �2.G/ D

G.

Proof. Suppose the theorem is false, i.e., �2.G/ < G. We consider cases p D 2 andp > 2 separately.

Case 1. Let p D 2. Set E D �2.G/ and hzi D Z.E/ so that jEj D 22nC1,n � 1, where n is the width of E. Let F be a subgroup of G containing E such thatjF W Ej D 2. To get a contradiction, one may assume, without loss of generality,that G D F . We use induction on n. Suppose that n D 1. Then E Š D8 or Q8. IfCG.E/ 6� E, then CG.E/�hzi contains elements of order� 4, a contradiction. HenceCG.E/ � E and then G is of maximal class. But then �2.G/ D G, a contradiction.We assume now n > 1. Since jEj � 25 and exp.E/ D 4,G has no cyclic subgroups

of index 2 and so G is not of maximal class. It follows that G has a normal four-subgroup R. We have R < E since �2.G/ D E. In particular, z 2 R and wemay set R D hz; ui for some involution u 2 .E � hzi/ so that CG.R/ D CG.u/.Since jE W CE .R/j D 2, it follows that CG.R/ covers G=E. By the structure of E,CE .u/ D hui � E0, where E0 is extraspecial of order 22.n�1/C1. Set F0 D CG.u/

so that jF0 W .hui � E0/j D 2 and consider the factor-group F0=hui D NF0 (barconvention), where j NF0 W NE0j D 2 and NE0 is extraspecial of width n � 1.


By induction, there is an element x 2 F0 � .hui � E0/ such that o. Nx/ � 4. Ifo. Nx/ D 2, then x2 2 hui and so o.x/ � 4, a contradiction. Hence o. Nx/ D 4 and sox4 2 hui but x2 62 hui. We have x2 2 CE .u/. If x2 is an involution, then o.x/ D 4, acontradiction. Hence x2 is an element of order 4 in CE .u/ and so x4 D z, contrary tox4 2 hui. This completes Case 1.

Case 2. Now let p > 2 and let E D �2.G/ be extraspecial. Since G > E, wehave exp.E/ D p2. We have E D E1 � � Em�1 � Em, where, in case m > 1,E1; : : : ; Em�1 are nonabelian of order p3 and exponent p and Em Š Mp3 (see �4).Then we have S D �1.E/ D �1.G/ D E1 : : : Em�1�1.Em/, where exp.S/ D p

and jE W S j D p (here we use the fact thatE is regular; see Theorems 7.1 and 7.2). Set�1.Em/ D R. We have CS .R/ D E1 � � Em�1R D S so R D Z.S/. It followsthat R is normal in G. Set C D CG.R/; then jG W C j D p, S < C and G D EC .Since S D �1.G/ and �1.G=S/ D E=S is of order p so G=S is cyclic since p > 2

and E=C is the unique subgroup of order p in G=S , we must have E=S � C=S . Thisis a contradiction since R 6� Z.E/. The proof is complete.

�56

Theorem of Ward on quaternion-free 2-groups

In this section we present the proof of the following important result of Ward [War]:

Theorem 56.1 (H.N. Ward [War]). Let G be a nonabelian quaternion-free 2-group.Then G has a characteristic subgroup of index 2.

Recall that a 2-group is quaternion-free if it has no sections isomorphic to Q8; thatproperty is inherited by sections. Note that the original proof of that theorem is basedessentially on properties of the Burnside group B.4; 2/ (this is a finite two-generatorgroup of exponent 4 of maximal order; by Burnside, jB.4; 2/j D 212). The offeredproof is shorter and elementary. However, we use ideas of [War]. The offered proof isdue to the second author.Let P 2 Syl2.G/ be the kernel of a Frobenius group G. If P is nonabelian, it has a

section isomorphic to Q8. Indeed, P has no characteristic subgroups of index 2 so theresult follows from Theorem 56.1. Let P 2 Syl2.G/, where G D Sz.22mC1/ is theSuzuki simple group. Since NG.P / is a Frobenius group, P has a section isomorphicto Q8, by what has been said.LetG be a nonabelian quaternion-free 2-group and letA � Aut.G/ be of odd order.

Then A has a fixed point on G=ˆ.G/ (use Maschke’s theorem).First we prove the following two auxiliary results.

Lemma 56.2. In a Q8-free 2-group X there are no elements x, y with o.x/ D 2k > 2

and o.y/ D 4 so that xy D x�1. If D � X and D Š D8, then CX .D/ is elementaryabelian.

Proof. If y2 D x2k�1

, then we have hx; yi Š Q2kC1 . If hxi \ hyi D f1g, then we

have hx; yi=hx2k�1

y2i Š Q2kC1 . In both cases hx; yi is not Q8-free, a contradiction.Suppose that D � X , where D D ha; t j a4 D t2 D 1; at D a�1i. If v is an elementof order 4 in CX .D/, then o.tv/ D 4 and tv inverts a, contrary to what has just beenproved. Thus, CX .D/ must be elementary abelian.

Lemma 56.3. Let X be a Q8-free 2-group with ˆ.X/ � Z.X/. If a; b 2 X , o.a/ D

o.b/ D 4, and Œa; b� ¤ 1, then .ab/2 D 1.

Proof. Set c D Œa; b� so that c2 D Œa; b�2 D Œa2; b� D 1 and therefore c is a centralinvolution in X . Also, a2 and b2 are central involutions in X . Set W D ha2c; b2ci

�56 Theorem of Ward on quaternion-free 2-groups 135

and so W � Z.ha; bi/ is of order � 4 and exponent 2. In particular, a; b 62 W . Wecompute ab D aŒa; b� D ac D a�1.a2c/, ba D bŒb; a� D bc D b�1.b2c/.It follows from the above equalities that .aW /b D .aW /�1 and .bW /a D .bW /�1.

Since ha; bi=W is Q8-free, Lemma 56.2 implies that at least one of aW or bW isan involution. Hence a2 D b2c or b2 D a2c (indeed, W contains exactly threeinvolutions a2c, b2c, a2b2 and a2 ¤ a2b2 ¤ b2). In any case a2b2 D c. But then.ab/2 D a2b2Œb; a� D c2 D 1. and we are done.

Proof of Theorem 56.1. We proceed by induction on jGj. Let K > f1g be a minimalcharacteristic subgroup of G. If G=K is nonabelian, it has a characteristic subgroupH=K of index 2, by induction; then H is characteristic of index 2 in G. Therefore,we may assume that K D G0, i.e., G0 is the unique minimal characteristic subgroupof G; then G0 is elementary abelian. Since G0 \ Z.G/ is a nonidentity characteristicsubgroup of G, we get G0 � Z.G/. Thus, the group G is of class 2 and so for allx; y 2 G, we have Œx2; y� D Œx; y�2 D 1, which gives ˆ.G/ D Ã1.G/ � Z.G/; inparticular, G=Z.G/ is elementary abelian. The map x 7! x2m

is an endomorphism ofG if m � 2 since

.xy/2m

D x2m

y2m

Œy; x�.2m.2m�1//=2 D x2m

y2m

;(1)

where x; y 2 G. Let exp.G/ D 2nC1 .n � 1/. If n � 2, G is generated by theelements of order 2nC1. For, if x2n

D 1 and o.y/ D 2nC1, then, taking m D n in(1), we get o.xy/ D 2nC1 and x D .xy/y�1 is a product of two elements of order2nC1. If n D 1, we may also assume that G is generated by the elements of order4. Indeed, let H D H2.G/ be the subgroup generated by all elements of G of order4. Since H > f1g, then H < G implies jG W H j D 2 (Straus–Szekeres). As H ischaracteristic, this is a contradiction. Thus, we may assume that H D G, i.e., G isgenerated by elements of order 4 D 2nC1 again. We prove our theorem in six steps.

Step 1. Suppose that n D 1, i.e., exp.G/ D 4. Denote Z0 D �1.Z.G//. We claimthat if a1; a2; : : : ; am are elements of G of order 4 such that a1a2 : : : am 2 Z0, thena2

1a22 : : : a

2m D 1.

Indeed, let m be the smallest integer for which a1; : : : ; am exist satisfying the hy-pothesis but not the conclusion of the above statement. Then m > 1. In fact m � 3,for if a1a2 D z 2 Z0, then a1 D a�1

2 z so a21 D a�2

2 z2 D a�22 and a2

1a22 D 1 2 Z0.

Since Œai ; aj � 2 G0 � ˆ.G/ � �1.Z.G// D Z0, elements ai and aj commutemodulo Z0 so the hypothesis a1a2 : : : am 2 Z0 is independent of the ordering of ai ’s.We have for any i ¤ j , .aiaj /

2 D a2i a

2j Œaj ; ai � and so Œai ; aj � D a2

i a2j .aiaj /

2.

Suppose that Œai ; aj � D 1. If, in addition, .aiaj /2 ¤ 1, then o.aiaj / D 4 and

.aiaj /2 D a2

i a2j . Then the two terms ai and aj could be combined to .aiaj / and

m lowered by one, contrary to the assumption. Thus if Œai ; aj � D 1, then a2i a

2j D

.aiaj /2 D 1. If Œai ; aj � ¤ 1, then Lemma 56.3 implies .aiaj /

2 D 1. So in any event,Œai ; aj � D a2

i a2j .


Now let i; j; k be distinct. Then, by the last result, using the collecting process, weobtain, since Ã1.G/ � Z.G/,

.aiaj ak/2 D a2

i a2j a

2k Œai ; aj �Œai ; ak �Œaj ; ak � D a2

i a2j a

2k a2

i a2j a2

i a2k a2

j a2k

D a2i a

2j a

2k :

If then .aiajak/2 ¤ 1, ai ; aj ; ak can be combined to .aiaj ak/ and m lowered by 2.

Therefore a2i a

2j a

2k

D 1 for all triples i; j; k. In particular,m > 3, so all the squares a2i

coincide (for example, it follows from a21a

22a

23 D 1 D a2

2a23a

24 that a

21 D a2

4). But thenŒai ; aj � D a2

i a2j D a4

i D 1 so the elements ai ’s commute, and, since a1a2 : : : am 2

Z0, we get 1 D .a1a2 : : : am/2 D a2

1a22 : : : a

2m after all.

Step 2. Suppose, as in Step 1, that n D 1, i.e., exp.G/ D 4. Denote again Z0 D

�1.Z.G//. We claim that there is the unique homomorphism s W g 7! gs of G intoZ0 such that if o.a/ D 4, then as D a2. In addition, xs D 1 for every x 2 Z0. (gs iscalled the “artificial square”of g.) It remains to define s on involutions from G �Z0.For g 2 G, write g D a1a2 : : : am, where o.ai / D 4 (recall that G is generated

by elements of order 4, by the paragraph preceding Step 1). By Step 1, a21a

22 : : : a

2m

depends only on g, not of choice of a1; : : : ; am. Indeed, if, in addition, g D b1 : : : bt ,then a1 : : : amb

�1t : : : b�1

1 D 1 2 Z0 so, by Step 1, a21 : : : a

2mb

�2t : : : b�2

1 D 1 anda2

1 : : : a2m D b2

1 : : : b2t , justifying our claim. If one defines g

s D a21a

22 : : : a

2m, the re-

sults stated in the previous paragraph are direct consequences of Step 1. This definitionalso works in the case o.g/ D 4: since gs depends only on g we must have gs D g2.It follows that im.N/ D NG is characteristic in G.

Step 3. We define the endomorphism N of G by Nx D x2n

if n > 1 and Nx D xs (theartificial square, defined in Step 2) if n D 1; recall that exp.G/ D 2nC1. It is easy tocheck, using (1) with m D n if n > 1 and Step 1 if n D 1 that N is a homomorphism.We observe that, if n > 1, then the kernel of N equals �n.G/, which is < G, by (1)withm D n; in that case, exp. NG/ D 2. We claim that in any case Œa; b� 2 h Na; Nbi for alla; b 2 G. e have im.N/ � Ã1.G/ \�1.Z.G//.Let n D 1. We have .ab/2 D a2b2Œb; a�. Then, if o.a/ D o.b/ D o.ab/ D 4,

we get Œa; b� D a2b2.ab/2 D Na Nbab D Na2 Nb2 D 1 2 h Na; Nbi. If o.ab/ < 4, then,independent of the orders of a and b, Œa; b� D a2b2 2 h Na; Nbi. If o.a/ < 4, thenŒa; b� D b2.ab/2 2 h Na; Nbi. Similarly, the same inclusion is true, if o.b/ < 4. Thus,our claim is true for n D 1.Now let n > 1. Suppose first that Na ¤ 1, Nb ¤ 1, and Na Nb ¤ 1 (or, what is the same,

elements a; b; ab have the same order 2nC1 D exp.G/). Then ha4; b4i � Ã2.G/ <

ˆ.G/ � Z.G/, exponent of ha; bi=ha4; b4i is 4, and the orders of a and b equal 4modulo ha4; b4i. If Œa; b� 62 ha4; b4i, then Lemma 56.3 implies .ab/2 2 ha4; b4i.But taking 2n�1-th powers of the last inclusion and using (1), one obtains .ab/2

n

2

ha2nC1

; b2nC1

i D f1g since exp.G/ D 2nC1, or, what is the same, Na Nb D 1, contrary tothe assumption. Therefore, Œa; b� 2 ha4; b4i and since �1.ha

4; b4i/ D hNa; Nbi (recallthat exp. NG/ D 2) and G0 is elementary abelian, Œa; b� 2 h Na; Nbi.


Next let a be an involution and Nb ¤ 1 (this means that o.b/ D 2nC1). We mayassume that j NGj � 4 (otherwise the kernel �n.G/ of N is a characteristic subgroupof G of index 2). Hence, for some element c 2 G, we have Nc 62 h Nbi (otherwise,NG D h Nbi were of order 2); it follows that o.c/ D 2nC1. In G=hb2i (b2 2 Z.G/), then,some power of c leads to a central element of order 4. By the second part of Lemma56.2, G=hb2i has no subgroups isomorphic to D8. Hence, a and b commute modulohb2i, and we have Œa; b� 2 hb2i and since o.Œa; b�/ � 2 (recall that exp.G0/ D 2),Œa; b� 2 h Nbi.

Now let Na D 1, Nb ¤ 1 (this means that o.a/ � 2n, o.b/ D 2nC1), and let ahave the minimal order 2m .2 � m � n/ among such elements for which Na D 1

but Œa; b� 62 h Na; Nbi D h Nbi (case m D 1 was considered in the previous paragraph).Then ha2m�1

; Nbi � �1.ha4; b4i/, where ha4; b4i � Z.G/. If Œa; b� 62 ha4; b4i, then

by Lemma 56.3, .ab/2 2 ha4; b4i and so (taking again 2n�1-th powers) we get Nb D

1 Nb D Na Nb D ab D 1, contrary to the assumption of this paragraph. Hence, sinceo.Œa; b�/ � 2, we get Œa; b� 2 �1.ha

4; b4i/ and so Œa; b� 2 ha2m�1

; Nbi. Again letc 2 G be such that Nc 62 h Nbi. Since ac D Nc ¤ 1 and cb D Nc Nb ¤ 1, we have, bythe above, Œac; b� 2 h Nc; Nbi and Œc; b� 2 h Nc; Nbi. Then Œac; b� D Œa; b�Œc; b� impliesŒa; b� 2 h Nc; Nbi. Combining this with Œa; b� 2 ha2m�1

; Nbi, we get Œa; b� 2 h Nbi, whichis a contradiction, unless a2m�1

2 h Nc; Nbi. But in this case a2m�1

D z�2m�1

for somez 2 hb2; c2i � Z.G/. Then .az/2

m�1

D 1 and so o.az/ < 2m and Œa; b� D Œaz; b� 2

h Nbi, by minimality of m. This contradiction proves that for any a; b 2 G with Na D 1

and Nb ¤ 1, we have Œa; b� 2 h Nbi.

If Na D Nb ¤ 1, then Œa; b� D Œab; b� with ab D Na2 D .a2/2n

D a2nC1

D 1, so thatŒa; b� D Œab; b� 2 h Nbi, by the previous paragraph.

Finally, suppose that Na D Nb D 1. Let c 2 G be any element with Nc ¤ 1. ThenŒa; bc� 2 h Na; bci D h Nci and Œa; c� 2 h Na; Nci D h Nci, by the above. Hence Œa; bc� D

Œa; b�Œa; c� gives Œa; b� 2 h Nci. Since j NGj � 4, there is d 2 G with Nd 62 h Nci. Then,again, Œa; b� 2 h Nd i and so Œa; b� 2 h Nd i \ h Nci D f1g (recall that exp. NG/ D 2), asneeded. All possibilities have been considered and so Œa; b� 2 h Na; Nbi for all a; b 2 G.

Step 4. We claim now that G0 D NG. Obviously, G0 � NG since G0 is the uniqueminimal characteristic subgroup of G and NG is a nonidentity characteristic subgroupof G (this follows from the definition of NG; see Step 2). Note that if G0 ¤ NG, thenj NG W G0j � 4. Assume that this is false; then j NG W G0j D 2. We have NG Š G=K withcharacteristic subgroupK < G, the kernel of N. SinceG0K < G is characteristic inG,we get jG W G0Kj � 4. It follows j NG W NG0j D jG=K W .G0K=K/j D jG W G0Kj � 4,as claimed.

Suppose that G0 < NG. Then if Na 2 NG � G0, we have Œa; b� D 1 for all b 2 G.To see this, note first that if Na; Nb; ab, all involutions in h Na; Nbi, are not contained inG0, then Œa; b� 2 G0 \ hNa; Nbi (using Step 3), and that intersection equals f1g sinceh Na; Nbi D f1; Na; Nb; abg. Now suppose that Na 62 G0 but Nb 2 G0. Find c 2 G withNc 62 hG0; Nai (c exists since j NG W G0j � 4 and exp. NG/ D 2). Then ac 62 G0, so that


Œa; c� D 1 by what has just been proved. And, Œa; b� D Œa; b�Œa; c� D Œa; bc� 2 h Na; bci,by Step 3. But h Na; bci \ G0 D f1g. Indeed, Na; bc 62 G0, by the choice. It remainsto show, that the third involution Nabc D Na Nb Nc of h Na; bci is not contained in G0. Ifthis is false, then, since Nb 2 G0, we get ac D Na Nc D Na Nb Nc Nb 2 G0, contrary to whathas been proved. Hence Œa; b� D 1. Finally, if Na 62 G0, Nb 62 G0, and Na Nb 2 G0,then Œa; b� D Œa; ab� D 1 (the second equality follows, by the previous case). Thus,Œa; b� D 1 for all b 2 G. But G is generated by the elements a 2 G with Na 62 G0,and G would be abelian, a contradiction. Set T D hx 2 G j Nx 62 G0i. To justifyour claim, we must to prove that G D hT i. Indeed, G is generated by all elementsof order 2nC1 D exp.G/. Let Ng 2 G0 with Ng ¤ 1 and let Na 2 NG � G0. Thenalso ga D Ng Na 2 NG � G0 and Ng D ga Na so that g D gaat with Nt D 1. But theng D .ga/.at/ and ga 62 G0 and at D Na 62 G0, i.e., ga; at 2 T . Thus, G D hT i,as claimed.

Thus NG D G0. Let K be the kernel of N . Then NG D G=K is isomorphic to G0 andso jG0j D jG W Kj > 2; the last inequality holds in view of G is nonabelian and K ischaracteristic in G.

Step 5. Suppose that jG0j � 8. Then we obtain a characteristic subgroup of index 2in G as follows.(i) There exist a; b 2 G with jh Na; Nbij D 4 and Œa; b� D 1 (since exp. NG/ D 2, if such

a; b exist, then h Na; Nbi is a four-group).For, let a and b be such that jh Na; Nbij D 4 but Œa; b� ¤ 1 (we may assume that NG is

noncyclic). By Step 3, 1 ¤ Œa; b� 2 h Na; Nbi and so without loss of generality we maytake Œa; b� D Na. (If Œa; b� D Nb, then replace Œa; b� with Œb; a� D Œa; b��1 D Nb�1 D Nb.If Œa; b� D Na Nb, then replace Œa; b� with Œab; b� D Œa; b� D ab.) Now take c withNc 62 h Na; Nbi (such c exists in view of j NGj D jG0j � 8). Then Œa; bc� D Œa; b�Œa; c� D

NaŒa; c� 2 h Na; bci \ Nah Na; Nci D f1; Nag. If Œa; bc� D 1, then a and bc are the desiredelements. If Œa; bc� D Na.D Œa; b�/, then Œa; c� D 1 so a and c are the desired elements.

(ii) Suppose that Na ¤ 1, Nb ¤ Na, and Œa; b� D 1. Then Œb; x� 2 h Nbi for all x 2 G.For, (by Step 3) Œax; b� D Œx; b� 2 hax; Nbi \ h Nx; Nbi. Assume that this intersection

is not h Nbi. Then hax; Nbi D h Nx; Nbi is a four-group. Since Na ¤ 1, we get ax D Nb Nx soNa D Nb, contrary to the assumption. Thus, our intersection is h Nbi, as claimed.

(iii) Now let H be the set of elements a 2 G for which Œa; x� 2 h Nai for all x 2 G.We show thatH is a subgroup of G of index 2; it will clearly be characteristic.If a1; a2 2 H , then we have to show that a1a2 2 H . Suppose that Nb 62 h Na1; Na2i.

Then Œa1a2; b� D Œa1; b�Œa2; b� 2 ha1a2; Nbi \ h Na1ih Na2i D ha1a2i, as needed. Since Gis generated by the set of all such b’s (noting that NG D G=K is elementary abelian oforder jG0j � 8), we have Œa1a2; b� 2 ha1a2i for all b 2 G. Thus a1a2 2 H soH is acharacteristic subgroup of G. Obviously, G0 � Z.G/ � H .By (i) and (ii), there is an element a 2 H with Na ¤ 1. Indeed, by (i), there exist

a; b 2 G such that Na ¤ 1, Nb ¤ 1, Nb ¤ Na and Œa; b� D 1. Then, by (ii), b 2 H #,proving our claim. Suppose for such an a that Œa; b� D 1 for some b 2 G. If Nb ¤ Na,


then b 2 H by (ii); but if Nb D Na, then ba D Na Na D 1 ¤ Na and Œa; ba� D Œa; b� D 1

and so by (ii) again, ba 2 H . Since a 2 H , so b 2 H in any event. We have provedthat CG.a/ � H . So if b 2 G �H , then Œa; b� ¤ 1. But then Œa; b� D Na (as a 2 H

and o. Na/ D 2). Hence if H ¤ G, then jG W H j D 2. (Suppose there are b1; b2 2

G � H such that b1b2 2 G � H . Then Na D Œa; b1b2� D Œa; b1�Œa; b2� D Na Na D 1,a contradiction.)Assume that H D G. Then G is Dedekindian so abelian since it is Q8-free (see

Theorem 1.19).

Step 6. Finally suppose that jG0j D 4 (by the last paragraph of Step 4. jG0j > 2).It cannot be that K, the kernel of N , is central, for jG W Kj D 4 would then implyjG0j D 2, by Lemma 1.1. Thus there is c 2 K (or, what is the same, Nc D 1) anda 2 G with Œa; c� ¤ 1. By Step 3, Œa; c� 2 h Na; Nci D h Na; 1i D h Nai and so Œa; c� D Na.Suppose that Nb 62 h Nai. Then Œab; c� D Œa; c�Œb; c� D NaŒb; c� 2 habi \ Nah Nbi, by Step 3,and so Œab; c� D ab. Thus Œb; c� D abŒa; c��1 D ab Na�1 D Nb. Since G is generatedby all b 2 G with Nb 62 h Nai, this implies Œg; c� D Ng for all g 2 G. It follows thatjK W .K \ Z.G//j D 2. Indeed, assume that there are c1; c2 2 K � Z.G/ such thatc1c2 2 K � Z.G/ and Na ¤ 1. Then Na D Œa; c1c2� D Œa; c1�Œa; c2� D Na Na D 1, whichis a contradiction.Let a; b 2 G be such that h Na; Nbi D G0. If Œa; b� ¤ 1, we may assume that Œa; b� D

Na. (If Œa; b� D Na Nb, then take the elements ab; b so that Œab; b� D Œa; b� D ab. IfŒa; b� D b, then take the elements b; a so that Œb; a� D Œa; b��1 D Nb�1 D Nb.) ThenŒa; bc� D Œa; b�Œa; c� D Na Na D 1 and jh Na; bcij D 4 (with c 2 K � Z.G/ as inthe previous paragraph). Thus we may assume from the start that Œa; b� D 1. Now,G D ha; biK D ha; b; ciZ.G/. By the previous paragraph, jK W .K \ Z.G//j D 2.Therefore, Z.G/ < K because otherwise jG W Z.G/j D 4 and then jGj D 2jZ.G/jjG0j

(Lemma 1.1) would imply jG0j D 2, a contradiction (see the last paragraph of Step 4).Hence we have Z.G/ < K and since jK W Z.G/j D 2, it follows that jG W Z.G/j D 8.The fact Œa; b� D 1 implies that Z.G/ha; bi is an abelian maximal subgroup of G.Since jG W Z.G/j D 8, we see that Z.G/ha; bi is the unique abelian maximal subgroupofG and therefore it is a characteristic subgroup of index 2. The proof is complete.

�57

Nonabelian 2-groups all of whoseminimal nonabelian subgroups are isomorphic

and have exponent 4

By Proposition 10.28, a nonabelian p-group is generated by its minimal nonabeliansubgroups. Therefore it is natural to try to determine the structure of a p-group if thestructure of its minimal nonabelian subgroups is known. Here we classify the titlegroups and note that there are exactly five minimal nonabelian 2-groups of exponent 4(see Lemma 65.1). All results of this section are due to the second author.Recall that there are exactly five minimal nonabelian 2-groups of exponent 4: D8,

Q8, H2 D ha; b j a4 D b4 D 1; ab D a�1i, H16 D ha; t j a4 D t2 D 1; Œa; t � D

z; z2 D Œa; z� D Œt; z� D 1i and H32 D ha; b j a4 D b4 D 1; Œa; b� D z; z2 D

Œa; z� D Œb; z� D 1i.It is proved in Appendix 17 (Corollary A.17.3) that, if G is a nonabelian 2-group

all of whose minimal nonabelian subgroups are isomorphic to Q8, then G D Q � V ,whereQ Š Q2n , n � 3, and exp.V / � 2. In Theorem 10.33 was proved that if G is anonabelian 2-group all of whose minimal nonabelian subgroups are isomorphic to D8,then G is generalized dihedral. Here we consider the other three minimal nonabelian2-groups of exponent 4 and first prove the following two key lemmas.

Lemma 57.1. Let G be a nonabelian p-group and let A be a maximal abelian normalsubgroup ofG. Then for any x 2 G�A, there is a 2 A such that Œa; x� ¤ 1, Œa; x�p D

1, and Œa; x; x� D 1 which implies that ha; xi is minimal nonabelian. Therefore, G isgenerated by its minimal nonabelian subgroups.

Proof. Since CG.A/ D A, we have CA.x/ ¤ A and therefore hxiCA.x/ is a properabelian subgroup of hxiA. Let B be a subgroup of hxiA containing hxiCA.x/ as asubgroup of index p. Then j.A \ B/ W CA.x/j D p, CA.x/ � Z.B/ and B 0 � CA.x/.Let a 2 .A \ B/� CA.x/ so that ap 2 CA.x/. We get 1 D Œap; x� D Œa; x�p . On theother hand, Œa; x� 2 CA.x/ and so Œa; x; x� D 1. We get ha; xi0 D hŒa; x�i and so, byLemma 65.2(a), ha; xi is minimal nonabelian.

Lemma 57.2. Let G be a nonabelian 2-group all of whose minimal nonabelian sub-groups are of exponent 4 and let A be a maximal normal abelian subgroup of G. Thenall elements inG�A are of order� 4 and so either exp.A/ D 2 or exp.A/ D exp.G/.

�57 Minimal nonabelian subgroups are of exponent 4 141

If x 2 G � A with x2 2 A, then x inverts each element in Ã1.A/ and in A=�1.A/. Ifexp.G/ > 4, then either G=A is cyclic of order � 4 or G=A Š Q8.

Proof. By Lemma 57.1, all elements in G � A are of order � 4. Thus, exp.A/ D 2

or exp.A/ D exp.G/. Let x 2 G � A with x2 2 A. Then for each a 2 A wehave .aax/x D axax2

D axa D aax and so aax D w with w 2 CA.x/ andax D a�1w. We compute .xa/2 D xaxa D x2axa D x2w, where o.x2/ � 2.Since o.xa/ � 4, we have o.w/ � 2 and so x inverts each element of A=�1.A/.Finally, .a2/x D .ax/2 D .a�1w/2 D a�2 and so x inverts each element of Ã1.A/.If exp .A/ > 4, thenG=A does not possess a four-subgroup and soG=A is either cyclicof order at most 4 or G=A Š Q8.

Theorem 57.3. Let G be a nonabelian 2-group all of whose minimal nonabelian sub-groups are isomorphic to H2 D ha; b j a4 D b4 D 1; ab D a�1i. Then the followingholds:

(a) If G is of exponent � 8, then G has a unique abelian maximal subgroup A. Wehave exp.A/ � 8 and E D �1.A/ D �1.G/ D Z.G/ is of order � 4. Allelements in G �A are of order 4 and if v is one of them, then CA.v/ D E and vinverts ˆ.A/ and A=E.

(b) If G is of exponent 4, then G D K � V , where exp.V / � 2 and for the group Kwe have one of the following possibilities:

(b1) K Š H2 is of order 24;(b2) K is the minimal nonmetacyclic group of order 25 (see Theorem 66.1(d));(b3) K is a unique special group of order 26 with Z.K/ Š E4 in which every

maximal subgroup is minimal nonmetacyclic of order 25 (from (b2)):

K D ha; b; c; d j a4 D b4 D 1; c2 D a2b2; Œa; b� D 1;

ac D a�1; bc D a2b�1; d2 D a2;

ad D a�1b2; bd D b�1; Œc; d � D 1iI

(b4) K is a splitting extension of B D B1 � � Bm, m � 2, with a cyclicgroup hbi of order 4, where Bi Š C4, i D 1; 2; : : : ;m, and b inverts eachelement of B (and b2 centralizes B).

Proof. Since D8 is not a subgroup of G, �1.G/ is elementary abelian of order > 2

(because G is not generalized quaternion). Let x 2 G � �1.G/ with o.x/ D 4 andassume that there is a 2 �1.G/ such that Œa; x� ¤ 1. Then Œa; x� is an involution andwe have 1 D Œa; x2� D Œa; x�Œa; x�x so that Œa; x�x D Œa; x� and ha; xi is minimalnonabelian, a contradiction. Here we have used the fact that �1.B/ � Z.B/ for eachminimal nonabelian subgroup B � G, where B Š H2. Since ��

2.G/ D G, we haveproved that �1.G/ � Z.G/. Hence, if A is a maximal normal abelian subgroup of G,then �1.G/ D �1.A/ � Z.G/, G=A is elementary abelian, �1.A/ < A, and G � A

consists of elements of order 4.


Suppose that there is an element v of order 4 such that CG.v/ is nonabelian andlet H D ha; b j a4 D b4 D 1; ab D a�1i be a minimal nonabelian subgroup ofCG.v/, where we set a2 D z, b2 D u. First suppose that hvi \ H D f1g. Then.av/2 D a2v2 D zv2 62 H , .av/b D abv D a�1v D .av/z, so that hav; bi is thenonmetacyclic minimal nonabelian group of order 25 and exponent 4, a contradiction.Hence, v 62 H but v2 2 Z.H/ and so v2 2 fz; u; uzg. If v2 D z, then i D va is aninvolution, i 62 H and ib D iz and so i 62 Z.G/, a contradiction. If v2 D u, thenj D vb is an involution, j 62 H and aj D avb D ab D a�1 so that j 62 Z.G/,a contradiction. If v2 D uz, then .vb/2 D v2b2 D .uz/u D z, vb 62 H , andavb D ab D a�1 so that hvb; ai Š Q8, a contradiction. We have proved that thecentralizer of each element of order 4 is abelian. In particular, for each x 2 G � A,CA.x/ D �1.A/ and �1.A/ D Z.G/.If exp.A/ > 4, then Lemma 57.2 implies that jG W Aj D 2 and we have obtained

groups in part (a) of our theorem. Indeed, since x 2 G � A inverts A=�1.A/ andexp.A/ � 8, we have jG0j > 2 and so, using Lemma 1.1, we see that A is a uniqueabelian maximal subgroup of G (otherwise, jG W Z.G/j D 4).From now on we assume that exp.G/ D 4. In that case ˆ.G/ � �1.A/ D Z.G/

and jˆ.G/j � 4 (since jˆ.H2/j D 4). If x; y are elements of order 4 in G withŒx; y� ¤ 1, then Œx; y� is an involution in Z.G/ and so hx; yi is minimal nonabelianand hx; yi Š H2 implies that in case x2 ¤ y2 we have yx 2 fy�1; yx2g.Considering G=ˆ.G/, we get G D K � V , where exp.V / � 2 and �1.K/ D

Z.K/ D ˆ.K/ D ˆ.G/. It is easy to determine the structure ofG in case jˆ.G/j D 4.Our groupK has exactly three involutions and Z.K/ Š E4 is noncyclic. By the resultsstated in the introduction to �82, K has a metacyclic normal subgroup M such thatK=M is elementary abelian of order � 4. Since in our case K is of exponent 4,we have jM j � 24 and so jKj � 26. If jKj � 24, then K Š H2. Suppose thatjKj D 25. In this case K is nonmetacyclic (since exp.K/ D 4). If K is not minimalnonmetacyclic, then Theorem 66.1 implies that K must possess a subgroup which isisomorphic to E8 or Q8. This is not the case and so K is minimal nonmetacyclic oforder 25. Finally, assume that jKj D 26. Each maximal subgroup of K is minimalnonmetacyclic of order 25 and such a group K is unique according to [CIS, Theorem2 and Remarks] so K is special with Z.K/ Š E4. We have obtained the groups statedin parts (b1), (b2) and (b3).It remains to treat the case jˆ.G/j > 4. We note that jA W �1.A/j D jÃ1.A/j

and let x 2 G � A. Consider any element y 2 A with y2 62 hx2i and suppose thatyx ¤ y�1 so that yx D yx2. Assume that there is v 2 A with v2 62 hx2; y2i. Sinceyx D yvx (and .vx/2 2 fv2; x2g and so .vx/2 ¤ y2), we get x2 D .vx/2 whichgives vx D v�1. Since .vy/2 62 hx2; y2i, we also get .vy/x D .vy/�1. Hence,yx D y�1 for all y 2 A with y2 62 hx2i and so x inverts A. (If z 2 A with z2 D x2,then x inverts zy and so x also inverts z. But in that case hz; xi Š Q8 which cannothappen.) We have proved that in case jA W �1.A/j � 8, each element x 2 G � A

inverts A and so jG W Aj D 2 and G is as in (b4).


Assume jA W �1.A/j D jÃ1.A/j D 4 so that A D hy; zi�1.A/ and Ã1.A/ D

hy2; z2i. Since jˆ.G/j > 4, there is u 2 G � A such that u2 62 hy2; z2i. By thearguments in the previous paragraph, u inverts A, and so Ahui is as in (b4). Supposethat Ahui ¤ G. Then there is x 2 G � .Ahui/ such that Ã1.Ahxi/ D Ã1.A/ and wemay assume that x2 D z2. If yx D y�1, then xu 2 G � A and xu centralizes y, acontradiction. It follows that yx D yx2. From yx D yzx and x2 ¤ .zx/2 (noting thathz; xi Š H2) follows that y2 D .zx/2 D Œz; x�. We compute .yz/x D yx2zŒz; x� D

yx2zy2 D y�1z�1 D .yz/�1. But then xu centralizes yz, a contradiction. Hence,we must have G D Ahui.Finally, suppose jA W �1.A/j D 2 so that A D hyi�1.A/. Let x 2 G � A, where

we may assume that x2 ¤ y2. (Indeed, we have hx; yi Š H2 and so if x2 D y2,then we replace x with x0 D xy, where .x0/2 ¤ y2. We may assume that yx D yx2

(otherwise, if each element in G � A inverts A, then jG W Aj D 2 and G would be asin (b1)). Let u 2 G � hA; xi with u2 62 hx2; y2i. Then we have yu 2 fy�1; yu2g.First suppose that yu D y�1. Then yxu D .yx2/u D y.y2x2/. But .xu/2 isequal either x2u2 (if Œu; x� D 1) or u2 or x2 (if Œu; x� ¤ 1 and so hu; xi Š H2)and therefore in any case .xu/2 ¤ y2. Since hy; xui Š H2, we get yxu D yy2

or yxu D y.xu/2. Since yxu D y.y2x2/, the only possibility is .xu/2 D y2x2

which is a contradiction. Hence, we must have the second possibility yu D yu2 andwe get yxu D .yx2/u D y.u2x2/ and therefore (because u2x2 D y2 is excluded)x2u2 D .xu/2 which implies Œu; x� D 1. In that caseA1 D hx; ui�1.A/ is an abeliannormal subgroup of G with jA1 W �1.A/j D 4 which leads to one of the former cases,when we consider a maximal normal abelian subgroup A� (instead of A) containingA1. Our theorem is proved.

Theorem 57.4. Let G be a nonabelian 2-group all of whose minimal nonabelian sub-groups are isomorphic to H16 D ha; t j a4 D t2 D 1; Œa; t � D z; z2 D Œa; z� D

Œt; z� D 1i. Then �1.G/ is a self-centralizing elementary abelian subgroup and G isof exponent 4. Moreover, the centralizer of any element of order 4 is abelian of type.4; 2; : : : ; 2/.

Proof. Since D8 is not a subgroup of G, �1.G/ is elementary abelian (of order > 2).We choose a maximal normal abelian subgroup A of G so that �1.G/ � A. UsingLemma 57.2, we see that all elements inG�A are of order 4which impliesˆ.G/ � A.By Lemma 57.1, we have also CG.�1.A// D A.Let v be an element of order 4 such that CG.v/ is nonabelian and let H D ha; t j

a4 D t2 D 1; Œa; t � D z; z2 D Œa; z� D Œt; z� D 1i be a minimal nonabelian subgroupof CG.v/. Assume that hvi \ H D f1g so that hH;vi D H � hvi is of order 26.Since .vt/2 D v2, it follows that hvti \H D f1g and hvti normalizes H . Considerthe subgroup ha; vti. Since Œa; vt � D Œa; t � D z and z commutes with a and vt , wehave ha; vti0 D hzi and so S D ha; vti is minimal nonabelian of order 25 (sinceS \ H D ha; zi Š C4 � C2 and S covers hH;vi=H ), a contradiction. Hence wemust have hvi \ H D hv2i, where v2 2 Z.H/ and so v2 2 fz; a2; a2zg. Suppose


that v2 D z so that .vt/2 D z and Œa; vt � D z which implies that ha; vti is metacyclicminimal nonabelian of order 24, a contradiction. Suppose that v2 D a2 so that va isan involution and Œva; t � D z ¤ 1, a contradiction. Similarly, if v2 D a2z, then atvis an involution and Œatv; t � D z ¤ 1, a contradiction. We have proved that for eachelement v of order 4, CG.v/ is abelian.Let x; y 2 G be such that x 2 G � A, Œx; y2� D 1 D Œx2; y� and Œx; y� ¤ 1. Then

hx; yi Š H16 and so either y or xy is an involution. Indeed, c D Œx; y� 2 A sinceG=A is abelian. We get 1 D Œx; y2� D Œx; y�Œx; y�y , 1 D Œx2; y� D Œx; y�x Œx; y�, andso cx D c�1, cy D c�1. Suppose o.c/ > 2 and let c0 be an element of order 4 inhci so that cx

0 D .c0/�1 D c0c

20 , hc0; xi0 D hc2

0i and therefore hc0; xi is a metacyclicminimal nonabelian subgroup, a contradiction. Hence, c D Œx; y� is an involutioncommuting with x and y so that hx; yi is a minimal nonabelian subgroup isomorphictoH16.Suppose that A ¤ �1.A/. Take an element x 2 G � A and an element y of

order 4 in A. Since CG.y/ is abelian, Œx; y� ¤ 1. By Lemma 57.2, x inverts eachelement in A=�1.A/ and so Œx; y� 2 �1.A/. We have Œx2; y� D 1 and Œx; y2� D

Œx; y�Œx; y�y D Œx; y�2 D 1. By the above, hx; yi Š H16 and so xy must be aninvolution, a contradiction. We have proved that A D �1.A/ and so exp.G/ D 4.For each a 2 G � A, we have CG.a/ D ha;CA.a/i. Suppose that this is false. Let

b 2 CG.a/ � .Ahai/ so that ha; bi Š C4 � C4 and ha; bi \ A D ha2; b2i Š E4.Indeed, if a2 D b2, then ab 2 G � A and ab is an involution, a contradiction. SetA0 D CA.a/ so that A0 ¤ A and A0 D CA.b/ (since CG.a/ and CG.b/ are abelian).Let t 2 A � A0 and consider the subgroup ha; bti. We have Œa; bt � ¤ 1, o.bt/ D 4

and since CA.bt/ D CA.b/ D CA.a/ D A0, we have a2; .bt/2 2 A0 and thereforeŒa; .bt/2� D Œa2; bt � D 1. By the above, ha; bti Š H16 which implies that abt(2 G � A) is an involution, a contradiction.

Theorem 57.5. Let G be a nonabelian 2-group all of whose minimal nonabelian sub-groups are isomorphic to H32 D ha; b j a4 D b4 D 1; Œa; b� D z; z2 D Œa; z� D

Œb; z� D 1i. Then �1.G/ � Z.G/ and G is of exponent 4 and class 2.

Proof. In exactly the same way as in the first paragraph of the proof of Theorem57.3, we show that if A is a maximal normal abelian subgroup of G, then �1.G/ D

�1.A/ � Z.G/, �1.A/ < A and G �A consists of elements of order 4. Suppose thatexp.A/ > 4. By Lemma 57.2, jG W Aj D 2 and if x 2 G � A, then o.x/ D 4 and xinverts each element in Ã1.A/. Let y 2 Ã1.A/ with o.y/ D 4. Then yx D y�1 sothat hx; yi is metacyclic minimal nonabelian (of order 8 or 16), a contradiction. Henceexp.A/ D 4 and so ˆ.G/ � �1.G/ � Z.G/.

In conclusion we classify the nonabelian 2-groups all of whose A1-subgroups areisomorphic to M16.

Theorem 57.6. Let G be a nonabelian 2-group all of whose minimal nonabelian sub-groups are isomorphic to M16 D ha; t j a8 D t2 D 1; at D a5i. Then E D �1.G/


is elementary abelian and if A is a maximal normal abelian subgroup of G containingE, then A is of type .2s ; 2; : : : ; 2/, s � 2, jG W Aj � 4, each element x in G � A is oforder 8, jE W CE .x/j D 2, x inverts each element in A=E and in Ã1.A/ and we havethe following possibilities:

(a) s > 2 in which case jG W Aj D 2 and G=E Š Q2s is generalized quaternion oforder 2s;

(b) s D 2 in which case either G Š M16 � V with exp.V / � 2 or G=E Š Q8 andjG W Aj D 4.

Proof. Let G be a nonabelian 2-group all of whose minimal nonabelian subgroups areisomorphic to M16. Since G does not possess dihedral subgroups, E D �1.G/ iselementary abelian. Let A be a maximal normal abelian subgroup of G containing Eso that E D �1.A/. By Lemma 57.1, each element in G � A is of order 4 or 8.Suppose that x 2 G � A and o.x/ D 4. Then x2 2 E and exp.Ehxi/ D 4. It

follows that Ehxi is abelian. But then Ahxi is nonabelian and E � Z.Ahxi/. This isa contradiction since in this case M16 cannot be a subgroup of Ahxi. We have provedthat all elements in G � A are of order 8 and so G=A is elementary abelian (G ismetabelian) and exp.A/ � 4.We want to determine the structure of A. Let x be a fixed element in G � A so

that o.x/ D 8, x2 2 �2.A/ � E and we set z D x4. By Lemma 57.1, there is asubgroup M Š M16 in Ahxi which contains hxi as a subgroup of index 2. Since alleight elements inM �A are of order 8, we haveM \A D M \�2.A/ Š C4 �C2 andM \E Š E4. Let t be an involution .M \E/�hzi so that tx D tz andM 0 D hzi. Setb D x2t so that b2 D z and bx D .x2t /x D .x2t /z D bz D b�1. It follows that eachelement xs with s 2 A is of order 8 and inverts the cyclic subgroup hbi of order 4. Thisforces hxsi\hbi D hzi. Indeed, if hxsi\hbi D f1g (hxsi cannot contain hbi since xsinverts b), then hb; xsi is metacyclic minimal nonabelian of order 25, a contradiction.We compute .xs/2 D xsxs D x2sxs and z D .xs/4 D x4.sxs/2 D z.sxs/2 and so.sxs/2 D 1 and therefore sxs D w 2 E. Thus, sx D s�1w with w 2 E which showsthat x inverts each element of A=E. Further, .s2/x D .sx/2 D .s�1w/2 D s�2 whichshows that x also inverts each element in Ã1.A/.Assume that there is an element a 2 �2.A/�E such that a2 ¤ z. First suppose that

x centralizes a. In that case, .ab/x D abx D ab�1 D .ab/z and .ab/2 D a2b2 D

a2z ¤ z. Hence hab; xi with habi \ hxi D f1g is minimal nonabelian of order 25,a contradiction. Now, suppose that ax D a� with 1 ¤ � 2 E so that Œa; x� D �.But a D ax2

D .a�/x D ax�x D a��x and so �x D � which gives h�i D ha; xi0.It follows that ha; xi is minimal nonabelian of order � 25 because hai \ hxi D f1g.We have proved that �2.A/ is of type .4; 2; : : : ; 2/ which forces that A is of type.2s; 2; : : : ; 2/, s � 2.It is easy to see that jE W CE .x/j D 2. Indeed, set E0 D CE .x/ and suppose

jE W E0j � 4. Since x induces on E an involutory automorphism, we know that xcentralizes E=E0 and so ŒE; x� � E0 and jŒE; x�j D jE=E0j, where E0 � hzi. In


that case, there is e 2 E �E0 such that Œe; x� 2 E0 � hzi. Then hx; ei0 D hŒe; x�i andtherefore hx; ei is nonmetacyclic minimal nonabelian of order 25, a contradiction. Wehave proved that CE .x/ is a hyperplane of E for each x 2 G � A.First suppose that s > 2 or equivalently, exp.A/ > 4. We know that each element

x 2 G � A inverts Ã1.A/ and here exp.Ã1.A// � 4. This forces jG W Aj D 2 notingthat G=A is elementary abelian. Since A=E Š C2s�1 and jG=Ej D 2s , s � 3, G=Ecannot be cyclic (otherwise, an element y 2 G � A would be of order 2sC1, contraryto the fact that all elements in G � A are of order 8). On the other hand, �2.A/=E isa unique subgroup of order 2 in G=E because�2.A/ D �2.G/. We have proved thatG=E Š Q2s .It remains to consider the case s D 2 so that A D �2.A/ D �2.G/ and jA W

Ej D 2 since A is of type .4; 2; : : : ; 2/. For any x 2 G � A, hxi covers A=E andhx4i D Ã1.A/. LetM Š M16 be a minimal nonabelian subgroup in Ahxi containinghxi, where we have again used Lemma 57.1. Since jE W CE .x/j D 2, we get Ahxi D

M � V with V � E. If Ahxi D G, we are done. Suppose that jG W Aj � 4 andwe note that G=A is elementary abelian. Hence G=E is not cyclic and the fact that�2.G/ D A shows that A=E is a unique subgroup of order 2 in G=E. It follows thatG=E is generalized quaternion. But then d.G=E/ D 2 and soG=Amust be elementaryabelian of order 4 which forces that G=E Š Q8 and jG W Aj D 4.

�58

Non-Dedekindian p-groupsall of whose nonnormal subgroupsof the same order are conjugate

A non-Dedekindian p-group in which all nonnormal subgroups of the same order areconjugate is termed a CO-group. The second author classified CO-groups (see Theo-rem 58.3); this solves Problem 1261. Below we offer another proof of that nice resultwhich is due to the first author. A p-group G is termed a CCO-group if it is notDedekindian and all its nonnormal cyclic subgroups of the same order are conjugate.Obviously, a CO-group is a CCO-group.

Lemma 58.1. Let G D ha; bi be a minimal nonabelian p-group as in Lemma 65.1(see also Exercise 1.8a). If all nonnormal cyclic subgroups of G of the same order areconjugate, then G Š Mpt .

Proof. Write A D hai and B D hbi. Let H 6E G; then H \ G0 D f1g and HG D

H �G0.

Let G D ha; b j apm

D bpn

D cp D 1; Œa; b� D c; Œa; c� D Œb; c� D 1i andassume thatm � n. Let�n.A/ D hxi. Then B and hbxi of the same order are neitherG-invariant nor conjugate since �1.BG/ ¤ �1.hbxiG/ so G is not a CCO-group.Let G D ha; b j apm

D bpn

D 1; ab D a1Cpm�1

i. If n D 1, then G Š MpmC1

satisfies the condition. Now let n > 1. If n � m, set �n.A/ D hxi; then B and hbxi

of the same order pn are neither G-invariant nor conjugate. Now assume that n > m.Then B and habi of the same order pn are neither G-invariant nor conjugate. Indeed,.ab/p

m

D bpm

so o.ab/ D pn, habi \ A D f1g, and we conclude that habi is notnormal in G (otherwise, G D ha; abi is abelian). Since �1.habiG/ ¤ �1.BG/, habi

and B are not conjugate in G so G is not a CCO-group.

Lemma 58.2. If a p-group G is a CCO-group and jG0j D p, then G Š Mpt .

Proof. In view of Lemma 58.1, one may assume that G has a proper minimal non-abelian subgroup, say B; then B GG since B 0 D G0. In that case, G D B � C whereC D CG.B/ (Lemma 4.2) and we conclude that B is either Dedekindian or CCO-group; in the first case B Š Q8, in the second case B Š Mpt (Lemma 58.1). The sizeof every conjugate class of non-G-invariant cyclic subgroups equals p.


(i) Let B D hx; y j x4 D 1; y2 D x2; xy D x3i Š Q8. We get exp.C / > 2 sinceG is not Dedekindian. Let L D hli � C be cyclic of order 4. WriteH D B �L; thenH GG. If B \L D Z.B/, thenH has exactly 6 > 2 nonnormal subgroups of order 2so they are not conjugate in G, a contradiction. Thus, B \L D f1g; thenH D B �L

has 3 > 2 distinct nonnormal cyclic subgroups hxli, hyli, hxyli of the same order4 so they are not conjugate, a contradiction. Thus, B has no subgroups isomorphicto Q8.

(ii) Now let B D hu; v j upt�1

D vp D 1; uv D u1Cpt�2

i Š Mpt . Assume thatX D hxi < G of order p is such that X 6� B; then X is not conjugate with hvi soXGG and F D B�X has exactlyp2 > p noncentral so non-G-invariant subgroups oforder p; then G is not a CCO-group, a contradiction. Thus, �1.G/ D �1.B/ Š Ep2

so C has exactly one subgroup of order p; in that case C is cyclic, by (i). Then thereis a cyclic W D hwi � C such that W 6� B and wp 2 Z.B/. Let A < B be cyclicof index p. Then AW is noncyclic abelian with cyclic subgroup A of index p so, bybasic theorem on abelian groups, AW D A�Y , where jY j D p and Y 6� B has orderp, contrary to what has just been proved.

Theorem 58.3 (Janko). If a p-group G is a CO-group, then G Š Mpt .

Proof (Berkovich). We use induction on jGj. In view of Lemma 58.2, one may assumethat jG0j > p. By Theorem 1.23, there is K D G0 \ Z.G/ of order p such that G=Kis non-Dedekindian. Since G=K is a CO-group, we get G=K Š Mpt , by induction.Since K < G0 � ˆ.G/, we get d.G/ D d.G=K/ D 2. Let A=K and B=K betwo distinct cyclic subgroups of index p in G=K; then A and B are abelian maximalsubgroups of G. It follows that A \B D Z.G/ so jG0j D p (Lemma 1.1), contrary tothe assumption.

Corollary 58.4. Suppose that all nonnormal subgroups of the same order of a nilpotentnon-Dedekindian group G are conjugate. Then G D P � C , where P 2 Sylp.G/ isnon-Dedekindian and P Š Mpn , C is cyclic.

Theorem 58.5 (Berkovich). If G is a non-Dedekindian p-group all of whose nonnor-mal cyclic subgroups of the same order belong to the same conjugate class of size p,then G Š Mpt .

Proof. We use induction on jGj. In view of Lemma 58.1, one may assume that jG0j >

p.Assume that G is a 2-group of maximal class. Then jGj > 8. If G Š Q2n n > 3,

then G has > 2 nonnormal cyclic subgroups of order 4. If G 6Š Q2n , then G has > 2

nonnormal subgroups of order 2. Thus, G is not a 2-group of maximal class.

By Theorem 1.23, there is in G0 \ Z.G/ a subgroup R of order p such that G=Ris non-Dedekindian. Let X=R < G=R be nonnormal cyclic. Assume that X is non-cyclic. Then X D Xi � R, where Xi is cyclic of index p in X , i D 1; : : : ; p. Thesubgroup Xi is nonnormal in G (otherwise, X D XiR G G). By hypothesis, the

�58 Nonnormal subgroups of the same order are conjugate 149

subgroups X1; : : : ; Xp form the conjugate class inG soX D X1 : : : Xp GG, a contra-diction. Thus, X must be cyclic. By hypothesis, there are in G exactly p nonnormalcyclic subgroups of order jX j. Therefore, there are exactly p nonnormal cyclic sub-groups of order jX=Rj in G=R so G=R satisfies the hypothesis. By induction, thenon-Dedekindian group G=R Š Mpn . Then G=R has two distinct cyclic subgroupsA=R and B=R of index p. In that case, A and B are two distinct abelian maximalsubgroups of G so A \ B D Z.G/. Then jG0j D 1

pjG W Z.G/j D p (Lemma 1.1),

contrary to the assumption.

�59

p-groups with few nonnormal subgroups

By Lemma 58.1, if all nonnormal subgroups of a minimal nonabelian p-group G areconjugate, then G Š Mpt . Of course, Theorem 59.1 follows from Theorem 58.3.

Theorem 59.1 ([Sch3]). If all nonnormal subgroups of a p-group G are conjugate,then G Š MpnC1 .

Proof (compare with [Sch3]). Let H < G be nonnormal; then H is cyclic since it isnot generated by their (G-invariant) maximal subgroups. Set jH j D ps . If s D 1,then G Š MpnC1 (Theorem 1.23). Now we assume that s > 1. We use induction onjGj. We have R D �1.H/ GG. By induction, G=R Š Mpn; then jH=Rj D p. SinceR � ˆ.H/ < ˆ.G/, we get d.G/ D d.G=R/ D 2. Let U=R and V=R be distinctcyclic subgroups of index p in G=R; then U and V are abelian subgroups of index pin G so ˆ.G/ D U \ V D Z.G/ and G is minimal nonabelian. Then, by the remarkpreceding the theorem, G Š MpnC1 .

Given m > 1, let H2;m D ha; b j ap2

D bpm

D 1; ab D a1Cpi be a metacyclicminimal nonabelian group of order p2Cm.We suggest to the reader, using the above approach, to prove the following fairly

deep

Theorem 59.2 ([Sch4]). If a group G is nilpotent and has exactly two conjugateclasses of non-invariant subgroups, then one of the following holds:

(a) G Š D8.

(b) G Š Q16.

(c) p D 2 and G Š H2;m.

(d) G Š Mpn � Cq , where a prime q ¤ p.

Note that the proof presented in [Sch4], is not full. For the proof of this theorem,see [Ber26].

�60

The structure of the Burnside group of order 212

I. N. Sanov [Sano] has proved that each finitely generated group of exponent 4 is fi-nite. W. Burnside [Bur4] has shown that each two-generated group of exponent 4 is ahomomorphic image of a certain group G D B.4; 2/ of order at most 212 which wecall the Burnside group.In this section, written by the second author, we determine completely the structure

of the Burnside groupG by presenting this group in a very convenient form fromwhicheach structural question about that group can be easily answered which is importantfor applications (Theorem D). For example, we show that ˆ.G/ D �1.G/ is a special2-group of order 210 with Z.ˆ.G// Š E25 . The group G has exactly 12 maximalelementary abelian subgroups. Each of them is of order 26 and the intersection of anytwo of them is equal Z.ˆ.G//. If M;N;K are the maximal subgroups of G, thend.M/ D d.N / D d.K/ D 3, cl.M/ D cl.N / D cl.K/ D 4 and M 0 D N 0 D K 0 isabelian of order 27 and type .4; 4; 2; 2; 2/. The subgroup G0 is of order 28 and class2, d.G0/ D 5, �1.G

0/ D Z.ˆ.G// D Z.G0/, G00 Š E4, G00 � Z.G/ and Z.G/ Š E8.Next, jAut.G/j D 221 3 (Theorem C).

Theorem A (see [Hup, p. 300]). Let G be a group of exponent 4 generated with anelement v of order 4 and an involution t . Then G0 D hŒv�1; t �; Œv; t �i is abelian oforder � 8 and so jGj � 26.

Proof. We have .tv/3 D .tv/�1 D v�1t and .tv�1/3 D .tv�1/�1 D vt hence

Œv; t �Œv�1; t � D .v�1tvt/.vtv�1t / D v�1.tv/3v2t

D v�1.v�1t /v2t D .v2t /2 D Œv2; t �;

Œv�1; t �Œv; t � D .vtv�1t /.v�1tvt/ D v.tv�1/3v2t

D v.vt/v2t D .v2t /2 D Œv2; t �;

and so:

Œv; t �Œv�1; t � D Œv�1; t �Œv; t � D Œv2; t � D .v2t /2:(1)

Hence the subgroup P D hŒv�1; t �; Œv; t �i is abelian of exponent � 4 and order � 8

since Œv�1; t �Œv; t � is of order � 2 and so P cannot be isomorphic to C4 � C4. Also,P � G0 and it is easy to see that P is normal in G. Indeed, from 1 D Œvi ; t2� D


Œvi ; t �Œvi ; t �t follows Œvi ; t �t D Œvi ; t ��1 for each integer i (mod 4). From 1 D

Œv�1v; t � D Œv�1; t �vŒv; t � follows Œv�1; t �v D Œv; t ��1 and from Œv2; t � D Œv; t �vŒv; t �

follows (using (1)) Œv; t �v D Œv2; t �Œv; t ��1 D Œv�1; t �. Since G=P is abelian, we getG0 D hŒv�1; t �; Œv; t �i and so jGj � 26.

Theorem B (see [Hup, p. 300]). Let G D ha; bi be a group of exponent 4 such thatŒa2; b2� D 1. Then jGj � 29 and ha2iG D ha2; .a2/b

�1

; .a2/b�1a�1

i is elemen-tary abelian of order � 8 (where .a2/b

�1a�1b�1

D a2.a2/b�1

.a2/b�1a�1

). Simi-larly, hb2iG D hb2; .b2/a

�1

; .b2/a�1b�1

i is elementary abelian of order � 8 (where.b2/a

�1b�1a�1

D b2.b2/a�1

.b2/a�1b�1

).

Proof. Since Œa2; b2� D .a2b2/2 D 1, we get a2b2a2b2 D 1 and a2 D b2a2b2 andso

ba2b3 D b.b2a2b2/b3 D b3a2b:(�)

Now,

.a2.a2/b�1

/2 D .a2ba2b3/2 D a2.ba2b3/ .a2ba2b3/

D a2b3a2b a2ba2b b2 D a2b3.a3b/3b2 D a2b3.a2b/�1b2

D a2b3b�1a2b2 D .a2b2/2 D 1;

which implies Œa2; .a2/b�1

� D 1 and so also Œa2; .a2/b�1a�1

� D 1.Further, using (�), baba D a3b3a3b3 (which is a consequence of .ba/4 D 1) and

.a3b/3 D .a3b/�1 D b3a, we get:

.a2/b�1a�1b�1

D baba ab3a3b3 D a3b3a3b3ab3a3b3 D a2.ab3a3b3/2

D a2..baba3/�1/2 D a2.baba3/2 D a2 baba3 baba3

D a2b.a2a3/ba3b.a3a2/ba3 D a2ba2.a3b/3b3a2ba3

D a2ba2.b3a/b3a2ba3 D a2ba2b3a.b3a2b/a3

D a2.a2/b�1

a.ba2b3/a3 D a2.a2/b�1

.a2/b�1a�1

:

We have obtained .a2/b�1a�1b�1

D a2.a2/b�1

.a2/b�1a�1

and so each elementconjugate to a2 is contained in ha2; .a2/b

�1

; .a2/b�1a�1

i which gives ha2iG D

ha2; .a2/b�1

; .a2/b�1a�1

i. Further, a2 commutes with .a2/b�1

and .a2/b�1a�1

andso a2 2 Z.ha2iG/. But then .a2/b

�1

and .a2/b�1a�1

are also contained in Z.ha2iG/

and so ha2iG is elementary abelian of order � 8. By Theorem A, G=ha2iG is of order� 26 and so jGj � 29. The second half of the theorem is obtained by interchanging aand b.

Theorem C. Let G be the group given with: G D ha; b j a4 D b4 D .ab/4 D

.a2b/4 D .ab2/4 D .a�1b/4 D .a2b2/4 D .bab/4 D .aba/4 D 1i. Then G is of

�60 The structure of the Burnside group of order 212 153

order 212 and exponent 4 and so G is the Burnside group. The group G has exactlyk.G/ D 88 conjugate classes. The automorphism group of G is of order 221 3.

Proof. This theorem (apart from the last statement) is proved by using a computer(W. Lempken at the University of Essen).We may define our group G also in the form: G D ha; b j x4 D 1 for all x 2 Gi.

Then each map ˛ of fa; bg into G such that ha˛; b˛i D G induces an automorphismof G. Hence jAut.G/j D jAut.G=ˆ.G/jjˆ.G/j2 D 6 220.

The Burnside group G is given in Theorem C in terms of generators a; b and 9 rela-tions. However, from this presentation it is extremely difficult to pin down the structureof all subgroups and factor-groups of G. Therefore we give in the next theorem thegroup G in terms of 12 generators and 78 relations which will be very convenient toanswer any structural question about that group. We shall also deduce many importantproperties of the Burnside group G by using this new presentation. In the sequel, weuse from Theorem C only the fact that the Burnside group G is of order 212.

TheoremD (Complete determination of the structure of the Burnside group). LetG D

ha; bi be the Burnside group of order 212 as defined in Theorem C. Then G has thefollowing properties:

(a) ˆ.G/ D �1.G/ is a special group of order 210 with Z.ˆ.G// D .ˆ.G//0 D

ˆ.ˆ.G// Š E25 .

(b) If M;N;K are maximal subgroups of G, then they are of class 4 and they aregenerated by three elements but not by two elements. In addition, M 0 D N 0 D K 0 isabelian of order 27 and type .4; 4; 2; 2; 2/, Z.G/ Š E8, Z.G/ D Z.M/ < Z.ˆ.G//,K3.M/ Š E24 , K3.M/ < Z.ˆ.G//, K4.M/ Š E4, K4.M/ < Z.G/, and an auto-morphism of order 3 of G acts transitively on fM;N;Kg.

(c) The commutator group G0 is of order 28 and class 2,G=G0 Š C4 �C4,�1.G0/ D

Z.ˆ.G// D Z.G0/, ˆ.G0/ D Z.G/, G00 � Z.G/, G00 Š E4, and M 0 D K3.G/

is an abelian maximal subgroup of G0. In addition, K4.G/ D Z.ˆ.G//, E4 Š

K5.G/ D K4.M/ � Z.G/, whereM is maximal in G, and so G is of class 5. Finally,Z2.G/ D Z.ˆ.G// Š E25 , Z3.G/ D K3.G/ D M 0 (abelian of type .4; 4; 2; 2; 2/),and Z4.G/ D ˆ.G/.

(d) The groupG has exactly 12 maximal elementary abelian subgroups. Each of themis of order 26 and the intersection of any two of them is equal Z.ˆ.G// Š E25 . Hencethe number of involutions in G is 31C 12 32 D 415 and each abelian subgroup of Gis of rank � 6.

(e) The special group ˆ.G/ D �1.G/ is generated with five involutions d; e;m; n; yand we may set Z.ˆ.G// D hc; g; h; i; j i Š E25 so that:

Œd; e� D ch; Œd;m� D c; Œd; n� D g; Œd; y� D icgh; Œe;m� D h;

Œe; n� D cgh; Œe; y� D ig; Œm; n� D cg; Œm; y� D jc; Œn; y� D jg:


We have G D ha; bi with a2 D d , b2 D m, .a�1b�1/2 D y, where the action of a�1

and b�1 on ˆ.G/ is given with:

ca�1

D g; ga�1

D c; ha�1

D cgh; ia�1

D ich; j a�1

D jgh;

da�1

D d; ea�1

D ei; ma�1

D n; na�1

D mc; ya�1

D demnychij;

cb�1

D h; gb�1

D cgh; hb�1

D c; ib�1

D igh; j b�1

D jcg;

db�1

D e; eb�1

D dc; mb�1

D m; nb�1

D nj; yb�1

D demnygi:

We have determined completely the structure of our group G. We have here M D

ˆ.G/hai, N D ˆ.G/hbi, K D ˆ.G/ha�1b�1i, whereM 0 D N 0 D K 0 D K3.G/ D

Z.ˆ.G//hde; mni, K3.M/ D hc; g; h; ii, K5.G/ D K4.M/ D hcg; chi, G0 D

M 0hemyi, G00 D hcg; chi, and Z.G/ D Z.M/ D hcg; ch; cij i Š E8. Also,hciG D hc; g; hi Š E8, ha2iG D hd; e; i; g; hi and hb2iG D hm;n; j; g; hi aresubgroups of orders 26 which have its commutator subgroup of order 2.

(f) The factor-groupH D G=Z.ˆ.G// (of order 27) is the group of maximal possibleorder having the exponent 4, being generated with two elements, and having the prop-erty that the squares of any two elements commute. We have ˆ.H/ D �1.H/ Š E25 ,Z.H/ Š E4, Z.H/ < H 0 Š E23 , and ŒH;H 0� D Z.H/ so that H is of class 3. Theclass number k.H/ ofH is 26.

Proof. Let G D ha; bi be the Burnside group (as defined in Theorem C) so that G isthe free group of exponent 4 generated with two elements. We have jGj D 212 and wewant to determine the exact structure of G.If Œa2; b2� D 1, then by Theorem B we get jGj � 29, a contradiction. Hence

Œa2; b2� D .a2b2/2 D c is an involution and ha2; b2i Š D8 with Z.ha2; b2i/ D hci

so that a and b are elements of order 4 and c commutes with a2 and b2.We first determine the structure of hciG . We set h D cb�1

and claim that Œc; h� D 1.Indeed, since b2 D b3b3 and .b3a2/3 D .b3a2/�1 D a2b, we get:

s1 D ch D ccb�1

D .a2b2/2b.a2b2/2b3 D a2b2a2b2 ba2b2a2b2b3

D a2.b3b3/a2b3a2.b3b3/a2b D a2b3.b3a2/.b3a2/.b3a2/a2b3a2b

D a2b3 a2b a2b3a2b D .a2b3a2b/2

and so

s1 D ch D .a2b3a2b/2 D .a2.a2/b/2:(2)

Hence .ch/2 D 1 and so Œc; h� D 1. Also, sb�1

1 D .ccb�1

/b�1

D cb�1

cb2

D cb�1

c D

hc D ch D s1, and so s1 commutes with b.


It is possible to show that s1 also commutes with a. Using (2),

a2b3ab3a2 D a.ab3 ab3 ab3/ba D a.ab3/3ba D a.ab3/�1ba D aba3ba;

.a3bab/2 D .b3a3b3a/�2 D .b3a3b3a/2;

b3a3b3a3b3 D .b3a3/3a D .b3a3/�1a D aba:

we get (noting that a2 D a3a3):

s1sa�1b1 D a2b3a2b a2b3.a2b b3a a2/b3a2b a2b3a2b a3b

D a2b3a2b.a2b3ab3a2/ba2b3a2ba3b

D a2b3.a2b aba3ba b/a2b3a2ba3b

D a2b3a3.a3bab/2a2b3a2ba3b

D a2b3a3.b3a3b3a b3a3b3a/a2b3a2ba3b

D a2.b3a3b3a3b3 a b3a3b3a3b3/a2ba3b

D a2 aba a aba a2ba3b D .a3b/4 D 1:

Hence s1 also commutes with a�1b which implies that s1 D ch D ccb�1

is containedin Z.G/. Applying the automorphism ˛ which interchanges a and b (noting that c˛ D

Œb2; a2� D Œa2; b2� D c), we get Œc; ca�1

� D 1 and s2 D cca�1

2 Z.G/. We setca�1

D g so that hch; cgi � Z.G/, where hcg; chi is elementary abelian of order � 4.Since c commutes with h and g, we get ch cg D hg and so .hg/2 D 1 and thereforeh also commutes with g and hc; g; hi is elementary abelian of order � 8.We have

hb�1

D cb2

D c; ga�1

D ca2

D c;

ha�1

D .c ch/a�1

D ca�1

ch D gch D cgh;

gb�1

D .c cg/b�1

D cb�1

cg D hcg D cgh;

which implies that hc; g; hi is a normal subgroup of G and so hc; g; hi D hciG . ByTheorem B, G=hciG is of order � 29. Since jGj D 212, we have hciG Š E8 andG=hciG is of order 29. Also, hcg; chi Š E4 and hcg; chi � Z.G/. We have alsodetermined the action of a and b on hc; g; hi Š E8.We determine now the structure of ha2iG . Set a2 D d , e D .a2/b

�1

D db�1

,f D .a2/b

�1a�1

D ea�1

and S D hc; g; hihd; e; f i. By Theorem B, S=hc; g; hi isa normal elementary abelian subgroup of order � 8 of G=hc; g; hi with ha2iG � S .By Theorem A, jG=S j � 26. Since G=hc; g; hi is of order 29, we have S=hc; g; hi Š

E8 and jG=S j D 26. We note that hs1; s2i is a maximal subgroup of hc; g; hi andhs1; s2i Š Z.G/.


We know that a2 D d centralizes c and so d centralizes hc; s1; s2i D hciG .From Œc; a2� D 1 follows Œcb�1

; .a2/b�1

� D 1 and so Œh; e� D 1 which gives thate centralizes hh; s1; s2i D hciG . From Œc; e� D 1 follows Œca�1

; ea�1

� D 1 and soŒg; f � D 1 which gives that f centralizes hg; s1; s2i D hciG . We have proved thathc; g; hi � Z.S/ and so S is of class � 2.From (2) follows s1 D Œa2; .a2/b � and conjugating this relation with b�1 gives

s1 D Œ.a2/b�1

; a2� so that (since s1 D ch is an involution) Œe; d � D Œd; e� D s1 D ch.Conjugating the last relation with a�1 we get Œea�1

; d � D s1 and so Œf; d � D Œd; f � D

ch. By Theorem B, f b�1

D def l for some l 2 hc; g; hi. From Œd; f � D s1 followsconjugating with b�1:

Œdb�1

; f b�1

� D s1 and so Œe; def l� D Œe; d �Œe; f � D s1;

which implies s1Œe; f � D s1 and Œe; f � D 1. We get Œd; ef � D Œd; e�Œd; f � D s1s1 D

1 and so ef 2 Z.hd; e; f i/ which gives:

hd; e; f i D hef i � hd; ei Š C2 � D8 with hd; e; f i0 D hs1i;

S D hc; g; hi � hd; e; f i; where hc; g; hi \ hd; e; f i D hs1i;

so that S 0 D hs1i and Z.S/ D hc; g; h; ef i Š E24 .It remains to show that ha2iG D S . For that purpose we have to determine exactly

the element f b�1

. First we note that c commutes with a2 and b2. From Œa2; b2� D c

follows a2b2a2b2 D c and a2 D cb2a2b2 and so

ba2b3 D b cb2a2b2 b3 D .bcb3/.b b2a2b2 b3/ D cb�1

.b3a2b/;

which gives:

ba2b3 D h.b3a2b/:(��)

We compute:

f b�1

D .a2/b�1a�1b�1

D baba2b3a3b3

(using baba D a3b3a3b3 which follows from .baba/.baba/ D 1)

D a3b3a3b3 a b3a3b3 D a2.ab3a3b3/.ab3a3b3/

D a2.ab3a3b3/2 D a2.baba3/�2 D a2.baba3/2

D a2baba3baba3 D a2b.a2a3/ba3b.a3a2/ba3

D a2ba2.a3b/.a3b/.a3b/b3a2ba3


(since .a3b/3 D .a3b/�1 D b3a and using (��))

D a2ba2 b3a .b3a2b/a3 D a2.ba2b3/a hba2b3a3

D a2.ba2b3/.aha3/.ab a2 b3a3/

D a2.a2/b�1

ha�1

.a2/b�1a�1

D deha�1

f D def cgh;

where we have used ha�1

D cgh (which was proved before) and cgh 2 Z.S/.We have obtained the desired relation f b�1

D defcgh. We compute f a�1

D

ea2

D ed D eŒe; d � D ech and so

f b�1a�1

D .defcgh/a�1

D df .ech/gc.cgh/ D dfec D defc:

Now, hd D a2; e; f i � ha2iG and so Œd; e� D ch 2 ha2iG . From f b�1

D def cgh

follows that cgh 2 ha2iG and from f b�1a�1

D def c follows that c 2 ha2iG . Hencehc; g; hi � ha2iG and so S D hd; e; f ihc; g; hi D ha2iG .Let ˛ be the automorphism of order 2 of G such that a˛ D b and b˛ D a. Note

that c˛ D Œb2; a2� D Œa2; b2� D c and g˛ D h, h˛ D g so that .hciG/˛ D hciG D

hc; g; hi. Set hb2iG D T so that T D S˛, jT j D 26, T > hc; g; hi, T=hc; g; hi Š E8.We have U D ST D ha2; b2iG and G=U is generated with two involutions and sojG=U j � 23. Since jU j � 29, we must have jG=U j D 23 and then jU j D 29,S \ T D hc; g; hi, and G=U Š D8. Set b2 D m, .b2/a

�1

D n, .b2/a�1b�1

D p

so that m D d˛ , n D e˛ , p D f ˛ and ˛ transports the relations for S into therelations for T . We get hm;n; pi D hnpi � hm;ni Š C2 � D8, where Œn; p� D 1,Œm; n� D Œm; p� D s2 D cg and T D hm;n; pi � hc; g; hi with hm;n; pi \ hc; g; hi D

hs2i, Z.T / D hc; g; h; npi Š Z24 , T 0 D hs2i. Also, from f b�1

D defcgh followspa�1

D mnpcgh.Since hc; g; hi � Z.S/ and hc; g; hi � Z.T /, we have hc; g; hi � Z.U /. But

ˆ.U / � S \ T D hc; g; hi (noting that U=S Š U=T Š E8) and so U is of class 2.We shall determine completely the structure of U D ST D ha2; b2iG . This is doneby the following computation (using all the previous relations):

f a�1

D .a2/b�1a2

D ea2

D ed D eŒe; d � D ech;

dm D .a2/b2

D a2c D dc;

f m D f b2

D .f b�1

/b�1

D .defcgh/b�1

D e dc defcgh h cgh c D f h;

en D eab2a�1

D .ea�1

/a2b2a�1

D .f d /ma�1

D .fch/ma�1

D .f h c h/a�1

D .fc/a�1

D echg;

dn D dma�1

D dama�1

D .a2/ma�1

D dma�1

D .dc/a�1

D dca�1

D dg;


and

dp D dnb�1

D dbnb�1

D .db2

/b�1nb�1

D .dc/b�1nb�1

D .eh/nb�1

D .echg h/b�1

D .ecg/b�1

D dc h cgh D dg;

em D eb2

D .eb�1

/b�1

D .dc/b�1

D eh;

ep D ebnb�1

D .eb2

/b�1nb�1

D .em/b�1nb�1

D .eh/b�1nb�1

D .dc c/nb�1

D dnb�1

D .dg/b�1

D ecgh;

f n D f ama�1

D .f a�1

/a2ma�1

D .ech/dma�1

D .ech c h/ma�1

D ema�1

D .eh/a�1

D f cgh;

f p D f bnb�1

D .f b2

/b�1nb�1

D .f m/b�1nb�1

D .f h/b�1nb�1

D .defcgh c/nb�1

D .defgh/nb�1

D .dg echg fcgh g h/b�1

D .def h/b�1

D e dc defcgh c D fcgh:

From the above relations we get at once the desired commutators which determinecompletely the structure of U D ST :

Œd;m� D c; Œd; n� D g; Œd; p� D g; Œe;m� D h; Œe; n� D cgh;

Œe; p� D cgh; Œf;m� D h; Œf; n� D cgh; Œf; p� D cgh:

In particular, we see that U 0 D ˆ.U / D hc; g; hi. We know that hc; g; hi � Z.U /,ef 2 Z.S/ and np 2 Z.T /. In addition, we compute (noting that U is of class 2):

Œef;m� D Œe;m�Œf;m� D h h D 1; Œef; n� D 1; Œef; p� D 1;

Œd; np� D 1; Œe; np� D 1; Œf; np� D 1;

and so ef; np 2 Z.U /. Therefore Z.U / � hc; g; h; ef; npi.We show that we have in fact Z.U / D hc; g; h; ef; npi Š E25 . Indeed, we set

U1 D hc; g; hihd; e;m; ni, U2 D hef; npi Š E4 so that U D U1 � U2, whereU 0

1 D ˆ.U1/ D hc; g; hi � Z.U1/ and we claim that hc; g; hi D Z.U1/. Suppose thatw D d˛eˇm�nı 2 Z.U1/ .˛; ˇ; �; ı D 0; 1/ which implies

1 D Œw; d � D Œe; d �ˇ Œm; d �� Œn; d �ı D .ch/ˇc�gı D cˇC�gıhˇ

and so ˇ D � D ı D 0 and w D d˛ . From 1 D Œw; e� D Œd; e�˛ D c˛h˛ follows˛ D 0 and so w D 1. We have proved that Z.U1/ D hc; g; hi and so Z.U / D

hc; g; h; ef; npi.The factor-group G=U Š D8 is generated with involutions a�1U and b�1U so that

x D a�1b�1 must be an element of order 4 with hxi \ U D f1g. Set y D x2 so


that U hyi=U D Z.G=U / D ˆ.G=U /. It follows that ˆ.G/ D U hyi. By TheoremA, the set G � ˆ.G/ cannot contain involutions and so (since �1.U / D U ) we have�1.G/ D ˆ.G/.We have hciG D hc; g; hi D hc; s1; s2i, where hs1; s2i � Z.G/, and so CG.c/ D

CG.hc; g; hi/ which shows that CG.c/ is a normal subgroup of G. On the other hand,CG.c/ � U and jG W CG.c/j D 4 (since the four conjugates of c are contained inhc; g; hi � hs1; s2i), which implies that CG.c/ D U hyi D ˆ.G/ and so cy D c andhc; g; hi � Z.ˆ.G//.We determine now the action of G=U on U by giving the actions of the elements

a�1 and b�1 (we use a�1 and b�1 rather than a and b because of some technicalreasons) on the elements c; g; h; d; e; f;m; n; p of U . We get (by using all previousrelations):

ca�1

D g; ga�1

D ca2

D c;

ha�1

D cgh; da�1

D d;

ea�1

D f; f a�1

D ..a2/b�1

/a2

D ed D eŒe; d � D ech;

ma�1

D n; na�1

D .b2/a2

D b2Œb2; a2� D mc;

pa�1

D mnpcgh; cb�1

D h;

gb�1

D cgh; hb�1

D cb2

D c;

db�1

D e; eb�1

D db2

D dm D dŒd;m� D dc;

f b�1

D defcgh; mb�1

D m;

nb�1

D p; pb�1

D ..b2/a�1

/b2

D .ma�1

/m D nm D nŒn;m� D ncg:

Since y D x2 D .a�1b�1/2, we get from the above relations the action of y on U .We know that y centralizes hc; g; hi and we get:

dy D defcgh; ey D fg; f y D eg; my D mnpc; ny D pg; py D ng:

Since .ef /y D ef and .np/y D np, we get that Z.U / D hc; g; h; ef; npi � Z.ˆ.G//.From the above relations we also get:

Œd; y� D .ef /cgh; Œe; y� D .ef /g; Œf; y� D .ef /g;

Œm; y� D .np/c; Œn; y� D .np/g; Œp; y� D .np/g:

Our relations also show that ˆ.G/=Z.U / is abelian and so elementary abelian. Onthe other hand, we see that ef; np 2 .ˆ.G//0 and so .ˆ.G//0 D Z.U /. If Z.ˆ.G// >Z.U /, then ˆ.G/ D U � Z.ˆ.G// which implies that .ˆ.G//0 D U 0 D hc; g; hi, a


contradiction. Thus Z.ˆ.G// D Z.U / Š E25 and so ˆ.G/ is a special group of order210, as claimed in part (a) of our theorem.It remains to determine the elements ya�1

and yb�1

. This will complete the deter-mination of the action of a�1 and b�1 on the special group ˆ.G/.Since x D a�1b�1 and y D x2, we get

ya�1

D .x2/a�1

D .xa�1

/2 D .b�1a�1/2 D b�1a�1b�1a�1

D b�1.a�1b�1a�1b�1/b D b�1yb D yb D .yb2

/b�1

D .ym/b�1

D .Œm; y�y/b�1

D ..np/cy/b�1

D .y.np/c/b�1

D yb�1

p ncg h

D yb�1

.np/cgh;

and so ya�1

D yb�1

.np/cgh. On the other hand, xa D a�1.a�1b�1/a D a2b�1a D

a2b2.ba/ D a2b2.a�1b�1/�1 D dmx�1, and so

ya D .x2/a D .xa/2 D dmx�1dmx�1 D dm.dm/xx2 D dm.dm/xy:

Conjugating the last relation with a�1, we get da�1

ma�1

.dm/a�1b�1a�1

ya�1

D y,and so using our previous relations we get:

d n .f mnpcgh/ ya�1

D y; ya�1

D .pnmf nd/ycgh:

Using our commutator relations in ˆ.G/, we get pnmf nd D demng.ef /.np/, andso ya�1

D demnych.ef /.np/. Putting the last relation in (3), we get also yb�1

D

demnyg.ef /.We have determined completely the structure of the Burnside group G D ha; bi.

We set ef D i and np D j so that Z.ˆ.G// D hc; g; h; i; j i Š E25 and the specialgroup ˆ.G/ is generated with five involutions d; e;m; n; y with a2 D d , b2 D m,.a�1b�1/2 D y and the commutator relations and the action of a�1 and b�1 onˆ.G/as given in part (e) of our theorem.In the rest of the proof we shall use only this presentation of the group G as given

in part (e) of our theorem. We show now that Z.G/ D hcg; ch; cij i Š E8. Indeed,G=U Š D8 and Z.G=U / D U hyi=U and so Z.G/ � U hyi D ˆ.G/. It follows thatZ.G/ � Z.ˆ.G//. On the other hand, hcg; chi � Z.G/. We compute:

.cij /a�1

D g ich jgh D cij; .cij /b�1

D h igh jcg D cij;

and so cij 2 Z.G/. But hi; j i is a complement of hcg; ch; cij i in Z.ˆ.G// and

ia�1

D ich; j a�1

D jgh; .ij /a�1

D ijcg;

which shows that hi; j i \ Z.G/ D f1g and therefore Z.G/ D hcg; ch; cij i Š E8.


We examine now the non-trivial cosets of Z.ˆ.G// in ˆ.G/. Each such coset con-sists either of involutions or only of elements of order 4. Using the defining relationsforˆ.G/, we find out that there are exactly 12 such cosets which consist of involutionsand they are the cosets with representatives from the set:

V D fd; e; m; n; y; dem; den; dey; dmn; emn; mny; demnyg:

Also we see that if v;w 2 V , v ¤ w, then o.vw/ D 4. This shows that G hasexactly 12 maximal elementary abelian subgroups. Each of them is of order 26 andthe intersection of any two of them is equal Z.ˆ.G//.In order to determine the structure of various subgroups ofG which contain the sub-

group Z.ˆ.G//, it is convenient first to determine the structure of H D G=Z.ˆ.G//of order 27. Since Z.ˆ.G// D .ˆ.G//0, H is obviously the group of maximal pos-sible order having the exponent 4, being generated with two elements, and havingthe property that the squares of any two elements commute. The images of elementsa; b; d; e;m; n; y of G in H D G=Z.ˆ.G// we denote again with the same symbols.Then we have E D �1.H/ D ˆ.H/ D hd; e;m; n; yi Š E25 , H D ha; bi witha2 D d , b2 D m, .a3b3/2 D y, and the action of a and b on E is given with:

da D d; ea D e; ma D n; na D m; ya D demny;

db D e; eb D d; mb D m; nb D n; yb D demny:

We have CE .a/ D hd; e;mni, CE .b/ D hde;m; ni, CE .ab/ D hde;mn; yi, Z.H/ D

hde;mni Š E4, .Ehai/0 D .Ehbi/0 D .Ehabi/0 D hde;mni D Z.H/, Ehai=Z.H/,Ehbi=Z.H/, Ehabi=Z.H/ are abelian groups of type .4; 2; 2/. It follows that themaximal subgroups of H are of rank 3. Also, it is easy to see from these informationthat the number of conjugate classes inH is k.H/ D 26. Indeed, 4 elements in Z.H/give 4 classes of size 1, 12 elements in dZ.H/ [mZ.H/ [ yZ.H/ give 6 classes ofsize 2, the remaining 16 elements in E give 4 classes of size 4, and 96 D 27 � 25

elements inH �E give 12 classes of size 8 each. We compute:

y D .a�1b�1/2 D a�1b�1a�1b�1 D .a�1b�1ab/ b�1a�1a�1b�1

D Œa; b�b�1a2b�1 D Œa; b� db b2 D Œa; b�em;

so that Œa; b� D emy. This gives H 0 D hde;mn; emyi Š E8 and H=H 0 Š C4 � C4.Since

.emy/a D e n demny D dmy D .emy/de;

.emy/b D d m demny D eny D .emy/mn;

it follows that ŒH;H 0� D Z.H/ D hde;mni, H is of class 3, and H=ŒH;H 0� isthe non-metacyclic minimal nonabelian group of order 25 and exponent 4. We haveobtained the results stated in part (f) of our theorem.


We return now to our group G and consider the structure of maximal subgroupsM D ˆ.G/hai, N D ˆ.G/hbi, K D ˆ.G/ha�1b�1i. By the preceding para-graph, M 0 D N 0 D K 0 D Z.ˆ.G//hde; mni and M=M 0 is an abelian group oftype .4; 2; 2/ so that d.M/ D 3. Let be the automorphism of G of order 3 in-duced with a D b, b D a�1b�1. Then acts transitively on fM;N;Kg andtherefore it is enough to determine the structure of M . By the preceding paragraph,G0 D Z.ˆ.G//hde; mn; emyi D M 0hemyi, G=G0 Š C4 �C4, andM 0=Z.ˆ.G// D

Z.H/ D K3.H/,H D G=Z.ˆ.G// and thereforeM 0 D Z.ˆ.G//hde; mni is a char-acteristic subgroup of G. It follows from d.G/ D 2 that G0=K3.G/ is cyclic andG0 D

K3.G/hŒa; b�i. But 1 D .ab/4 a4b4Œb; a�6 .mod K3.G// so that Œb; a�2 2 K3.G/

and jG0=K3.G/j D 2. This gives that K3.G/ D Z.ˆ.G//hde; mni D M 0.We compute:

Œde;mn� D Œd;m�Œd; n�Œe;m�Œe; n� D c g h cgh D 1;

.de/2 D d2e2Œe; d � D ch; .mn/2 D cg; .demn/2 D gh;

which shows that M 0 D K3.G/ is abelian of type .4; 4; 2; 2; 2/ with ˆ.M 0/ D

hcg; chi Š E4. Since CG.c/ D ˆ.G/, we have Z.M/ � Z.ˆ.G//. But Chi;j i.a�1/ D

f1g and so Z.M/ D Z.G/ D hcg; ch; cij i, where we have used the fact that hi; j i is acomplement of Z.G/ in Z.ˆ.G//. Since

.de/a�1

D .de/i; .de/y D .de/ch; .de/n D .de/ch; .mn/a�1

D .mn/g;

.mn/y D .mn/cg; .mn/n D .mn/cg; ia�1

D ich; j a�1

D jgh;

we get that hc; g; h; ii Š E24 is normal inM ,M 0=hc; g; h; ii � Z.M=hc; g; h; ii/ andK3.M/ D hc; g; h; ii. Finally, hc; g; h; ii � Z.ˆ.G//, ca�1

D c.cg/, ga�1

D g.cg/,ha�1

D h.cg/, ia�1

D i.ch/ so that K4.M/ D hcg; chi � Z.G/ andM is of class 4.We determine now the structure of G0 D Z.ˆ.G//hde; mn; emyi, where G=G0 Š

C4 �C4. We know that K3.G/ D Z.ˆ.G//hde; mni is an abelian maximal subgroupof G0, where�1.K3.G// D Z.ˆ.G//. But we see that the elements:

emy; de emy D dmy; mn emy eny; de mn emy dny .mod Z.ˆ.G///

are all of order 4 and so �1.G0/ D Z.ˆ.G//. We compute Œde; emy� D ch and

Œmn; emy� D cg and so G00 D hcg; chi � Z.G/. From jG0j D 2jZ.G0/jjG00j followsjZ.G0/j D 25 and so Z.G0/ D Z.ˆ.G///. Also, .de/2 D ch, .mn/2 D cg and.emy/2 D cghij , which gives ˆ.G0/ D hcg; ch; cghij i D hcg; ch; cij i D Z.G/.We have M 0 D K3.G/ and K3.M/ D ŒM;M 0� D ŒM;K3.G/� D hc; g; h; ii,

.mn/b�1

D .mn/j , and so K4.G/ � hc; g; h; i; j i D Z.ˆ.G//. On the otherhand, we know that K3.G/=Z.ˆ.G// D Z.G=Z.ˆ.G/// D Z.H/ and so K4.G/ D

Z.ˆ.G//. We have ib�1

D i.gh/, j a�1

D j.gh/, j b�1

D j.cg/, and so we see thatK5.G/ D hcg; chi � Z.G/, K6.G/ D f1g, and G is of class 5.Finally, we see that Z2.G/ D Z.ˆ.G// Š E25 , Z3.G/ D K3.G/ D M 0 (abelian of

type .4; 4; 2; 2; 2/), and Z4.G/ D ˆ.G/. Our theorem is proved.

�61

Groups of exponent 4

generated by three involutions

It is known that a group generated with two non-commuting involutions is dihedral.But the problem to determine groups generated with three involutions is already verydifficult. In this section we classify groups of exponent 4 which are generated withthree involutions.

Theorem 61.1. Let G be a group of exponent 4 with d.G/ D 3 so that G is generatedwith involutions a; b; c and at least one of ab; ac; bc is an involution. Then jGj � 26

and one of the following holds:

(a) G Š C2 � D8, where jGj D 24.

(b) G is a splitting extension of the abelian groupH D hv;w j v4 D w4 D Œv;w� D

1; v2 D w2i of type .4; 2/ with the four-group ha; bi, where va D w, vb D v�1,wb D w�1. We have jGj D 25 and G D ha; b; ci with c D bv.

(c) G is a splitting extension of the nonmetacyclic minimal nonabelian group H D

hv;w j v4 D w4 D .wv/2 D Œw2; v� D 1i of order 24 with the four-groupha; bi, where va D w, vb D v�1, wb D w�1. We have jGj D 26 and G D

ha; b; ci with c D bv. The group G is isomorphic to a Sylow 2-subgroup of thealternating group A8.

Proof. Let G be a group of exponent 4 with d.G/ D 3 so that G is generated withthree involutions a; b; c. If a; b; c pairwise commute, then G Š E8, a contradiction. Ifa commutes with b and c but Œb; c� ¤ 1, then G D hai � hb; ci Š C2 � D8.Assume now that Œa; b� D .ab/2 D 1 but G is not isomorphic to C2 � D8. This

implies Œb; c� ¤ 1 and Œa; c� ¤ 1. Set v D bc so that o.v/ D 4 and D D hb; ci D

hb; vi Š D8 with hv2i D Z.D/. Also, ha; ci Š D8 and so o.ac/ D 4.Set w D va. Since b inverts v and a centralizes b, it follows that b inverts hwi.

Also, wa D v and so the four-group ha; bi normalizes the subgroup H D hv;wi.Hence G D H ha; bi. We compute

.ac/2 D .aca/c D cac D .bv/abv D .bwb/v D w�1v

and so o.w�1v/ D 2. From .av/4 D 1 follows .av/2 D .ava/v D vav D wv and


.wv/2 D 1. The last relation gives

1 D wvwv D w2.w�1vw�1/w2v D w2.w�1vw�1v/v�1w2v

D w2.w�1v/2v�1w2v D Œw2; v�:

If hwi D hvi, thenH D hvi and G D Dhai. There is an involution i 2 ha; bi � hbi

which centralizes hvi and so G D hii �D, contrary to our assumption.We have hvi ¤ hwi and since Œw2; v� D 1, we get w2 2 Z.H/. Consider NH D

H=hw2i (bar convention) so that NH D h Nw; Nvi. Suppose v2 D w2 which implies thatNH is generated with two distinct involutions Nv and Nw. We know that .wv/2 D 1 andso wv D Nw Nv is also an involution. Hence NH Š E4. Suppose that H is nonabelian.Since hvi and hwi are two distinct cyclic subgroups of order 4 inH , we getH Š Q8.But we know that .wv/2 D 1, a contradiction. HenceH is abelian of type .4; 2/. Wehave G D H ha; bi and so the group G is a splitting extension of the abelian groupH D hv;w j v4 D w4 D Œv;w� D 1; v2 D w2i of type .4; 2/ with the four-groupha; bi, where va D w, vb D v�1, wb D w�1. We get jGj D 25 and G D ha; b; ci

with c D bv.Assume now that v2 ¤ w2 so that hvi \ hwi D f1g. We have o. Nw/ D 2, o. Nv/ D 4

and o. Nw Nv/ D 2 (since .wv/2 D 1 ). Hence NH Š D8 and so jH j D 24. The groupH isnot of maximal class sinceH possesses the abelian subgroup hw2i�hvi of type .4; 2/.HenceH is of class 2 and therefore 1 D Œw2; v� D Œw; v�2 D Œw; v2�, so that Œw; v� isan involution. Since Œw; v� 2 Z.H/, we get that H=hŒw; v�i is abelian. Consequently,H 0 D hŒw; v�i and so H is minimal nonabelian with Z.H/ D hw2; v2i Š E4. Also,1 D .wv/2 D w2v2Œv;w�, and so Œv;w� D Œw; v� D w2v2. Since wv 62 Z.H/ D

hw2; v2i, we get hw2; v2; wvi Š E8 and so H is non-metacyclic. The group G is asplitting extension of the non-metacyclic minimal nonabelian groupH D hv;w j v4 D

w4 D .wv/2 D Œw2; v� D 1i of order 24 with the four-group ha; bi, where va D w,vb D v�1, wb D w�1. We have ha; bi\H D f1g and so jGj D 26 andG D ha; b; ci

with c D bv and Z.G/ D hv2w2i D h.bc/2..bc/2/ai. Indeed, it is easy to see thatthe involutions in ha; bi Š E4 induces on H only outer automorphisms. No elementx 2 H could invert an element y of order 4 inH since Œw; v� D w2v2 is not a squareinH . Also, hyxi \ hyi � hy2i since y2 2 ˆ.H/ D Z.H/.The group G is isomorphic to a Sylow 2-subgroup of the alternating group A8. Set

D D hv; bi Š D8 so that .v2/a D w2 and therefore v2 62 DG . No subgroup of order2 in D � hvi is normal in D. Hence DG D f1g and so G has a faithful transitivepermutation representation of degree 8. If we set:

a D .1; 2/.3; 4/.5; 6/.7; 8/; b D .4; 5/.3; 6/; c D .2; 3/.6; 7/;

then we see that the permutations a; b; c satisfy all the above relations for G and

.bc/2..bc/2/a D .1; 8/.2; 7/.3; 6/.4; 5/ ¤ 1

and so the obtained permutation representation is faithful. Since these permutationsare even, we are done.

�61 Groups of exponent 4 generated by three involutions 165

Theorem 61.2. Let G be the group given with

G D ha; b; c j a2 D b2 D c2 D .ab/4 D .bc/4 D .ca/4 D 1;

.abc/4 D .c.ab/2/4 D .b.ac/2/4 D .acab/4 D 1i:

Then G is of order 210 and exponent 4. Hence each group of exponent 4 which isgenerated with three involutions is a epimorphic image of the above group of order 210.

Proof. This theorem is proved by using a computer (W. Lempken of the University ofEssen).

In what follows we shall determine the structure of the group G D ha; b; ci of order210 as defined in Theorem 61.2 so that G is the free group of exponent 4 generatedwith three involutions a; b; c.We set .ab/2 D d , d c D e, ea D f , and f b D g. We shall determine the

structure of the normal subgroup S D hd iG � hd; e; f; gi, where d; e; f; g are someinvolutions in ˆ.G/. Since ha; bi Š D8, the involution d D Œa; b� commutes with aand b. From the above follows ec D d , f a D e and gb D f . We compute

.de/2 D .abab cababc/2 D .ababc ababc/2 D .ababc/4 D 1;

and so Œd; e� D 1. From the last relation we obtain (conjugating with the element aand then with the element b): Œd; ea� D Œd; f � D 1, Œd; f b� D Œd; g� D 1. Furthercomputation gives (noting that .ab/2 D .ba/2 and .bac/3 D .bac/�1 D cab):

.ef /2 D .e aea/2 D .ea/4 D 1;

.fg/2 D .f bf b/2 D .f b/4 D 1;

.eg/2 D .cbabac bacbabacab/2 D .cba.bac bac bac/cbacab/2

D .cba cab cba cab/2 D .cbacab/4 D 1;

and so Œe; f � D 1, Œf; g� D 1 and Œe; g� D 1. We have proved that the subgrouphd; e; f; gi is elementary abelian of order � 24.It is possible to show that the subgroup hd; e; f; gi is normal in G and so we get

S D hd iG D hd; e; f; gi. Indeed, we compute (noting that .ca/3 D .ca/�1 D ac,.ab/3 D ba, .cbab/3 D .cbab/�1 D babc):

ff c D acababca cacababcac D acabab.ca/3babcac D acabab ac babcac

D ac.ababab/bcbabcac D ac ba bcbabcac

D a.cbab/.cbab/.cbab/babac D a babc babac

D .ab/2c.ba/2c D dd c D de;


and so ff c D de which gives f c D def . Also we get:

ga D .d c/aba D eaba D eabab�b D .e.ab/2

/b D .ed /b D eb ;

and so ga D eb .We compute (noting that .ab/2 D .ba/2 and bcbcb D .bc/3c D .bc/�1c D cbc):

eeb D caba.bc bcb/abacb D caba.cbc/abacb D .cabacb/2;

and similarly (noting that baca baca ba D .baca/3ac D .baca/�1ac D acabac),

fg D acba.baca bacaba/bcab D acba acabac bcab D .acbcab/2

D .acbcab/�2 D .bacbca/2:

Using the above relations we get (noting that .ac/2 D .ca/2, .bca/2 D .bca/�2 D

.acb/2, .cab/2 D .bac/2, and .b.ac/2/3 D .b.ac/2/�1 D .ac/2b):

eebfg D .cabacb cabacb/.bacbca bacbca/

D cabac.bca bca cab cab/acbca D cabac.acb/2.bac/2acbca

D cabacacbacbbacbacacbcaD ca .b.ac/2 b.ac/2 b.ac/2/bca

D ca .ac/2b bca D ca.ca/2ca D .ca/4 D 1;

and so eb D efg D ga.Finally, we get (using .ab/3 D ba):

gce D cbacbabac.abc cabab/c D cba cba bac.ab/3c D .cba/2.bac/2

D .cba/�2.bac/�2 D .abc/2.cab/2 D abcabccabcab

D abc.ab/2cab D .ab/2 bac.ab/2cab D dg;

and so gc D deg. We have proved by now:

da D d; ea D f; f a D e; ga D efg;

db D d; eb D efg; f b D g; gb D f;

d c D e; ec D d; f c D def; gc D deg:

This shows that hd; e; f; gi is normal in G and so S D hd; e; f; gi D hd iG is elemen-tary abelian of order � 24. By Theorem 61.1, G=S is of order � 26. Since jGj D 210

(Theorem 61.2), we have S Š E24 and jG=S j D 26.From the above results we get at once that hef; egi � Z.G/. The subgroup hd; ei

is a complement of hef; egi in S and Z.G/ \ hd; ei D f1g. This gives Z.G/ \ S D

hef; egi.

�61 Groups of exponent 4 generated by three involutions 167

Define the automorphism of G induced with a D b, b D c, c D a. Set

d D h D .bc/2; e D i D ..bc/2/a;

f D j D ..bc/2/ab; g D k D ..bc/2/abc

so that T D hhiG D hh; i; j; ki Š E24 and transports the above relations for theelements in S into the relations for the elements in T . In particular, ha D i , ia D h,hb D h, ib D j , hc D h, ic D ijk.We want to determine S\T . For that purpose we compute (noting that .bc/3 D cb,

.cba/2 D .abc/2, .ca/3 D ac, .bc/3 D cb):

ef ik D cababc acaba.bca abcbc/a cbabcbcabc

D cababcacaba.cb acba/bcbcabc D cababcacaba abcabc bcbcabc

D cabab.ca/3.bc/3abc D cabab ac cb abc D c.ab/4c D 1;

and similarly (noting that we have .bc/2 D .cb/2, .cbab/2 D .babc/2, .ab/3 D ba,.bac/3 D cab):

egjk D cababc bacababcab bacbcbab cbabcbcabc

D cababcbacaba.cbabcbab/cbcabc

D cababcbac.aba bab/cbabc cbcabc D cababc.bac ba cbac/abc

D cababc cab abc D c.ab/4c D 1;

which gives ef D ik, eg D jk and so S\T � hef; egi Š E4 and ib D j D i.ij / D

i.fg/, ic D i.jk/ D i.eg/. We have jST j � 26. By Theorem 61.1, jG=ST j � 24

and so the fact that jGj D 210 gives U D ST is of order 26 and G=U Š C2 � D8.This implies that y D .ca/2 62 U and S \ T D hef; egi D hik; jki � Z.G/. Notethat hh; ii is a complement of S \ T in T and we compute:

dh D dbcbc D d.fg/; d i D dabcbca D d.fg/;

eh D ebcbc D e.fg/; ei D eabcbca D e.fg/;

and so Œd; h� D Œd; i � D Œe; h� D Œe; i � D fg.We have U 0 � S \ T � Z.G/ and so U is of class 2 and, by the above, U 0 D hfgi

and U D hd; e; f; gihh; ii � ˆ.G/. Since y D .ca/2 62 U and G=G0 is generatedwith involutions, we have ˆ.G/ D G0 D �1.ˆ.G// D U hyi.We note that fg; eg; ef 2 Z.G/,

hg D ghŒh; g� D ghŒh; e.eg/� D ghŒh; e� D ghfg


and Œd; hi � D Œd; h�Œd; i � D .fg/2 D 1 so

yb D .caca/b D ch ad ch ad D cahadchad D caidchad D cacicd chad

D caci.eg/ehad D cacighad D .caca/iagahad D y h.ef /gi d

D y.ef /.hg/id D y.ef /gh.fg/id D y.ef /.fg/g.hi/d D yedhi;

and hence we have yb D ydehi , ya D y and yc D y.Set fg D z1, eg D z2 so that E4 Š hz1; z2i � Z.G/,

dy D d caca D dz1z2; ey D ez1z2; hy D hz2; iy D iz2;

Z.U / D hz1; z2; de; hii Š E24 , U 0 D hz1i.Obviously we have Z.ˆ.G// � Z.U / and since .de/y D de, .hi/y D hi , we get

hde; hii � Z.ˆ.G// and so Z.ˆ.G// D Z.U /. By the above, z1; z2 2 .ˆ.G//0 andsince the factor-group ˆ.G/=hz1; z2i is abelian, we get G00 D .ˆ.G//0 D hz1; z2i.Also, hde; hii is a complement of hz1; z2i in Z.ˆ.G//, Z.G/ � ˆ.G/ and .de/a D

df , .hi/b D hj , .dehi/a D df hi which all imply that hz1; z2i D Z.G/.Our relations give at once that K3.G/ D Z.ˆ.G// D hz1; z2; de; hii Š E24 and

K4.G/ D hz1; z2i D Z.G/ so that G is of class 4. We have proved the following

Theorem 61.3. Let G be the free group of exponent 4 which is generated with threeinvolutions a; b; c. Then G is of order 210 and has the following properties:

(a) ˆ.G/ D G0 D �1.G0/ is of order 27 and class 2 with Z.ˆ.G// Š E24 and

.ˆ.G//0 D G00 D ˆ.ˆ.G// Š E4.

(b) We have K3.G/ D ŒG;G0� D Z.ˆ.G// and K4.G/ D .ˆ.G//0 D Z.G/ so thatG is of class 4.

(c) The groupˆ.G/ is generated with five involutions d; e; h; i; y so that .ˆ.G//0 D

hz1; z2i Š E4 and Œd; e� D Œh; i � D 1, Œd; h� D Œd; i � D Œe; h� D Œe; i � D z1,Œd; y� D Œe; y� D z1z2, Œh; y� D Œi; y� D z2, where hz1; z2i D Z.G/ andZ.ˆ.G// D hz1; z2; de; hii Š E24 . We have G D ha; b; ci with .ab/2 D d ,.bc/2 D h, .ca/2 D y, where the action of a; b; c on ˆ.G/ is given with:

da D d; ea D ez1z2; ha D i; ia D h; ya D y;

db D d; eb D ez1; hb D h;

d c D e; ec D d; hc D h; ic D iz2; yc D y:

�62

Groups with large normal closuresof nonnormal cyclic subgroups

Let H be a nonnormal subgroup of a p-group; then jG W HG j � p, where HG is anormal closure ofH in G. Therefore, it is natural to classify the p-groups G such thatjG W HG j D p for all nonnormal subgroups H < G. Such G are classified by thesecond author but all proofs in this section are due to the first author.

Theorem 62.1 (Z. Janko). Let G be a non-Dedekindian p-group. Suppose that jG W

HG j D p for all nonnormal cyclic subgroups H of G. Then one and only one of thefollowing holds:

(a) jGj D p3.

(b) G D ha; b j ap2

D bp2

D 1; ab D a1Cpi is the unique nonabelian metacyclicgroup of order p4 and exponent p2.

(c) G is a 2-group of maximal class.

(d) p D 2, G D H Z is a semidirect product with cyclic kernel Z of order> 4 and a cyclic complement H of order 4. Write R D �1.H/. We haveZ.G/ D R � �1.Z/ Š E4, G=R is of maximal class not isomorphic withgeneralized quaternion group, and G=G0 is abelian of type .4; 2/. We haveG D

hh, z j h4 D z2n

D 1, n > 2, zh D z�1C�2n�1

, � D 0; 1i.

It is trivial that groups (a), (b) and (c) of Theorem 62.1 satisfy the hypothesis. Forexample, if N G G is of index > 2, where G is 2-group of maximal class, then N �

ˆ.G/, and this implies that groups of (c) satisfy the hypothesis.Now let G be as in Theorem 62.1(d) and let X < G be nonnormal cyclic. Let us

prove that jG W XGj D 2. It follows from �1.G/ D Z.G/ that jX j > 2. The groupG has no proper subgroups which are generalized quaternion (Proposition 10.19(a))hence �1.G/ < XG (otherwise, XG is cyclic so X G G). If �1.H/ D R < X ,thenX=R is a nonnormal cyclic subgroup of the 2-group of maximal classG=R (sincej.G=R/ W .G0=R/j D 4) so jG W XG j D 2, by (c). Now we assume that R 6� X .Suppose that XR D X �R 6E G; then XR=R 6E G=R so jG W .XR/Gj D 2. We have.XR/G D XGR D XG since R < �1.G/ < XG , so jG W XGj D 2. Now assumethat XR G G. Then XR=R � T=R, where T=R is a cyclic subgroup of index 2 inG=R. We have T D V � R and ˆ.V / D ˆ.T / G G so all cyclic subgroups of T of


order 2k for all k < n, where jV j D 2n (n > 2), are G-invariant, so jX j D 2n; thenjG W X j D 4. In that case, XG D T so jG W XGj D 2, as required.

Lemma 62.2. Let G be a non-Dedekindian minimal nonabelian p-group of order >p3. If jG W HG j D p for all nonnormal H < G, then G is the unique nonabelianmetacyclic group of order p4 and exponent p2 which we denote Hp;2.

Proof. Let H < G be nonnormal cyclic; then HG D H � G0 so jG W H j D p2. Ifthe nonmetacyclic G is as in Lemma 65.1(a), then o.a/ D o.b/ D p so jGj D p3,contrary to the hypothesis. Thus, G is metacyclic as in Lemma 65.1(b) so G D B A,where A D hai GG, B D hbi and jAj D p2, by the above. Assume that n > 2. Write�2.B/ D hxi, F D hxai; then jF j D p2. We have .xa/b D xa1Cp 62 F so F is notG-invariant; then FG D F �G0 has index pn�1 > p in G, a final contradiction.

Proof of Theorem 62.1. LetH < G be nonnormal cyclic. SinceHG � Hˆ.G/ < G,we get jG W Hˆ.G/j D p so H 6� ˆ.G/ and d.G/ D 2. Hence all cyclic subgroupsof ˆ.G/ are G-invariant so ˆ.G/ is Dedekindian. Since ˆ.G/ has no G-invariantsubgroups isomorphic to Q8 (Lemma 1.4), it is abelian. If jG0j D p, thenG is minimalnonabelian (Lemma 65.2(a)) so G is as in Theorem 62.1(b), by Lemma 62.2. Nowsuppose that jG0j > p so G is not minimal nonabelian.

(i) Let ˆ.G/ be cyclic. If p > 2, then ˆ.G/ D Z.G/ and jG0j D p (Theorem 4.4),a contradiction. Thus, p D 2; then ˆ.G/ D Ã1.G/ so G has a cyclic subgroup ofindex 2 hence G is of maximal class. Now suppose that ˆ.G/ is noncyclic.

(ii) Let p > 2. Suppose that ˆ.G/ D U � V and U > f1g is cyclic (by (i),V > f1g and, by the above, U;V are G-invariant). Take x 2 G. By induction,Œx;ˆ.G/� � U \ V D f1g so ˆ.G/ � Z.G/. Then G is minimal nonabelian, acontradiction.

(iii) Thus, p D 2 and ˆ.G/ is noncyclic abelian. Assume that G is not metacyclic.Then, by Schreier inequality (see Appendix 25) and Corollary 36.6, there isM 2 �1

such that d.M/ D 3. Write NG D G=ˆ.M/; then NG of order 24 is neither abelian (inview of d.G/ D 2 < d.M/) nor of maximal class so there is NL < NM of order 2 suchthat NL 6� Z. NG/. Since NG is not of maximal class, we get j NG0j D 2, by Taussky, soNL

NG D NL � NG0 has index 4 in NG, a contradiction (our hypothesis is inherited by non-Dedekindian epimorphic images of G). Thus, G is metacyclic so G=Z is cyclic for acyclicZGG. SinceG is not of maximal class, we get�1.G/ Š E4 so�1.G/ � Z.G/in view of G=�1.G/ > 2. Since G is not of maximal class, we get jG=G0j � 8

(Taussky). Let R < G0 be of index 2; then G=R is minimal nonabelian and non-Dedekindian since d.G=R/ D 2 and jG=R/ > 8 (Lemma 62.2). Therefore, by Lemma62.2, G=R Š H2;2 soG=G0 is abelian of type .4; 2/. It follows that jG W Zj D 4 sinceZ < G0.

LetM < G be minimal nonabelian; then jM j > 8 (Proposition 10.19) so M is asin Lemma 65.1(b); thenM D H1 Z1 is a semidirect product with cyclic factors Z1

�62 Groups with large normal closures of nonnormal cyclic subgroups 171

andH1 andH1 is not normal inM so in G. However,M 0 D �1.Z1/ < G0 and G0 is

cyclic soH1 \G0 D f1g. Since 2 D jG W HG1 j, we getHG

1 D H1 G0; then jH1j D 4

since�1.G/ � Z.G/, andG=G0 is abelian of type .4; 2/,H1\Z D f1g sinceG0 < Z

and Z is cyclic. It follows that G D H1 Z, a semidirect product. Since Ã1.H1/ GG

as a subgroup of ˆ.G/, then A D Ã1.H1/ � Z is abelian of index 2 in G so 4 D12jG W G0j D jZ.G/j (Lemma 1.1). Since jG=Ã1.H1/ W .G=Ã1.H1//

0j D 4, the groupG=Ã1.H/ is of maximal class (Taussky); moreover, it is not generalized quaternionsince G=�1.G/ D .H1�1.G/=�1.G// .Z�1.G/=�1.G// is a semidirect product.Next, jZj > 4 (otherwise, G is as in (b)). If G=�1.G/ is generalized quaternion, thenj�2.G/j D 8 so G is a group of Lemma 42.1(c); then Z.G/ is cyclic, a contradiction.Thus, G is as stated in (d).

�63

Groups all of whose cyclic subgroupsof composite orders are normal

Definition 1. If all cyclic subgroups of a p-group G of composite orders are normal,then G is said to be a Kp-group.

The property Kp is inherited by sections. Below we classify Kp-groups and provethat the property (Kp) is equivalent to the following condition: Whenever f1g < B <

A � G, where A is cyclic, then NG.B/ D NG.A/ [Kaz1]. All proofs, apart of theproof of Supplement to Theorems 63.1 and 63.3, are due to the first author.

Proposition 63.1. If G is a nonabelian Kp-group, p > 2, and exp.G/ > p, thenjG0j D p.

Proof. Let Z < G be cyclic of order p2 and C0 D �1.Z/; then C0 G G. Assumethat K=C0 < G=C0 is nonnormal of order p; then K is abelian of type .p; p/. SetH D KZ; then jH j D p3. Since p > 2, H has exactly p cyclic subgroups of orderp2 and they generateH soH E G. Then �1.H/ D K GG, a contradiction.

Lemma 63.2. Let a 2-group G be a K2-group. Then:

(a) If L GG is of order 2 and G=L Š Q8, then L is a direct factor of G.

(b) If T < G is cyclic of order 2e , e > 1, then G=T 6Š Q8.

Proof. (a) By Theorem 1.2, G has no cyclic subgroups of index 2 so L 6� G0, byTaussky’s theorem. Then, if Z=L < G=L is cyclic of order 4, then Z is abelian oftype .4; 2/ and �1.Z/ D Z.G/ so all subgroups of order 2 are normal in G, and weconclude that G is Dedekindian. Then, by Theorem 1.20, L is a direct factor of G.

(b) Assume that G=T Š Q8. We use induction on jGj. Let L < T be of order 2;then jT=Lj D 2, by induction, so, by (a),G=L D .T=L/�.Q=L/, whereQ=L Š Q8.By (a), Q D L � Q1, where Q1 Š Q8. Since Q1 Š Q8 is generated by cyclicsubgroups of order 4, it is normal in G so G D T �Q1. Since all subgroups of order2 are normal in G, G is Dedekindian, contrary to Theorem 1.20.

Recall that the Hp-subgroup Hp.G/ of a groupG is defined as hx 2 G j o.x/ ¤ pi.By Burnside, a nontrivial H2-subgroup has index 2 in G. If G is a (Kp)-group, thenHp.G/ centralizesG0. Note that if jG W H2.G/j D 2, then exp.Z.G// D 2.

�63 Groups all of whose cyclic subgroups of composite orders are normal 173

Theorem 63.3 ([Kaz1]). If G is a nonabelian K2-group, then either (a) jG0j D 2 or(b) H2.G/ has index 2 in G.

Proof. Suppose that G is a counterexample of minimal order. As we noticed, H2.G/

centralizes G0. Since jG W H2.G/j � 2, G0 < H2.G/. If jG W H2.G/j D 2, wehave case (b). Now we assume that G D H2.G/. Then, by what has just been said,cl.G/ D 2.

(i) Let jG0j > 4. If C < G0 is of order 2, then, by induction, G=C has an abeliansubgroup A=C of index 2 such that all elements in .G=C / � .A=C / are involutionssince j.G=C /0j D jG0=C j > 2. Let x 2 G � A; then x2 2 A. Assume that o.x/ > 2;thenX D hxiGG. In that case, jXC=C j D 2 so x2 2 C andG=C D .A=C /�.X=C /

is abelian so C D G0 is of order 2, which is not the case. Thus, if jG0j > 4, we obtaincase (b). It remains to consider the case where G0 is either cyclic of order 4 or abelianof type .2; 2/.

(ii) Let G0 D hci Š C4. We have c D Œx; y� for some x; y 2 G. Set H D hx; yi;then H 0 D G0 D hci is a central subgroup of order 4. We see that (a nonabelianK2-group)H of composite exponent is not a group from the conclusion (indeed, sinceexp.Z.H// > 2, H coincides with its H2-subgroup; in addition, jH 0j D 4 > 2);therefore, H D G, by induction. If x; y are involutions, then G is dihedral of class 2so jG0j D 2, a contradiction. Therefore, one may assume that o.x/ > 2; in that case,G D hxihyi is metacyclic. Next, G has no cyclic subgroups of index 2 (otherwise,sinceG is of class 2, jG0j D 2). Then�1.G/ is of type .2; 2/ andG has no nonabeliansubgroups of order 8 (this follows from Proposition 10.19). Therefore, if K=�1.G/

is a subgroup of order 2 in G=�1.G/ then K is abelian of type .4; 2/ so it is normalin G since K D hy j o.y/ D 4i. It follows that G=�1.G/ is nonabelian Dedekindian(nonabelian since G0 is cyclic of order 4 > 2). Then G=�1.G/ Š Q8 since it ismetacyclic (Theorem 1.20). By Taussky’s theorem, G=G0 is abelian of type .4; 2/.Since G0 � Z.G/, G is minimal nonabelian. In that case, jG0j D 2, a contradiction.

(iii) It remains to consider the case where G0 Š E4. Let Z < G be cyclic of order2e D exp.G/; then G=Z is nonabelian since G0 — Z. By Lemma 63.2(b), G=Z hasno subgroups isomorphic to Q8. It follows from Theorem 1.20 that G=Z contains anonnormal subgroup L=Z of order 2. Since L 6E G, the subgroup L is not generatedby cyclic subgroups of composite orders so L is dihedral, by Theorem 1.2 (indeed,jL W Zj D 2 and Z is cyclic). Since L0 is cyclic and exp.G0/ D 2, we get jL0j D 2

hence jLj D 8, jZj D 4 and e D 2. Since G D H2.G/ is nonabelian, there aretwo cyclic subgroups, say U and V , of order 4 that generate a nonabelian subgroupQ D UV . In that case, Q Š Q8 and Q G G. Since G0 6� Q, the quotient groupG=Q is nonabelian so it contains a cyclic subgroup M=Q of order 4 .D exp.G//.Set M D hx;Qi; then X D hxi is of order 4 and M D Q � X . Since M is notDedekindian (Theorem 1.20) and all subgroups of order 2 are normal inM , it containsa nonnormal cyclic subgroup of order 4, a final contradiction.


Supplement to Theorems 63.1 and 63.3 ([Kaz1]). Let G be a (Kp)-group. If thereholds exp.G/ > p and jG0j D p, then ˆ.G/ is cyclic.

Proof (Kazarin, personal communication). Assume that ˆ.G/ is noncyclic; then wehave jGj > p3 and exp.G/ � exp.G=G0/ > p.

(i) If x; y 2 G, then 1 D Œx; y�p D Œx; yp � so yp 2 Z.G/ andˆ.G/ D G0Ã1.G/ �

Z.G/.

(ii) Let C < G be cyclic of order > p. Assume that G0 6� C ; then G=C isnonabelian, C \ G0 D f1g so ŒC;G� � C \ G0 D f1g and hence C � Z.G/.By Lemma 63.2(b), G=C has no subgroups isomorphic to Q8 so it is not Dedekindian.Therefore, there is a nonnormal B=C < G=C of order p; B is abelian of type .jC j; p/

since C � Z.G/. Then B GG since B D hx 2 B j o.x/ > pi, contrary to the choiceof B . Thus, every cyclic subgroup of G of composite order contains G0.

(iii) Let G be minimal nonabelian. Then G=G0 D haG0i � hbG0i is abelian oftype .pm; pn/ with m � n; then m > 1. Let o.aG0/ D pm, o.bG0/ D pn. By(ii), G0 � hai so G0 is not a maximal cyclic subgroup of G so G is metacyclic andG D hbi hai (Lemma 65.1). By (ii), n D 1 so ˆ.G/ is cyclic.

(iv) If x; y 2 G are elements of composite orders, thenH D hx; yi is either abelianor minimal nonabelian and in both cases it contains a cyclic subgroup of index p.Indeed, suppose that H is abelian. Let o.x/ � o.y/. Then H D hxi � hzi so, by(ii), o.z/ D p, i.e., H has a cyclic subgroup hxi of index p. Now suppose that His nonabelian. Since hxi G G, H is metacyclic. Since ˆ.H/ � Z.H/, by (i), andH=ˆ.H/ Š Ep2 , H is minimal nonabelian. By (iii), H has a cyclic subgroup ofindex p so ˆ.H/ is cyclic. Thus, in any case, hxp ; ypi is cyclic for all x; y 2 G.

Let x 2 G be of maximal order. Then, by (iv), Ã1.G/ � hxi so, by (ii), ˆ.G/ D

Ã1.G/G0 � hxi whence ˆ.G/ is cyclic.

Thus, Theorems 63.1 and 63.3 can be formulated as follows.

Theorem 63.4 ([Kaz1]). If G is a Kp-group, then one of the following holds:

(a) G is abelian,

(b) exp.G/ D p,

(c) jG0j D p and ˆ.G/ is cyclic,

(d) p D 2 and jG W H2.G/j D 2.

The above four groups satisfy the hypothesis.

Definition 2. A p-group G is said to be an Np-group if every cyclic subgroup A ofcomposite order satisfies the following condition: Whenever f1g < B < A, thenNG.B/ D NG.A/. (Clearly, subgroups of Np-groups areNp-groups.)

Theorem 63.5 ([Kaz]). Let G be a nonabelian p-group, exp.G/ > p > 2. Then thefollowing assertions are equivalent:


(a) G is an Np-group,

(b) G is a Kp-group.

Proof. Obviously, (b) implies (a). Now suppose that an Np-group G is not a Kp-group. Then G has a nonnormal cyclic subgroup A of composite order so, in view ofTheorem 1.2, G has no cyclic subgroups of index p. By hypothesis,

(i) Every nonidentity subgroup of A is not normal in G, i.e., AG D f1g.

Suppose that G is a counterexample of minimal order. In that case, every propersubgroup of G of composite exponent is a Kp-group so the normalizer of every non-normal cyclic subgroup ofG of composite order is maximal inG since this normalizeris an Kp-group; thus N D NG.A/ 2 �1. Then, by Proposition 63.1, jN 0j � p. LetM 2 �1 � fN g; then A 6� M (otherwise, A is normal inM so in G). If jA\M j > p,then, by induction, NG.M \A/ � MN D G so AG � A\M > f1g, contrary to (i).Thus jM \ Aj D p so jAj D p2. Thus

(ii) Let A be a nonnormal cyclic subgroup of G of composite order. Then jAj D p2

and A is contained in exactly one maximal subgroup of G.

Suppose that jG0j D p. Then G0 6� A so G0A D G0 � A G G. In that case,ˆ.G0 �A/ > f1g is contained inA and normal inG, contrary to (i). Thus, if jG0j D p.the theorem is true. In what follows we assume that

(iii) jG0j > p.

(iv) G has at most one abelian subgroup of index p. Indeed, otherwise, we havejG W Z.G/j D p2 so jG0j D p, by Lemma 1.1.

Let R � N be of composite exponent. Since jN 0j � p and p > 2, R is regular.It follows that R is generated by cyclic subgroups of composite orders so R is nor-mal in N (by induction, N is a Kp-group). It follows that N=A is abelian since it isDedekindian and p > 2 so N 0 < A. By (i), N 0 D f1g so N is abelian. Thus, by (iv),

(v) N is the unique abelian maximal subgroup of G. Moreover, the normalizer ofany nonnormal cyclic subgroup of composite order is an abelian maximal subgroup ofG so coincides with N .

It follows that

(vi) All nonnormal cyclic subgroups of orders > p in G are contained in N .

Let B G G be cyclic of order > p. Assume that B 6� N , or, what is the same,B does not normalize A; then B G G, by (vi). Then, AB is not a Kp-group so it isnot an Np-group and we get, by induction, AB D G. Thus, G is metacyclic. By (i),A \ B D f1g, and, obviously, CN .B/ D Z.G/ (recall that N is the unique abelianmaximal subgroup of G). Note that jB W .B \ N/j D p and B \ N � Z.G/.Since AG D f1g, we get Z.G/ < B so B \ N D Z.G/. Then G=Z.G/ is abelianof type .p2; p/. We have Z.G/ D ˆ.B/ � ˆ.G/. Let U=Z.G/ and V=Z.G/ bedistinct cyclic subgroups of order p2 in G=Z.G/. Then U and V are distinct abelianmaximal subgroups of G, contrary to (v). Thus, B < N so all cyclic subgroups of


composite orders are contained in N , i.e., N D Hp.G/. This is a contradiction sincehG �N i D hG � Hp.G/i D G. Thus,

(vii) All cyclic subgroups of G of composite orders are contained in N D Hp.G/.

Let B G G be normal cyclic of composite order. Take x 2 G � N and set H D

hx;Bi; then H is metacyclic. It follows from p > 2 that H is generated by its cyclicsubgroups of maximal order soH � N , by (vii), contrary to the choice of x. Thus,

(viii) All cyclic subgroups of composite orders are not normal in G. We see that thenormalizer of every cyclic subgroup of composite order is abelian so coincides withN . It follows that N is the unique maximal subgroup of G of composite exponent, by(ii). Moreover,�2.N / D N , by (ii), so exp.G/ D p2.

Let M 2 �1 � fN g; then exp.M/ D p, by (viii). In that case N contains theelementary abelian subgroup N \ M of index p. Then N D A � E, where E iselementary abelian. In that case,ˆ.N/ D ˆ.A/ > f1g is characteristic inN so normalin G, contrary to (i). Thus, all cyclic subgroups of composite orders are normal in Gso G is a Kp-group.

Theorem 63.6 ([Kaz1]). Let G be a nonabelian 2-group, and let exp.G/ > 2. Thenthe following assertions are equivalent:

(a) G is a N2-group,

(b) G is a K2-group.

Proof. Clearly, (b) implies (a). It remains to prove the reverse implication. SupposethatG is a counterexample of minimal order. ThenG has a nonnormal cyclic subgroupB D hbi of composite order. As above,

(�) jX j D 4 and XG D f1g, where X < G is nonnormal cyclic of order > 2.

Let B < N , where jG W N j D 2. Then N is a K2-group, by induction, so N D

NG.B/ and N is the unique maximal subgroup of G containing B . Let a 2 G � N

be such that o.a/ is as large as possible; then Ba ¤ B . If �1.Ba/ D �1.B/, then

�1.B/a D �1.B/ and NG.�1.B// � hN;ai D G, contrary to (�). Thus,�1.B

a/ ¤

�1.B/ so B \ Ba D f1g. Since B;Ba � N , these two subgroups are normal inN so H D BBa D B � Ba is abelian of type .4; 4/. Since B 6E ha;Bi, we getG D ha;Bi D ha;H i, by induction. Set A D hai; then H G G since a2 2 N ,G D AH , G=H is cyclic. Since H is abelian, A \H G G. If o.a/ D 2, then G is aK2-group (indeed, then H is an H2-subgroup of G). In what follows we assume thato.a/ > 2.If A \ H > f1g, then A G G since A \ H G G (see (�)), and so, by induction,

G D BA is a semidirect product with kernel A since B \ A D f1g, by (�). In thatcase,G D B A is metacyclic.Suppose that A \ H D f1g. Then A is not normal in G since G is nonabelian

so CG.H/ D H , by (�). In that case, the abelian subgroup NG.A/ D A � Z.G/ is


maximal in G, by induction, so Z.G/ is a subgroup of index 2 inH ; then B \Z.G/ >f1g, contrary to (�). Thus, it remains to consider the following two cases.

(i) Let G D B A, a semidirect product with kernel A D hai Š C2n (n > 1),B D hbi Š C4. As above, N D NG.B/, a maximal subgroup of G, is abelian oftype .22; 2n�1/, CG.A/ D A and N \ A D Z.G/ has index 2 in A. Then G=Z.G/is abelian of type .4; 2/ so G is minimal nonabelian. In that case, Z.G/ D ˆ.G/ andf1g < B \ Z.G/ � BG , contrary to (�).

(ii) Let G D A H , a semidirect product with abelian kernel H of type .4; 4/,A D hai Š C4, H D B � Ba, where B D hbi Š C4. Set NG.A/ D M ; thenjG W M j D 2, M \H D Z.G/ is abelian of type .4; 2/ so BG � B \ Z.G/ > f1g,contrary to (�).

It follows from the above that epimorphic images of Np-groups are Np-groups orhave exponent � p.Let G be a p-group with exp.G/ > p. If every subgroup of composite exponent

is normal in a p-group G, then either jG0j � p or G Š D24 [Man18]. Obviously, Gis a Kp-group. Suppose that jG0j > p. Then p D 2 and A D H2.G/ has index 2 inG, by Theorem 63.4. If A is cyclic, we are done, by Theorem 1.2. Assume that A isnot cyclic. Since G0 > f1g, we get exp.A/ > 2. Let Z be a cyclic subgroup of order4 in A. By hypothesis, all subgroups are normal in G=Z, i.e., G=Z is Dedekindian.By Lemma 63.2(b), G=Z has no subgroups isomorphic to Q8 so G=Z is abelian. LetZ1 ¤ Z be another cyclic subgroup of order 4 in A (see Theorem 1.17(b)). ThenG=Z1 is abelian, by what has just been proved. It follows that G0 � Z \ Z1, sojG0j � 2.

Proposition 63.7 ([Li3] (compare with Theorem 58.5)). Suppose that all nonnormalcyclic subgroups of a nonabelian p-group G, p > 2, are conjugate. Then G Š Mpn .

Proof. Let T < G be a nonnormal cyclic subgroup of G. Set jT j D pt . LetK be theset of all nonnormal cyclic subgroups of G; then jKj is a power of p.

(i) Let t D 1.

(i1) Assume that exp.G/ D p. Set jGj D pm, jZ.G/j D pz . Then jKj Dpm�pz

p�1D pz.1C pC C pm�z�1/. Since m� z � 1 > 0, jKj is not a power of p

so not all members of the setK are conjugate in G, a contradiction.

(i2) Thus, exp.G/ > p. Then G is a Kp-group so, by Theorem 63.1, jG0j D p, andG is regular since p > 2. Let j�1.G/j D pk and j�1.Z.G//j D p. Then the number

of nonnormal subgroups of order p inG equals pk�p�

p�1 D p.1CpCpk��1/, whichis a power of p, so k D C 1. Since jG0j D p, it follows that there are exactly psubgroups conjugate with T inG. It follows that D 1 so k D 2. Then jG=Ã1.G/j D

j�1.G/j D p2 so G is metacyclic. It follows that G is minimal nonabelian (Lemma65.2(a)). Then, by Lemma 58.1, G Š Mpn .


(ii) Now let t > 1. Set T0 D �1.T /; then T0 � TG G G so T0 is contained inall nonnormal cyclic subgroups of G. Set NG D G=T0. Let NU < NG be nonnormalcyclic. If U is noncyclic, then U D T0 � U0 and U0 is not normal in G. ThenT0 — U0, contrary to what has just been said. Thus, U is cyclic so conjugate with T .It follows that all nonnormal cyclic subgroups of NG are conjugate, i.e., NG satisfies thehypothesis. Therefore, by induction, NG Š Mpn . We also have T0 � ˆ.T / � ˆ.G/

so d.G/ D d. NG/ D 2. If A=T0; B=T0 < G=T0 be two distinct cyclic subgroups ofindex p, then A;B 2 �1 are distinct abelian so A \ B D Z.G/, and we conclude thatG is minimal nonabelian. By Lemma 58.1, G Š Mpn . However, all nonnormal cyclicsubgroups of Mpn have order p < pt , so case t > 1 is impossible.

Exercise 1. Suppose that all nonnormal cyclic subgroups of a nonabelian 2-group G,are conjugate. Then G Š M2n , i.e., Theorem 63.1 is also true for p D 2.

Exercise 2. Let a group G be nonabelian of exponent p. Then the number of nonnor-mal subgroups of order p in G is not a power of p.

There are in the book a number of results on cyclic subgroups of p-groups.

Problem. Classify the non-Dedekindian p-groups in which any two nonnormal cyclicsubgroups of the same order are conjugate.

�64

p-groups generated by elements of given order

1o. In what follows, G is a group of order pm, ��k.G/ D hx 2 G j o.x/ D pki,

where pk � exp.G/. Then ��k.G/ � �k.G/ and �

�k.G/ is the least subgroup H

of G subjecting ck.H/ D ck.G/. We haveP1

iD0 '.pi /ci .G/ D jGj, where '.�/ is

Euler’s totient function. Given k > 0, set solk.G/ D jfx 2 G j o.x/ � pkgj. Thensolk.G/ D

PkiD0 '.p

i /ci .G/. For definition of Ls- and U2-groups, see ��17, 18, 67.A 2-group G is said to be a Un-group, if there is E2n Š R G G such that G=R is

of maximal class with cyclic subgroup T=R of index 2 and �1.T / D R. It is easy toshow that R contains all normal elementary abelian subgroups of G and all elementsin the set G � T have orders � 8. A subgroup R is called the kernel of G.For the sake of completeness, we collected in Lemma 64.1 some known facts. We

use this lemma until end of the book.

Lemma 64.1. Let G be a p-group.

(a) (Theorems 7.1 and 7.2) If G is regular of exponent pe and k � e, then we haveexp.�k.G// D pk . All p-groups of class < p are regular. Groups of exponent p areregular. Regular 2-groups are abelian.

(b) (Theorem 12.1) If G has no normal subgroups of order pp and exponent p, itis either absolutely regular or of maximal class. If an irregular p-group G has anabsolutely regular maximal subgroup H , then G is either of maximal class or G D

H�1.G/, where j�1.G/j D pp .

(c) (Theorems 9.5 and 9.6) A p-group of maximal class and order > pp is irregular.A p-group of maximal class and order pm contains only one normal subgroup of orderpi for 1 � i � m � 2.

(d) (Theorem 9.6, Lemma 12.3) A p-group of maximal class and order > ppC1 hasno normal subgroups of order pp and exponent p. For suchG, we have c1.G/ 1C

pC C pp�2 .mod pp/ and c2.G/ pp�2 .mod pp�1/. If p2 < pk � exp.G/,then ck.G/ D 1 if p D 2 (see Lemma 64.10, below) and pp�1 divides ck.G/ if p > 2.

(e) (Theorem 9.6) An irregular p-group G of maximal class has an absolutely regularsubgroup G1 of index p such that j�n.G1/j D p.p�1/n for pn < exp.G1/ D exp.G/.If, in addition, jGj > ppC1, then Z.G1/ is noncyclic (G1 is called the fundamentalsubgroup of G); all other maximal subgroups of G are of maximal class. If 2 < k �

exp.G/, then ��k.G/ � G1.


(f) (Theorems 13.2 and 13.5) If G is neither absolutely regular nor of maximal class,then c1.G/ 1CpC Cpp�1 .mod pp/ and, for k > 1, ck.G/ 0 .mod pp�1/.The number of subgroups of order pp and exponent p in G is 1 .mod p/.

(g) (Theorem 13.19) If G is an irregular p-group of maximal class and H � G is oforder > pp , thenH is either absolutely regular or of maximal class.

(h) If A G G is of order pp�1 and exponent p, G is a p-group of maximal class andR < G is of order pp and exponent p, then A < R. Next, A is contained in everyirregular subgroup of G.

(i) (Proposition 10.17) If B � G is nonabelian of order p3 such that CG.B/ < B ,then G is of maximal class.

(j) (Corollary 18.7) Suppose that an irregular p-group G is not of maximal class,k > 2. Then ck.G/ 0 .mod pp/, unless G is an Lp- or U2-group. In particular, inthat case, solk.G/ 0 .mod ppC2/.

(k) Let C < H < G, where G is a 2-group which is not of maximal class, jH W C j D

2, C is cyclic. Then the number of subgroups of G of order jH j, that are of maximalclass, is even.

(l) (Theorems 41.1 and 66.1) If G is minimal nonmetacyclic, then one of the followingholds: (i)G is of order p3 and exponent p, (ii)G is of order 34 and class 3, j�1.G/j D

9, (iii) G D C2 �Q8, (iv) G D Q8 �C4 Š D8 �C8 of order 24, (v)G is special groupof order 25 with jZ.G/j D 22. Thus, if �2.G/ is metacyclic then G is metacyclic.

(m) (Lemma 44.1, Corollary 44.6, Theorem 9.11) A p-group G is metacyclic if one ofthe following quotient groups is metacyclic: G=ˆ.G0/K3.G/, G=ˆ.G0/, G=K3.G/.If p > 2 and jG=Ã1.G/ � p2, then G is metacyclic (Huppert).

(n) (Theorem 44.5) A 2-group G is metacyclic if G and all its maximal subgroups aretwo-generator.

(o) (Corollary 44.9) If G=Ã2.G/ is metacyclic then G is metacyclic.

(p) ([Bla5, Theorem 4.2]) If a p-group G, p > 2, and all its maximal subgroups aretwo-generator, then G is either metacyclic or Ã1.G/ D K3.G/ is of index p3 in G (inthe last case, jG W G0j D p2).

(q) (Tuan; see Lemma 1.1) If G is nonabelian and A < G is abelian of index p, thenjGj D pjG0jjZ.G/j.

(r) (Lemma 4.2). If jG0j D p, then G D .A1 � � As/Z.G/, where A1; : : : ; As areminimal nonabelian; then G=Z.G/ is elementary abelian.

(s) (Taussky; Proposition 1.6) If G is a nonabelian 2-group with jG W G0j D 4, thenG is of maximal class, i.e, dihedral, semidihedral or generalized quaternion.

(t) If a nonabelian p-group G has a cyclic subgroup of index p, then G is eitherMpn

or p D 2 and G is of maximal class (see (s)).

�64 p-groups generated by elements of given order 181

(u) (see Exercise 1.69) If U;V 2 �1 are distinct, then jG0 W U 0V 0j � p.

(v) (Lemma 1.4) LetM be a G-invariant subgroup of ˆ.G/. If Z.M/ is cyclic,M isalso cyclic. In particular,M cannot be nonabelian of order p3.

(w) (Exercise 9.13) If a p-group G of maximal class, p > 2, has a subgroup H withd.H/ > p � 1, then G is isomorphic to a Sylow p-subgroup of Sp2 .

(x) (�5) c1.G/ 1C p .mod p2/ and, for k > 1, ck.G/ 0 .mod p/, unless G iseither cyclic or a 2-group of maximal class. Under the last exceptions,G has a normalabelian subgroup of type .p; p/.

(y) If hU 0 j U 2 �1i < G0, then d.G/ D 2.

Let us prove parts (g), (h), (y) of Lemma 64.1.

Proof of Lemma 64.1(g). Let jGj > ppC1. If jGj D ppC2, the result follows fromTheorem 9.6. Now let jGj > ppC2 and H — G1, where G1 is the fundamentalsubgroup of G. Let H < M 2 �1; thenM.¤ G1/ is of maximal class (Lemma J(e))andM1 D M \ G1 D ˆ.G/ is the fundamental subgroup ofM since jM j > ppC1.Then, by induction on jGj, H is of maximal class sinceH 6� M1.

Proof of Lemma 64.1(h). We have A D �1.ˆ.G//. One may assume that jGj >

ppC1 (if jGj D ppC1, then A D ˆ.G/). Let R < G be either of order pp andexponent p or irregular. Since j�1.R \G1/j D pp�1, the result follows.

Proof of Lemma 64.1(y). Set D D hU 0 j U 2 �1i. Since, by hypothesis, D < G0,there exist x; y 2 G such that Œx; y� 62 D. Assume that H D hx; yi � M 2 �1; thenM 0 6� D, a contradiction. Thus, H D G.

2o. We begin with the following

Definition 1 (Ito). Let G be a group of order pm. Then G is called one-stepped ifthere exist m elements x1; : : : ; xm of order p in G such that jhx1; : : : ; xi ij D pi fori D 1; : : : ;m. We consider the identity group f1g as one-stepped.

Lemma 64.2. A p-group G is one-stepped if and only if �1.G/ D G.

Proof. We have to prove that if �1.G/ D G, then G is one-stepped. Let H � G

be one-stepped of maximal order. Assume that H < G. Set N D NG.H/. IfN � H has an element x of order p, then H1 D hx;H i is one-stepped of order> jH j, a contradiction. Thus, H D �1.N /, i.e., H is characteristic in N . ThenN D G > H D �1.G/ D G, a contradiction.

Lemma 64.3. Suppose that A < G are one-stepped p-groups and jG W Aj D pn.Then there are x1; : : : ; xn 2 G � A of order p such that jAi W Ai�1j D p, whereA0 D A, Ai D hA; x1; : : : ; xi i for i D 1; : : : ; n.


Proof. Suppose that Ai has constructed. Assuming that Ai < G, we have to constructAiC1. Set Ni D NG.Ai /. As in the proof of Lemma 64.2, there is xiC1 2 Ni �Ai oforder p. Set AiC1 D hxiC1; Ai i. Then An D G.

A series A D A0 < A1 < < An D G, constructed in Lemma 64.3, we call theIto series or (Ito 1-series; see below) connecting A D A0 with G.

3o. In this subsection we give some estimates of the order of a p-groupG D ��k.G/

in terms of ck.G/. For p-groups of maximal class and order > ppC1 these estimatesare best possible.Set �.G/ D Œ 1

pp c1.G/� and, for k > 1, set k.G/ D Œ 1pp�1 ck.G/�.

Theorem 64.4. Suppose that a one-stepped p-groupG is not a group of maximal class.Then jGj � ppC1C�.G/.

Proof. We use induction on jGj. One may assume that jGj D pm > ppC1 (other-wise, there is nothing to prove). If G is regular, then exp.G/ D p (Lemma 64.1(a)),c1.G/ D 1C p C C pm�1, and we have to check only that

m � 1C p C

�

1

pp .1C p C C pm�1/

�

D 1C p C .1C p C C pm�p�1/:

This is true since 1C p C C pm�p�1 � 1Cm � p � 1 D m � p.Now let G be irregular. Then there is R G G of order pp and exponent p (Lemma

64.1(b)) and x 2 G � R of order p. Set A D A0 D hx;Ri. Let A D A0 < A1 <

< An D G be an Ito series connecting A D A0 with G. Then jGj D jA0j pn D

ppC1Cn so it suffices to prove that n � �.G/. For i D 0; 1; : : : ; n � 1 we havec1.AiC1/ > c1.Ai/ and, since AiC1 is not of maximal class in view of R G Ai andjAiC1j > ppC1 (Lemma 64.1(d)), we get c1.AiC1/ c1.Ai / .mod pp/ (Lemma64.1(f)), so

c1.G/ � c1.A0/C n pp � .1C p C C pp�1/C n pp :

However, by Lemma 64.1(f), c1.G/ D 1CpC Cpp�1C�.G/pp so n � �.G/.

Corollary 64.5. If a p-group G is not of maximal class, then j�1.G/j � ppC1C�.G/

so, if p D 2 and c1.G/ D 4k C 3, then j�1.G/j � 23Ck .

Let G be one-stepped of maximal class and order pm > ppC1 and let G1 < G bethe fundamental subgroup. Then R D �1.G1/ G G is of order pp�1 and exponentp. Let x 2 G � R be of order p; then H D hx;Ri is of order pp and exponent p(Lemma 64.1(a)). By Lemma 64.1(g), NG.H/ is of maximal class and order ppC1

so there are exactly jG W NG.H/j D pm�p�1 conjugates with H in G. All theseconjugates contain exactly

c1.R/C .c1.H/ � c1.R//pm�p�1 D 1C p C C pp�2 C pm�2


distinct subgroups of order p and they generateM 2 �1 (take into account the aboveformula!). Since G is one-stepped, there is y 2 G � M of order p. Setting H1 D

hy;Ri, we, as above, obtain at least pm�2 new subgroups of order p that together withR generateM1 2 �1 � fM g. Let �G be the number of one-stepped members in theset �1. Then (see Lemma 12.3) we get

1C p C C pp�2 C �.G/ pp D c1.G/ � �G pm�2:

It follows that pm � ppC2

�G �.G/. Since G is one-stepped, p � �G > 1 so we obtain

Theorem 64.6. Let G be a one-stepped p-group of maximal class and order > ppC1.Then jGj � ppC1 �.G/ or, what is the same. jGj � p c0

1.G/, where c01.G/ is the

number of subgroups of order p not contained in G1.

Definition 2. A p-group G is said to be k-stepped if it has elements x1; : : : ; xt , all oforder pk , such that G D hx1; : : : ; xt i and jAi W Ai�1j > 1, where Ai D hx1; : : : ; xi i

and xi 2 NG.Ai�1/ for i D 1; : : : ; t . If ��k.A/ D A D A0 < G and there are

elements x1; : : : ; xn, all of order pk , such that, denoting Ai D hA0; x1; : : : ; xi i, wehave xi 2 NG.Ai�1/ and An D G, then the series A0 < A1 < < An D G is saidto be an Ito k-series connecting A0 with G.

If A0 < A1 < < An D G is an Ito k-series connecting A0 with G, thenjG W A0j � pkn. As in Lemma 64.2, G is k-stepped if and only if ��

k.G/ D G. If

A D A0 is a k-stepped subgroup of a k-stepped p-group G, then there exists an Itok-series A0 < A1 < < An < G connecting A0 with G.

Lemma 64.7. LetRGG be of order pp and exponent p and letG=R be cyclic of order> p. Then exp.�j .G// D pj for all j with pj � pe D exp.G/. Next,��

e .G/ D G,i.e.,G is e-stepped. If G=R is of order pe, then ce.G/ D pp . If G=R is of order pe�1

(in that case �1.G/ D R), then ce.G/ D pp�1.

Proof. It suffices to show that exp.�1.G// D p. However, this follows from Lem-ma 17.4(c).

Let k > 2 and suppose that a p-group G D ��k.G/ is not absolutely regular (by

Theorem 13.19, G is not of maximal class); then there is R G G of order pp andexponent p (Lemma 64.1(b)). Take in G an element x0 of order pk and set A0 D

hx0; Ri; then ppCk�1 � jA0j � jRjpk D ppCk . By Lemma 64.7, A0 is k-stepped.Let A0 < A1 < < An D G be an Ito k-series connecting A0 with G. Sinceck.A0/ � pp�1 and ck.A1/ > ck.A0/ � pp�1, A1 is not an Lp-group. Next, ifp D 2 and k > 3, then A1 is not a U2-group. Then, if p > 2, then ck.A1/ � pp

(Lemma 64.1(j)), and this is also true for p D 2. It follows that ck.AiC1/ > ck.Ai/,and, by Lemma 64.1(j), ck.AiC1/ ck.Ai / .mod pp/, i D 1; : : : ; n� 1, so we havek.G/p

p�1 D ck.G/ � ck.A1/ C .n � 1/pp � npp , hence k.G/ � np, and we

conclude that jGj � jA0jpkn � ppCkCk

h

1p

�k.G/i

. Thus, we have


Theorem 64.8. Suppose that k > 2 and a k-stepped p-group G of order pm is neitherabsolutely regular nor of maximal class. Then m � p C k C k

pk.G/.

Corollary 64.9. Let k > 2 and let a p-group G be neither absolutely regular nor of

maximal class. Then j��k.G/j � p

pCkCk�h

1p

�k.G/i

.

The case k D 2 we consider for p-groups of maximal class only.Let G D ��

k.G/ be a p-group of maximal class, k > 1, jGj D pm � p2p and

p > 2; then k D 2 (Lemma 64.1(e)) and exp.G/ > p2. Let c02.G/ be the number

of cyclic subgroups of order p2 in G not contained in G1. Set R D �1.G1/, whereG1 < G is fundamental; then jRj D pp�1. Let x 2 G � G1 be of order p2. SetA D hx;Ri; then A is absolutely regular of order pp since xp 2 Z.G1/ (Theorem13.19). Set N D NG.A/. Since N 6� G1 and jN j > pp , N is of maximal class(Lemma 64.1(g)). Since A 6E G, we get N < G so A is not characteristic in Nand hence ��

2.N / > A. Then jN W Aj D p (every normal subgroup of index > p

in N is characteristic) so there are exactly jG W N j D pm�p�1 subgroups conjugatewith A in G; their intersection equals R so their contribution in c02.G/ is equal topm�p�1c2.A/ D pm�p�1pp�2 D pm�3. These G-conjugates of A generate M 2

�1 (every normal subgroup of index� p2 in G is contained in ˆ.G/ < G1). We haveG1 \ M D ˆ.G/. Since m � 2p, we get �2.G1/ � ˆ.G/. Since G is 2-stepped,there is y 2 G � M of order p2; by the above, also y 2 G � G1. Set A1 D hy;Ri;then A\A1 D R. As above, the G-conjugates of A1 generateM1 2 �1 � fM g; theirintersection equals R again. Let G be the number of 2-stepped members in the set

�1. Then c02.G/ D G pm�3 so jGj D pm D p3

G c0

2.G/. If p D 2, then G Š Q2m

since ��2.G/ D G, and equality is attained. Thus we have

Supplement to Theorem 64.8. If G is a 2-stepped p-group of maximal class and

order � p2p , then jGj D p3

G c0

2.G/, where G is the number of 2-stepped membersin the set �1.

Theorem 64.10. Let a p-group G, c1.G/ D 1C p C p2 and exp.�1.G// > p. Thenp D 2, �1.G/ D D8 � C2 is of order 24 and one of the following holds:

(a) G D D � C , where D8 Š D, C is cyclic of order � 4, D \ C D Z.D/.

(b) G D DC , whereD8 Š D GG, C is nonnormal cyclic of index 4 inG,D\C D

Z.D/ and CG.D/ D Z.G/ has index 2 in C , G=CG.D/ Š D8.

(c) G D DQ, where D8 Š D G G, Q is a nonnormal generalized quaterniongroup of index 4 in G, D \ Q D Z.D/, CG.D/ is a cyclic subgroup of index2 in Q, G=CG.D/ Š D8. If L is a cyclic subgroup of order 4 in Q such thatL 6� CG.D/, thenDL Š SD24 .

Proof. Let p > 2. Let Ep2 Š R GG and let x 2 G �R be of order p in G �R; thenH D hx;Ri is of order p3 and exponent p. Since c1.G/ D c1.H/, �1.G/ D H isof order p3 and exponent p, a contradiction. For p D 2, the assertion coincides withTheorem 43.9.


4o. In this subsection we study p-groups G with small ck.G/.

Proposition 64.11. Let G be a p-group, p > 2, with �1.G/ D G, c1.G/ D 1C p C

C pp and exp.G/ > p. Then:

(a) G is irregular of order ppC2; all members of the set �1 have exponent p2.

(b) d.G/ D 3, ˆ.G/ D G0 is of order pp�1 and exponent p, cl.G/ D p.

(c) G=Ã1.G/ is of order ppC1 and exponent p, Ã1.G/ D Kp.G/.

(d) c2.G/ D pp.

(e) �1 D fM1; : : : ;Mp2 ; T1; : : : ; TpC1g, whereM1; : : : ;Mp2 are of maximal class,T1; : : : ; TpC1 are not generated by two elements so regular and have expo-nent p2. Next, �.G/ D

TpC1iD1 Ti has exponent p2 and index p2 in G, where

�.G/=K3.G/ D Z.G=K3.G//.

(f) G has exactly pC 1 subgroups of order pp and exponent p, all those subgroupscontain ˆ.G/ so normal in G.

(g) Exactly p2 subgroups of G of order pp , containing ˆ.G/, have exponent p2.

(h) If L is a subgroup of order pp and exponent p in G, then exactly p maximalsubgroups of G, containing L, are of maximal class.

(i) Let S 2 �2 be of exponent p2. If S ¤ �.G/ (see (e)), then S is contained inexactly p irregular members of the set �1.

Proof. By Theorems 12.3 and 7.2(b), G is neither of maximal class nor absolutelyregular so there is R GG of order pp and exponent p (Theorem 12.1(a)).

(a) By Theorem 64.1(a), G is irregular. By Theorem 64.4, jGj � ppC1C�.G/ D

ppC2, since �.G/ D Œ 1pp c1.G/� D 1; here Œx� is the integer part of x 2 R. Since

�1.G/ D G, we get G=R Š Ep2 so exp.G/ D p2. We also have jGj D ppC2. As-sume that exp.H/ D p for someH 2 �1. Then c1.H/ D c1.G/ so .G D/ �1.G/ D

H has exponent p, a contradiction. Thus, all members of the set �1 have exponentp2.

(b) If x 2 G �R is of order p, then exp.hx;Ri/ > p, by (a), soM D hx;Ri 2 �1

is of maximal class (Theorem 7.1); then d.G/ D 3 and ˆ.G/ D G0 D ˆ.M/ D M 0

has exponent p, by (a) and Theorem 12.12(a), and we conclude that cl.G/ D p.

(c) follows from Theorem 12.12(b).

(d) follows from (a) and the hypothesis. Indeed, c2.G/ D jGj�.p�1/c1.G/�1

'.p2/D pp .

(e) The first assertion follows from Theorem 12.12(c). Now assume exp.�.G// D

p. Then c2.Ti / D 1p.p�1/

.jTi j � j�.G/j/ D pp�1 for i D 1; : : : ; p C 1 so c2.G/ DPpC1

iD1 c2.Ti / D .p C 1/pp�1 > pp, contrary to (d). Thus, exp.�.G// D p2.

(f) Assume that S < G of order pp and exponent p is not normal in G; thenˆ.G/ D G0 6� S so H D Sˆ.G/ 2 �1. It follows from Theorem 7.2(b) and (a) that


H is irregular so it is of maximal class; in that case,ˆ.G/ D ˆ.H/ (compare orders!)so H D S 62 �1, a contradiction. If t is the number of subgroups of order pp andexponent p in G, then

1CpC Cpp�1Cpp D c1.G/ D c1.ˆ.G//Ctpp�1 D 1CpC Cpp�2Ctpp�1

so t D p C 1.

(g) follows from (a), (b) and (f).

(h,i) By (e), the intersection of two distinct regular maximal subgroups of G co-incides with �.G/. Let L ¤ �.G/ be a normal subgroup of index p2 in G; then Lis contained in at most one regular maximal subgroup of G and G=L Š Ep2 since�1.G/ D G. LetD be aG-invariant subgroup of index p2 in L. Set C D CG.L=D/.Let L < H � C , where H 2 �1; then H is regular. It follows that L is contained inexactly one regular member of the set �1, and the proof is complete.

Exercise 1. Suppose thatG D �1.G/ is an irregular p-group containing exactly pC1

subgroups of order pp and exponent p. Then c1.G/ D 1C p C C pp. If p D 2,then �1.G/ D Q8 � C4.

Exercise 2. Let G be a p-group of maximal class, p > 2. Then �1.G/ D G or��

2.G/ D G. (Hint. �1.G/��2.G/ D �2.G/ D G. Since any two normal subgroups

of G of distinct orders are incident, we are done.)

Proposition 64.12. Suppose that a p-group G, p > 2, is such that ck.G/ D pp�2 fork > 1. Then one of the following holds:

(i) G is regular with cyclic G=�k�1.G/ and j�k�1.G/j D ppCk�3.

(ii) k D 2 and G is of maximal class and order ppC1 having exactly p subgroupsof order pp and exponent p.

Proof. (i) Let G be regular and j�k.G/j D pa, j�k�1.G/j D pb . Then

pp�2 D ck.G/ Dj�k.G/j � j�k�1.G/j

pk�1.p � 1/Dpa�b � 1

p � 1pb�kC1:

It follows that a � b D 1 so j�k.G/=�k�1.G/j D p, and we conclude that thequotient group G=�k�1.G/ is cyclic. Then p�2 D b�kC1 so j�k�1.G/j D pb D

ppCk�3.

(ii) Now let G be irregular. By Lemma 64.1(f), G is of maximal class. Assumethat jGj > ppC1. Let G1 be the fundamental subgroup of G; then j�k�1.G1/j D

p.k�1/.p�1/. If k > 2, then

ck.G1/ Dj�k.G1/j � j�k�1.G1/j

'.pk/�p.k�1/.p�1/C1 � p.k�1/.p�1/

.p � 1/pk�1;

p.k�1/.p�2/ > pp�2 D ck.G/;


a contradiction. Thus, k D 2. Then j�2.G1/j � p.p�1/C2 D ppC1 so

c2.G1/ �ppC1 � pp�1

p.p � 1/D pp�2.p C 1/ > pp�2 D c2.G/;

again a contradiction. Thus, jGj D ppC1. The second assertion in (ii) is checkedeasily.

�65

A2-groups

A p-group G is said to be an An-group if it contains a nonabelian subgroup of indexpn�1 but all its subgroups of index pn are abelian.

Lemma 65.1 (= Exercise 1.8a (Redei)). Let G be a minimal nonabelian p-group (=A1-group). Then G D ha; bi and one of the following holds:

(a) apm

D bpn

D cp D 1, Œa; b� D c, Œa; c� D Œb; c� D 1, jGj D pmCnC1,G D hbi .hai � hci/ D hai .hbi � hci/ is nonmetacyclic.

(b) apm

D bpn

D 1, ab D a1Cpm�1

, jGj D pmCn and G D hbi hai is metacyclic.

(c) a4 D 1, a2 D b2, ab D a�1, G Š Q8.

We have jG0j D p andH=Ã1.H/j � p3 forH � G. If jGj > p3 and j�1.G/j � p2,then G is metacyclic. The group G is nonmetacyclic if and only if G0 is a maximalcyclic subgroup of G. Next, jG=Ã1.G/j � p3 with equality if and only if G is from(a) and p > 2. If, in (a), u 2 G �ˆ.G/, then hui is not normal in G.

Suppose that G0 < L < G, where L is cyclic of order p2. By Theorem 6.1,L=G0 � C=G0, where C=G0 is a cyclic direct factor of G=G0; in particular, G=C iscyclic. It remains to show that C is cyclic. We get G0 D ˆ.L/ � ˆ.C/. It followsthat C=ˆ.C / as an epimorphic image of a cyclic group C=ˆ.L/, is cyclic so C iscyclic and G is metacyclic. The last assertion of Lemma 65.1 is a partial case of thefollowing general fact. If a nonmetacyclic G D hu; vi, then hui 6E G.The aim of this section is to clear up the subgroup and normal structure of A2-

groups. Using comparatively easy arguments, we obtain a lot of information on A2-groups which is difficult to read out from their defining relations. Some results whichwe shall prove in the sequel, are contained in the thesis of Lev Kazarin (unpublished),however, the proofs presented below, as a rule, are new. The length of [She], alsodevoted to classification of A2-groups, indicates the degree of the difficulty of thisproblem.Every nonabelian group of order p4 is either an A1- or A2-group. Therefore, in

what follows, we assume that jGj D pm > p4. A2-groups belong to a wider class ofp-groups all of whose nonabelian subgroups are normal.

Lemma 65.2. Let G be a nonabelian p-group. Then:

(a) If G0 � �1.Z.G// and d.G/ D 2, then G is an A1-group.

�65 A2-groups 189

(b) If G is metacyclic with jG0j D p2, then G is an A2-group.

(c) If G has two distinct abelian maximal subgroups, then jG0j D p.

(d) Suppose that jH 0j � p for all H 2 �1. Then jG0j � p3 and G0 is abelian. IfF;H 2 �1 are distinct and F 0 � H 0, then jG0j � p2. If jG0j D p3, then G hasno abelian maximal subgroups, and if, in addition, d.G/ > 2, then d.G/ D 3

and Ep3 Š G0 � Z.G/.

(e) If G0 \ Z.G/ is cyclic and jH 0j � p for all H 2 �1, then jG0j � p2. If, inaddition, G0 6� Z.G/, then d.G/ D 2 and G=.G0 \ Z.G// is an A1-group.

Proof. (a) If a; b 2 G, then 1 D Œa; b�p D Œa; bp� so Ã1.G/ � Z.G/. Then ˆ.G/ D

G0Ã1.G/ � Z.G/ so Z.G/ D ˆ.G/ has index p2 in G. Hence, all members of theset �1 are abelian so G is anA1-group.

(b) By Lemma 65.1, G is not an A1-group. Let L D Ã1.G0/ and H 2 �1; then

L < G0 < H . By (a), G=L is an A1-group so H=L is abelian; then, again by (a), His either abelian or an A1-group. Thus, G is anA2-group.

(c) follows from Lemma 64.1(u).

(d) Let F;H 2 �1 be distinct. Then jG0j � pjF 0H 0j � p3 (Lemma 64.1(u)). If,in addition, F 0 � H 0, then jG0j � p2. Now let jG0j D p3. Since F 0 6� H 0 andH 0 6� F 0, �1 has no abelian members. Let, in addition, d.G/ > 2. Then A0 ¤ B 0

for distinct A;B 2 �1 so c1.G0/ � j�1j � 1 C p C p2, and we get d.G0/ D 3,

Ep3 Š G0 � Z.G/. Therefore, since 1C pC p2 D c1.G0/ D j�1j, we get d.G/ D 3.

(e) We have D D hH 0 j H 2 �1i � �1.G0 \ Z.G// so jDj D p since �1.G

0 \

Z.G// is cyclic. Then G=D is either abelian or A1-group so j.G=D/0j � p (Lemma65.1), and we get jG0j � p2. Let, in addition, G0 6� Z.G/. Then G=D is nonabelianso it is anA1-group. Since D < ˆ.G/, we get d.G/ D d.G=D/ D 2.

Corollary 65.3. A metacyclic p-group G is an A2-group if an only if jG0j D p2.

Remarks. 1. Let us prove that if a p-group G has the cyclic derived subgroup G0, thenall elements of G0 are commutators. We have G0 D hŒx; y�i for some x; y 2 G. Onemay assume that G D hx; yi. By Lemma 64.1(u), there is H 2 �1 such that H 0 D

Ã1.G0/. By induction, all elements of H 0 are commutators. By [BZ, Theorem 3.27],

all generators of G0, i.e., members of the set G0 � H 0, are commutators, completingthe proof.

2. Let us prove that if G of order p4 and exponent p has no nontrivial direct factors,it is of class 3. Let H be an A1-subgroup of G; then jH j D p3. If Z.G/ 6� H , thenG D H � C , where C < Z.G/ is of order p such that C 6� H . Thus, Z.G/ < H

so Z.G/ is of order p. Since CG.H/ D Z.G/, we get CG.H/ < H . Then G is ofmaximal class (Proposition 10.17).

Lemma 65.4. Let G be an A2-group of order pm > p4.

(a) If K � G, then jK=Ã1.K/j � p4, and this estimate is best possible.


(b) If K < G with jK=Ã1.K/j D p4, then Ep4 Š �1.K/ GG and, if�1.K/ � H 2

�1, then H is abelian. If, in addition, jG0j > p, then jG0j D p2, jG W Z.G/j D p3

and G=�1.K/ is cyclic.

(c) If K G G and G=K is generated by subgroups of index p2, then K � Z.G/. Inparticular, if d.G/ D 3, then ˆ.G/ � Z.G/.

(d) Let jG0j D p. Then either G D H � C , jC j D p or G D H �Z, whereH 2 �1

is an A1-subgroup and Z is cyclic of order p2.

(e) Ã2.G/ � Z.G/. If p > 2 and G is not metacyclic, then Ã1.G/ � Z.G/ andjG=Z.G/j � p3.

(f) Let jG=Ã1.G/j D p4, m > 5 and G0 6� Z.G/. Then p > 2, G0 Š Ep2 ,cl.G=Ã1.G// D 3, Ã1.G/ � Z.G/ is cyclic, G=G0 Š Cpm�3 � Cp so d.G/ D 2. IfL=G0 is a direct factor of G=G0 of order p, then L Š Ep3 and G D Z L is a semidi-rect product of L with a cyclic subgroup Z of order pm�3 hence Ã1.G/ D Ã1.Z/,�1.G/ D L � �1.Ã1.G// Š Ep4 , G0 \ Ã1.G/ D f1g and G is an L4-group (see��17, 18). Next, G=.G0 \ Z.G// is an A1-group.

(g) If jG0j D p3 and d.G/ D 2, then p > 2 and jGj D p5.

(h) If G=Z.G/ is nonabelian, then G=.G0 \ Z.G// is an A1-group.

(i) Let G be metacyclic. Then G=Z.G/ is of order � p4 and exponent p2. If thenumber jG=Z.G/j D p3, then p D 2 and G=Z.G/ Š D8. Now suppose thatjG=Z.G/j D p4. If p D 2, then G=Z.G/ is abelian of type .4; 4/. If p > 2, thenG=Z.G/ can be abelian or nonabelian of exponent p2 (see examples following thelemma)

(j) If G has no normal elementary abelian subgroups of order p3, then it is eithermetacyclic or minimal nonmetacyclic.

Proof. (a) In view of Lemma 65.1, one may assume that K < G. Suppose thatjK=Ã1.K/j > p3. Then K is not a subgroup of an A1-group (Lemma 65.1) so itis abelian. Considering K \ A, where A 2 �1 is nonabelian, we get j�1.K/j � p4

(Lemma 65.1), and we are done since jK=Ã1.K/j D j�1.K/j. If A2-group G D

U � C , where U is a nonmetacyclic A1-group of order > p3 and jC j D p, then�1.G/ is of order p4 and exponent p.

(b) Let jK=Ã1.K/j D p4 for K < G. Then every member of the set �1 containingK is abelian (Lemma 65.1) so Ep4 Š �1.K/ G G. Let, in addition, jG0j > p. ThenG=�1.K/ is cyclic since the set �1 has exactly one abelian member (Lemma 65.2(c)).By Lemma 65.2(d), jG0j D p2 so jG W Z.G/j D p3 (Lemma 64.1(q)).

(c) Let R=K < G=K be of index p2 in G=K. Then R is abelian so R � CG.K/

and K � Z.G/ since G D hR < G j K < R; jG W Rj D p2i. Now the secondassertion follows.

(d) This follows from Lemma 64.1(r).

�65 A2-groups 191

(e) If there isK GG such that G=K is of order p3 and exponent p, thenK � Z.G/,by (c), so we are done since Ã2.G/ � K. Now assume that such K does not exist.Then G=Ã1.G/ Š Ep2 . If Ã1.G/ is cyclic, then p D 2, m D 4 and so Ã2.G/ D

Z.G/ (Lemma 64.1(t)). Now let Ã1.G/ be noncyclic. Then Ã1.G/ has a G-invariantsubgroup T such that Ã1.G/=T Š Ep2 . In that case, Ã2.G/ � T and the resultfollows from (c).

(f) Since d.G/ � 3, we get p > 2. By (c), Ã1.G/ < Z.G/. Next, G=Ã1.G/

has an abelian subgroup A=Ã1.G/ of index p; d.A/ � d.A=Ã1.G// D 3 so Ais abelian (Lemma 65.1). Then jG0j D p2 (Lemma 64.1(u)) so jG=Z.G/j D p3

(Lemma 64.1(q)). By Corollary 65.3, G0 Š Ep2 since G is nonmetacyclic. By theabove, A=Ã1.G/ is the unique abelian maximal subgroup of G=Ã1.G/, and we con-clude that d.G=Ã1.G// D 2; then d.G/ D 2 since Ã1.G/ < ˆ.G/. It follows thatcl.G=Ã1.G// D 3 (Remark 2). Thus, G=Z.G/ is nonabelian so G0 6� Z.G/. Then, by(c),G=G0 is not generated by subgroups of index p2 so it is abelian of type .pm�3; p/.Let L=G0 be a direct factor of G=G0 of order p; then L is abelian of order p3 sincejG W Lj D pm�3 > p2. We have G=L Š Cpm�3 so there exists a cyclic Z < G withG D Z L (semidirect product) since Z \L D f1g in view of jG=Ã1.G/j D p4. Wealso see that Ã1.Z/ D Ã1.G/ (compare indices!) so exp.L/ D p, L Š Ep3 . ThenG0 \Ã1.G/ � L\Ã1.Z/ D f1g. Since a Sylow p-subgroup of Aut.L/ is nonabelianof order p3 and exponent p, we get �1.G/ D L � �1.Ã1.G// Š Ep4 (recall thatm > 5). By remark, preceding the lemma, cl.G=Ã1.G// D 3. Next, the last assertionfollows from Lemma 65.2(a).

(g) All members of the set �1 areA1-groups (Lemma 65.2(d)) so they are generatedby two elements (Lemma 65.1). By Lemma 64.1(n), p > 2. In that case, jG W G0j D

p2 (Lemma 64.1(p)) so jGj D jG W G0jjG0j D p2p3 D p5.

(h) Since G0 6� Z.G/, we get jG0j � p2, and, by (c), d.G/ D 2. Then T D

G0 \ Z.G/ has index p in G0. Indeed, this is true, if jG0j D p2. If jG0j D p3, thereare nonabelian F;H 2 �1 such that F 0 ¤ H 0 (Lemma 64.1(u)) so F 0H 0 D T hasindex p in G0. Then G=T is anA1-group since jG=T /0j D p and d.G/ D 2 (Lemma65.2(a,e)).

(i) Since G is not an A1-group, we have jG W Z.G/j > p2. Assume that jG W

Z.G/j D p3. Then G=Z.G/ has exactly one cyclic subgroup of index p (other-wise, jG W Z.G/j D p2) so G=Z.G/ Š D8. Since Ã2.G/ � Z.G/, by (e), weget jG=Z.G/j � p4. Suppose that p D 2 and NG D G=Z.G/ is nonabelian of order 24.Then NG D ha; b j a4 D b4 D 1; ab D a�1i. Set NU D hb2i. Then G=U Š D8, and soU � Z.G/, by (c), a contradiction. Thus, in that case,G=Z.G/ is abelian.

(j) Let H 2 �1. Since �1.H/ G G, we get j�1.H/j � p2 so H is metacyclic(Lemma 65.1).

Not all groups of Lemma 65.4(d) are A2-groups. Indeed, let G D H � C , whereH D ha; b j a4 D b4 D 1; ab D a�1i, C D hc j c4 D 1i;H \ C D �1.C /. Thegroup G is anA2-group if and only if c2 D a2.


The group G D ha; b j a8 D b4 D 1; Œa; b� D a2i is a metacyclic A2-group withG=Z.G/ Š D8.The group G D ha; b j apn

D bp2

D 1; Œa; b� D apn�2

i. n > 3, is metacyclic withG=Z.G/ Š Cp2 � Cp2 .

Let p > 2 and G D ha; b j ap3

D bp2

D 1; Œa; b� D api. Then G0 D hapi,Z.G/ D hap2

i, G=Z.G/ is nonabelian of order p4. The above three groups realize allpossibilities of Lemma 65.4(i).Let G be a metacyclic 2-group with G=Z.G/ Š D8. Then jG0j D 4 (Lemma

64.1(u)) so G is anA2-group (Corollary 65.3).If G is a metacyclic p-group such that G=Z.G/ is of order p4 and exponent p2,

then G is an A2-group. Indeed, if H 2 �1, then H=Z.G/ is abelian of type .p2; p/

soH is either abelian or minimal nonabelian since d.H/ D 2.

Lemma 65.5. Let an A2-group G of order > p4 have no abelian maximal subgroups,and d.G/ D 2. Then:

(a) If p D 2, then G is metacyclic.

(b) If p > 2 and G is nonmetacyclic, then jGj D p5. K3.G/ D Ã1.G/ D Z.G/ isof order p2 and all subgroups of index p2 in G contain Z.G/.

Proof. (a) follows from Lemmas 65.1 and 64.1(n).

(b) The quotient group G=Ã1.G/ is nonabelian of order p3 and exponent p,Ã1.G/ D K3.G/, jG W G0j D p2 (Lemma 64.1(p)) and Ã1.G/ D Z.G/ (Lemma65.4(c)). If T < G is of index p2, then T Z.G/ is abelian so Z.G/ < T . Next, wehave jGj D jG W G0jjG0j � p5 so jGj D p5.

Lemma 65.6. Let G be a group of order pm > p4 and all members of the set �1 aregenerated by two elements.

(a) If Z.G/ D ˆ.G/ has index p3 in G, then G is an A2-group.

(b) If d.G/ D 2 and Z.G/ D Ã1.G/ has index p3 in G, then G is an A2-group.

Proof. (a) IfH 2 �1, then d.H/ D 2 hence all maximal subgroups ofH are membersof the set �2 so abelian, andH is either abelian or minimal nonabelian.

(b) Given H 2 �1, it follows from d.H/ D 2 that all maximal subgroups of Hcontain Z.G/ (since Z.G/ � ˆ.G/) so abelian. Thus, again H is either abelian orminimal nonabelian.

Theorem 65.7. Suppose that G is an A2-group of order > p4.

(a) If G0 is cyclic of order > p, then G is metacyclic and jG0j D p2.

(b) Suppose that jG0j D p3. Then every subgroup of index p2 in G contains Z.G/.If d.G/ D 2, then p > 2 and jG W Z.G/j D p3. If d.G/ D 3, then ˆ.G/ D

Z.G/ and G0 Š Ep3 . If p D 2, then d.G/ D 3 so G0 � Z.G/ is elementaryabelian.

�65 A2-groups 193

(c) If p D 2, G is nonmetacyclic and D 2 �1 is abelian, then G0 � Z.G/.

(d) If G is nonmetacyclic, then G0 is elementary abelian.

(e) If G is nonmetacyclic, then jG=Z.G/j � p3.

(f) If G is nonmetacyclic and A;B 2 �1 are distinct nonabelian, then Z.A/ D

Z.B/ � Z.G/.

(g) If p D 2 and G=Z.G/ Š D8, then G is metacyclic.

(h) Suppose that G is nonmetacyclic and the set �1 has exactly one abelian memberA; then jG0j D p2. If G0 6� Z.G/, then p > 2 and G=Z.G/ is nonabelian oforder p3 and exponent p. If G0 � Z.G/, then G=Z.G/ Š Ep3 .

Proof. (a) Since G0 is cyclic, all nonabelian maximal subgroups of G have the samederived subgroup D �1.G

0/. Then, by Lemma 64.1(u), jG0j D p2. Let a nonabelianF 2 �1. Since F 0 < G0 < F , then F is metacyclic (Lemma 65.1). If all membersof the set �1 are metacyclic, then G is also metacyclic (Theorem 69.1). Supposethat there is a nonmetacyclic A 2 �1; then A is abelian. We have �1.A/ Š Ep3 sinced.A/ D 3: F \A is metacyclic of index p inA. Since all members of the set �1 whichcontain �1.A/, are nonmetacyclic so abelian, G=�1.A/ is cyclic. Then G0 < �1.A/

soG0 Š Ep2 , a contradiction. Thus, A does not exist so, by what has been said before,G is metacyclic.

(b) If T < G is of index p2, then T Z.G/ D T and jG W Z.G/j > p2 since T isabelian and the set �1 has no abelian members (Lemma 65.2(d)); therefore, ifH 2 �1,then Z.G/ � ˆ.H/ � ˆ.G/. Suppose that d.G/ D 2. Since G is nonmetacyclic(Corollary 65.3 and (a)), we get p > 2 (Lemma 64.1(n) since all members of the set �1

are two-generator) and jG=Ã1.G/j � p3 (Lemma 64.1(m)), Ã1.G/ � Z.G/ (Lemma65.4(c)) and jG W Z.G/j D p3. Now let d.G/ D 3. Then G0 � ˆ.G/ � Z.G/(Lemma 65.4(c)) so ˆ.G/ D Z.G/ (Lemma 64.1(q)), exp.G0/ � exp.G=Z.G// D p,and we have G0 Š Ep3 . Let, in addition, p D 2. Then d.G/ D 3 since all members ofthe set �1 are two generator (Lemma 64.1(n)), and Œx; y�2 D Œx; y2� D 1 for x; y 2 G,so exp.G0/ D 2 since G0 is abelian (Burnside).

(c) Assuming G0 6� Z.G/, we get jG0j D 4 (Lemma 64.1(u)), G0 Š E4, by (a), andd.G/ D 2 (Lemma 65.4(c)). By Lemma 65.2(c), D is the unique abelian member ofthe set �1. Since G=Z.G/ is nonabelian of order 8 (Lemma 64.1(q)), it has exactlyone cyclic subgroup of index 2 so G=Z.G/ Š D8. LetH 2 �1 � fDg; thenH \D D

ˆ.G/. If a 2 H � ˆ.G/ and x 2 D �ˆ.G/, then G D ha; xi. We have o.Œa; x�/ D

2.D exp.G0// and hŒa; x�i 6E G (otherwise, G=hŒa; x�i is abelian so jG0j D 2). Next,a2 2 D so

1 D Œa2; x� D a�2.x�1ax/2 D a�2.aŒa; x�/2 D a�2a2Œa; x�2ŒŒa; x�; a� D ŒŒa; x�; a�

since ha; Œa; x�i.� H/ has class at most 2 (Lemma 65.1) and exp.G0/ D 2. Thena centralizes G0 D hG0 \ Z.G/; Œa; x�i. It follows from G0 < D that CG.G

0/ �

ha;Di D G, contrary to the assumption.


(d) Assume that exp.G0/ > p; then G0 is abelian (Lemma 64.1(v)) of type .p2; p/

since G0 is noncyclic, by (a). By Lemma 65.1, H 0 � �1.G0/ < G0 for all H 2 �1

so d.G/ D 2 (Lemma 64.1(y)). By Lemma 65.2(d), all members of the set �1 asA1-groups are two-generator so p > 2 and K3.G/ D Ã1.G/ has index p3 inG hencejG W G0j D p2 (Corollary 36.6). Since Ã1.G/ � Z.G/ (Lemma 65.4(c)), we getÃ1.G/ D Z.G/. There is H 2 �1 such that H 0 ¤ Ã1.G

0/. Then .G=H 0/0 D G0=H 0

is cyclic of order p2 so G=H 0 is metacyclic, by (a). This is a contradiction sinceG=Z.G/ is nonmetacyclic andH 0 � Z.G/.

(e) Assume that jG=Z.G/j > p3. Then p D 2, exp.G=Z.G// D 4 (Lemma65.4(e)), G has no epimorphic images isomorphic to D8 or E8 (Lemma 65.4(c)) sod.G/ D 2. By Lemma 64.1(n), there exists A 2 �1 with d.A/ > 2; then A is abelian(Lemma 65.1). Let B 2 �1 be nonabelian; then Z.B/ D ˆ.B/ < ˆ.G/ < A soCG.Z.B// � AB D G. It follows that jG W Z.G/j � jG W Z.B/j D jG W BjjB W

Z.B/j D 8, a contradiction.

(f) By (e), G=Z.G/j � p3. Assume that Z.A/ ¤ Z.B/. Then Z.A/Z.B/.� ˆ.G//

has index p2 in G so d.G/ D 2 and Z.G/ < ˆ.G/ (< since A is nonabelian); more-over, jG=Z.G/ D p3. It follows that Z.A/ D Z.G/. Similarly, Z.B/ D Z.G/. ThenZ.A/ D Z.B/, and G is not a counterexample. (Clearly, (e) follows from (f).)

(g) Assume that G=Z.G/ Š D8 and G is nonmetacyclic. Then G0 6� Z.G/. IfD=Z.G/ is the cyclic subgroup of order 4 in G=Z.G/, then D 2 �1 is abelian. Then,by (c), G0 � Z.G/, a contradiction.

(h) By Lemma 65.4(d), jG0j > p so jG0j D p2 (Lemma 64.1(u)). First assume thatG0 6� Z.G/. By (c), p > 2. By (e) and Lemma 65.4(e), G=Z.G/ is nonabelian oforder p3 and exponent p. Now let G0 � Z.G/. By Lemma 64.1(q), jG W Z.G/j D

pjG0j D p3 so G=Z.G/ Š Ep3 .

Probably, (a) and (d) are most important parts of Theorem 65.7. We use those partsin the study of A3- andA4-groups in �72.

Theorem 65.8. Let a two-generator A2-group G be a 2-group. Then G is metacyclic.

Proof. Assume that an A2-group G is a nonmetacyclic two-generator 2-group. Thenthere is A 2 �1 with d.A/ > 2 (Lemma 64.1(n)) so d.A/ D 3, by Schreier’s theorem(Appendix 25). Since A is not an A1-group, it is abelian. Let �1 D fA;M;N g; thenM and N are nonabelian (ifM is abelian, then G is an A1-group) so A1-subgroups.We have Z.M/ D ˆ.M/ < ˆ.G/ < A so CG.Z.M// � AM D G whence Z.M/ D

Z.G/ since Z.G/ < ˆ.G/. Similarly, Z.N / D Z.G/. Since a nonelementary abeliangroup G=Z.G/ of order 8 has two four-subgroups, we get G=Z.G/ Š D8. By Lemma65.4(c), if U G G is such that G=U Š D8, then U D Z.G/. Since jG0j > 2 (Lemma65.2(a)), A is the unique abelian member of the set �1 (Lemma 65.2(c)).Write NG D G=Ã1.A/. Then NG is nonabelian of order 16 since d.A/ D 3 > 2 D

d.G/. Since exp. NG/ D 4, NG is not of maximal class. Then, by Proposition 10.17, NG

�65 A2-groups 195

has no nonabelian subgroups of order 8 (otherwise, d.G/ > 2) so it is an A1-group.It follows that �1. NG/ D NA (Lemma 65.3) so Z. NG/ Š E4. Let NR < Z. NG/ be oforder 2 and NR ¤ NG0; then NG= NR Š D8 since a nonabelian group NG= NR has a subgroupNA= NR Š E4. By the previous paragraph, R D Z.G/. Let B=R < G=R be cyclic oforder 4; then B 2 �1 � fAg is abelian, a contradiction.

Lemma 65.9. Let G be a nonmetacyclic p-group and R < G0 be G-invariant of orderp. If G=R is minimal nonabelian and all nonabelian maximal subgroups of G aretwo-generator, then G is an A2-group and p > 2.

Proof. By hypothesis, jG0j > p so G is not an A1-group. We also have d.G/ D

d.G=R/ D 2. Therefore, in view of Theorem 65.8, we have to prove that G is anA2-group. Let M 2 �1 be nonabelian. Then d.M/ D 2 and M=R (as a maximalsubgroup of G=R) is abelian which implies M 0 D R. Hence M is an A1-group(Lemma 65.2(a)) and so G is anA2-group.

Theorem 65.10 ([CP] (for p D 2), [ZAX]). Suppose that a nonabelian p-group Gis neither minimal nonabelian nor metacyclic nor minimal nonmetacyclic. If all non-abelian subgroups of G are metacyclic, then one and only one of the following holds:

(a) G D M � C , whereM 6Š Q8 is a metacyclic A1-group and jC j D p.

(b) p > 2, d.G/ D 2, G D �1.G/C , where �1.G/ Š Ep3 , C is a cyclic subgroupof index p2 in G, CG D Ã1.C / D Z.G/ is of index p3 in G.

Proof. Let us check that groups of (a) and (b) satisfy the hypothesis. This is clearfor groups from (a). Now let G be as in (b) and V 2 �1. If �1.G/ � V , thenV D �1.G/Ã1.C / so V is abelian since�1.G/ is abelian and Ã1.C / D Z.G/. Nowassume that �1.G/ 6� V . Then �1.V / D V \�1.G/ Š Ep2 and V=�1.V / is cyclichence V has a cyclic subgroup of index p and so metacyclic.Now, assuming that G satisfies the hypothesis, we have to prove that G is either as

in (a) or in (b). By hypothesis, there are in G two maximal subgroups M and A suchthatM is nonabelian so metacyclic and A is nonmetacyclic so abelian; then d.A/ > 2and d.G/ � d.M/C 1 D 2C 1 D 3. SinceM \A is a metacyclic maximal subgroupof A, we get d.A/ D 3. Set E D �1.A/; then Ep3 Š E G G. By the productformula, G D ME soM \E Š Ep2 . All maximal subgroups of G containing E, arenonmetacyclic so abelian, hence d.G=E/ D d.M=.M \E// � 2 (Exercise 1.6(a) andLemma 65.1). In what follows, A;M and E are such as defined in this paragraph.Let d.G=E/ D 2. Then there is a maximal subgroup B=E < G=E with B ¤ A so

E � A \ B D Z.G/ since B is abelian. If x 2 E � M , then G D M � X , whereX D hxi. Let a noncyclic N < M be maximal. Then N � X is abelian (otherwise,d.N / � 2 so d.N �X/ � 3). Thus,M is a metacyclicA1-group so G is as in (a).Now suppose that G=E is cyclic; then G0 < E.

(i) Let jG=Ej D p; then jM j D p3. If CG.M/ < M , then G is of maximalclass (Proposition 1.17) and p > 2 since G is not metacyclic, and E is the unique


abelian maximal subgroup of G (Lemma 65.2(c)) so exp.G/ D p2 since M 2 �1

is metacyclic. If E D Z.G/ � L, then LG D f1g so G is isomorphic to a subgroupof exponent p2 of a Sylow p-subgroup of the symmetric group Sp2; then G has asubgroup ¤ A of order p3 and exponent p which is nonabelian and nonmetacyclic, acontradiction. Thus, G D MZ.G/. If Z.G/ is noncyclic, then G D M �L is as in (a)(in that case,M 6Š Q8 since G is not minimal nonmetacyclic). If Z.G/ is cyclic, then,since jZ.G/j D p2, we get G D EZ.G/, by the product formula, so G is abelian, acontradiction.

(ii) Now suppose that G=E is cyclic of order > p; then G0 < E so jG0j � p2. Wehave �1.G=�1.G// < A=�1.G/ so �1.G/ D �1.A/ D E. SinceM=.M \ E/ D

M=�1.M/ Š G=E is cyclic, we get M Š Mpn , n > 3, since M is nonabelianand has a cyclic subgroup of index p (Theorem 1.2). Thus, all nonabelian maximalsubgroups of G are isomorphic to Mpn . (It follows that G is an A2-group so one canuse the classification of A2-groups, however we prefer to present independent, moreelementary, proof.)Let d.G/ D 2; then Z.G/ < ˆ.G/ and, since G is not an A1-group, we get

G0 6� Z.G/ (Lemma 65.2(a)) so G0 Š Ep2 ; then cl.G/ D 3, A is the unique abelianmaximal subgroup of G and jG W Z.G/j D pjG0j D p3 (Lemma 64.1(q)). SinceG0 < M , we get G0 D �1.M/ so M=G0 is a cyclic subgroup of index p of theabelian group G=G0. Since Z.G/ < M , then Z.G/ D Z.M/ (compare indices!) soZ.G/ is cyclic. Assume that there is a cyclic U=Z.G/ of index p in G=Z.G/. ThenU is abelian and metacyclic so U ¤ A, a contradiction. Thus, exp.G=Z.G// D p soG=Z.G/ is nonabelian of exponent p; then p > 2. We have Z.G/ < C < M , whereC is cyclic of index p in M . Since jG W C j D p2, we get G D EC , and C is notnormal in G since G0 is noncyclic. Thus, G is as in (b).Now we assume that d.G/ D 3. Then G=G0 has no cyclic subgroups of index

p so jG0j D p, and we get jG W Z.G/j D pjG0j D p2 (Lemma 64.1(q)). SincejA W Z.G/j D p and d.A/ D 3, the subgroup Z.G/ is noncyclic. By what has beenproved already, M Š Mpn . In that case, �1.Z.G// 6� M since Z.M/ is cyclic. IfL < Z.G/ of order p is not contained inM , then G D M � L so G is as in (a).

Exercise 1. Let G be a nonabelian p-group and d.G/ > 2. Prove that all members ofthe set �2 are abelian if and only if d.G/ D 3 and ˆ.G/ � Z.G/.

Exercise 2 (Janko). LetA < G be a maximal normal abelian subgroup of a nonabelianp-group G. Prove that, for x 2 G � A, there exists a 2 A such that hx; ai is an A1-group.

Exercise 3. Let a p-groupG be neither abelian nor minimal nonabelian and letA1.G/

be the set of all minimal nonabelian subgroups ofG. Prove that ifG is not generated byany proper subset of the setA1.G/, then p D 2, G is anA2-group and jA1.G/j D 2.(Hint. Use Lemma 76.5. Try to give an independent proof.)

Problem. Classify the p-groups all of whose nonnormal subgroups are abelian.

�66

A new proof of Blackburn’s theoremon minimal nonmetacyclic 2-groups

Here we present a new proof, due to Janko, of Blackburn’s theorem on minimal non-metacyclic 2-groups avoiding his tedious calculations. It is difficult to estimate thecrucial role of this theorem in our book. Only case p D 2 is considered below since,in case p > 2, the original proof is easy but nonelementary.

Theorem 66.1 (Blackburn). Suppose thatG is a minimal nonmetacyclic 2-group. ThenG is one of the following groups:

(a) E8.

(b) The direct product C2 � Q8.

(c) The central product Q8 � C4 Š D8 � C4 of order 24.

(d) The group G D ha; b; ci with a4 D b4 D Œa; b� D 1, c2 D a2, ac D ab2,bc D ba2, where G is special of order 25 with exp.G/ D 4, �1.G/ D G0 D

Z.G/ D ˆ.G/ D ha2; b2i Š E4, M D hai � hbi Š C4 � C4 is the uniqueabelian maximal subgroup of G, and all other six maximal subgroups of G areisomorphic to X D hx; y jx4 D y4 D 1; xy D x�1i (which is the metacyclicminimal nonabelian group of order 24 and exponent 4).

Proof [Jan8]. Suppose that G is a minimal nonmetacyclic 2-group. Then Theorem44.5 implies d.G/ D 3. If jGj � 23, then ˆ.G/ D f1g and so G Š E8. Clearly, if Gis abelian then a consideration of �1.G/ shows that G D �1.G/ Š E8.Suppose that jGj D 24. Since d.G/ D 3, G is neither abelian nor minimal non-

abelian. Let Q be a nonabelian subgroup of order 8 in G. Since G is not of maximalclass, Proposition 10.17 implies G D Q � Z with jZj D 4 and Q \ Z D Z.Q/.If Z Š E4, then Q Š Q8 (because in case Q Š D8 the group G would contain a(proper) subgroup isomorphic to E8 which is nonmetacyclic). Thus, G Š C2 �Q8. IfZ Š C4, then G Š Q8 � C4 Š D8 � C4 with Q8 \ C4 D Z.Q/.In what follows we assume that jGj � 25. Assume that G contains a nonabelian

subgroup D of order 23. If CG.D/ � D, then, by Proposition 10.17, G is of maximalclass and so G is metacyclic, a contradiction. Hence CG.D/ 6� D and take in CG.D/

a subgroup U of order 4 containing Z.D/. Then jD � U j D 24 and D � U is notmetacyclic since d.D � U/ D 3, a contradiction. We have proved that every subgroupof order 8 in G is abelian.


Now suppose that jGj D 25. Since d.G/ D 3, it follows that jˆ.G/j D 4. Assumeˆ.G/ 6� Z.G/. Then there exists an element x 2 G � CG.ˆ.G//; then hˆ.G/; xi

is nonabelian of order 8 since x2 2 ˆ.G/, contrary to the result of the previous para-graph. Hence ˆ.G/ � Z.G/. On the other hand, G0 � ˆ.G/ and suppose thatjG0j D 2. Then G=G0 is abelian and minimal nonmetacyclic of order 24, contrary tothe result of the first paragraph of the proof. Hence G0 D ˆ.G/. For any x; y 2 G,we have Œx; y�2 D Œx2; y� D 1 since G is of class 2, and so ˆ.G/ D G0 Š E4. Inparticular, exp.G/ D 4 and �1.G/ D ˆ.G/. Assume that Z.G/ > ˆ.G/ so thatjZ.G/j D 23 (recall that jGj D 25) and G has an abelian subgroup of index 2. Butthen Lemma 64.1(q) gives jGj D 2jZ.G/jjG0j D 26, a contradiction. We have provedthat�1.G/ D G0 D Z.G/ D ˆ.G/ Š E4, and so G is a special group of order 25. Infact, we shall show that the structure of G is uniquely determined.There are elements x1; x2 2 G � G0 such that Œx1; x2� D t1, where t1 is an in-

volution in G0. Then X D hx1; x2i is a maximal subgroup of G since it is non-abelian so of order 24, and X 0 D ht1i since X=ht1i is abelian. Since G=ht1i isnonabelian in view of jG0j D 4, there is x3 2 G � X such that (interchangingx1 and x2 if necessary) Œx1; x3� D t2 2 G0 � ht1i. We have ht1; t2i D G0 andG D hX;x3i D hx1; x2; G

0; x3i D hx1; x2; x3i.For Œx2; x3� we have one of the following possibilities:

(1) Œx2; x3� D 1. In that case, hx2; x3; G0i is an abelian maximal subgroup of G.

(2) Œx2; x3� D t1 and then Œx1x3; x2� D Œx1; x2�Œx3; x2� D t1t1 D 1. In that case,hx1x3; x2; G

0i is an abelian maximal subgroup of G.

(3) Œx2; x3� D t2 and then Œx1x2; x3� D Œx1; x3�Œx2; x3� D t2t2 D 1. In that case,hx1x2; x3; G


(4) Œx2; x3� D t1t2 and then Œx1x2; x2x3� D Œx1; x2x3�Œx2; x2x3� D t1t2 t1t2 D 1.In that case hx1x2; x2x3; G


We see that at least one maximal subgroup M of G is abelian and so (being meta-cyclic of exponent 4)M Š C4 � C4. Since jZ.G/j D 4,M is also the unique abelianmaximal subgroup of G, by Lemma 64.1(q).Take an element c 2 G �M so that c2 is an involution (inG0). Such element exists

(otherwise, hc;G0i Š E8 since exp.G/ D 4; this argument shows that �1.G/ D G0).Since Z.G/ D G0, we get CM .c/ D G0 and c stabilizes the chain M > G0 > f1g.Let a 2 M �G0 be an element (of order 4) with a2 D c2 (recall that every involutionin M is a square of an element from M ). If c normalizes hai, then ha; ci Š Q8 isnonabelian, a contradiction. Hence ac D at , where t 2 G0 � hai D G0 � ha2i. Letb 2 M � G0 be such that b2 D t . We get M D hai � hbi and ac D ab2. SinceŒb; c� 2 G0 � hb2i (as above), it follows that either bc D ba2 or bc D b.a2b2/.However, if bc D b.a2b2/, then .cb/2 D cbcb D c2bcb D a2b.a2b2/b D 1. Thisis a contradiction since cb 62 G0 and �1.G/ D G0. Hence bc D ba2 and the structureof G is uniquely determined as stated in the theorem.


It remains to show that jGj < 26. Assume that G is a minimal counterexample.Then, considering an appropriate quotient group G=N , where N is a subgroup oforder 2 inG0 \Z.G/, we get jGj D 26. Let N be a subgroup of order 2 in G0 \Z.G/.Then G=N is isomorphic to the group (d) of our theorem. Set W=N D .G=N/0 D

ˆ.G=N/ Š E4 so that W D G0 D ˆ.G/ is of order 8. Then W is abelian, byBurnside; moreover, W is abelian of type .4; 2/ since W is metacyclic and W=N Š

E4. We get N D ˆ.W / D Ã1.W / D hzi and setW0 D �1.W /.It is easily seen that W0 6� Z.G/. Indeed, if W0 � Z.G/, then we consider a

subgroup N0 of order 2 in W0 Š E4 with N0 \ N D f1g so that W=N0 Š C4.On the other hand, G=N0 is also isomorphic to the group (d) of our theorem andtherefore .G=N0/

0 Š E4 coincides with W=N0 Š C4, a contradiction. We haveexp.G/ � 23 becauseG=W Š E8 and exp.W / D 4. Also, Z.G/ � N and Z.G/ � W

since Z.G=N/ D W=N . It follows that either Z.G/ D N or Z.G/ Š C4 withN < Z.G/ < W .Set C D CG.W0/ so that jG W C j D 2. If x 2 G � C , then x2 2 ˆ.G/ D W .

If x2 2 W0, then hx; W0i Š D8 is nonabelian of order 8, a contradiction. Thusx2 2 W �W0 is an element of order 4 and so all elements in G �C are of order 8. Inparticular, exp.G/ D 23. Since C D CG.W0/ is metacyclic, there are no involutionsin C �W0 and so W0 D �1.G/.By the structure of G=N , G has exactly six maximal subgroups M1;M2; : : : ;M6

such that Mi=N Š X .i D 1; 2; : : : ; 6/, where X D hx; y jx4 D y4 D 1; xy D

x�1i is the metacyclic minimal nonabelian group of order 24 and exponent 4. For theseventh maximal subgroupM7 of G we haveM7=N Š C4 � C4. HenceM7 is eitherabelian of type .23; 22/ orM 0

7 D N in which caseM7 is minimal nonabelian (Lemma65.2(a)) with exp.M7/ D 23. In both casesW0 � Z.M7/ and soM7 D C D CG.W0/

(indeed, if M7 is minimal nonabelian, W0 D �1.M7/ and M7=W0 is noncyclic soW0 � Z.M7/).We shall determine the structure of Mi , i 2 f1; 2; : : : ; 6g. We have ˆ.Mi / D W

since Mi=W Š E4 and Mi is metacyclic. Also, M 0i � W and M 0

i > f1g is cyclicsinceMi is metacyclic. Suppose jM 0

i j D 2. The caseM 0i D N is not possible since

in that case Mi=N would be abelian. Hence M 0i ¤ N and so W0 D M 0

i � N . Butthen CG.W0/ � MiM7 D G hence W0 � Z.Mi /, contrary to what has been provedalready. HenceM 0

i is cyclic of order 4 and W > M 0i > N . Since CG.W0/ D M7, we

have W0 6� Z.Mi / for i < 7. BecauseMi , i < 7, is metacyclic and exp.Mi / D 23,Mi possesses a cyclic normal subgroup Zi > M 0

i with jZi j D 23 andMi=Zi Š C4.If Mi splits over Zi , then Mi D ZiUi with Zi \ Ui D f1g and Ui Š C4. Butthen �1.Ui / � W0 and �1.Ui / centralizes Zi since Aut.Zi / Š E4. In that caseW0 � Z.Mi /, a contradiction. HenceMi does not split over Zi . Let ai 2 Mi �Zi besuch that hai i coversMi=Zi . Since exp.Mi / D 23, we have hai i \ Zi D N D hzi.Then a2

i is of order 4, a2i 2 W �W0, and a2

i 62 M 0i since a

2i 62 Zi , by the choice, and

M 0i < Zi . Set hzi i D Zi so that z

ai

i D z�1i z�, � D 0; 1 (recall that z 2 N #/. We

have �2.Mi / D W and so Mi is uniquely determined according to Lemma 42.1. In


particular, we may assume � D 0 and we have Z.Mi / D ha2i i < W . We see thatM 0

i

and Z.Mi / are two distinct cyclic subgroups of order 4 in W .On the other hand, c2.W / D 2 and let W1 and W2 be two cyclic subgroups of

order 4 in W . For a given i < 7, one of these two subgroups is equal to Z.Mi /

and the other one is equal to M 0i . Therefore, we have (interchanging W1 and W2 if

necessary) W1 D Z.Mi / D Z.Mj / for certain i ¤ j , i; j 2 f1; 2; : : : ; 6g. We getCG.W1/ � hMi ;Mj i D G and so W1 D Z.G/. In that case W1 D Z.Mi / andW2 D M 0

i for all i < 7. It follows that all seven maximal subgroups of G=W2 areabelian so G=W2 is abelian. Then E4 < G

0 � W2 Š C4, a final contradiction.

For another proof of Theorem 66.1, see Theorem 69.1.

Lemma 66.2 (= Theorem 1.25). Suppose that all nonnormal subgroups of a non-Dede-kindian 2-group G have order 2. Then one of the following holds: (a) G Š M2n;(b) G D Z �G0, where Z is cyclic and G0 D D8; (c) G D D8 � Q8.

Exercise 1. Classify the p-groups all of whose subgroups of index p2 are absolutelyregular.

Exercise 2 ([Pas]). Let G be a non-Dedekindian 2-group all of whose nonnormal sub-groups have the same order 2n > 2. Then either G is metacyclic or n D 2.

Solution. As in Supplement to Theorem 1.25, E4 Š �1.G/ � Z.G/, unlessQ Š Q16,and all nonnormal subgroups of G are cyclic. Suppose that G is nonmetacyclic. Itfollows from Theorem 66.1 that ifM is a minimal nonmetacyclic subgroup of G thenM is either the group of Theorem 66.1(d) orM Š Q8 �C2 or is the group of Theorem66.1(d). In the first case,M has a subgroup X Š ha; b j a4 D b4 D 1; ab D a�1i.Since X has a nonnormal subgroup hbi of order 4, we get n D 2. In the second case,G has a subgroup Q Š Q8, and Q G G since Q is noncyclic. Set C D CG.Q/.Assume that C has a cyclic subgroup Z of order 4. If Q \ Z > f1g, then QC D

Q �Z Š D8 �Z has a nonnormal subgroup of order 2 (see Appendix 16, Subsection1o), a contradiction, since n > 1. If Q \ Z D f1g, then Q � Z has a nonnormalcyclic subgroup of order 4 (Theorem 1.20) since �1.Q � Z/ � Z.G/, and we getn D 2. Thus, one may assume that C is elementary abelian. Next,G=C is isomorphicto a 2-subgroup of Aut.Q/ Š S4 that contains a subgroup which is isomorphic toQ=Z.Q/ Š E4. If G=C Š E4, then G D Q � C D Q � E, where E is a subgroupof index 2 in C ; in that case G is Dedekindian, a contradiction. Thus, G=C Š D8.If x 2 G � C is such that hxC i is not normal in G=C , then the subgroup hxi is notnormal in G. Since o.x2/ � exp.C / D 2 and n > 1, we get o.x/ D 4 so n D 2.

Exercise 3. Classify the groups of Exercise 2 containing a subgroup Q Š Q8.

Exercise 4. If a 2-group of Exercise 2 is metacyclic, then either G Š Q16 or G D

ha; b j a2m

D b2n

D 1; ab D a1C2m�1

i, n � m (in the second case, G is minimalnonabelian).


Exercise 5. Suppose that a nonmetacyclic 2-group G of Exercise 2 has a subgroupsuch as the group of Theorem 66.1(d). If Z is a nonnormal subgroup of G of order 4then, all nonnormal subgroups of G=�1.Z/ have the same order 2.

Solution. Clearly, Z is cyclic. Suppose that H=�1.Z/ 6E G=�1.Z/; then H 6E G.Since G is nonmetacyclic, we get jH j D 4 (Exercise 2) so jH=Zj D 2.

Exercise 6. Suppose that a 2-groupG of Exercise 2 has a nonnormal (cyclic) subgroupZ. Then G=ZG is such as in Lemma 66.2. If G=ZG Š M2t , then G is metacyclic.

Hint. It remains prove that if G=ZG Š M2n , then G is metacyclic. Assume thatthis is false. Then jZj D 4, by Exercise 2. The group G=ZG has two distinct cyclicsubgroups A=ZG and B=ZG of index 2. The subgroups A and B are abelian soA \ B D Z.G/. Since ZG D ˆ.Z/ � ˆ.G/, we get dG/ D d.G=ZG/ D 2, and weconclude that G is minimal nonabelian. As above, �1.G/ Š E4 so G is metacyclicand G is not a counterexample.

Exercise 7. Study the non-Dedekindian 2-groups G all of whose nonnormal abeliansubgroups are cyclic of the same order 2n > 2. (The groups Q2n , n > 3, satisfy thiscondition.)

Hint. We have �1.Z.G// D �1.G/ Š E4, unless G is a generalized quater-nion group (indeed, given U < G nonnormal cyclic, let us consider the abelianU�1.Z.G//). All cyclic subgroups of G=�1.G/ are normal so this group is Dedekin-dian. Assume that G=�1.G/ is nonabelian. Then G=�1.G/ contains a subgroupL=�1.G/ Š Q8 (Theorem 1.20). Since all subgroups of L containing �1.G/, areabelian, we get jL W Z.L/j D 4 so jL0j D 2. It follows that L0 < �1.G/, a contra-diction since L=�1.G/ is nonabelian. Thus, G=�1.G/ is abelian. Now assume thatG has a subgroup Q Š Q8 and set C D CG.Q/. If G is metacyclic, it is generalizedquaternion (Proposition 1.19). The subgroup C is elementary abelian of order 4 sinceG is not of maximal class (Proposition 1.17). It follows that QC D Q � L, wherejLj D 2. Since all noncyclic abelian subgroups of QC are normal in G and QC isgenerated by these subgroups, we getQCGG. It follows thatG=C Š E4 sinceG=C isisomorphic to a subgroup of Aut.Q/ Š S4 containing a subgroup Š Q=Z.Q/ Š E4.In that case,G D Q�E, whereE is a subgroup of index 2 in C , soG is Dedekindian,contrary to the hypothesis. Thus, G has no subgroups isomorphic to Q8. It followsthat ifM is a minimal nonmetacyclic subgroup of G, thenM is isomorphic to a groupof Lemma 66.1(d).

Problem. Classify the nonmetacyclic 2-groups all of whose minimal nonmetacyclicsubgroups have order 16.

�67

Determination of U2-groups

Recall that a 2-group G is said to be a U2-group if it contains a normal four-subgroupR such that G=R is of maximal class and, provided T=R is cyclic of index 2 in G=R,then �1.T / D R (see ��17, 18). The subgroup R is said to be the kernel of G. Thekernel R is the unique normal four-subgroup of G.

Theorem 67.1 (Janko [Jan8]). Let G be a U2-group with d.G/ D 3. Then there isM 2 �1 of maximal class, and one of the following holds:

(a) G D M � C2;

(b) G D M � C4 withM \ C4 D Z.M/;

(c) G D M hui, where u is an involution inducing on M an involutory “outer”automorphism such that CM .u/ D M1 is of maximal class, jM W M1j D 2, andfor each x 2 M �M1, xu D xz with hzi D Z.M/ (in particular, z2 D 1).

Proof. Let G be a U2-group with the kernel R and d.G/ D 3. Then G possesses amaximal subgroupM such thatR 6� M since, by hypothesis, R 6� ˆ.G/. In that case,M \ R D hzi is of order 2 andM=hzi Š G=R soM is nonabelian. Next,M has noG-invariant abelian subgroups of type .2; 2/ since R 6� M soM is of maximal class(Lemma 1.4). Since jM j > 23,M=hzi Š G=R is dihedral.Let u 2 R � hzi. If R � Z.G/, then G D hui � M and we are done. Suppose

R 6� Z.G/ and setM1 D CM .R/ D CM .u/ so that jM W M1j D 2. IfM1 is cyclic,then take an element v of order 4 in M1 and x 2 M � M1. Since R# D fu; z; uzg

and u and z are not G-conjugate, we have .uv/x D uxvx D .uz/vx D .uz/v�1 D

.uz/.vz/ D uv and so C D huvi Š C4 and C � Z.G/ since hM �M1i D M andCM D G. In this caseG D M �C with C \M D hzi D Z.M/ and we have obtainedthe group from (b). Now suppose thatM1 is of maximal class. Then u induces onMan outer involutory automorphism such thatM1 D CM .u/ is a maximal subgroup ofM and for each x 2 M �M1, xu D xz.

In the rest of this section we assume that G is a U2-group with the kernel R andd.G/ D 2. Then G=R is of maximal class and order 2n, n � 3, and ˆ.G/ � R.Let T=R (of order 2n�1) be a cyclic subgroup of index 2 in G=R and let hai be acyclic subgroup of T which covers T=R. Since �1.T / D R, hai is of order 2n andhai \R D hzi is of order 2. Hence T is either abelian of type .2n; 2/ or T Š M2nC1 .

�67 Determination of U2-groups 203

In any case, ˆ.T / D ha2i, z 2 ha2i, ha2i is normal in G, and z 2 Z.G/. Letv D a2n�2

so that o.v/ D 4, v2 D z, A D Rhvi D �2.T / is abelian of type.4; 2/, and hvi D A \ ˆ.T / is normal in G. Hence A=R D Z.G=R/ and so for eachx 2 G � T , x2 2 A since G=R is of maximal class. Since Ã1.G/ D ˆ.G/ D Rha2i

is of exponent 2n�1, we get hai 6� ˆ.G/ so, if hai is normal in G, G is metacyclic.If G=R Š Q2n , then for each x 2 G � T , x2 2 A � R. All elements in A � R

are of order 4 and so o.x/ D 8. In this case �2.G/ D �2.T / D A and so G ismetacyclic and is isomorphic to a group (c) of Lemma 42.1. Therefore, we may assumein the sequel that G=R 6Š Q2n . In particular, T=R is the unique cyclic subgroupof index 2 in G=R. Conversely, assume that G is metacyclic. Then G0 is cyclic,G0 � ˆ.G/ D Rha2i, jG0 \ Rj D 2, and G0 covers .Rha2i/=R. Thus jG0j D 2n�1.Let S be a cyclic normal subgroup of G such that G0 < S and jS W G0j D 2. Butthen jS j D 2n and .RS/=R is a cyclic subgroup of index 2 in G=R. The uniquenessof T=R implies RS D T . But cn.T / D 2, and so hai is normal in G.We have proved that (under our assumptions) G is metacyclic if and only if hai is

normal G. We shall use the above notation in the rest of this section.

Theorem 67.2. Let G be a metacyclic U2-group with the kernel R. Then one of thefollowing holds.

(a) G D ha; b j a2n

D b4 D 1; ab D a�1z�; � D 0; 1; z D a2n�1

; n � 3i,where R D hb2; zi D Z.G/ and G=R Š D2n .

(b) G is isomorphic to a group (c) of Lemma 42.1 and hereG=R Š Q2n and Z.G/ Š

C4.

(c) G D ha; b j a2n

D b4 D 1; ab D a�1C2n�2

; z D a2n�1

; n � 4i, whereR D hb2; zi, G=R Š SD2n , and Z.G/ D hzi Š C2.

Proof. Suppose thatG is a metacyclic U2-group with the kernelR, whereG=R 6Š Q2n

(for the case G=R Š Q2n , see the second paragraph following the proof of Theorem67.1). Then hai is normal in G, by the paragraph preceding the theorem. Since G isnot of maximal class, R D �1.G/.Assume that G=R is dihedral. Take x 2 G � T ; then x2 2 R so H D hx;Ri

is of order 8 and abelian (otherwise, G would be of maximal class). It follows thatCG.R/ � hG � T i D G. We have obtained the group (a).Suppose now that G=R Š SD2n , n � 4. There is b 2 G � T with b2 2 R. If

b2 2 hzi, then ha; bi is a maximal subgroup of G, by the product formula so R <

ˆ.G/ < ha; bi. However, this is a contradiction since we must have ha; bi=R D G=R

so that ha; bi D G. Thus b2 D u 2 R � hzi. We have, in view of the structure ofG=R that ab D a�1vz� .� D 0; 1/, where v D a2n�2

(recall that hai is normal inG). We have .a4/b D .ab/4 D .a�1vz�/4 D a�4, and so vb D v�1 since v 2 ha4i

(recall that hai is normal in G). We compute further ab2

D au D .a�1vz�/b D

.a�1vz�/�1v�1z� D av�2 D az, so au D az and therefore T Š M2nC1 . If � D 1,then we replace b with b0 D bu; then .b0/2 D b2 D u and ab0

D abu D .a�1vz/u D


a�1zvz D a�1v. Hence, we may assume from the start that ab D a�1v, and we haveobtained the group (c).

Theorem 67.3. Let G be a nonmetacyclic U2-group with d.G/ D 2 and the kernel R.Then one of the following holds:

(a) G D ha; b j a2n

D 1; a2n�1

D z; b2 D z�; � D 0; 1; ab D a�1u; u2 D

Œu; a� D Œu; b� D 1; n � 3i; where R D hu; zi D Z.G/ and G=R Š D2n .

(b) G D ha; t j a2n

D t2 D 1; a2n�1

D z; at D a�1u; u2 D 1; Œu; a� D

Œu; t � D z; n � 3i, where R D hu; zi and G=R Š D2n . If n D 3, thenZ.G/ D ha2i Š C4 and if n > 3, then Z.G/ D hzi Š C2.

(c) G D ha; t j a2n

D t2 D 1; a2n�1

D z; a2n�2

D v; at D a�1vu; u2 D

Œu; a� D 1; ut D uz; n � 4i, where R D hu; zi, G=R Š SD2n , and Z.G/ D

hvui Š C4.

(d) G D ha; b j a2n

D 1; a2n�1

D z; a2n�2

D v; b2 D z�; � D 0; 1; ab D

a�1vu; u2 D Œu; b� D 1; ua D uz; n � 4i, where R D hu; zi, G=R Š SD2n

and Z.G/ D hzi Š C2.

Proof. Suppose that G is a nonmetacyclic U2-group with the kernel R and d.G/ D 2.In that case G=R 6Š Q2n (see the proof of Theorem 67.2) so G=R has the uniquecyclic subgroup T=R of index 2. In that case, T is either abelian of type .2n; 2/ orT Š M2nC1 . Let hai be a cyclic subgroup of index 2 in T ; then hai is not normalin G (see the paragraph preceding Theorem 67.2). Let z be an involution in hai; thenz 2 Z.G/.

(i) Suppose that G=R Š D2n and T is abelian. If x 2 G � T , then x2 2 R. SinceÃ1.G/ D ˆ.G/ � R and Ã1.T / does not contain R, there is b0 2 G � T with b2

0 D

u0 2 R � hzi and so R � Z.G/. But hai is not normal in G and so ab0 D a�1u0z�

.� D 0; 1/. We have .b0a/2 D b0ab0a D b2

0ab0a D u0a

�1u0z�a D z� . Hence,

setting b0a D b and u0z� D u, we have b2 D z� and ab D a�1u. We have obtained

group (a).

(ii) Now suppose that G=R Š D2n and T Š M2nC1 . In that case, R 6� Z.G/.Since �1.G=R/ D G=R, it follows that there is a subgroup H=R of order 2 such thatH 6� T and H is nonabelian. Then H Š D8. Let t 2 H � R be an involution. Thenat D a�1u with u 2 R�hzi, ut D uz, and ua D uz. We have obtained the group (b).

(iii) It remains to consider the case where G=R Š SD2n , n � 4. IfQ=R is the gen-eralized quaternion subgroup of index 2 in G=R, then .T \Q/=R is a cyclic subgroupof index 2 in Q=R and �1.T \ Q/ D R. Thus Q is a U2-group with the kernel R.Since Q=R is generalized quaternion, Q is metacyclic in view of �2.Q/ D �2.T /

is of order 8 (see Lemma 42.1), and the same lemma implies that Z.Q/ Š C4. Inparticular, R 6� Z.G/.Suppose, in addition, that T is abelian. Then CG.R/ D T . If D=R is the dihedral

subgroup of index 2 in G=R, then �1.D \ T / D R and so D is a U2-group with

�67 Determination of U2-groups 205

R 6� Z.D/. In particular, there is an element x 2 D � T such that x2 2 R andRhxi Š D8. It follows that there is an involution t 2 D � T acting faithfully on R(see (ii)) and, since hai is not normal in G, we get at D a�1vu with u 2 R � hzi,ut D uz, Œu; a� D 1. We have obtained the group (c).Finally, suppose that T is nonabelian, i.e., T Š M2nC1 so that hai acts non-trivially

on R. Then, in view of R 6� Z.G/, M D CG.R/ is a maximal subgroup of G,where M=R is noncyclic so dihedral, T \ M D ha2iR, ˆ.T / D ha2i is normal inG, �1.T \M/ D R, and soM is a U2-group. Since R � Z.M/, M=R is dihedral.Indeed, if Q=R is the generalized quaternion subgroup of index 2 in G=R, then weknow thatQ is metacyclic. Since d.G/ D d.T / D d.Q/ D 2 andG is, by assumption,nonmetacyclic, Theorem 44.5 implies that d.M/ D 3. Hence M is isomorphic to agroup (a) of Theorem 67.1. In particular, there is b 2 M � T such that b2 2 hzi.We set b2 D z� , � D 0; 1. Finally, b centralizes R, hai acts non-trivially on R, andab D a�1vu with u 2 R � hzi. We have obtained the group (d).

Corollary 67.4 (Berkovich). Suppose that a 2-group G of rank three has order 2nC2

and class n > 2. Then one of the following holds:

(a) G D H1 � C , where H1 is of maximal class and jC j D 2 (three groups); allfour subgroups of maximal class and index 2 are isomorphic.

(b) G D H1 �C , whereH1 is of maximal class and C D Z.G/ is cyclic of order 4,C \H1 D Z.H1/.

In what follows we assume that Z.G/ D Z.H1/ so Z.G/ is of order 2.

(c) G D ha; b; x j a2n

D b2 D x2 D 1; ab D a�1; ax D a1C2n�1

; bx D

a2n�1

bi. HereH1 D ha; bi Š D2nC1 .

(d) G D ha; b; x j a2n

D x2 D 1; a2n�1

D b2; ab D a�1; ax D a1C2n�1

; bx D

a2n�1

bi. HereH1 D ha; bi Š Q2nC1 .

(e) G D ha; b; x j a2n

D b2 D x2 D 1; ab D a�1C2n�1

; ax D a1C2n�1

; bx D

a2n�1

bi. HereH1 D ha; bi Š SD2nC1 .

Proof. Let K D K3.G/; then jG0 W Kj D 2. We have jZ.G=K/j D 4 and G=K is notminimal nonabelian. Let H=K < G=K be minimal nonabelian; then G D K�.G/,where �.G/=K D Z.G=K/. It follows from Theorem 1.40 that cl.H/ D cl.G/ soH is of maximal class. By Lemma 1.4, G has a normal subgroup R of type .2; 2/.Since jH j > 8, we get R 6� H so G D HR. Since G=R Š H=.H \ R/ Š D2n , weconclude that G is a U2-group. Now the result follows from Theorem 67.1.

Exercise. Let G be a group of order pm, cl.G/ D m � 2 and G=G0 Š Ep3 . Let L bea G-invariant subgroup of index p in G0. LetH=L < G=L be nonabelian of index p.Prove thatH is of maximal class.

�68

Characterization of groups of prime exponent

Recall that the set † D fAi gn1 of subgroups of a group G is a partition of G if G D

Sn1 Ai and Ai \Aj D f1g for i ¤ j . In what follows we assume that† is a nontrivial

partition of a group G (this means that n > 1 and Ai > f1g for all i ). SubgroupsAi are called components of †. If all components of † have equal order, then G issaid to be equally partitioned [Isa8]. It follows from Cauchy’s lemma that, for equallypartitioned G, we have �.A1/ D �.G/.In this section we characterize groups of prime exponent as equally partitioned

groups proving the following Isaacs’ result [Isa8]: a group G is equally partitionedby a nontrivial partition † if and only if G is of prime exponent. Note that this theo-rem is not a consequence of known results on partitioned groups.

Lemma 68.1. Let † be a nontrivial partition of a group G and x; y 2 G# with xy D

yx. If x and y lie in different components of †, then o.x/ D o.y/ is a prime.

Proof. Suppose that o.x/ < o.y/. Then .xy/o.x/ D yo.x/ ¤ 1, and so xy and y lie inthe sameH 2 †. Then x D .xy/y�1 2 H , which is not the case. Thus, o.x/ D o.y/.If n > 1 is a proper divisor of o.x/, the nonidentity commuting elements xn and yof different orders lie in different components of †, contrary to what has just beenproved. Thus o.x/ D o.y/ is a prime.

If a component H 2 † does not contain Z.G/, then exp.H/ D p for some primep. Indeed, if x 2 H # and z 2 Z.G/ �H , then x and xz commute and lie in differentcomponents of † so o.x/ D o.xz/ is a prime, say p (Lemma 68.1); then o.z/ D p.Assume that y 2 H # is of prime order q ¤ p. In that case, o.y/ D o.yz/ D q soo.z/ D q ¤ p, a contradiction. It follows that, if a p-groupG of exponent> p admitsa nontrivial partition † and Z.G/ � H 2 †, thenH is the unique component of † ofcomposite exponent soH contains the Hudges subgroup Hp.G/ of G.

Lemma 68.2 ([Isa8]). Let † be a nontrivial partition of an equally partitioned groupG and ¿ ¤ X � G#. Then there existsH 2 † such that Xz 6� H for all z 2 G.

Proof. Suppose that the lemma is false and for each H 2 †, choose XH , a G-conjugate of X , with XH � H . Let NH D NH .XH / so that H contains at leastjH W NH j conjugates of X . Let N D NG.X/, g D jGj, h D jH j. We have

jG W N j D jG W NG.XH /j � jG W NH j D jG W H j jH W NH j

�68 Characterization of groups of prime exponent 207

(the first equality follows since X and XH are conjugate inG), and hence jH W NH j �hg

jG W N j. Since g�1h�1

� gh

D g�hh.h�1/

> 0, we get gh< g�1

h�1D j†j. Now jG W N j is

the number of conjugates of X in G and thus, since † is a partition, we get

jG W N j �X

H2†

jH W NH j D j†jjH W NH j � j†j h

gjG W N j;

so gh

� j†j D g�1h�1

, a contradiction.

Lemma 68.3 ([Isa8]). Let † be a nontrivial partition of an equally partitioned groupG. Then every element of G# has a prime order.

Proof. Suppose that x 2 G has a composite order and let K D Kx be the conjugacyG-class of x. By Lemma 68.2, there existsH 2 † such that xz 62 H for all z 2 G sothat K \H D ¿. By Lemma 68.1, no element of H # centralizes any element of K.Thus, H acts semi-regularly by conjugation on K and hence jH j divides jKj. Since1 2 H and 1 62 K it follows that jH j < jKj.Now pick F 2 † with x 2 F . Since jF j D jH j < jKj, the set K � F is

nonempty, and, obviously, this set is F -invariant. By Lemma 68.1, F acts on K � F

semi-regularly via conjugation so that jH j D jF j divides jK � F j > 0. Since K D

.K�F /[ .K\F / is a partition, jF j divides jK\F j < jF j, which is a contradiction.Thus elements of G# have prime orders.

Exercise 1. Suppose that U is a nontrivial normal p-subgroup of G, where p is thelargest prime divisor of jGj. Assume that every element of G# has prime order and letP 2 Sylp.G/. Then either P D G or jG W P j is prime (and so P GG).

Solution. Take q 2 �.G/�fpg andQ 2 Sylq.G/. ThenQU is a Frobenius group withkernel U so jQj D q. It follows that P G G. Indeed, if r is a minimal prime divisorof jGj, then jGjr D r so G is r-nilpotent (Burnside). Applying this to the r-normalcomplement of G, we at last prove our claim. If P < G, then G is a Frobenius groupwith kernel P and complement of square free order without elements of compositeorder. It follows that p0-Hall subgroup of G is of prime order [BZ, Chapter 10].

Exercise 2. Suppose that every element of G# is of prime order. Let P 2 Sylp.G/,where p is the largest prime divisor of G. Then P is a TI-set, i.e., P \ P x D f1g forx 2 G � NG.P /, unless P E G.

Solution. Assume that f1g < D D P \ P x , where P ¤ P x and jDj is as largeas possible. Applying Exercise 1 to NG.D/ whose Sylow p-subgroup is not normal(Burnside), we obtain a contradiction.

Now we are ready to prove the main result of this section.

Theorem 68.4 ([Isa8]). Let† be a nontrivial partition of an equally partitioned groupG. Then exp.G/ D p for some prime p.


Proof. One may assume that j�.G/j > 1 (Lemma 68.3). Let p be the largest primedivisor of jGj and P 2 Sylp.G/. By Lemma 68.3 and Exercise 2, P is a TI-set sinceP 6E G. By Lemma 68.3 and Exercise 1, NG.P / D C P , where either C D f1g orjC j D q, a prime.In the first case, G D P H is a Frobenius group with kernel H and complement

P . Then jP j D p, a prime. In that case,G is not equally partitioned (by Lemma 68.2:take there X D P #).Thus, NG.P / D C P is a Frobenius group with kernel P and complement C of

prime order q ¤ p.Let jGj D g, jP j D pb , jH j D h for H 2 † and p˛ D hp, the p-part of h. By

Lemma 68.2, given x 2 P #, there exists F 2 † such that xz 62 F for all z 2 G, so˛ < b since jF j D jH j D h (Sylow).Since P is a TI-set, jP \H j D p˛ for all H 2 † such that P \ H ¤ f1g (note

that a Sylow p-subgroup of any component H of † is also a TI-subset inH ).By Lemma 68.2, one may choose H 2 † with H \ C x D f1g for all x 2 G. Take

P0 2 Sylp.H/. We may assume that P0 < P . Since P is a TI-set, we must haveNH .P0/ � NG.P / D C P . It follows that NH .P0/ D P0 C0, where p − jC0j.Assume that y 2 C #0 . Then f1g < P0 � P \ P y so P y D P and so C0 is conjugatewith C in NG.P /, by Sylow’s theorem, contrary to the choice of H . It follows thatC0 D f1g so NH .P0/ D P0. By assumption, �.g/ D �.h/ so, to complete the proof,it suffices to prove thatH is a p-group. Assume that this is false. Then H D P0 Q0

is a Frobenius group with kernel Q0 and complement P0 since P0 is a TI-subgroupin H (see Frobenius’ theorem in [BZ, Chapter 10]). Since P0 has only one subgroupof order p and its exponent is p, we get jP0j D p, i.e., ˛ D 1. Since Q0 is nilpotent(Thompson), it is a q-group for some prime q, by Lemma 68.3. Then G is a fp; qg-group, and so G is solvable, by Burnside’s two-prime theorem. We have f1g < Q1 D

Oq.G/ since P is a nonnormal TI-subgroup of G (in view of the existence of H ).Since P Q1 is a Frobenius group, we get jP j D p D jP0j, a contradiction since1 D ˛ < b. Thus, H is a p-subgroup.

For more elementary proof of Theorem 68.4, see [Isa8].

Exercise 3. Suppose that a 2-group G of exponent > 2 admits a nontrivial partition.Then G D Ahbi, where o.b/ D 2 and b inverts A. (Hint. Use Lemma 68.1 and theparagraph following it.)

Problem 1 (Isaacs). Does there exist a group partitioned by proper subgroups of equalorder not all of which are abelian?

Problem 2. Let † be a nontrivial partition of G. Study the structure of G if orders ofcomponents of † form a chain under divisibility.

Problem 3. Classify all nontrivial partitions of elementary abelian p-groups.

�69

Elementary proofs of some Blackburn’s theorems

In this section we present elementary proofs of some Blackburn’s theorems; theseproofs are due to the first author.

1o. Blackburn has classified minimal nonmetacyclic p-groups. For a simpler proofin the case p D 2, due to the second author, see Theorem 66.1. Here we offer anotherelementary proof for all p.The p-groups, p > 2, without normal elementary abelian subgroups of order p3

were classified by Blackburn (see Theorem 13.7). The proof of Theorem 13.7 wasbased on the deep theorem 12.1(a). Here we offer another proof based on Theorem12.12 which is elementary for p D 3 (see Remark 1.)

Remark 1. The proof of Theorem 12.12(a) is elementary. Now let p D 3. In the proofof part (c) of that theorem we use the following fact: If G is a 3-group of maximalclass and order 3m > 33, then jG=Ã1.G/j D 33. It suffices to prove this for m D 4.By Lemma 64.1(m), jG=Ã1.G/j > 32, since G is nonmetacyclic. It is known thatgroups of exponent 3 are of class � 2, so jG=Ã1.G/j D 33.

Similarly, using Theorem 9.5. it is easy to show that, if a nonabelian group G ofexponent 3 is generated by two elements, then its order is 33.

Theorem 69.1. If G is a minimal nonmetacyclic p-group, then, one of the followingholds: (a) G is of order p3 and exponent p; (b) G is a group of maximal class andorder 34; (c) p D 2 and jGj � 25; if jGj D 25, then G is an A2-group.

Proof. We use induction on jGj. One may assume that jGj > p3. Clearly, we havejG=Ã1.G/j � p3 (otherwise, all maximal subgroups of G are nonmetacyclic).

(i) Let p > 2. Obviously, G has no subgroups of order p3 and exponent p. Then Gis irregular (otherwise, jG=Ã1.G/j D j�1.G/j D p2 and G is metacyclic, by Lemma64.1(m)). Therefore, if jGj D p4, then cl.G/ D 3 so p D 3 (Lemma 64.1(a)). Nowsuppose that jGj > p4. Let R � Ã1.G/ \ Z.G/ be of order p. By Lemma 64.1(m),G=R is nonmetacyclic so minimal nonmetacyclic; then p D 3 and G=R is of maximalclass and order 34, by induction. We get jGj D 35 and jG W Ã1.G/j D 33 (Remark 1).LetH=R be maximal in G=R; thenH=R is either abelian of type .9; 3/ or isomorphicto M33 . Since H is metacyclic of order 34, jH 0j � 3 so H is either abelian or A1-group (Lemma 65.2(a)). Since R � ˆ.H/ � ˆ.G/, we conclude that all members of


the set �1 are either abelian or A1-groups so G is an A2-group. By Lemma 64.1(p),jG W G0j D 32 hence G0 Š E33 (Theorem 65.7(d)) so nonmetacyclic, a contradiction.

(ii) Now assume that p D 2; then d.G/ D 3, by Lemma 64.1(n). Since all maximalsubgroups of an (abelian) groupG=G0 are 2-generator and d.G=G0/ D 3, we concludethat G=G0 Š E8 so G0 D ˆ.G/. TakeH 2 �1.

Set NG D G=K3.G/; then j NGj � 24. Since d. NG/ D 3, NG is neither metacyclic noran A1-group. Next, NG0 � Z. NG/ since cl. NG/ D 2. In view of j NH W .Z. NG/ \ NH/j �

j NH W NG0j D 4 and d. NH/ D 2, NH is either abelian or A1-group (Lemma 65.2(a)). Itfollows that NG is a (nonmetacyclic)A2-group. By Theorem 65.7(d), NG0 is elementaryabelian; in particular, j NG0j � 4 ( NG0 is metacyclic!) and exp. NG/ D 4. Therefore, wehave 22 � j NG0j D 1

8j NGj so j NGj � 25.

Suppose that j NGj D 25; then j NG0j D 4. In that case, Z. NG/ D NG0 .Š E4/, byLemmas 4.9 and 64.1(q) (indeed, if Z. NG/ > NG0, then j NG0j D 2). Then all (noncyclic!)subgroups NA of NG of order 8 contain Z. NG/ D NG0 (otherwise, d. NAZ. NG// > 2, which isnot the case since NAZ. NG/ < NG/ so NA G NG. Since NG0 D ˆ. NG/, we conclude that NG isspecial.

Assume that K D K3.G/ > f1g. By the above, NG D G=K is anA2-group of order� 25.

Suppose that jGj D 25. Assume that G is not an A2-group. Then G contains anonabelian subgroup H of order 8. By Lemma 64.1(i), CG.H/ contains a subgroupF of order 4 since G is not of maximal class. Taking F so that H \ F D Z.H/, weget H � F 2 �1 and d.H � F / D 3, a contradiction. Thus, if jGj D 25, then G is aspecialA2-group.

Suppose that j NGj D 24. We claim that this is impossible. Let R be a G-invariantsubgroup of index 2 in K; then cl.G=R/ D 3. One may assume that R D f1g: thenjGj D 25, contrary to the previous paragraph.

It remains to consider the case where K > f1g and jG=Kj D 25. We claim thatthis is impossible. Let R be a G-invariant subgroup of index 2 in K. To obtain acontradiction, one may assume that R D f1g; then jGj D 26. Since cl.G/ D 3 andd.G/ D 3, G is not an A2-group (Lemma 65.4(c)). It follows that K D K3.G/

is the unique minimal normal subgroup of G. Therefore, there exists a nonabeliansubgroup H of index 4 in G. We have K < H (otherwise, d.K � H/ > 2). SinceE4 Š Z.G=K/ < H=K , H is not of maximal class. There are the following twopossibilities for (the metacyclic subgroup) H : eitherH Š M24 or H D ha; b j a4 D

b4 D 1; ab D a�1i. Since NG0 < NH , we conclude thatH GG.

Assume that H Š M24 . We have c3.H/ D 2 and H G G so, if L is a cyclicsubgroup of H of index 2, then jG W NG.L/j � 2. Let H < M � NG.L/ withjM W H j D 2. There are two possibilities forM=L: M=L 2 fC4;E4g. IfM=L Š C4,then CM .L/ D L sinceH is the unique subgroup of order 16 inM containing L andH is nonabelian. Then M=L Š C4 is a subgroup of Aut.L/ Š E4, a contradiction.Now suppose that M=L Š E4. Let F=L be a subgroup of order 2 in M=L with

�69 Elementary proofs of some Blackburn’s theorems 211

F=L ¤ H=L. SinceK < L < F , F=K G G=K (see the end of the second paragraphof (ii)). We haveG0 D H\F D L is cyclic. This is a contradiction sinceG0=K Š E4.Now assume that H D ha; b j a4 D b4 D 1; ab D a�1i. Since H has exactly

two nonidentity squares (a2 2 K and b2) and K is characteristic inH , we see that allsubgroups of order 2 are characteristic in H so normal in G. This is a contradictionsince K is the unique normal subgroup of G of order 2. The proof is complete.

The groups of Theorem 69.1(c) are described in Theorem 66.1.

2o. Here we offer another proof of Theorem 13.7. As follows from the proof ofTheorem 13.7, it suffices to prove Theorem 69.3, below. First we prove the following

Lemma 69.2. Let G be a p-group, p > 2 and c1.G/ D 1 C p. Then G is eithermetacyclic or a 3-group of maximal class.

Proof. Let G be a counterexample of minimal order. Then every proper subgroup ofG is either metacyclic or a 3-group of maximal class, by induction. Let H � G beminimal nonmetacyclic; thenH is a 3-group of maximal class and order 34 (Theorem69.1) soH < G. Let B be a nonabelian subgroup of order 33 inH (B exists sinceHis not minimal nonabelian); then CG.B/ 6� B (Lemma 64.1(i)). Let U be a subgroupof order 9 in CG.B/ with B \ U D Z.B/; then BU < G is neither metacyclic nor ofmaximal class, a contradiction.

It follows from Lemma 69.2 the following new

Proof of the last assertion of Lemma 64.1(m) = Theorem 9.11. We have to prove that,if p > 2 and jG=Ã1.G/j � p2, thenG is metacyclic. Indeed, in that case,G is regular,by Lemma 64.1(a). If jG=Ã1.G/j D p, then G is cyclic. Now let jG=Ã1.G/j D p2;then j�1.G/j D p2 (Lemma 64.1(a)) so c1.G/ D 1Cp. By Lemma 69.2, G is eithermetacyclic or an irregular 3-group of maximal class. In the second case, however,jG=Ã1.G/j D 33, by Remark 1.

Remark 2. If a p-group G has an absolutely regular maximal subgroup A and irreg-ular subgroup M of maximal class, then G is of maximal class. This coincides withProposition 12.13.

Theorem 69.3. If a p-group G, p > 2. has no normal subgroups of order p3 andexponent p. then G is either metacyclic or a 3-group of maximal class.

Proof. Suppose that G is a counterexample of minimal order. Then G is irregular(Lemma 64.1(a,m)). By Theorem 10.4, G has no elementary abelian subgroups oforder p3 so it has no subgroups of order p4 and exponent p. Let Ep2 Š R G G

(Lemma 64.1(x)). By Lemma 69.2, T D CG.R/ is metacyclic since �1.T / D R,and so T 2 �1. By Lemma 69.2 again, there exists x 2 G � R of order p; thenB D hx;Ri is of order p3 and exponent p (Lemma 64.1(a)). Since B 6E G, there


exists y 2 NG.B/ � B of order p (see the proof of Lemma 64.2). Set H D hy;Bi;then exp.H/ > p so p D 3, H is of maximal class and order 34 (Lemma 64.1(a)).Existence of T andH shows: G is of maximal class (Remark 2).

We are ready to prove Blackburn’s Theorem 13.7.

Theorem 69.4 (Blackburn). Suppose that a group G of order pm, m > 3, p > 2, hasno subgroups Š Ep3 . If G is neither metacyclic nor a 3-group of maximal class, thenG D UE, where E D �1.G/ is nonabelian of order p3 and exponent p, U is cyclic,Z.G/ � C .

Proof. By Theorem 69.3, G has a normal (nonabelian) subgroup E of order p3 andexponent p. By Theorem 10.4, G has no elementary abelian subgroups of order p3 soit has no subgroups of order p4 and exponent p. Therefore, setting C D CG.E/ wesee that subgroups of order p in C lie in C \E D Z.E/ so C is cyclic. Next, C GG

and G=C is isomorphic to a p-subgroup of Aut.E/ (which is nonabelian of order p3

and exponent p) containing a subgroup EC=C Š E=.E\C/ Š Ep2 . IfG=C Š Ep2 ,then G D E � C , and we are done since G is of class 2 so�1.G/ D E.Now let jG=C j D p3; then G=C is nonabelian of order p3 and exponent p. Let

Ep2 Š R < E be normal in G and set F D CG.R/; then F 2 �1 since E 6� CG.R/.Since �1.F / D R, F is metacyclic (Lemma 69.2). Assume that there is x 2 G � E

of order p. ThenH D hx;Ei is of exponent > p so p D 3 andH is of maximal classand order 34. Existence of F and H shows that G is of maximal class (Remark 2), acontradiction. Thus, �1.G/ D E. Let U=C < G=C be a subgroup of order p suchthat U=C 6� CE=C . Since �1.U / D U \ E D Z.E/, U has only one subgroup oforder p, namely Z.E/, so U is cyclic; clearly, G D EU . Since CG.C / � EU D G,the proof is complete.

Corollary 69.5. Let A be a p-group of operators of a nonmetacyclic p-group G, p >2, jGj > p3. If all proper A-invariant subgroups of G of order p3 are metacyclic,then G is a 3-group of maximal class.

Indeed, by Theorem 10.4,G has no elementary abelian subgroups of order p3. Nowthe result follows from Theorem 69.4.

Exercise 1. Let G be a noncyclic p-group. Suppose that for each H G G, there ish 2 H such that hhx j x 2 Gi D H . Prove that G is of maximal class.

Solution. We use induction on jGj. One may assume that G is nonabelian and jGj >

p3. Clearly, jG W G0j D p2. Indeed, if this is not true, let H=G0 < G=G0 be of type.p; p/. Then there is no h 2 H such that hhx j x 2 Gi D H . Next, the factors of thelower central series of G apart of the first one, are cyclic and the factors of the uppercentral series of G apart of the last one, are cyclic. One may assume that jGj > p3.Let R be a minimal normal subgroup of G. Then, by induction, G=R is of maximalclass. Assuming that G is not of maximal class, we conclude that Z.G/ Š Cp2 and,

�69 Elementary proofs of some Blackburn’s theorems 213

by Lemma 1.4, G has a normal abelian subgroup L of type .p; p/. We have R < L soZ.G=R/ Š Ep2 , and G=R is not of maximal class, a contradiction.

Exercise 2. If a 3-groupG of order> 34 is nonmetacyclic but all its minimal nonmeta-cyclic subgroups have the same order 34, then G is of maximal class with �1.G/ Š

E9. (Hint. Use Theorem 69.4.)

Exercise 3. If G is a minimal nonmetacyclic group of order 25, then (a) G is specialwith jG0j D 4, (b) the set �1 has exactly one abelian member.

Solution. (a) By Theorem 69.1, G=G0 Š E8 so G0 D ˆ.G/. Assume that G has anonabelian subgroupH of order 8. SinceG is not of maximal class, we get CG.H/ 6�

H (Lemma 64.1(i)). If F is a subgroup of order 4 in CG.H/ with Z.H/ < F , thend.HF / D 3, a contradiction. Thus, G is an A2-group so G0 Š E4 (Lemma 65.7(d)).We have CG.G

0/ � hH j H 2 �2i so G0 � Z.G/. If G0 < Z.G/, then by Lemma64.1(q), jG0j D 2, a contradiction. It follows that G is special.(b) Assume that the set �1 has no abelian members. Let Zi , i D 1; 2; 3, be all

subgroups of order 2 in G and let Ai D fT 2 �1 j T 0 D Zi g. Since G=Zi has exactlythree abelian subgroups of index 2, we have jAi j D 3, i D 1; 2; 3, and assume that allAi are non-empty. Obviously, the setsA1;A2;A3 are pairwise disjoint, a contradictionsince jA1 [ A2 [ A3j D 3 C 3 C 3 D 9 > j�1j. Thus, �1 has exactly one abelianmember since jG0j D 4.

�70

Non-2-generator p-groups all of whosemaximal subgroups are 2-generator

We begin with a study of p-groups of the title by proving the following five theorems.We use results on A2-groups from �65. The following theorem is essential in deter-mination of A2-groups (see �71). For p > 2, Theorem 70.1 was proved by Blackburn(however, another proof is presented below).

Theorem 70.1. Let G be a nonabelian p-group with d.G/ D 3. Suppose that allmembers of the set �1 are generated by two elements. If G is of class 2, then it is anA2-group of order � p6 and one of the following holds:

(a) jGj D p4 and either G D E � C , where E is nonabelian of order p3, C Š Cp2 ,E \ C D Z.E/ or G D Q � Z withQ Š Q8 and jZj D 2.

(b) G D ha; b; c j a4 D b4 D Œa; b� D 1; c2 D a2; ac D ab2; bc D ba2i isthe minimal nonmetacyclic group of order 25, G is special, �1.G/ D G0 D Z.G/ D

ˆ.G/ D ha2; b2i Š E4, ha; bi Š C4 �C4 is the unique abelian maximal subgroup ofG. All subgroups of G of order 8 contain Z.G/ D G0.

(c) jGj D p5, Ep2 Š ˆ.G/ D G0 D Z.G/ < �1.G/ Š Ep3 so G is special,Exactly p2 members of the set �1, not containing �1.G/, are metacyclic and exactlyone of them is abelian and p C 1 other members of the set �1, containing �1.G/, arenonmetacyclic minimal nonabelian. More precisely, G D ha; b; ci with

ap D bp2

D cp2

D Œb; c� D 1; Œa; b� D x; Œa; c� D y; bp D x˛yˇ ;

cp D x�yı ; xp D yp D Œa; x� D Œa; y� D Œb; x� D Œb; y� D Œc; x� D Œc; y� D 1;

where in case p D 2 we have ˛ D 0, ˇ D � D ı D 1, and in case p > 2,the number 4ˇ� C .ı � ˛/2 is a quadratic non-residue mod p. Here G0 D hx; yi,�1.G/ D G0 � hai, and hb; ci Š Cp2 � Cp2 is the unique abelian member of the set�1. All subgroups of G of order p3 contain Z.G/ D G0.

(d) p D 2, G D ha; b; ci is of order 26 and

a4 D b4 D c4 D 1; Œa; b� D c2; Œa; c� D b2c2; Œb; c� D a2b2;

Œa2; b� D Œa2; c� D Œb2; a� D Œb2; c� D Œc2; a� D Œc2; b� D 1;


where G0 D ha2; b2; c2i D Z.G/ D ˆ.G/ D �1.G/ Š E23 , so G is special and allmembers of the set �1 are nonmetacyclic minimal nonabelian. If H < G is of order24, thenH � G0.

Proof. Let G satisfy the assumptions of the theorem; then G is not an A1-group(Lemma 65.1) so jGj > p3. Since G=G0 is of rank 3 and all maximal subgroups ofG=G0 are two-generator, we get G=G0 Š Ep3 and so G0 D ˆ.G/. For any a; b 2 G,Œa; b�p D Œap; b� D 1 since G is of class 2, and so G0 is elementary abelian. IfjGj D p4 and E < G is an A1-subgroup, then G D EZ.G/ (Lemma 64.1(i)), so Gis from (a).From now on we assume that jGj � p5 and so jG0j D 1

p3 jGj � p2. If the

set �1 has an abelian member, then jGj D pjZ.G/jjG0j (Lemma 64.1(q)) impliesjG W Z.G/j � p3. If the set �1 has no abelian members, then jG W Z.G/j � p3 again.Since jG W G0j D p3 and G0 � Z.G/, we get Z.G/ D G0 D ˆ.G/ so G is specialand exp.G/ � p2. Let H 2 �1. Then jH W Z.G/j D p2 D pd.H/ so Z.G/ D ˆ.H/

andH is either abelian or an A1-subgroup (Lemma 65.2(a)). Thus G is anA2-groupso jG0j � p3 (Lemma 65.2(d)) and jGj D jG W G0jjG0j � p6. Since G has a minimalnonabelian subgroup H 2 �1, then exp.H/ � p2 so exp.G/ D p2.

(i) Assume first that jGj D p5; then G0 Š Ep2 . If each H 2 �1 is metacyclic,then G is minimal nonmetacyclic. By Lemma 64.1(l), p D 2 and G is the uniquelydetermined group of order 25 given in (b) (see Theorem 66.1).In what follows we assume that G is not minimal nonmetacyclic. Let H 2 �1

be nonmetacyclic. Since d.H/ D 2, H must be an A1-group and so H D ha; b j

ap2

D bp D 1; Œa; b� D c; cp D Œc; a� D Œc; b� D 1i; we have E D �1.H/ D

hap; b; ci Š Ep3 (Lemma 65.1). Assume that there is x 2 G � E of order p. ThenD D hx;Ei 2 �1 is neither abelian (since d.D/ D 2) nor an A1-group (Lemma65.1), a contradiction. Thus, E D �1.G/ Š Ep3 . Let T1=E; : : : ; TpC1=E be allmaximal subgroups of G=E. Then T1; : : : ; TpC1 are nonmetacyclicA1-groups (sinced.Ti / D 2) isomorphic to H (Lemma 65.1). Suppose that M 2 �1 does not containE. Then �1.M/ D E \M Š Ep2 and soM is metacyclic since it is either abelianor anA1-group (Lemma 65.1).Let X < G and jX j D p3 .> p2 D exp.G//; then X is noncyclic. If Z.G/ 6�

X then d.XZ.G// > 2 and XZ.G/ 2 �1, a contradiction. Hence .G0 D/Z.G/ < X

and so X is abelian and normal in G.Let G D hNa; Nb; Nci; then G0 D hŒ Na; Nb�; Œ Na; Nc�; Œ Nb; Nc�i. Hence G0.Š Ep2/ is generated

by two of these elements, say Nx D Œ Na; Nb� and Ny D Œ Na; Nc�. If Œ Nb; Nc� D Nx Ny� , letb D Nb Na�� , c D Nc Na . We compute (recall that G is of class 2)

Œb; c� D Œ Nb Na�� ; Nc Na � D Œ Nb; Nc�Œ Nb; Na� Œ Na; Nc�� D Nx Ny� Nx� Ny�� D 1:

Then A D hb; c;G0i 2 �1 is abelian and so A � E D �1.G/ (otherwise, d.A/ D 3).Hence A D hbi � hci Š Cp2 � Cp2 with �1.A/ D G0 D Ã1.A/. Take an element


a 2 E � A (of order p) so that G D ha; b; ci. Setting x D Œa; b�; y D Œa; c�, we getG0 D hx; yi D hbp; cpi since x ¤ 1 ¤ y ¤ x. Set bp D x˛yˇ , cp D x�yı .The subgroup L D ha; b�c�; G0i 2 �1 for any integers �; �, unless � �

0 .mod p/ since a; b; c independent modulo G0. Since, for such � and �, we haveŒa; b�c�� D x�y� ¤ 1, L is nonabelian, In that case, d.L/ D 2 so ˆ.L/ D G0 D

Z.L/, and G0 is generated by elements .b�c�/p D x˛�C��yˇ�Cı� and Œa; b�c�� D

x�y�, which are linear independent since G0 Š Ep2 . Henceˇ

ˇ

ˇ

˛�C�� ˇ�Cı��

ˇ

ˇ

ˇ

0

.mod p/ only if � � 0 .mod p/. This gives

ˇ�2 C .ı � ˛/�� 2 0 only if � � 0 .mod p/:(1)

From (1) we get ˇ 6 0 .mod p/ (otherwise, � D 1, � D 0 is a nontrivial solutionof (1)) and � 6 0 .mod p/ (otherwise, � D 0, � D 1 is a nontrivial solution of (1)).

(i1) Assume that p > 2. We compute

.2ˇ� C .ı � ˛/�/2 � .4ˇ� C .ı � ˛/2/�2 D 4ˇ.ˇ�2 C .ı � ˛/�� 2/:(2)

Using (1), we see (because ˇ 6 0 .mod p/) that the right hand side of (2) vanishes.mod p/ only if � � 0 .mod p/, and so we have the same for the left hand sideof (2):

.2ˇ� C .ı � ˛/�/2 .4ˇ� C .ı � ˛/2/�2 .mod p/

only if � � 0 .mod p/:(3)

Hence 4ˇ� C .ı � ˛/2 is a quadratic non-residue .mod p/. Indeed, if 4ˇ� C .ı �

˛/2 2 .mod p/, then we set � D 1 in (3) and solve the congruence .2ˇ� C .ı �

˛//2 2 .mod p/ for � , which gives a contradiction (since for � D 1 we haveobtained a nontrivial solution of (3) in � and �).

(i2) Suppose that p D 2. Then ˇ � 1 .mod 2/ and so relation (1) becomes

�2 C .ı C ˛/��C �2 0 .mod 2/ only if � � 0 .mod 2/:(4)

If ı C ˛ 0 .mod 2/, then � � 1 .mod 2/ satisfies (4), a contradiction. Henceı C ˛ 1 .mod 2/ and we get b2 D x˛y, c2 D xy1C˛ . If ˛ 0 .mod 2/, then(recall that o.x/ D o.y/ D 2)

b2 D y; c2 D xy:(5)

If ˛ 1 .mod 2/, then

b2 D xy; c2 D x:(6)

However, interchanging b and c and also x and y, we obtain from (6) the relations (5).Hence we may assume from the start that ˛ D 0, ˇ D � D ı D 1 and our group G isuniquely determined. We have obtained the groups from (c).


(ii) Now suppose that jGj D p6 so that ˆ.G/ D G0 Š Ep3 . Each H 2 �1 isnonmetacyclic and minimal nonabelian since it is 2-generator, and so one may setH D hb; c j bp2

D cp2

D 1; Œb; c� D z; zp D Œb; z� D Œc; z� D 1i, where ˆ.H/ D

hbp; cp; zi D ˆ.G/ D G0 Š Ep3 , and �1.H/ D ˆ.H/ implies �1.G/ D G0 D

ˆ.G/ D Z.G/ and we again conclude that G is special.

(ii1) First assume that p > 2. Then Ã1.H/ D hbp; cpi and H 0 D hzi, where z 62

Ã1.H/. All p2 CpC1members of the set �1 are isomorphic toH (Lemma 65.1) andtheir derived subgroups (of order p) are pairwise distinct (Lemma 65.2(d)). Since Gis regular and jG W Ã1.G/j D p3 (if jG=Ã1.G/j > p

3, then G=Ã1.G/ has a maximalsubgroup that is not generated by two elements), we getÃ1.G/ D G0 D fxp j x 2 Gg

(Lemma 64.1(a)). In particular, there exists a 2 G � H with ap D z. Set bp D x,cp D y. We have G D ha;H i D ha; b; ci and G0 D hx; y; zi (by Lemma 65.2(d), allelements of G0 are commutators). Set Œa; b� D x˛yˇ z� , Œa; c� D x�yız�, and recallthat Œb; c� D z. We shall construct a maximal subgroup K of G with d.K/ > 2. To dothis we shall do some purely number-theoretic computations using heavily the oddnessof p. First, we solve the equation

t2 �2.�2 � 4ˇ/C 2��.�� 2.ı � ˛//C �2.�2 C 4�/ .mod p/(�)

for t , � , � with � , � not both 0 .mod p/. If r D �2 C 4� 0 .mod p/, we set� D 1, � D 0. If r D �2 C4� 6 0 .mod p/, we put � D 1 and set ��2.ı�˛/ D s,�2 � 4ˇ D u, so that (�) can be rewritten in the simplified form

t2 r.� C r�1s/2 � r�1s2 C u .mod p/;(��)

where r , s, u are constant numbers mod p and � , t run through all integers mod p. Theleft hand side of (��) runs through 1C.p�1/=2 D .pC1/=2 distinct values (quadraticresidues) mod p. Now, � C r�1s runs through all numbers mod p, .� C r�1s/2 runsthrough .p C 1/=2 distinct values (quadratic residues) mod p so that the right handside of .��/ runs through .p C 1/=2 distinct values (mod p) (take into account that�r�1s2Cu is a constant number modulo p). Since .pC1/=2C.pC1/=2 D pC1 > p,it follows that there are values for t and � which satisfy (��), as required.Since at least one of the integers � , � is not a multiple of p, we can solve t

2.� N� � N��/ C ��C �� .mod p/ for N�, N�. Substituting this into .�/, we obtain (afterreforming)

.��C �� C N�� N�/.� N� � N��/C �.ˇ�C ı�/ � �.˛�C ��/ 0 .mod p/;

or in the determinant form:ˇ

ˇ

ˇ

ˇ

ˇ

ˇ

��C �� C N�� N� ˛�C �� ˇ�C ı�

1 N� N�

0 � �

ˇ

ˇ

ˇ

ˇ

ˇ

ˇ

0 .mod p/:


We now define the elements a� D ab N�cN� , b� D b�c� ofG, where p does not divide

at least one of the integers � , �. Let K D ha�; b�; G0i; then K 2 �1. It follows fromˆ.K/ D Ã1.K/K

0 that ˆ.K/ D hŒa�; b��; .a�/p; .b�/pi. We have

Œa�; b�� D z��C��C N�� N�x˛�C��yˇ�Cı� ; .a�/p D zx N�yN� ; .b�/p D x�y� ;

where x, y, z are determined in the first paragraph of this part (ii1). From the van-ishing of the above determinant, we see that “linear independent” elements x, y,z cannot be expressed as “linear combinations” of Œa�; b��, .a�/p , .b�/p and soˆ.K/ < G0 D hx; y; zi so jK W ˆ.K/j > p2 or, what is the same, d.K/ > 2,contrary to the hypothesis of the theorem.

(ii2) Now assume that p D 2. Let G0x, G0y be two distinct involutions in G=G0.Then X D hx;G0i and Y D hy;G0i are distinct abelian subgroups of type .4; 2; 2/.We have ˆ.X/ D hx2i and ˆ.Y / D hy2i. Assume that x2 D y2. Then X=ˆ.X/ Š

E8 Š Y=ˆ.Y / D Y=ˆ.X/ and X=ˆ.X/ ¤ Y=ˆ.X/. This is a contradiction since�1.G=ˆ.X// Š E8. Thus, x2 ¤ y2. Hence, the seven nontrivial elements in G=G0

produce seven pairwise distinct squares in G0 which are the seven involutions in G0.In particular, each involution in G0 is a square in G.Let H 2 �1 be fixed. We may set H D ha; b j a4 D b4 D 1; Œa; b� D z; z2 D

Œa; z� D Œb; z� D 1i (Lemma 65.1(a)). Set a2 D x, b2 D y so thatG0 D hxi�hyi�hzi

and .ab/2 D a2bab D xbzb D xb2z D xyz. There exists c 2 G � H suchthat c2 D z (see the previous paragraph; note that z is not a square in H ). ThenG D hH; ci D ha; b; ci. We claim that ac ¤ ca. Assume that this is false. ThenCG.c/ � ha;Z.G/; ci, and the subgroup at the right hand side is abelian of index � 2

in G, a contradiction since the set �1 has no abelian members (Lemma 65.2(d)). Itfollows that ha; ci is nonabelian. Since G is an A2-group, we get ha; ci 2 �1. Sinceˆ.ha; ci/ D ha2 D x; c2 D z; Œa; c�i D G0, we must have Œa; c� 2 G0 � hx; zi

and so Œa; c� D x˛yzˇ . Similarly, considering the maximal subgroup hb; ci, we getŒb; c� 2 G0 � hy; zi and so Œb; c� D xy�zı .Consider the subgroup K D ha; bci. Assume that a bc D bc a. It follows

bca D ab c D baz c so ca D cz D cc2 D c�1 and ha; ci has order 16 so is not amember of the set �1, a contradiction. Since G is anA2-group and d.G/ D 3, we getK 2 �1. We have

ˆ.K/ D ha2 D x; .bc/2 D xy1C�z1Cı ; Œa; bc� D x˛yz1Cˇ i Š E23 ;

and since ˆ.K/ D G0 D hx; y; zi, we haveˇ

ˇ

ˇ

ˇ

ˇ

ˇ

1 0 0

1 1C � 1C ı

˛ 1 1C ˇ

ˇ

ˇ

ˇ

ˇ

ˇ

ˇ

D 1

and so

ˇ C � C ˇ� C ı D 1:(7)


Considering the maximal subgroup L D hab; ci, we get

ˆ.L/ D hxyz; z; x1C˛y1C�zˇCıi Š E23 ;

and soˇ

ˇ

ˇ

ˇ

ˇ

ˇ

1 1 1

0 0 1

1C ˛ 1C � ˇ C ı

ˇ

ˇ

ˇ

ˇ

ˇ

ˇ

D 1;

which gives

˛ C � D 1:(8)

ConsideringM D hb; aci 2 �1, we get ˆ.M/ D hy; x1C˛yz1Cˇ ; xy�z1Cı i so

˛ C ı C ˛ı C ˇ D 1:(9)

Finally, consider the maximal subgroup N D hab; aci. We get

ˆ.N/ D h.ab/2 D xyz; .ac/2 D x1C˛yz1Cˇ ; Œab; ac� D x1C˛y1C�z1CˇCıi;

and so

˛� C ˛ı C ˇ� D 1:(10)

Relations (7–10) have exactly two solutions:

(A) ˛ D 0, ˇ D 1, � D 1, ı D 0, which implies Œa; c� D yz, Œb; c� D xy, and

(B) ˛ D 1, ˇ D 0, � D 0, ı D 1, which implies Œa; c� D xy, Œb; c� D xz.

However, interchanging a and b and also a2 D x and b2 D y, we obtain fromthe solution (A) the solution (B). Hence we may assume from the start that we havesolution (A) and so the groupG of order 26 is uniquely determined as stated in part (d).Assume that F is a subgroup of G of order 24 and F 6� G0. We have FG0 < G so

M D FG0 2 �1. But thenM=.F \ G0/ Š E23 so d.M/ � 3, a contradiction. Thus,G0 is contained in all subgroups of G of order 24. The proof is complete.

Theorem 70.2. Suppose that all maximal subgroups of a p-group G are generated bytwo elements but d.G/ D 3. If G is of class> 2, then p D 2, jGj � 27, andG=K3.G/

is isomorphic to the group of order 26 of Theorem 70.1(d). If, in addition, jGj D 27,then the class of G is 3 and G is uniquely determined, namely, G D ha; b; ci, where

a4 D b4 D c4 D k2 D Œa; k� D Œb; k� D Œc; k� D 1;

Œa; b� D c2; Œa; c� D b2c2; Œb; c� D a2b2; Œa2; b� D k; Œa2; c� D k;

Œb2; a� D 1; Œb2; c� D k; Œc2; a� D k; Œc2; b� D 1:

Here hki D K3.G/ D ŒG;G0� D Z.G/ is of order 2, ha2; b2; c2; ki D G0 D ˆ.G/ D

�1.G/ Š E24 , and G exists as a subgroup of the alternating group A16.


Proof. Suppose thatG is a p-group of class> 2, d.G/ > 2, and all members of the set�1 are generated by two elements. Then K3.G/ ¤ f1g andG=K3.G/ is isomorphic toa group of Theorem 70.1. In particular, jGj � p5. Let jGj D p5 and assume that Ghas a nonabelian subgroup S of order p3. Since d.G/ D 3, G is neither of maximalclass nor an A1-group. Hence CG.S/ 6� S (Lemma 64.1(i)) and let F > Z.S/be a subgroup of order p2 in CG.S/ containing Z.S/. Then SF D S � F 2 �1

and d.SF / D 3, a contradiction. Thus S does not exist so G is an A2-group. Butthen jG=ˆ.G/j D p3 and so ˆ.G/ � Z.G/ (Lemma 65.4(c)), so G is of class 2, acontradiction. Thus, jGj � p6.

(i) Let jGj D p6. If jG=K3.G/j D p4, then take in K3.G/ a G-invariant subgroupL of index p. But then jG=Lj D p5 and G=L satisfies the hypothesis, contrary tothe previous paragraph. Hence jG=K3.G/j D p5 and so G=K3.G/ is isomorphicto a group (b) or (c) of Theorem 70.1. Also, K D K3.G/ is of order p. We haveG0=K D Z.G=K/ Š Ep2 and G=G0 Š Ep3 .The subgroup K is the unique minimal normal subgroup of G. Indeed, let L be a

minimal normal subgroup of G distinct from K. Then L < Z.G/ and so L < G0. Wehave d.G=L/ D 3 and each maximal subgroup of G=L is generated by two elements.By the above, G=L is of class 2 and so ŒG;G0� � L. This is a contradiction sinceŒG;G0� D K and K \ L D f1g.The group G is not an A2-group (otherwise, d.G/ D 3 would imply G0 � Z.G/,

by Lemma 65.4(c)). Therefore G has a nonabelian subgroup H of order p4. We haveH � K (otherwise, KH D K � H 2 �1 and d.KH/ > 2). By Theorem 70.1,applied to G=K, H > G0 and so H is normal in G. Also, G=K is an A2-group andsoH=K Š Ep2 H 0 D K.Assume that H is not minimal nonabelian. Then H (of class 2) contains a non-

abelian subgroup B of order p3, and we have d.H/ > 2 sinceH D BCG.B/ (Lemma64.1(i)) soH=H 0 D H=K Š Ep3 , G=K is isomorphic to a group (c) of Theorem 70.1and H=K D �1.G=K/, i.e., H is the uniquely determined subgroup of index p2 inG. Let H < M 2 �1. Since d.M/ D 2, we get ˆ.M/ D ˆ.G/ D G0 so allmaximal subgroups of M are normal in G. Any nonabelian p-group has exactly 0,1 or p C 1 abelian subgroups of index p (Exercise 1.6(a)) so M has a nonabelianmaximal subgroup F ¤ H . As above, K < F . Since H is the unique A2-subgroupof index p2 in G, by the above, F is an A1-subgroup and F is normal in G. SinceF=K ¤ H=K D �1.G=K/, the subgroup F=K is abelian of type .p2; p/. Assumethat F is not metacyclic. If p > 2, then Ã1.F / is a normal subgroup of G of order p(Theorem 9.11), distinct fromK, contrary to what has proved already. Now let p D 2.Then Z.F / Š E4 (Z.F / is noncyclic since Z.F / D ˆ.G/ D Ã1.F /) and F 0 D K.Let U D CG.Z.F //; then U 2 �1 since Z.G/ is cyclic. Let K1 ¤ K be a subgroupof order 2 in Z.F /. Then U=K1 contains a nonabelian subgroup F=K1 of order 8.Since d.U=K1/ D 2, it follows that U=K1 is of maximal class (Lemma 64.1(i)). ByTaussky’s theorem, K1 6� U 0 so U 0 Š C4. We have F < U so F 0 < U 0 < F and U 0

is cyclic of order 4. It follows that F is metacyclic (Lemma 65.1). Thus, all subgroups


of G of index p2 which are different from H , are either abelian or metacyclic. Weconclude that if G has an A2-subgroup of index p2, then G has a normal metacyclicminimal nonabelian subgroup F of the same index.Let p > 2. It follows from G0 < F that G0 is abelian of type .p2; p/. As we know,

Ã1.G/ D ˆ.G/ D G0. By Theorem 13.7 (or Theorem 69.4),G has a normal subgroupE Š Ep3 . It follows from the structure of G=K that E�1.H/=K/ � �1.G=K/ D

H=K soE � �1.H/, and we conclude thatM=E is cyclic of order p2 (indeed,M hasat most one abelian maximal subgroup so, in view of p > 2, it has at least p � 1 � 2

nonabelian metacyclic maximal subgroups). Then CM .E/ > E since p2 does notdivide exp.Aut.E//. Since H is the unique maximal subgroup of M containing E,we get H � CM .E/ so H is abelian, which is a contradiction. Thus, if p > 2, allnonabelian subgroups of G of index p2 areA1-groups.Now let p D 2. Assume that exp.F / D 4, where F ¤ H is metacyclic of order 16.

Then all subgroups of order 2 in F are characteristic so normal in G, a contradictionsince Z.G/ is cyclic. Thus, F Š M24 . This case we shall consider in (i1).(i1) Assume that p D 2 and F Š M24 . We have c3.F / D 2 and F G G. Therefore,

if L is a cyclic subgroup of F of index 2, then jG W NG.L/j � 2. Let F < M �

NG.L/ with jM W F j D 2. IfM=L Š C4, then CM .L/ D L since F=L is the uniquesubgroup of order 2 in M=L. But in this case M=L is isomorphic to a subgroup ofAut.L/ Š E4, a contradiction. Hence, we haveM=L Š E4. Let R=L be a subgroupof order 2 inM=L with R=L ¤ F=L. Since K < L < R and jR=Kj D 23, Theorem70.1 forces R > G0. But then R \ F D L � G0 and so G0 is cyclic. This is acontradiction since G0=K Š E4.Thus, if p D 2, then G has no nonabelian metacyclic subgroups of order 24. It

follows that then G has noA2-subgroups of index 4 (otherwise, by the above, it has anonabelian metacyclic subgroup of order 24).(i2) Assume that p D 2 and H is a nonmetacyclic A1-group of exponent 4. Then

ˆ.H/ D Z.H/ D L �K, where jLj D 2, Z.H/ is normal in G and H=Z.H/ Š E4.Set U D CG.L/ D CG.Z.H//. Then jG W U j D 2 since L 6� Z.G/ (recallingthat K is the unique minimal normal subgroup of G). In that case H=L is a (proper)nonabelian subgroup of order 8 in U=L which implies that U=L is not an A1-group.By hypothesis, d.U / D 2 and so L < ˆ.U / and CU .H=L/ < H=L. It followsthat U=L is of maximal class (Lemma 64.1(i)), and so .U=L/0 Š C4. By Taussky’stheorem, L 6� U 0 so U 0 Š C4. However, H < U and H 0 < U 0 < H so H ismetacyclic (Lemma 65.1), contrary to the assumption.Thus, if p D 2, then jGj > 26 (otherwise, G is anA2-group so of class 2).(i3) Now we assume that p > 2. Then, as we have been proved already, all non-

abelian subgroups of G of order p4 are either abelian or A1-groups. Let H be anonabelian subgroup of index p2 in G.Assume thatH is a nonmetacyclicA1-group. ThenH=K is abelian of type .p2; p/

and Ã1.H/ > f1g (see Lemma 65.1) is a normal subgroup of G not containing K D

H 0, a contradiction since K D K3.G/ is the unique minimal normal subgroup of G.


Hence,H is metacyclic,H=K is abelian of type .p2; p/ and NG D G=K is isomorphicto a group of Theorem 70.1(c). Set NE D �1. NG/ so that NE Š Ep3 and E (the inverseimage of NE inG) of order p4, is abelian since it is not anA1-group in view of d.E/ �

3 (see the paragraph, preceding (i1)). Since NH � Z. NG/ and Z. NG/ < �1. NG/, we getNE \ NH D Z. NG/ and jH \Ej D p3. ButH is metacyclic and soH \E is abelian oftype .p2; p/ and therefore E is abelian of type .p2; p; p/. Set E1 D �1.E/ so thatE1 Š Ep3 ,E1 \H Š Ep2 , and E1 > K. The quotient group G=E1 is noncyclic sinceNG= NE Š Ep2 is its epimorphic image. Let A=E1 and B=E1 be two distinct subgroupsof order p in G=E1 so that .AB/=E1 Š Ep2 . Note that Theorem 70.1(c) implies thatNA, NB being order p3 are normal in NG. Since A and B are nonmetacyclic of order p4,they must be abelian. Therefore CG.E1/ D AB 2 �1 (note that E1 6� Z.G/ becauseof Z.G/ is cyclic). Since d.AB/ D 2, AB is nonabelian. But AB has two distinctabelianmaximal subgroups and so j.AB/0j D 2 (Lemma 65.2(c)), and we get .AB/0 D

K. It follows that AB is a nonmetacyclic A1-group with Z.AB/ D ˆ.AB/ D E1

(Lemma 65.2(a)). Then Ã1.AB/ Š Ep2 is normal in G and Ã1.AB/ \ K D f1g,a contradiction since K is the unique minimal normal subgroup of G and Ã1.AB/ isnormal in G.Thus, jGj > p6 for all p.(ii) Let jGj � p7. As above, set K D K3.G/; it follows from Theorem 70.1 that

jG=Kj � p6. Assume that jG=Kj � p5. Take a G-invariant subgroup L < K

with jK W Lj D p. We have jG=Lj � p6 and G=L satisfies the assumptions of ourtheorem. By (i), we have a contradiction since L > f1g. Hence jG=Kj D p6; thenp D 2 (Theorem 70.1), and G=K is isomorphic to the group of Theorem 70.1(d).Let, in addition, jGj D 27 so that jKj D 2. Assume that L is another minimal

normal subgroup of G. Then jG=Lj D 26 so, by (i), G=L is of class 2. Then G, as asubgroup of .G=K/� .G=L/, is also of class 2, a contradiction. Thus,K is the uniqueminimal normal subgroup of G; in particular, Z.G/ is cyclic.LetH be an arbitrary subgroup of order 25 in G. Since exp.G/ � 23 (by the above

and Theorem 70.1), H is not cyclic. If K 6� H , then KH D K � H 2 �1 andd.KH/ � 3, a contradiction. HenceH > K and so, by Theorem 70.1(d), H > G0 D

ˆ.G/ and H is normal in G. Since G=K is an A2-group, H=K is abelian in view ofj.G=K/ W .H=K/j D 4.Set G0 D W . By Theorem 70.1(d), �1.G=K/ D W=K D Z.G=K/ Š E23 . If

A=K < W=K is of order 2, then A is normal in G soW � CG.A/ since jAut.A/j2 D

2. Thus, W D hA j K < A < W; jAj D 4i � Z.W / so W is abelian. It follows thatW 2 fE24 ;C4 � C2 � C2g.By Theorem 70.1(d), G D ha; b; ci and a; b; c are not involutions modulo K since

ˆ.G=K/ D �1.G=K/ so that hx;W i is of order 25 and class � 2 for x 2 G � W

since hx;W i=K is abelian (this fact will be used in the sequel) and ha2; b2; c2;Ki D

G0 .D W /,

Œa; b� c2; Œa; c� b2c2; Œb; c� a2b2 .mod K/:(��)


Also, for any x; y 2 G,

Œx; y�2 D Œy; x��2 D Œy; x�2; Œx2; y��1 D Œx2; y�; Œx2; y�1� D Œx2; y�;

Œx2; y� D x�2y�1x2y D x�2.y�1xy/2 D x�2.xŒx; y�/2

D x�2x2Œx; y�2ŒŒx; y�; x� D Œx; y�2Œx; y; x�:

Using the above identity and the relations for G=K (see (��)), we get

Œa2; c� D Œa; c�2ŒŒa; c�; a� D Œa; c�2Œb2c2k; a� D Œa; c�2Œb2; a�Œc2; a�;

where k 2 K � Z.G/ and so

Œa2; c� D Œa; c�2Œb2; a�Œc2; a�:(11)

In the same way we obtain the following relations:

Œb2; c� D Œb; c�2Œb; c; b� D Œb; c�2Œa2b2k; b� D Œb; c�2Œa2; b�;(12)

Œb2; a� D Œb; a�2Œb; a; b� D Œa; b�2Œc2k; b� D Œa; b�2Œc2; b�;(13)

Œc2; a� D Œc; a�2Œc; a; c� D Œc; a�2Œb2c2k; c� D Œa; c�2Œb2; c�;(14)

Œa2; b� D Œa; b�2Œa; b; a� D Œa; b�2Œc2k; a� D Œa; b�2Œc2; a�;(15)

Œc2; b� D Œc; b�2Œc; b; c� D Œc; b�2Œa2b2k; c� D Œb; c�2Œa2; c�Œb2; c�:(16)

For any x; y; z 2 G we get (noting that W is abelian and Œx; y�2� 2 Œx;Ã1.G/� D

Œx;G0� � K � Z.G/)

Œx; y�1� D Œx; yy�2� D Œx; y�2�Œx; y�y�2

D Œx; y�2�Œx; y�

sinceG0 is abelian and y�2 2 Ã1.G/ D G0, and so Œx; y�1; z� D ŒŒx; y�2�Œx; y�; z� D

Œx; y; z�. Hence from the Hall–Witt identity Œa; b�1; c�bŒb; c�1; a�c Œc; a�1; b�a D 1

we get Œa; b; c�bŒb; c; a�c Œc; a; b�a D 1 so Œc2; c�bŒa2b2; a�c Œb�2c�2; b�a D 1, by.��/, or, taking into account that Œb2; a�; Œc2; b� 2 K � Z.G/, we get

Œb2; a�Œc2; b� D 1; or, what is the same, Œb2; a� D Œc2; b�;(17)

It follows from (17) and (13) that Œa; b�2 D 1 so Œa2; b� D Œc2; a�, by (15). It followsfrom (15) and (12) that

Œc2; a� D Œa; c�2Œb2; c� D Œa; c�2Œb; c�2Œa2; b� D Œa; c�2Œb; c�2Œc2; a�

so, canceling, we get Œa; c�2 D Œb; c�2. Then, by (16), (12), (14), (11) we get

Œc2; b� D Œb; c�2Œa2; c�Œb2; c� D Œa2; b�Œa2; c� D Œc2; a�Œa2; c�

D Œa; c�2Œb2; a�Œc2; a�Œc2; a� D Œa; c�2Œc2; b�


so Œa; c�2 D 1. Thus, Œa; b�2 D Œa; c�2 D Œb; c�2 D 1 and one can rewrite (11–17) asfollows:

Œb2; a� D Œa2; b�Œa2; c�;(18)

Œb2; c� D Œa2; b�;(19)

Œc2; a� D Œa2; b�;(20)

Œc2; b� D Œa2; b�Œa2; c�:(21)

In particular, since the abelian group W D G0 D hŒa; b�; Œa; c�; Œb; c�;Ki is gener-ated by elements of order 2, we get W Š E24 . Since K D hki is the unique minimalnormal subgroup of G andK � Z.G/ � W , we get Z.G/ D K. If Œa2; b� D Œa2; c� D

1, then the relations (18–21) and (��) imply that W D ha2; b2; c2;Ki � Z.G/,a contradiction. It follows that we have exactly three possibilities for the values ofŒa2; b� and Œa2; c� (below K D hk j k2 D 1i):

Œa2; b� D k; Œa2; c� D kI(G1)

Œa2; b� D k; Œa2; c� D 1I(G2)

Œa2; b� D 1; Œa2; c� D k:(G3)

In all three cases we have (see (��)): Œa; b� c2, Œa; c� b2c2, Œb; c� a2b2

.mod K/.In case (G1) we substitute Œa2; b� D k, Œa2; c� D k in relations (18–21) and obtain

Œb2; a� D 1; Œb2; c� D k; Œc2; a� D k; Œc2; b� D 1:

If Œa; b� D c2k, then we replace a with a0 D a�1 D aa2 and get

Œa0; b� D Œaa2; b� D Œa; b�a2

Œa2; b� D Œa; b�k D c2k2 D c2;

and so writing again a instead of a0, we may assume from the start that

Œa; b� D c2:(˛)

If Œa; c� D b2c2k, then we replace c with c0 D c�1 so that the relation (˛) remainsvalid in view of o.c/ D 4, and we get

Œa; c0� D Œa; cc2� D Œa; c2�Œa; c�c2

D kŒa; c� D k2b2c2 D b2.c0/2;

and so we may assume from the start that

Œa; c� D b2c2:(ˇ)

Finally, in case Œb; c� D a2b2k, we replace b with b0 D b�1. Then the relation (˛)remains unchanged: Œa; b0� D Œa; bb2� D Œa; b2�Œa; b�b

2

D Œa; b�, and also the relation


(ˇ) remains unchanged and we get

Œb0; c� D Œbb2; c� D Œb; c�b2

Œb2; c� D Œb; c�k D a2b2k2 D a2.b0/2;

and so we may assume from the start that

Œb; c� D a2b2:(� )

The structure of G D G1 is uniquely determined. If we set (identify)

a D .1; 2/.3; 13; 7; 9/.4; 10; 8; 11/.5; 6/.12; 16/.14; 15/;

b D .1; 3/.2; 9/.4; 14; 8; 12/.5; 7/.6; 13/.10; 16; 11; 15/;

c D .1; 4/.2; 10; 6; 11/.3; 12/.5; 8/.7; 14/.9; 15; 13; 16/;

we see that the permutations a; b; c satisfy all relations for G1. Since

k D Œa2; b� D .1; 5/.2; 6/.3; 7/.4; 8/.9; 13/.10; 11/.12; 14/.15; 16/

and hki D Z.G/, we see that our permutations a; b; c induce a faithful permutationrepresentation from G1 into the alternating group A16.In case (G2) we substitute Œa2; b� D k, Œa2; c� D 1 in relations (18–21) and obtain

Œb2; a� D k; Œb2; c� D k; Œc2; a� D k; Œc2; b� D k:

If Œa; b� D c2k, then we replace a with a�1. If Œa; c� D b2c2k, then we replace c withc�1. Finally, in case Œb; c� D a2b2k, we replace a with a�1 and b with b�1. Doing so,we can assume (as before) from the start that Œa; b� D c2, Œa; c� D b2c2, Œb; c� D a2b2.The structure of G D G2 is uniquely determined. Similarly, the permutations:

a D .1; 2/.3; 9; 7; 10/.13; 16; 14; 15/.4; 12/.5; 6/.8; 11/;

b D .1; 3/.5; 7/.2; 9; 6; 10/.11; 15/.12; 16/.4; 13; 8; 14/;

c D .1; 4/.2; 11; 6; 12/.5; 8/.9; 16/.10; 15/.3; 13; 7; 14/;

induce a faithful permutation representation from G2 into A16.In case (G3) we substitute Œa2; b� D 1, Œa2; c� D k in the relations (18) to (21) and

obtain Œb2; a� D k, Œb2; c� D 1, Œc2; a� D 1, Œc2; b� D k. If Œa; b� D c2k, then wereplace b with b�1. If Œa; c� D b2c2k, then we replace a with a�1. Finally, in caseŒb; c� D a2b2k, we replace c with c�1. Doing so, we can assume from the start thatŒa; b� D c2, Œa; c� D b2c2, Œb; c� D a2b2. The structure of G D G3 is uniquelydetermined. The permutations:

a D .1; 2/.4; 14; 8; 11/.12; 15; 13; 16/.3; 10/.5; 6/.7; 9/;

b D .1; 3/.4; 13/.2; 9; 6; 10/.5; 7/.8; 12/.11; 16; 14; 15/;

c D .1; 4/.3; 12; 7; 13/.2; 11/.5; 8/.6; 14/.9; 16; 10; 15/;

induce a faithful permutation representation from G3 into A16.


Considering G1; G2; G3 as above subgroups of A16, we see that the permutation.2; 4; 11; 6; 8; 12; 9; 7; 14; 10/.3; 13/ conjugates G1 onto G2, and the permutation.2; 11; 8; 6; 14; 7; 10; 4/.3; 9; 13; 12/ conjugates G1 onto G3. In particular, G2 andG3 are isomorphic to G D G1. Since K < ˆ.G/, G satisfies the hypothesis sinceG=K does. The proof is complete.

Thus, groups of the title are classified for p > 2 (see also [Bla1, Theorem 3.1]).

Theorem 70.3. Let G be a p-group with d.G/ > 2, whose maximal subgroups aregenerated by two elements. If G is of class 3, then p D 2, 27 � jGj � 28 andG=K3.G/ is isomorphic to the group of order 26 of Theorem 70.1(d). If jGj D 27,then jK3.G/j D 2 and G is isomorphic to the group of Theorem 70.2. If jGj D 28,then K3.G/ D Z.G/ Š E4, G0 D ˆ.G/ D �1.G/ Š E25 and such groups exist.

Proof. Suppose that all maximal subgroups of a p-group G are generated by twoelements but d.G/ D 3. Assume, in addition, that G is of class 3.By Theorem 70.2, p D 2, jGj � 27, f1g < K D K3.G/ � Z.G/, and G=K is

isomorphic to the group of order 26 of Theorem 70.1(d). It follows that

G0 D ˆ.G/; G=G0 Š E8; G0=K Š E8; ŒG0; G� D K;

�1.G=K/ D Z.G=K/ D G0=K:

Let x 2 G0 and g 2 G. Then Œx; g� 2 K � Z.G/ and so Œx; g�2 D Œx2; g� D 1

since x2 2 K. Hence the abelian group K is generated by involutions and so K iselementary abelian. Let S be a maximal subgroup in K. Then G=S (of order 27)is isomorphic to the group of Theorem 70.2. In particular, Z.G=S/ D K=S and soK D Z.G/. Also, G0=S is elementary abelian. Hence ˆ.G0/ � S for any maximalsubgroup S of K. Since K is elementary abelian and noncyclic, we get ˆ.G0/ D f1g

and so�1.G/ D G0 is elementary abelian.We have G D ha; b; ci, ha2; b2; c2iK D G0 and for each maximal subgroup S of

K, G=S is isomorphic to the group of Theorem 70.2. Suppose that jKj � 23. Thenthere exists a maximal subgroup S ofK which contains hŒa2; b�; Œa2; c�i. But then a2S

is contained in Z.G=S/. This is a contradiction since a2 62 K and Z.G=S/ D K=S .We have proved that jKj � 4.A group G of order 28 of Theorem 70.3 exists. For example, such a group is

G D ha; b; c j a4 D b4 D c4 D k2 D l2 D m2 D klm D 1;

Œa; k� D Œb; k� D Œc; k� D 1; Œa; l � D Œb; l � D Œc; l � D 1; Œa; b� D c2;

Œa; c� D b2c2; Œb; c� D a2b2; Œa2; b� D m; Œa2; c� D k;

Œb2; a� D l; Œb2; c� D m; Œc2; a� D m; Œc2; b� D li:


Here hk; li D Z.G/ D K3.G/ Š E4 and G0 D ˆ.G/ D ha2; b2; c2; k; li D

�1.G/ Š E25 . We get a faithful permutation representation of G by setting:

a D .1; 2/.3; 21; 22; 15/.4; 23; 10; 18/.5; 8/.6; 11/.7; 17/.9; 14; 12; 26/

.13; 16; 25; 27/.19; 30; 20; 29/.24; 32; 31; 28/;

b D .1; 3/.2; 14; 11; 15/.4; 24; 25; 20/.5; 9/.6; 12/.7; 22/.8; 21; 17; 26/

.10; 19; 13; 31/.16; 29; 27; 28/.18; 32; 23; 30/;

c D .1; 4/.2; 16; 17; 18/.3; 19; 12; 20/.5; 10/.6; 13/.7; 25/.8; 27; 11; 23/

.9; 24; 22; 31/.14; 32; 21; 30/.15; 28; 26; 29/:

In this way we see that the group G exists as a subgroup of A32.In view of K < ˆ.G/, G satisfies the hypothesis since G=K does.

Theorem 70.4. Let G be a p-group with d.G/ > 2 all of whose maximal subgroupsare generated by two elements. If G is of class > 2, then p D 2 and one of thefollowing holds:

(i) G=K4.G/ is isomorphic to the uniquely determined group of order 27 given inTheorem 70.2.

(ii) H D G=K4.G/ is isomorphic to one of the groups of order 28 described in Theo-rem 70.3 and more precisely,H D ha; b; ci with

a4 D b4 D c4 D k2 D l2 D m2 D klm D 1; Œa; k� D Œb; k� D Œc; k� D 1;

Œa; l � D Œb; l � D Œc; l � D 1; Œa; b� D c2k�; Œa; c� D b2c2k�; Œb; c� D a2b2;

Œa2; b� D m; Œa2; c� D k; Œb2; a� D l; Œb2; c� D m; Œc2; a� D m; Œc2; b� D l;

where � D 0; 1. These two groups (for � D 0 and � D 1) exist as subgroups of thealternating group A32 and they are not isomorphic. We have in both cases K3.H/ D

Z.H/ D hk; li Š E4 and H 0 D ˆ.H/ D �1.H/ D ha2; b2; c2; k; li Š E25 . Incase � D 1, H possesses an automorphism of order 7 which acts fixed-point-freelyon H=Z.H/ and so H=Z.H/ is isomorphic to the Suzuki group of order 26 (given inTheorem 70.1(d)).

Proof. We may assume that K4.G/ D f1g so that G is of class 3. Using Theorems70.2 and 70.3, we see thatG is either isomorphic to the group of order 27 given in The-orem 70.2 or a group of order 28 described in Theorem 70.3. It remains to determinecompletely the structure of G in the second case. In that case G=K3.G/ is the specialgroup of order 26 isomorphic to the group of Theorem 70.1(d), K3.G/ D Z.G/ Š E4,and G0 D ˆ.G/ D �1.G/ Š E25 . We have G D ha; b; ci, where a; b; c are el-ements of order 4 in G � G0 so that ha2; b2; c2i � K3.G/ D G0 and Œa; b� c2,Œa; c� b2c2, Œb; c� a2b2 .mod K3.G//. Also, for each maximal subgroup S of


K3.G/, jS j D 2 and G=S is isomorphic to the group of order 27 of Theorem 70.2. Inparticular, Z.G=S/ D Z.G/=S is of order 2.

We claim that each of the commutators Œa2; b�, Œa2; c�, Œb2; a�, Œb2; c�, Œc2; a�, Œc2; b�

are distinct from 1. Indeed, if (for example) Œa2; b� D 1, then we consider a maximalsubgroup S� of Z.G/ which contains Œa2; c�. Then a2S� 2 Z.G=S�/, contrary to thefact that Z.G=S�/ D Z.G/=S� (and noting that a2 62 Z.G/).

Set Œa2; b� D m and Œa2; c� D k. If m D k, then considering G=hmi, we havea2hmi 2 Z.G=hmi/, a contradiction. It follows that k ¤ m and so K3.G/ D hk;mi

since K3.G/ Š E4. Set l D km so that k, l , m are the three distinct involutions inZ.G/ D K3.G/. We compute (with some z 2 Z.G/):

Œb2; c� D Œbb; c� D Œb; c�bŒb; c� D .a2b2z/ba2b2z D .a2/ba2

D Œb; a2� D Œa2; b� D m;

Œc2; a� D Œc; a�cŒc; a� D .b2c2z/cb2c2z D .b2/cb2 D Œc; b2� D m;

Œc2; b� D Œc; b�cŒc; b� D .a2b2z/ca2b2z D .a2/ca2.b2/cb2

D Œc; a2�Œc; b2� D km D l;

Œb2; a� D Œb; a�bŒb; a� D .c2z/bc2z D .c2/bc2 D Œb; c2� D l:

Hence we have obtained the following relations:

Œa2; b� D m; Œa2; c� D k; Œb2; a� D l;

Œb2; c� D m; Œc2; a� D m; Œc2; b� D l;(˛�)

where Z.G/ D K3.G/ D hk; li Š E4 and m D kl . Also, we may set

Œa; b�c2 D kplq; Œa; c�b2c2 D kr ls; Œb; c�a2b2 D kt lu;(˛��)

where p, q, r , s, t , u are some integers mod 2.

For any x 2 G � G0, Œx;G0� � Z.G/ and so hx;G0i is of class � 2. This fact willbe used many times. For any x; y 2 G �G0 and any g1; g2 2 G0, we have

Œ.xg1/2; yg2� D Œ.xg1/

2; y� D Œx2g21Œg1; x�; y� D Œx2; y�:

Therefore, if we replace a; b; c with a0 D aa2.qCsCu/, b0 D bb2.sCu/, c0 D cc2s,respectively, then the relations (˛�) remain preserved. However, the relations (˛��)


are simplified, as the following computation shows:

Œa0; b0�.c0/2 D Œaa2.qCsCu/; bb2.sCu/�c2

D Œa; bb2.sCu/�a2.qCsCu/

Œa2.qCsCu/; bb2.sCu/�c2

D Œa; bb2.sCu/�Œa2.qCsCu/; bb2.sCu/�c2;

Œa; b2.sCu/�Œa; b�Œa2.qCsCu/; b2.sCu/�Œa2.qCsCu/; b�c2

D Œa; b2�sCuŒa; b�Œa2; b�qCsCuc2

D .Œa; b�c2/lsCumqCsCu D kplqlsCukqCsCulqCsCu D kpCqCsCu;

and we get in a similar way Œa0; c0�.b0/2.c0/2 D kqCrCu, Œb0; c0�.a0/2.b0/2 D ksCtCu.Writing again a, b, c (instead of a0, b0, c0), we see that the relations (˛��) are

simplified into:

Œa; b�c2 D k�; Œa; c�b2c2 D k�; Œb; c�a2b2 D k� ;(˛��)

where �, �, � are some integers mod 2. In this way we have obtained eight possibilitiesG1; G2; : : : ; G8 for the structure of G depending on the values of these integers:

� D 0; � D 0; � D 0I(G1)

� D 0; � D 0; � D 1I(G2)

� D 0; � D 1; � D 0I(G3)

� D 0; � D 1; � D 1I(G4)

� D 1; � D 0; � D 0I(G5)

� D 1; � D 0; � D 1I(G6)

� D 1; � D 1; � D 1I(G7)

� D 1; � D 1; � D 0:(G8)

We replace now a; b; c with

a� D bb2; b� D c; c� D aca2b2l;(ˇ�)

respectively, and verify first that the relations (˛�) remain unchanged. For example:

Œ.a�/2; c�� D Œb2; aca2b2l � D Œb2; ac� D Œb2; c�Œb2; a�c D mlc D ml D k:

We want to see what happens with the relations (˛��) for arbitrary integers �, �, �.mod 2/. First we compute:

.ac/2 D a.ca/c D a.acŒc; a�/c D a.acb2c2k�/c D a2.cb2/c3k�

D a2.b2cm/c3k� D a2b2k�C1l;


which gives

.c�/2 D .aca2b2l/2 D .ac/2.a2b2l/2Œa2b2l; ac� D a2b2k�C1l Œa2b2; c�Œa2b2; a�c

D a2b2k�C1l Œa2; c�Œb2; c�Œb2; a�c D a2b2k�C1l:

Then it is easy to transform the relations (˛��):

Œa�; b��.c�/2 D Œbb2; c�a2b2k�C1l D Œb; c�b2

Œb2; c�a2b2k�C1l

D .Œb; c�a2b2/mk�C1l D k�mk�C1l D k�C� ;

and we get in a similar way:

Œa�; c��.b�/2.c�/2 D k�C�C�C1

andŒb�; c��.a�/2.b�/2 D k�C1:

These results show that our group Gi defined with the constants �, �, � in the re-lations (˛��) is isomorphic with the group Gj defined with the constants � C �,� C �C � C 1, �C 1. In particular, the group G1 defined with � D � D � D 0 is iso-morphic to the group G4 defined with the constants 0; 1; 1. Further, the group G4 with� D 0, � D 1, � D 1 is isomorphic with G3 defined with the constants 0 ,1, 0 and thisone is isomorphic with G5. Then G5 is isomorphic with G2, G2 is isomorphic withG6 and G6 is isomorphic with G7. However, our replacement (ˇ�) sends G8 onto G8

and we verify that in fact (ˇ�) induces an automorphism of order 7 on G8 which actsfixed-point-freely on G8=Z.G8/. HenceG=Z.G/ is isomorphic to the Suzuki group oforder 26 given in Theorem 70.1(d).The group G1 is exactly the group occurring as an example in the proof of Theo-

rem 70.3, where we have found a faithful permutation representation of G1 of degree32 (with even permutations). We get a faithful permutation representation of G8 bysetting:

a D .1; 2/.3; 14; 21; 25/.4; 18; 10; 27/.5; 8/.6; 11/.7; 17/

.9; 24; 12; 15/.13; 26; 23; 16/.19; 30; 20; 29/.22; 32; 31; 28/I

b D .1; 3/.2; 14; 11; 15/.4; 22; 23; 20/.5; 9/.6; 12/.7; 21/

.8; 24; 17; 25/.10; 19; 13; 31/.16; 29; 26; 28/.18; 32; 27; 30/I

c D .1; 4/.2; 16; 17; 18/.3; 19; 12; 20/.5; 10/.6; 13/.7; 23/

.8; 26; 11; 27/.9; 22; 21; 31/.14; 32; 24; 30/.15; 28; 25; 29/:

In this way, we see that also the group G8 exists as a subgroup of A32.Finally, considering G1 and G8 as subgroups of A32, W. Lempken (Institut fur

experimentelle Mathematik, Essen) has shown that the groups G1 and G8 are not iso-morphic.


Let G be a 2-group of class > 3 with d.G/ > 2 but d.H/ D 2 for all H 2 �1.Then, it is possible to prove that we must have jGj � 29 and we exhibit here anexample of such a group G of order 29 and class 4, where G D ha; b; ci with thefollowing defining relations:

a8 D b4 D c8 D 1; a4 D c4 D u; Œa; u� D Œb; u� D Œc; u� D 1;

Œa2; b� D m; Œa2; c� D k; m2 D k2 D 1; km D l; l2 D 1;

Œb2; a� D lu; Œb2; c� D um; Œc2; a� D um; Œc2; b� D l; Œa; b� D c2;

Œa; c� D b2c2; Œb; c� D a2b2; Œa2; b2� D 1; Œa2; c2� D 1; Œb2; c2� D u;

Œm; a� D 1; Œm; b� D 1; Œm; c� D u; Œk; a� D 1; Œk; b� D u;

Œk; c� D 1; Œl; a� D 1; Œl; b� D u; Œl; c� D u:

Here G0 D ha2; b2; c2; k; li is of order 26, K4.G/ D Z.G/ D hui is of order 2,K3.G/ D hk; l; ui Š E8, and Z.G0/ D ha2; k; li is abelian of type .4; 2; 2/. Finally,we have shown that this group G exists as a subgroup of the symmetric group ofdegree 64.We can improve Theorem 70.4 in case (i).

Theorem 70.5. Let G be a p-group with d.G/ > 2 all of whose maximal subgroupsare generated by two elements. If G is of class > 2, then p D 2, and if G=K4.G/ isisomorphic to the group of order 27 given in Theorem 70.2 (case (i) of Theorem 70.4),then K4.G/ D f1g.

Proof. Assume that K4.G/ ¤ f1g. To get a contradiction we may assume jK4.G/j

D 2 (by considering a suitable quotient group of G). Then G=K4.G/ is isomorphic tothe group of order 27 of Theorem 70.2. We may set G D ha; b; ci, where

a4 b4 c4 k2 1; Œa; k� Œb; k� Œc; k� 1; Œa; b� c2;

Œa; c� b2c2; Œb; c� a2b2; Œa2; b� k;

Œa2; c� k; Œb2; a� 1; Œb2; c� k;

Œc2; a� k; Œc2; b� 1 .mod K4.G//:

Also we may set K4.G/ D hui � Z.G/. Then K3.G/ D hk; ui, W D G0 D ˆ.G/ D

ha2; b2; c2; k; ui is of order 25 and class � 2 sinceW=K4.G/ Š E24 .Since CG.K3.G// is a subgroup of index 2 in G, we get K3.G/ � Z.W /. From

Œa2; b� k .mod K4.G// with k 62 K4.G/ follows Œa2; b� D k0 with k0 2 K3.G/ �

K4.G/. From a4 1 .mod K4.G// follows a4 2 K4.G/ � Z.G/ and so

1 D Œa4; b� D Œa2a2; b� D Œa2; b�a2

Œa2; b� D ka2

0 k0 D k20 ;

since a2 2 W and K3.G/ D hk0; ui � Z.W /. Hence k0 is an involution and so wehave proved that K3.G/ Š E4.


Using Œc; b2� D kui and Œc2; b� D uj , where i; j are some integers (mod 2), wecompute

Œc2; b2� D Œcc; b2� D Œc; b2�cŒc; b2� D .kui /ckui D kck D c�1kck D Œc; k�;

since k is an involution. On the other hand, we get

Œc2; b2� D Œc2; bb� D Œc2; b�Œc2; b�b D uj .uj /b D .uj /2 D 1;

and so Œc; k� D 1.Using Œa; b2� 2 K4.G/ � Z.G/ and Œa2; b� D kui with an integer i .mod 2/, we

compute

Œa2; b2� D Œaa; b2� D Œa; b2�aŒa; b2� D Œa; b2�2 D 1;

Œa2; b2� D Œa2; bb� D Œa2; b�Œa2; b�b D kui.kui /b D kkb D Œk; b�;

and so Œk; b� D 1.Finally, using Œa; c2� D kuj and Œa2; c� D kul with some integers j , l .mod 2/,

we get

Œa2; c2� D Œaa; c2� D Œa; c2�aŒa; c2� D .kuj /akuj D kak D Œa; k�;

Œa2; c2� D Œa2; cc� D Œa2; c�Œa2; c�c D kul.kul /c D kkc D Œk; c�:

Hence (by the above) Œa; k� D Œk; c� D 1. We have obtained Œk; a� D Œk; b� D Œk; c� D

1, and so K3.G/ D hk; ui � Z.G/. This is a contradiction since G was of class 4.

The group G of Theorem 70.2 of order 27 is an A4-group since ha2; bi is a non-abelian subgroup of order 24 (this subgroup is minimal nonabelian nonmetacyclic) andG has no nonabelian subgroups of order 8. Obviously, D8 is not a subgroup ofG since�1.G/ D E24 D G0. Also,Q D Q8 is not a subgroup of G (otherwise,Q\G0 D C2

and so S D G0Q is a maximal subgroup of G and so d.S/ D 2 so ˆ.S/ D G0 and,by Taussky’s theorem, S is of maximal class, a contradiction).Both groups G of Theorem 70.4 of order 28 are A5-groups since ha2; bi is a non-

abelian subgroup of order 24 (this subgroup is minimal nonabelian nonmetacyclic) andG has no nonabelian subgroups of order 8. Obviously, D8 is not a subgroup ofG since�1.G/ D E25 D G0. Also,Q D Q8 is not a subgroup of G (otherwise,Q\G0 D C2

and so T D G0Q is a maximal subgroup of G hence d.T / D 2; then ˆ.T / D G0 so,by Taussky’s theorem, T is of maximal class, a contradiction).The above two paragraphs show that if a 2-group G is such that d.G/ D 3 but

d.H/ D 2 for all H 2 �1 and the class of G is > 2, then G is an An-group withn < 5 if and only if G is the unique group of order 27 from Theorem 70.2, and this Gis anA4-group.

�71

Determination ofA2-groups

We recall that a p-group G is called an An-group if every subgroup of index pn isabelian but at least one subgroup of index pn�1 is nonabelian. Here we determineA2-groups of order > p4 up to isomorphism in terms of generators and relations.This determination will follow five propositions covering allA2-groups.

Proposition 71.1. Suppose that G is an A2-group of order > p4. Then jG0j D p ifand only if G has at least two distinct abelian maximal subgroups and in that case oneof the following holds:

(i) G D H � Cp , whereH is minimal nonabelian.

(ii) G D ha; b; ci, where

apm

D bpn

D cp2

D 1; m � n � 1; m � 2; Œa; b� D d; cp D d;

Œa; d � D Œb; d � D Œa; c� D Œb; c� D 1:

Here H D ha; bi is nonmetacyclic minimal nonabelian, H 0 D G0 D hd i, G D

H � C , where C D hci is cyclic of order p2 andH \ C D hd i D hcpi D H 0.

Proof. Let G be an A2-group of order > p4 with jG0j D p. By Lemmas 65.2(c)and 65.4(d), jG0j D p if and only if the set �1 has at least two abelian members.In that case the set �1 has a nonabelian member H so that G D HZ.G/, whereZ.G/ \ H D Z.H/. We have jH j > p3, H 0 D G0, Z.H/ D ˆ.H/ D ˆ.G/, andjH W Z.H/j D p2 sinceH is anA1-group. Clearly, jZ.G/ W Z.H/j D p.Suppose that there is an element c 2 Z.G/�H of order p. Then G D H � hci and

it is clear that all such groups areA2-groups.We assume in the sequel that�1.Z.G// � H ; then d.Z.H// D d.Z.G// so jZ.H/ W

ˆ.Z.H//j D jZ.G/ W ˆ.Z.G//j D pjZ.H/ W ˆ.Z.G//j (note that ˆ.Z.H// <ˆ.Z.G// � Z.H/ since Z.H/ is maximal in Z.G/) and so jˆ.Z.G// W ˆ.Z.H//j D

p. Hence, if Z.G/=ˆ.Z.G// Š Epd , then Z.G/=ˆ.Z.H// Š Cp2 � Epd�1. Thisimplies that for each c 2 Z.G/ � Z.H/, we have cp 2 Z.H/ �ˆ.Z.H//.

(i) Assume, in addition, that H is metacyclic; then, by Lemma 65.1(b), we haveH D ha; b j apm

D bpn

D 1, m � 2, n � 1, ab D a1Cpm�1

i, whereH 0 D hapm�1

i

and jH j D pmCn, m C n > 3, jGj D pmCnC1. We have Z.H/ D hapi � hbpi D

ˆ.H/ and for each c 2 Z.G/ �H , cp 2 hap; bpi � hap2

; bp2

i since hap2

; bp2

i D

ˆ.Z.H//.


If n D 1, then H Š MpmC1 , m > 2, and Z.H/ D hapi is cyclic. In this case,replacing an element c 2 Z.G/�Z.H/ with ci , i 6 0 .mod p/, if necessary, we mayassume that cp D a�p and so .ca/p D cpap D 1. Since .ca/b D cab D .ca/apm�1

,it follows that hca; bi is nonabelian of order p3. This is a contradiction, since jGj �

p5 and G is anA2-group. We have proved that n � 2.Suppose that there exists an element x 2 G � H of order p. Since d.Z.G// D

d.Z.H//, we get x 62 Z.G/ and so Œa; x� ¤ 1 or Œb; x� ¤ 1. If Œa; x� ¤ 1, then (notingthat hai is normal in G) ha; xi is nonabelian of order pmC1. This is a contradictionsince jGj D pmCnC1 and n � 2. Hence we must have Œa; x� D 1 and Œb; x� ¤ 1. Inthat case hŒb; x�i D G0 D H 0 and so H� D hb; xi is a nonmetacyclic A1-group oforder pnC2. HenceH� must be a maximal subgroup of G containing Z.H/ D ˆ.G/

but not containing Z.G/ (every member of the set �1, containing Z.G/, is abelian!),and so jGj D pnC3. We get m D 2 and G D H�Z.G/, whereH� is a nonmetacyclicA1-group. This case will be considered in part (ii) of the proof.Hence assume that �1.G/ � H . By the paragraph, preceding (i), for each c 2

Z.G/�Z.H/, we get cp 2 Z.H/�ˆ.Z.H//, where Z.H/ D ˆ.H/ D Ã1.H/ sinceH is metacyclic. If there is h 2 H such that cp D hp, then the abelian subgrouphh; ci is not cyclic since hhi and hci are two distinct cyclic subgroups of hh; ci of thesame order. Since hh; ci \ H D hhi, there is an element of order p in hh; ci � H ,contrary to �1.G/ � H . We conclude that hcpi is a maximal cyclic subgroup of H .Since cp 2 Z.H/, the quotient group H=hcpi is noncyclic so cp � ˆ.H/ D Ã1.H/

since H is metacyclic. It follows that not every element in Ã1.H/ is a p-th power ofan element in H so H is irregular so p D 2 since H is of class 2 (Lemma 64.1(a)),and c2 is not a square in H . Set d D Œa; b� ¤ 1, where d2 D 1, and compute forany integers i; j : .aibj /2 D a2ib2j Œbj ; ai � D a2ib2jd ij . If i or j is even, thena2ib2j D .aibj /2 is a square in H . Therefore c2 D a2kb2l , where both k and lare odd. Consider the nonabelian subgroup S D ha; b�lci (recall that c 2 Z.G/).We have .b�lc/2 D b�2lc2 D b�2la2kb2l D a2k , and so jS j D 2mC1. This is acontradiction, since jG W S j D p2 > p in view of jGj D 2mCnC1 and n � 2.

(ii) It remains to consider the case where H is a nonmetacyclic A1-group of order� p4. We set

H D ha; b j apm

D bpn

D 1; Œa; b� D d; dp D Œa; d � D Œb; d � D 1i;

where we may assume thatm � n � 1. Here jH j D pmCnC1 and so jGj D pmCnC2.Since jH j � p4, must be m � 2. We have Z.H/ D ˆ.H/ D hapi � hbpi � hd i andˆ.Z.H// D hap2

i � hbp2

i since o.d/ D p. Also note that hd i is a maximal cyclicsubgroup in H and for each c 2 Z.G/ � Z.H/, cp 2 Z.H/ � ˆ.Z.H//. Note thatH \ Z.G/ D Z.H/.We want to show that there exists c 2 Z.G/ � Z.H/ so that 1 ¤ cp 2 hd i.Assume first that n D 1 so that Z.H/ D hapi � hd i and for an element c 2

Z.G/ � Z.H/, cp D aipd j ; recall that o.c/ > p). If i 0 .mod p/, then j 6 0

�71 Determination ofA2-groups 235

.mod p/ (otherwise, o.c/ D p) and we may set cp D ai 0p2

d j . Then we compute.a�i 0pc/p D a�i 0p2

cp D d j ¤ 1, and we are done since a�i 0pc 2 Z.G/ � Z.H/ D

Z.G/ � H . If i 6 0 .mod p/, then S D ha�ic; bi is nonabelian and .a�ic/p D

a�ipcp D a�ipaipd j D d j . Hence p � o.a�ic/ � p2. If o.a�ic/ D p2, thenj 6 0 .mod p/ and ha�ici � hd j i D G0. In any case, jS j D p3. This is acontradiction since jG W S j � p2 and G is anA2-group.Thus, we may also assume that n � 2. We have for c 2 Z.G/ � Z.H/, cp D

aipbjpdk , where at least one of the integers i; j; k is 6 0 .mod p/ since cp 62

ˆ.Z.H//.If i 0 .mod p/ and j 0 .mod p/, then k 6 0 .mod p/ and ai ; bj 2 Z.H/ <

Z.G/ so

.a�ib�j c/p D a�ipb�jpcp D a�ipb�jpaipbjpdk D dk ¤ 1;

and we are done since a�ib�j c 2 Z.G/ � Z.H/.Now assume that one of the integers i; j is 0 .mod p/ and the other one is 6 0

.mod p/. Note that a and b occur symmetrically since we have m � 2 and n � 2.Interchanging a and b (if necessary), we may assume that i 6 0 .mod p/ but j 0

.mod p/. In that case T D ha�ib�j c; bi is nonabelian and since b�j ; c 2 Z.G/we get .a�ib�j c/p D a�ipb�jpcp D dk . Thus p � o.a�ib�j c/ � p2 and ifo.a�ib�j c/ D p2, then k 6 0 .mod p/ and ha�ib�j ci � hdki D G0. In any case,jT j D pnC2. This is a contradiction since jG W T j D pm > p.It remains to study the possibility i 6 0 .mod p/ and j 6 0 .mod p/. We con-

sider again the nonabelian subgroup T D ha�ib�j c; bi. Ifp > 2, then .a�ib�j c/p D

a�ipb�jpcp D dk . If p D 2, then

.a�ib�j c/2 D .a�ib�j /2c2 D a�2ib�2j Œb; a�ij c2

D a�2ib�2jd ij a2ib2jdk D d ij Ck:

In any case, we have jT j D pnC2, a contradiction since jG W T j D pm > p.We have proved that there exists c 2 Z.G/�H such that 1 ¤ cp 2 hd i. Replacing

c with another generator of hci, we may assume that cp D d . The structure of G isuniquely determined.

Proposition 71.2. Let G be a metacyclic A2-group of order > p4. Then G0 Š Cp2

and we have one of the following possibilities:

(a) G D ha; b j apm

D 1, m � 3, bpn

D a�pm�1

, n � 1, � D 0; 1, m C n � 5,ab D a1Cpm�2

i, where in case p D 2, m � 4.

(b) p D 2, G D ha; b j a8 D 1, b2n

D a4�, � D 0; 1, n � 2, ab D a�1C4� ,� D 0; 1i.

Proof. By Corollary 65.3, a metacyclic p-group G is anA2-group if and only if G0 Š

Cp2 . Let G0 be the subgroup of order p in G0; then G=G0 is an A1-group (Lemma


65.2(a)). But jG=G0j > p3 and so G=G0 is a “splitting” metacyclic group (Lemma65.1(b,c)). We have G D AB , where A � G0, A \ B D G0, A=G0 (of order � p2)and B=G0 are cyclic. Since G0 < G0 < A and G0 is cyclic, A does not split overG0. Hence A is a cyclic normal subgroup of order pm, m � 3, and we set A D hai.Let b 2 B be such that hbi covers A=G0 so that bpn

2 G0 D hapm�1

i, wherepn D jB=G0j, n � 1. Changing a generator of hai if necessary, one may assumebpn

D a�pm�1

with � D 0; 1. Now, hbi induces an automorphism group of order p2

on A so that ŒA;B� Š Cp2 . In that case our result follows at once.

Proposition 71.3. Let G be a nonmetacyclic A2-group of order > p4 possessing ex-actly one abelian maximal subgroup. Then G0 Š Ep2 and assume, in addition, thatG0 6� Z.G/. In that case p > 2, d.G/ D 2, K3.G/ is of order p, G=K3.G/ isnonmetacyclic minimal nonabelian, and one of the following holds:

(i) G D ha; bi, where

ap D bpn

D 1; n � 3; Œa; b� D c; cp D Œa; c� D 1; Œb; c� D bpn�1

:

Here jGj D pnC2, G0 D hc; bpn�1

i Š Ep2 , K3.G/ D hbpn�1

i, M D

ha; c; bpi 2 �1 is abelian of type .p; p; pn�1/, and all members of the set�1 � fM g are metacyclic.

(ii) G D ha; bi, where

ap D bpn

D 1; n � 2; Œa; b� D c; Œb; c� D d;

cp D dp D Œa; c� D Œd; a� D Œd; b� D 1:

Here jGj D pnC3, G0 D hc; d i Š Ep2 , K3.G/ D hd i, M D ha; c; d; bpi 2

�1 is abelian of type .p; p; p; pn�1/.

(iii) G D ha; bi, where

ap2

D bpn

D 1; n � 2; Œa; b� D c; Œc; b� D asp ; cp D Œa; c� D 1;

and s D 1 or s is a fixed quadratic non-residue mod p. Here jGj D pnC3,G0 D hc; api Š Ep2 , K3.G/ D hapi,M D ha; c; bpi 2 �1 is abelian of type.p; p2; pn�1/, and all members of the set �1 are nonmetacyclic.

Proof. Lemma 65.2(d,e), Lemma 65.4(c) and Theorem 65.7(c,d) imply thatG0 Š Ep2 ,d.G/ D 2 and p > 2. Since G0 6� Z.G/, we have jK3.G/j D p and G=K3.G/ is anonmetacyclicA1-group (Lemmas 65.2(e) and 64.1(m)).By the structure of G=K3.G/, there are cyclic subgroups A=K3.G/ > f1g and

B=K3.G/ > f1g of G=K3.G/ such that

hA;Bi D G; A \ G0 D B \G0 D A \ B D .G0A/\ B D .G0B/ \ A D K3.G/:


Let A D ha;K3.G/i and B D hb;K3.G/i. The commutator Œa; b� is of order p sinceab ¤ ba and exp.G0/ D p and hŒa; b�i\K3.G/ D f1g sinceG=K3.G/ is nonabelian.SinceG0 6� Z.G/, bothA andB cannot centralizeG0. To fix ideas, assume thatB doesnot centralize G0, which implies ŒB;G0� D K3.G/. Since G0B < G is nonabelian,we must have jG W .G0B/j D p. But .G0B/ \ A D K3.G/ implies jAj D p2, bythe product formula. If also A does not centralize G0, then G0A is nonabelian and soG0A 2 �1. In that case jGj D pjG0Aj D pp3 D p4, a contradiction. We have provedthat A centralizes G0. Let B1 � B be such that K3.G/ � B1 and jB W B1j D p. LetM 2 �1 be abelian. Since G0 < M and G0 6� Z.G/, we have M D CG.G

0/. Onthe other hand, A centralizes G0 and so A < M . Finally, B1 � ˆ.G/ < M and soM D G0AB1 (recall that jG W B1G

0j D p2) andA 6� B1G0 sinceG D hA;Bi). Since

jGj > p4 and jAG0j D p3, we have jB1j � p2 and so jBj � p3.

(i) Suppose first that B does not split over K3.G/, i.e., B is cyclic. Set B D hbi

so that o.b/ D pn with n � 3 and hbpn�1

i D K3.G/. The subgroup BG0 Š MpnC1

is maximal in G since .BG0/A D G, .BG0/ \ A D K3.G/ and jAj D p2. ByLemma 64.1(p), d.M/ > 2 (otherwise, all members of the set �1 are two-generatorwhich is not the case since a nonmetacyclicA1-group has a maximal subgroup that isnot generated by two elements; see Lemma 65.1(a)); moreover, d.M/ D 3 (considerM \ .BG0/) and so �1.M/ Š Ep3 . By the product formula, G D B�1.M/. Itfollows from jABj < jGj and hA;Bi D G that B is not normal in G. Therefore,there is a 2 �1.M/ that does not normalize B . One can take A D ha;K3.G/i. SetŒa; b� D c, where c is an element of order p in G0 � K3.G/. Since BG0 Š MpnC1 ,

one may assume that bc D b1Cpn�1

(if necessary, changing b by another element oforder pn in BG0). Thus we have obtained the group (i) of our proposition.

(ii) Suppose that B splits over K3.G/ so that B D K3.G/� hbi, where o.b/ D pn,n � 2. Consider first the subcase

(ii1) whereA also splits over K3.G/ so that A D K3.G/�hai with o.a/ D p. ThenŒa; b� D c, where c 2 G0 � K3.G/ is of order p and Œb; c� D d , where hd i D K3.G/.We have obtained the group (ii) of our proposition.

(ii2) It remains to consider the subcase where A does not split over K3.G/ so thatA D hai is cyclic of order p2 and hapi D K3.G/. We have again Œa; b� D c, wherec 2 G0 �K3.G/ is of order p and Œc; b� D asp , where s 6 0 .mod p/. For any i 6 0

.mod p/, we replace b with b0 D bi . Then Œa; b0� D Œa; bi � D Œa; b�i� D ci� D c0

with a suitable element � 2 K3.G/ � Z.G/. Hence Œc0; b0� D Œci�; bi � D Œc; b�i2

D

ai2sp D as0p, where s0 D i2s 6 0 .mod p/. It follows that we may assume (runningthrough all i 6 0 .mod p/) that either s0 D 1 or s0 is a (fixed) quadratic non-residuemod p. Writing again b; c; s instead of b0; c0; s0, we have obtained the groups stated inpart (iii) of our proposition.

Proposition 71.4. Let G be a nonmetacyclic A2-group of order > p4 possessing ex-actly one abelian maximal subgroup. Then G0 Š Ep2 and assume, in addition, that


G0 � Z.G/. In that case d.G/ D 3, Z.G/ D ˆ.G/, and one of the following holds:

(a) If G has no normal elementary abelian subgroups of order p3, then p D 2 and

G D ha; b; c j a4 D b4 D Œa; b� D 1; c2 D a2; ac D ab2; bc D ba2i

is the minimal nonmetacyclic group of order 25. HereG is a special group of exponent4,�1.G/ D G0 D Z.G/ D ˆ.G/ D ha2; b2i Š E4, C4 �C4 Š M D hai � hbi 2 �1

is abelian, and all other six members of the set �1 are metacyclic of exponent 4.

(b) If G has a normal elementary abelian subgroup E of order p3, then E D �1.G/

and one of the following holds:

(b1) G D ha; b; d i with

ap˛

D bp D dp2

D 1; ˛ � 2;

Œa; b� D dpa�p˛�1

; Œd; b� D 1; Œa; d � D ap˛�1

;

where � D 0 unless p D 2 and ˛ D 2 in which case � D 1. Here

jGj D p˛C3; G0 D hdp; ap˛�1

i Š Ep2 ;

E D �1.G/ D hb; dp ; ap˛�1

i 6� Z.G/;

and hb; d; api 2 �1 is abelian of type .p; p2; p˛�1/.

(b2) G D ha; b; ci with

ap˛

D bp2

D cp2

D Œb; c� D 1; ˛ � 1;

Œa; b� D x; Œa; c� D y; bp D x˛0

yˇ ; cp D x�yı ;

xp D yp D Œa; x� D Œa; y� D Œb; x� D Œb; y� D Œc; x� D Œc; y� D 1;

where in case p D 2, ˛0 D 0, ˇ D � D ı D 1, and in case p > 2,4ˇ� C .ı � ˛0/2 is a quadratic non-residue mod p. We have jGj D p˛C4,G0 D hx; yi Š Ep2 , E D �1.G/ D G0hap˛�1

i. Let A 2 �1 be abelian. Thenin case ˛ D 1, A D hb; ci Š Cp2 � Cp2 and E 6� Z.G/ and in case ˛ � 2,A D hap ; b; ci is abelian of type .p˛�1; p2; p2/ and E � Z.G/.

Proof. Let G be a nonmetacyclicA2-group of order > p4 having exactly one abelianmaximal subgroup. Then we have G0 Š Ep2 (Lemmas 64.1(q), 65.4(d) and Theorem65.7(a)), and we assume here, in addition, that G0 � Z.G/. It follows from Lemma65.2(a) that d.G/ D 3. Next, Z.G/ D ˆ.G/ (Lemma 65.4(c)), G=Z.G/ Š Ep3 , and ifH ¤ K 2 �1 are both nonabelian, then Z.H/ D Z.K/ D Z.G/ (Theorem 65.7(f)). Itfollows that if x; y 2 G, then jhx; yij � p o.x/o.y/ (since d.G/ D 3, hx; yi is eitherabelian or anA1-subgroup). Also, ifG has no normal elementary abelian subgroups oforder p3, then G is minimal nonmetacyclic (Lemma 65.4(j)). In the last case, p D 2,


jGj D 25, and G is the group stated in part (a) of our proposition (Lemma 64.1(l)and Theorem 66.1).From now on we assume that the group G has a normal subgroup E Š Ep3 . By

Lemma 65.4(k), either �1.G/ D E or E < �1.G/ D K Š Ep4 . Suppose thatthe second case occurs. Every maximal subgroup of G containing K must be abelian(Lemma 65.1) and G=K > f1g is cyclic (Lemma 65.4(b)). Let G D Khai; thenK \ hai D D is of order p. Indeed, jDj � p. Assume that D D f1g. Then, ifhai < T 2 �1, then �1.T / � h�1.hai/; T \ Ki is of order � p4, so T is abelian(Lemma 65.1). Since K 6� T , the set �1 has two distinct abelian members, contraryto the assumption. Thus, ps D jhaij � p2, D � Z.G/, and jGj D psC3. SinceKhapi is abelian, hapi � Z.G/. We have G0 Š Ep2 , G0 < K, and G0 � Z.G/ (byassumption). Since G=Z.G/ Š Ep3 and G0D � Z.G/, we get G0 > D (otherwise,d.G=G0/ D 2 < 3 D d.G/). Therefore, G0hai is an abelian normal subgroup of G oftype .ps ; p/, s � 2, and Z.G/ D G0hapi. We have CK.a/ < Z.G/ and so CK.a/ D

G0 and CG.a/ D G0hai. The subgroup G0hai has exactly p cyclic subgroups of orderps . Since jG W .G0hai/j D p2, we get NK.hai/ > G0. Let x 2 NK.hai/ � G0 sothat x (of order p) induces an automorphism of order p on hai. But then ha; xi is anonabelian subgroup of order psC1 so its index in G equals p2, a contradiction. Wehave proved that �1.G/ D E Š Ep3 . Since G0 < E, the quotient group G=E isabelian. Since d.G=G0/ D 3, G=E is also noncyclic.In what follows A denotes the unique abelian maximal subgroup of G.

(i) First assume thatE 6� ˆ.G/ .D Z.G//; then d.G=E/ D 2. IfA does not containE, then G D AE, �1.A/ D A \E D G0 Š Ep2 and so A is metacyclic. Hence eachmaximal subgroup of G is generated by two elements (Lemma 65.1) and d.G/ D 3.Since G is of class 2 (by hypothesis), Theorem 70.1 shows that G is isomorphic to agroup (b2) for ˛ D 1 in our proposition.We may assume that E < A and so each maximal subgroup of G which does not

containE is nonabelian metacyclic (Lemma 65.1, since all members of the set�1�fAg

are A1-groups). LetM be a nonabelian maximal subgroup of G containing E (thereare exactly p such subgroups since d.G=E/ D 2). Then M is a nonmetacyclic A1-group with Z.M/ D ˆ.M/ D ˆ.G/ D Z.G/ (Theorem 65.7(f)). By Lemma 65.1, wehave, since E 6� Z.M/ (otherwise, E � Z.G/, which is not the case since d.G=E/ D

2 < 3 D d.G=Z.G//),

M D ha; b j ap˛

D bp D 1; ˛ � 2; Œa; b� D c; cp D Œa; c� D Œb; c� D 1i;

where

hci D M 0; b 2 E �G0; M D hbi .hci � hai/;

Z.M/ D Z.G/ D hci � hapi; jGj D p˛C3; G0 D hap˛�1

i � hci:

In particular, G=E has a cyclic subgroup M=E of index p. Now, ha; ci .� G0/ isan abelian normal subgroup of type .p˛; p/ and it has exactly p cyclic subgroups of


index p which are not normal inM (see the last assertion of Lemma 65.1). Therefore,NM .hai/ D ha; ci and so N D NG.hai/ must cover G=M since N 6� M in viewof jG W N j D c˛.ha; ci/ D p D jG W M j (here c˛.X/ is the number of cyclicsubgroups of order p˛ in a p-group X). Thus N 2 �1 withM \ N D ha; ci and soE \ N D G0. Therefore N is nonabelian metacyclic, CG.a/ D ha; ci, N is an A1-group, and so it is “splitting” metacyclic (Lemma 65.1). We have f1g < N 0 < hai\G0

and so N 0 D hap˛�1

i � Ã1.N / (Lemma 65.1). It follows that Z.G/ D hap; ci D

Z.N / D ˆ.N/ D Ã1.N /, and so there exists d 2 N � M D N � ha; ci such thatdp 2 G0 � hap˛�1

i and o.d/ D p2 (note that�1.N / � ha; ci). We have N D ha; d i

and N=hai is cyclic. Next, CG.E/ D A.If ˛ > 2, then working in the abelian subgroup hd; ap˛�2

i Š Cp2 � Cp2 , we may

choose an element d 0 2 hd; ap˛�2

i �M such that .d 0/p D c and so we may assumefrom the start that dp D c D Œa; b�. If u 2 G is of order p2, then jhu;Eij D p4 sohu;Ei is abelian, and we conclude that u 2 CG.E/ D A. It follows that �2.G/ D

Ehu 2 G j o.u/ D p2i � A so Œd; b� D 1. Replacing d with d 0 D d i (i 6 0

.mod p/), we may assume that ad 0

D ad i

D a1Cp˛�1

. But then we also replaceb with b0 D bi and c with c0 D ci and obtain .d 0/p D .dp/i D ci D c0 D

Œa; b0� and the relation Œb0; d 0� D 1 remains valid. Writing again d; b; c instead ofd 0; b0; c0, respectively, we obtain exactly the relations stated in part (b1) for ˛ > 2 ofour proposition.It remains to investigate the case ˛ D 2, where jGj D p5, G0 D Z.G/ D ˆ.G/ Š

Ep2 , E D �1.G/ Š Ep3 , exp.G/ D p2. For each x 2 N � ha; ci, xp 2 G0 � hapi

(otherwise, ha; xi would be nonabelian of order p3; here N D NG.hai). Since G D

EN , we have A D E.A\N/, by the modular law. Note that A\N is abelian of type.p2; p/ so it is generated by elements of order p2. Since N D hai.A \ N/ (indeed,M D haiE is nonabelian so a 62 CG.E/ D A), there exists d 2 .A \ N/ ��1.A \

N/ such that ha; d i is nonabelian. It follows that dp 62 hapi (otherwise, ha; d i isnonabelian of order p3, contrary to the fact that G is an A2-group). Replacing dwith another generator of hd i if necessary, we may assume that ad D a1Cp. Setdp D cia�jp , where i 6 0 .mod p/. We replace b with b0 D bi and c with c0 D ci ,so that we get Œa; b0� D Œa; bi � D Œa; b�i D ci D c0 and dp D c0a�jp. Thus, we mayassume from the start that Œa; b� D c and c D ajpdp with j 2 Z.We shall determine the integer j .mod p/. Consider the nonabelian maximal sub-

group S D hb; ad �i, where � is any integer (recall that d.G/ D 3). The subgroupS is a nonmetacyclic (otherwise, b 2 Ã1.S/ D ˆ.S/) A1-group containing E withZ.S/ D Z.G/ D G0 D hap; dpi D h.ad �/p ; Œad � ; b�i. If p > 2, then

.ad �/p D ap.dp/� ; Œad � ; b� D Œa; b� D c D ajpdp;

and so we must have (since the displayed two elements are independent)ˇ

ˇ

ˇ

1 �j 1

ˇ

ˇ

ˇ

6 0

.mod p/, which gives 1 � j� 6 0 .mod p/ for every integer � . If j 6 0 .mod p/,then we can solve the congruence j� 1 .mod p/ for � and get a contradiction.


Hence j 0 .mod p/ and we may set in this case j D 0; then c D ajpdp D dp. Ifp D 2, then taking � D 1, we have

.ad/2 D a2d2Œd; a� D a2d2a2 D d2; Œad; b� D Œa; b� D c D a2jd2

(by the above, ad D a1Cp for any p), and we must haveˇ

ˇ

0 1j 1

ˇ

ˇ 6 0 .mod 2/, whichgives j 6 0 .mod 2/ and so we may set in this case j D 1; then c D a2d2. We haveobtained the relations stated in part (b1) for ˛ D 2 of our proposition.

(ii) Suppose that Ep3 Š E D �1.G/ � Z.G/ D ˆ.G/. LetH 2 �1 be nonabelian.Then E < H and soH is a nonmetacyclicA1-group:

H D ha; b j ap˛

D bpˇ

D 1; Œa; b� D c; cp D Œa; c� D Œb; c� D 1i;

where E D hap˛�1

; bpˇ�1

; ci. Since E � Z.G/ D Z.H/ D ˆ.H/ (Theorem65.7(f)), we must have ˛ � 2 and ˇ � 2 so jGj D pjH j D p˛CˇC2 � p6. LetG0 < G0 be of order p such that G0 ¤ H 0. Since .G=G0/

0 D G0=G0 is of order pand d.G=G0/ D 3, a nonabelian group G=G0 is an A2-group of order � p5 with thecommutator group of order p, whereH=G0 is anA1-subgroup. By Proposition 71.1,there exists d 2 G � H such that dp 2 G0. Hence o.d/ D p2 and d 62 Z.G/ sinced 62 H and �1.G/ D E < H . It follows that d does not centralize a or b. Withoutloss of generality, we assume that Œd; a� ¤ 1 and so nonabelian hd; ai 2 �1 in viewof d.G/ D 3. Since o.d/ D p2 and o.a/ D p˛ , we have jhd; aij � p˛C2C1 in viewof o.Œd; a�/ D p, and therefore jGj � p˛C4. Hence jGj D p˛CˇC2 � p˛C4 and soˇ � 2. This gives ˇ D 2 and jGj D p˛C4 with ˛ � 2. We have Z.G/ D Ehapi.Assume first that ˛ > 2; then jGj > p6. In that case, any two elements x; y 2 G

of orders � p2 commute (otherwise, hx; yi 2 �1, since d.G/ D 3, is nonabelianwith jhx; yij � p o.x/o.y/ � p5, a contradiction). Since Œb; d � D 1 in view ofo.b/ D o.d/ D p2, we get hb; d i D hbi � hd i. Assume that this is false. Thenhb; d i D hbi � hd1i with o.d1/ D p. Since b 2 H , d 62 H , we get d1 62 H , acontradiction: �1.G/ D E < H . Consider K D ha; d i, where K 2 �1 is nonabeliancontaining E (by assumption, E < ˆ.G/). Let G0 ¤ K 0 be a subgroup of order pin G0. Then applying Proposition 71.1 on G=G0, we see that there is b0 2 G � K

with bp0 2 G0 and o.b0/ D p2. Since Œd; b0� D 1 and there are no elements of order

p in G � K, we see (as before) that hd; b0i D hd i � hb0i. But G D ha; d; b0i,b0 62 Z.G/, and so Œa; b0� ¤ 1. Replacing H with H0 D ha; b0i (¤ K) and bwith b0, we may assume from the start that bp 2 G0. Hence, we may assume thatÃ1.hb; d i/ D �1.hb; d i/ D G0. Since jGj D p˛C4 and NG D G=G0 is generatedby Nb; Nd ; Na (bar convention) of orders at most p;p; p˛ , respectively, and j NGj D p˛C2,we get o. Na/ D p˛ which gives hai \ G0 D f1g and so hai \ hb; d i D f1g since�1.hb; d i/ D G0.Suppose that ˛ D 2; then jGj D p6, exp.G/ D p2,E D Z.G/ D ˆ.G/ D �1.G/.

Let X < G0 be of order p. Since d.G=X/ D 3 and .G=X/0 is of order p, G=X is anA2-group of order p5, and so we may apply again Proposition 71.1. There isM 2 �1


such that M=X is an A1-group and there is c 2 G � M such that cp 2 G0 andŒG; c� � X . If cp 2 X , then hci (of order p2; recall that �1.G/ < M ) is normal inG. Since c 62 Z.G/ D E, there is an element l of order p2 with Œl; c� ¤ 1 and sohŒl; c�i D hcpi. But then jhl; cij � o.l/o.c/ � p4, a contradiction. Thus cp 2 G0�X .We consider now the subgroup Y D hcpi of order p in G0 and apply Proposition 71.1to G=Y . As before, there is an element b 2 G such that bp 2 G0 � Y . It followshbp; cpi D G0. The abelian group hb; ci=G0 is generated by two elements of orderp and so jhb; cij � p4. Thus hb; ci is abelian and since hbi \ hci D f1g, we haveS D hb; ci Š Cp2 � Cp2 and S is normal in G. But, E D ˆ.G/ D G0Ã1.G/ and sothere is a 2 G such that ap 2 E � G0. Hence hai \ S D f1g and so Shai D G sinceo.a/ D p2.In any case (for all ˛ � 2) we have obtained the following configuration (where in

case ˛ > 2 we write c instead of d ). We have G D hai .hbi � hci/, with ap˛

D 1,˛ � 2, bp2

D cp2

D 1, Œb; c� D 1, where G is a semidirect product with kernelhbi � hci Š Cp2 � Cp2 , G0 D hbp; cpi, and A D hap ; b; ci is the unique abelianmaximal subgroup of G (of type .p˛�1; p2; p2/).Set Œa; b� D x, Œa; c� D y. Then G0 D hx; yi D hbp; cpi (� Z.G/) and so bp D

x˛0

yˇ and cp D x�yı , where, since bp and cp are independent, we getˇ

ˇ

ˇ

˛0 ˇ� ı

ˇ

ˇ

ˇ

6 0

.mod p/.The subgroup ha; b�c�i is nonabelian if and only if Œa; b�c�� D x�y� ¤ 1 which

occurs if and only if both � and � are not multiples of p. In that case L D ha; b�c�i

is a nonmetacyclic A1-subgroup of index p in G. Since hx�y�i D L0 and L0 is amaximal cyclic subgroup in L, we have G0 D h.b�c�/p ; x�y�i, where .b�c�/p D

.bp/�.cp/� D x˛0�C��yˇ�Cı�. Hence,ˇ

ˇ

ˇ

˛0�C�� ˇ�Cı��

ˇ

ˇ

ˇ

0 .mod p/ only if �

� 0 .mod p/. This gives

ˇ�2 C .ı � ˛0/�� 2 0 .mod p/(1)

only if � � 0 .mod p/. From (1) follows (setting � D 1, � D 0 or � D 0, � D 1)ˇ 6 0 .mod p/ and � 6 0 .mod p/.Suppose first p > 2. Then we compute (using (1))

.2ˇ�C.ı�˛0/�/2�.4ˇ�C.ı�˛0/2/�2 D 4ˇ.ˇ�2C.ı�˛0/��2/ 0 .mod p/

only if � � 0 .mod p/. Hence

.2ˇ� C .ı � ˛0/�/2 .4ˇ� C .ı � ˛0/2/�2 .mod p/(2)

only if � � 0 .mod p/. From (2) follows at once that 4ˇ� C .ı � ˛0/2 is aquadratic non-residue mod p. Indeed, if 4ˇ�C .ı�˛0/2 is a quadratic residue modulop, then we set � D 1 in (2) and then we solve the congruence .2ˇ� C .ı � ˛0//2

4ˇ� C .ı � ˛0/2 .mod p/ for � and get a contradiction.


Suppose now that p D 2. We know already that ˇ D � D 1. Then (1) yields

�2 C .ı C ˛0/��C �2 0 .mod 2/(3)

only if � � 0 .mod 2/. Setting � D � D 1 in (3), we get ı C ˛0 D 1. This givesb2 D x˛0

y and c2 D xy˛0C1, where ˛0 D 0; 1. Hence we have the following twopossibilities:

b2 D y; c2 D xy;(4)

b2 D xy; c2 D x:(5)

However, interchanging b and c and also x and y, we see that the relations Œa; b� D x

and Œa; c� D y remain unchanged but (4) goes onto (5). Hence, we may assume thatwe have relations (4) and so we get ˛0 D 0 and ˇ D � D ı D 1. Our group G isuniquely determined.

Proposition 71.5. Let G be a nonmetacyclic A2-group of order > p4 all of whosemaximal subgroups are nonabelian. Then we have one of the following possibilities:

(a) d.G/ D 3, p D 2, and G D ha; b; ci with

a4 D b4 D c4 D 1; Œa; b� D c2; Œa; c� D b2c2; Œb; c� D a2b2;

Œa2; b� D Œa2; c� D Œb2; a� D Œb2; c� D Œc2; a� D Œc2; b� D 1;

where G0 D ha2; b2; c2i D Z.G/ D ˆ.G/ D �1.G/ Š E23 , exp.G/ D 4 andall members of the set �1 are nonmetacyclic A1-groups. (From Theorem 70.4follows that G is isomorphic to the Suzuki 2-group of order 26.)

(b) d.G/ D 2, p > 2, G is of order p5:

G D ha; x j ap2

D xp2

D 1; Œa; x� D b; Œa; b� D y1; Œx; b� D y2;

bp D yp1 D y

p2 D Œa; y1� D Œx; y1� D Œa; y2� D Œx; y2� D 1;

ap D y˛1 y

ˇ2 ; x

p D y�1 y

ı2i;

where in case p > 3, 4ˇ� C .ı � ˛/2 is a quadratic non-residue mod p. HereG0 D hb; y1; y2i D �1.G/ Š Ep3 , Z.G/ D K3.G/ D Ã1.G/ D hy1; y2i Š

Ep2 .

Proof. Let G be a nonmetacyclic A2-group of order > p4 all of whose maximalsubgroups are nonabelian and therefore they are generated by two elements.Suppose first that d.G/ D 3. ThenG0 � Z.G/ (Lemma 65.4(a)). By Theorem 70.1,

G is a uniquely determined group of order 26 as stated in part (a) of our proposition.Now suppose that d.G/ D 2. If p D 2, then G is metacyclic (Lemma 64.1(n)), a

contradiction. Hence p > 2; then Ã1.G/ D K3.G/, jG W G0j D p2 (Lemma 64.1(m))


so jG0j D p3 and jGj D p5 (Lemma 65.2(d)). Since G=K3.G/ is generated by itssubgroups of index p2, K3.G/ � Z.G/ (Lemma 65.4(c)) and so G is of class 3. Wehave K3.G/ D Z.G/ Š Ep2 and G0 Š Ep3 (Lemma 65.7(d)).Each M 2 �1 is a nonmetacyclic A1-group with G0 D �1.M/ Š Ep3 and so

ˆ.G/ D G0 D �1.G/ and exp.G/ D p2 (Lemma 65.1). Let G D ha; xi; thena; x 2 G � ˆ.G/ D G � �1.G/ and therefore o.a/ D o.b/ D p2. We haveG0 D hŒa; x�;K3.G/i and so Œa; x� D b 2 G0 � K3.G/ which gives o.b/ D p. SetŒa; b� D y1 and Œx; b� D y2, where y1; y2 2 K3.G/ D Z.G/. But ha; biZ.G/ andhx; biZ.G/ are members of the set �1 and so they are nonabelian. Hence o.y1/ D

o.y2/ D p (Lemma 65.1) and ha; biZ.G/ D ha; bi, hx; biZ.G/ D hx; bi. Sinceha; bi and hx; bi are distinct members of the set �1, we have hy1i ¤ hy2i (Lemma65.2(d)). It follows hy1; y2i D Z.G/. Since ap; xp 2 Z.G/, we get ap D y˛

1 yˇ2 and

xp D y�1 y

ı2 .

Suppose that p > 3. ThenG is regular (Lemma 64.1(a)) and sinceG0 is elementaryabelian, we get that G is p-abelian, i.e., .lk/p D lpkp for all l; k 2 G (see �7). Sinceha; xi D G, we have in this case hap; xpi D Ã1.G/ D Z.G/ D hy1; y2i. Considerthe subgroup S D hb; a�x�i (S < G since b 2 G0; it follows that S is of class � 2).Obviously, a�x� 2 G � G0 if and only if not both � and � are divisible by p. In thatcase Œa�x� ; b� D Œa; b�� Œx; b�� D y

�1y

�2 ¤ 1, and so S is nonabelian; then S 2 �1

hence E < S . It follows that S is nonmetacyclic. Here we have used the fact thatŒa; b�; Œx; b� 2 Z.G/. Since Z.G/ D Z.S/ and S 0 is a maximal cyclic subgroup of S ,we have Z.S/ D Z.G/ D hy1; y2i D h.a�x�/p; Œa�x� ; b�i. We compute

.a�x�/p D .ap/�.xp/� D y˛�C��1 y

ˇ�Cı�2 and Œa�x� ; b� D y

�1y

�2 ;

and soˇ

ˇ

ˇ

ˇ

˛� C �� ˇ� C ı�

� �

ˇ

ˇ

ˇ

ˇ

0 .mod p/(�)

only if � � 0 .mod p/. From (�) we get (as in the proof of Proposition 71.4) that4ˇ� C .ı � ˛/2 is a quadratic non-residue modulo p. Our proposition is proved andallA2-groups are determined.

The classification of A2-groups is contained in Propositions 71.1–71.5. The sub-group structure of A2-groups is given in �65.It is impossible to compare our proof with Kazarin’s proof since the last one is

not published and only sketched in his PhD Thesis. The paper [She], also devoted toA2-groups, is not accessible. In any case, our proof is based on other ideas. Indeed,the above authors were forced to use tables [HS]. Besides, Kazarin used Nazarova-Roiter’s classification of p-groups with abelian subgroup of index p so his proof is notelementary. Instead of this, we use Theorem 70.1 which is new for p D 2.

Exercise 1. Let a p-group G D A�C , where C Š Cp2 , is anA2-group. Describe theintersection A \ C . Consider in detail the case when A is metacyclic.


Exercise 2. Classify the metacyclic A1-groups G of order 2n such that there is aninvolution x 2 G which is not square.

Exercise 3. Find theA2-groupsG such that jHG W H j D p for all nonnormalH < G.

Exercise 4. Find the A2-groups G such that exp.HG/ D exp.H/ for allH < G.

The following three theorems are due to the first author.

Theorem 71.6 ([BJ2]). Let G be a nonmetacyclic two-generator 2-group. Then thenumber of two-generator members of the set �1 is even.

Proof. By hypothesis, G is nonabelian. If A 2 �1, then Ã2.G/ � Ã1.A/ D ˆ.A/

so d.A=Ã2.G// D d.A/. Therefore, without loss of generality, one may assumethat Ã2.G/ D f1g; then exp.G/ D 4. Let �1.G/ D fA;B;C g. Since G is notmetacyclic, one may assume that d.C / D 3 (Appendix 25 and Theorem 44.5). As-sume that the theorem is false. Then we may assume that d.A/ D 2 and d.B/ D 3

(see Appendix 25). Let R be a G-invariant subgroup of index 2 in G0. Then G=Ris minimal nonabelian (Lemma 65.2(a)) and nonmetacyclic (Theorem 36.1). Sinceexp.G=R/ D 4, we get jG=Rj � 25 (Lemma 65.1). If jG=Rj D 25, the set �1 has notwo-generator members (Lemma 65.1), contrary to the assumption. Thus, jG=Rj D 24

so jG=G0j D 8. One may assume that C=R D �1.G=R/. Since B=R is abelian oftype .4; 2/ so two-generator and d.B/ D 3, we get R ¤ ˆ.B/. Clearly, ˆ.B/ is aG-invariant subgroup of index 8 in B . Write NG D G=ˆ.B/; then NG is not of maximalclass since it contains NB Š E8. It follows that cl. NG/ D 2. Since jG=G0j D 8, we getjG0 W ˆ.B/j D 2. Since G0 has only one G-invariant subgroup of index 2 (Remark36.1), we conclude that ˆ.B/ D R. This is a contradiction since B=R is abelian oftype .4; 2/ and exp.B=ˆ.B// D 2. Thus, the number of two-generator members ofthe set �1 is even, as was to be shown.

Theorem 71.7. Suppose that all nonabelian maximal subgroups of a 2-group G aremetacyclic and jGj > 24. Then one of the following holds: (a) G is abelian. (b) G ismetacyclic. (c) G is an A1-group. (d) G is minimal nonmetacyclic. (e) G D M � C ,whereM is a metacyclic A1-group and jC j D 2.

Proof. Assume that G is not of types (a)–(d). Then, by Theorem 71.6, d.G/ > 2

and there is in G a maximal subgroup A that is not two-generator; in that case, byhypothesis, A is abelian. By assumption, there is inG a nonabelian maximal subgroupM ; thenM is metacyclic so d.G/ D 3. SinceM\A is a metacyclic maximal subgroupof A, we get d.A/ D 3. Write E D �1.A/; then E GG and E Š E8. By hypothesis,all maximal subgroups of G that contain E, are abelian. Since G has at most threeabelian maximal subgroups, we get d.G=E/ � 2.

(i) Suppose that d.G=E/ D 2. Let B=E be a maximal subgroup of G=E andB ¤ A. Then B is abelian so A \ B D Z.G/ has index 4 in G; in that case, jG0j D 2

(Lemma 64.1(q)) and E � Z.G/. It follows from jM 0j D 2 and d.M/ D 2 thatM is


an A1-group. Let C < E be a subgroup of order 2 not contained in �1.M/ .Š E4/;then G D M � C is as stated in (e).

(ii) Now letG=E be cyclic; thenE 6� Z.G/ sinceG is nonabelian. Since d.G/ D 3,we get jG0j D 2. We have E D �1.G/ since �1.G/ < A in view of G=E iscyclic of order > 2. Let, as above, M is a nonabelian maximal subgroup of G. ByLemma 65.2(a),M is anA1-subgroup; moreover,M Š M2n , n > 3, since�1.M/ D

M \ E Š E4 and M=�1.M/ Š G=E is cyclic of order > 2, and we conclude thatZ.M/ is cyclic. If Z.G/ is noncyclic, then G D M � C , where jC j D 2, and G isas stated in (e). Next assume that Z.G/ is cyclic. We get jG W Z.G/j D 2jG0j D 4

(Lemma 64.1(q)). Then, by the product formula, G D EZ.G/ so G is abelian, a finalcontradiction.

Theorem 71.8. Suppose that a two-generator nonabelian p-group G contains anabelian maximal subgroup A. Set R D hxp j x 2 G � Ai. Then R � Z.G/ andG=R is of maximal class, unless G is an A1-group.

Proof. Recall that if A is an abelian maximal subgroup of a nonabelian p-group Gand jG W G0j D p2, then G is of maximal class (this is proved by induction withhelp of Lemma 64.1(q)). We have R � Z.G/ \ ˆ.G/ since CG.x

p/ � hx;Ai for allx 2 G�A. Write NG D G=R; then NG is noncyclic. Since all elements of the set NG� NA

have the same order p, we get�1. NG/ � h NG � NAi D NG. It follows that NG0 D ˆ. NG/ sothat NG= NG0 Š Ep2 since d. NG/ D d.G/ D 2. Suppose that NG0 D fN1g. Then R D ˆ.G/

so G=Z.G/ Š Ep2 . In that case, all maximal subgroups of G are abelian so G is anA1-group. Now suppose that NG0 > fN1g; then NG is nonabelian. In that case, as we havenoticed, NG is of maximal class.

Problems

Problem 1. Classify the p-groups G such that, whenever H < G is minimal non-abelian andH Š H1 < G, thenH1 D H .

Problem 2. Classify the p-groups all of whose A2-subgroups are isomorphic. More-over, study the p-groups all of whoseA1-subgroups of the same order are isomorphic.

Problem 3. Classify the non-Dedekindian p-groups all of whose nonnormal sub-groups are abelian.

Problem 4. Classify the p-groups all of whoseA2-subgroups are metacyclic.

Problem 5. Let G be a p-group all of whose maximal subgroups areA2-groups. Is ittrue that jGj is bounded?

Problem 6. Classify the p-groups with at most two A2-subgroups. (Note that p-groups with exactly one A2-subgroup are not classified.)


Problem 7. Describe the automorphism groups of all A1- andA2-groups.

Problem 8. Study the p-groups G such that, whenever A;B 2 �1 are distinct, thenA \ B is either abelian or minimal nonabelian.

Problem 9. Study the p-groups all of whoseA2-subgroups are not two-generator.

Problem 10. Study the subgroup structure of A � A, where A is anA1-group.

Problem 11. Classify the p-groups G which are lattice isomorphic with B , where Bis (i) anA1-group, (ii) anA2-group.

Problem 12. Study the p-groups G such that, whenever A < G is an A1-subgroup,then A < B � G, where B is anA2-subgroup.

Problem 13. Classify the p-groups of exponent > p all of whose A2-groups haveexponent p (then these subgroups have order p4).

Problem 14. Study the p-groups G such that ˆ.G/ is anA1-group (A2-group).

Problem 15. Study the p-groups G all of whose subgroups of fixed order pn areA2-groups. Is it possible to estimate jGj in terms of n?

Problem 16. Classify the p-groups with minimal nonabelian subgroup of index p.

�72

An-groups, n > 2

In this section we study An-groups with n > 2. A p-group G is said to be an An-group if all its subgroups of index pn are abelian but G has a nonabelian subgroup ofindex pn�1.Main results of this section are due to the first author.

1o. First we classifyAn-groups G with G0 Š Cpn , n > 1 (for n D 2, see Corollary65.3).

Proposition 72.1. (a) Let G be a metacyclic p-group with jG0j D pn. Then G is anAn-group.

(b) Let G be an An-group, n > 1, with cyclic derived subgroup G0 of order � pn.Then G is metacyclic and jG0j D pn.

Proof. We use induction on jGj.

(a) If n D 1, then G is an A1-group (Lemma 65.2(a)). Now suppose that n > 1.Let L D Ã1.G

0/.� ˆ.G//. By Lemma 65.2(a), G=L is minimal nonabelian so, ifH 2 �1, thenH=L is abelian, and we haveH 0 � L. By induction, H is anAs-groupwith s < n. By Lemma 64.1(u), there is F 2 �1 with jG0 W F 0j D p so F is anAn�1-group since jF 0j D pn�1. It follows that G is anAn-group.

(b) Set jG0j D ps , s � n. In view of Lemma 65.2(a,b) and Theorem 65.7(a), onemay assume that n > 2. Take H 2 �1. If H 0 D G0, then, by induction, H is an As-group, a contradiction since s � n. Therefore, if G0 D hŒx; y�i, then G D hx; yi, i.e.,d.G/ D 2. By Lemma 64.1(u), there is F 2 �1 with jG0 W F 0j D p so, by induction,F is anAs�1-group and so s�1 � n�1 sinceG is anAn-group. It follows that s D n

and F is metacyclic, by induction since it is an An�1-group with cyclic F 0 of orderpn�1 > p. Let L D Ã1.G

0/ and T D Ã2.G0/. IfH 2 �1, thenH=L is abelian since

G=L is an A1-group (Lemma 65.2(a)). By Lemma 65.1, G=T is not an A1-group.Then G=T has a maximal subgroup A=T that is nonabelian so .A=T /0 > f1g. SinceA0 and G0 are cyclic, we get T < A0 so, by the above, jA0j D pn�1. By induction,A is metacyclic. Then A=T is an A1-group (Lemma 65.2(a)). Thus, every maximalsubgroup of G=T is either abelian or an A1-group so G=T is an A2-group. Sincej.G=T /0j D p2, G=T is metacyclic, by Lemma 65.7(a). Now, by Lemma 64.1(o), Gis metacyclic since T � Ã2.G/.

�72 An-groups, n > 2 249

Let G be a nonabelian metacyclic group of order pm and pk D min fjAj j A <

G; A0 > f1gg. Then G is an Am�.k�1/-group (indeed, all subgroups of G of orderpk�1 are abelian andG has a nonabelian subgroupA of orderpk) so jG0j D pm�.k�1/

(Proposition 72.1(a)) whence jG=G0j D pk�1. In that case, G0 < Z < G, where Zis cyclic of index pk�2 in G. In particular, if k D 3, then G has a cyclic subgroupof index p3�2 D p so it is either a 2-group of maximal class or p > 2 and m D 3

(Lemma 64.1(t)). The last result coincides with Proposition 10.19.Let p > 2 and suppose that a metacyclic p-group G of order pm > p4 contains

a normal nonabelian subgroup H of order p4 and exponent p2. Then H D �2.G/

so all subgroups of G of order p3 are abelian. It follows that G is an Am�3-group sojG0j D pm�3 (Proposition 72.1) and G has a normal cyclic subgroup Z of index p2,G0 < Z, such thatG=Z is cyclic. We have Z.H/ Š Ep2 . LetL be a subgroup of orderp in Z.H/, L ¤ H 0. Then jG W CG.L/j � p. Since CG.L/=L contains a nonabeliansubgroup H=L of order p3, we get CG.L/=L D H=L (Lemma 64.1(t)). In that case,jG W H j D p so jGj D pjG W H j D p5, and Z.H/ 6� Z.G/ so Z.G/ is cyclic.Let G be a metacyclic p-group and Mp4 Š H < G. In that case, G has no cyclic

subgroup of index p (Lemma 64.1(t)). Then G is an Am�3-group so jG0j D pm�3

and G has a normal cyclic subgroup Z ¤ ˆ.G/ of index p2.

Exercise. Describe the nonabelian metacyclic 2-groups of order 2m > 25 withoutnormal subgroups of order 24 and exponent 4.

Solution. Suppose that G has no cyclic subgroups of index 2 (otherwise, G satisfiesthe condition). Then �1.G/ Š E4. Write NG D G=�1.G/. Then NG has no normalabelian subgroups of type .2; 2/ so it is of maximal class (if NG is cyclic, then G has acyclic subgroup of index 2 so Š M2m). Suppose that NG is dihedral. By Proposition10.17, all subgroups of G of order 8 are abelian. Since �1. NG/ D NG, it follows that�1.G/ � Z.G/. If G=�1.G/ is of maximal class, then G has no normal subgroupsof order 24 and exponent 4 since any such subgroup must contain �1.G/. In the caseunder consideration, G is a U2-group (see �67).

2o. In this subsection and subsection 3o we estimate jG0j, whereG is anAn-group,n D 3; 4.If a p-group G is an A3-group and U;V 2 �1 are distinct, then jG0j � pjU 0V 0j �

p7 (Lemmas 65.2(d) and 64.1(u)). Proposition 72.2 yields essentially better estimate,and that estimate is best possible.

Proposition 72.2. If a p-group G is an A3-group, then jG0j � p4. For p D 5, thereexists an A3-group G with jG0j D 54.

Proof. Assume that jG0j > p4. Take U 2 �1. Then jU 0j � p3 and, if jU 0j D p3,then U 0 Š Ep3 (Theorem 65.7(a,d)).

(i) We claim that G0 is regular. This is the case if jG W G0j > p2 since then G0 isabelian. Let jG W G0j D p2 and let G0 be nonabelian; then p > 2 (Taussky’s theorem)


and G0 is anA1-group. In that case, G0 is regular since cl.G/ � 2 (Lemmas 65.1 and64.1(a)); moreover, G0 is metacyclic (Theorem 44.12).

(ii) We claim that if H � G, then jH=Ã1.H/j � p5, and this inequality is strongif H < G. Indeed, there is an A2-subgroup A 2 �1. Then jA=Ã1.A/j � p4 (Lemma65.4(a)) so jG=Ã1.G/j � p5. TakeH < G such that jH=Ã1.H/j � p5. By Lemmas65.1 and 65.4(b), H is abelian so �1.H/ is elementary abelian of order � p5. Thenall members of the set �1 containing �1.H/, are abelian (Lemmas 65.1 and 65.4(b)).By assumption, jG0j > p4 > p so H is the unique member of the set �1 containing�1.H/ (Lemma 65.2(c)), and we conclude that G=�1.H/ is cyclic whence G0 <

�1.H/. Considering H \ F , where F 2 �1 is nonabelian, we get j�1.H/j � p5 sojG0j < p5, a contradiction. (If G D S.p3/�Ep2 , where S.p3/ is nonabelian of orderp3 and exponent p > 2, then G is anA3-group, jGj D p5 and exp.G/ D p.)

(iii) Assume that jG0j > p5. Let U;V 2 �1 be distinct; then jU 0V 0j � 1p

jG0j � p5

(Lemma 64.1(u)) so one may assume that jU 0j D p3; then jV 0j � p2. If jV 0j D p3,then exp.U 0V 0/ D p since exp.U 0/ D exp.V 0/ D p (Theorem 65.7(d), Lemma64.1(a) and (i)), and jU 0V 0j � p5, contrary to (ii). Thus, jV 0j D p2; then U 0 \ V 0 D

f1g, by the product formula. By (ii), since jU 0V 0j � p5, we get exp.U 0V 0/ > p so V 0

is cyclic; then V is metacyclic (Theorem 65.7(a)), a contradiction since Ep3 Š U 0 <

G0 � ˆ.G/ < V . Thus, jG0j D p5. Then the set �1 has no abelian members, byLemmas 65.2(d) and 64.1(u) (otherwise, if A 2 �1 is abelian andH 2 �1 � fAg, thenjH 0j � p3 so jG0j � pjA0H 0j � p4, contrary to the assumption).

(iv) Assume that there are distinct U;V 2 �1 such that exp.U 0/ D exp.V 0/ D p.Then, using (i) and (ii), we get: U 0V 0 is of order p4 and exponent p so U 0V 0 <

G0, and exp.G0/ D p2 since jG0j D p5. In that case, U=U 0V 0 and V=U 0V 0 arecyclic (Lemma 65.4(b)). Thus, the nonabelian group G=U 0V 0 has two distinct cyclicsubgroups U=U 0V 0 and V=U 0V 0 of index p so it is either ordinary quaternion orŠ Mpn (Lemma 64.1(t)). In the first case jG W G0j D 4 so G is a 2-group of maximalclass (Lemma 64.1(s)), a contradiction. Assume that G=U 0V 0 Š Mpn . Let H=U 0V 0

be the noncyclic subgroup of index p in G=U 0V 0. In that case, H is neither A1- norA2-group (Lemmas 65.1 and 65.4(b)) so H is abelian, a contradiction (see the lastsentence of part (iii)).

Now we are ready to complete the proof. By (iv), there is V 2 �1 such thatexp.V 0/ > p; then V is metacyclic and V 0 is cyclic of order p2 (Theorem 65.7(d,a)).TakeU 2 �1�fV g. SinceU 0 < V , we get j�1.U

0/j � p2 so jU 0j � p2 (Lemma 65.1and Theorem 65.7(a,d)). Then U 0 \ V 0 D f1g (Lemmas 65.2(d) and 64.1(u)). SinceU 0V 0 D U 0 � V 0 < V and V is metacyclic, U 0 Š Cp2 (Corollary 65.3 and Lemma64.1(u)). Thus, all members of the set �1 are metacyclic. Since G is nonmetacyclic(Proposition 72.1(a)), we get, jG0j � p2 (Lemma 64.1(l)), a contradiction.

Below we present an A3-group G with jG0j D p4 for p D 5 (this example wasconstructed by the second author).

�72 An-groups, n > 2 251

Let G be the group of order 56 with the following generators and relations:

hb; h j b25 D h25 D 1; Œh; b� D a; a5 D 1; Œa; b� D c; c5 D 1;

Œb; c� D d; d5 D 1; Œc; a� D Œh; d � D Œb; d � D Œh; c� D 1;

Œh; a� D s; s5 D 1; Œh; s� D Œb; s� D 1; d D h5; b5 D si:

Here G0 D ha; c; d; si Š E54 , K3.G/ D hc; d; si Š E53 , K4.G/ D hd i is of order5, and Z.G/ D hd; si Š E52 . All maximal subgroups of G are A2-groups and so Gis an A3-group. Finally, we have shown that this group G exists as a subgroup of thesymmetric group of degree 625.

In the following proposition we investigate A3-groups with derived subgroup oforder p4 in some detail. We show that then G0 is nonmetacyclic. Note that jGj � p6

with strong inequality if p D 2 (Taussky’s theorem).

Proposition 72.3. Suppose that a p-group G is an A3-group with jG0j D p4.

(a) G0 is abelian.

(b) If exp.G0/ D p, then p > 2, G=G0 is abelian of type .pn; p/ (n � 1) andG0 Š Ep4 . If, in addition, n > 1, then �1.G/ D G0.

(c) G0 is not metacyclic.

Proof. By Proposition 72.1, G0 is noncyclic so G is not metacyclic.

(i) Assume that G0 is metacyclic. Let us prove that then p D 2, the set �1 hasonly one nonmetacyclic member and G has no normal elementary abelian subgroupsof order 8. Since the set �1 has a nonmetacyclic member (Lemma 64.1(l)), there isU 2 �1 such that exp.U 0/ � p (Theorem 65.7(a,d)); then jU 0j � j�1.G

0/j D p2 soU 0 � Z.G0/ (indeed, every G-invariant subgroup L of order � p2 in G0 is containedin Z.G0/: consider CG.L/). Take V 2 �1 � fU g. By Lemma 64.1(u), jG0 W U 0V 0j �

p so jU 0V 0j � p3 hence exp.U 0V 0/ > p since G0 is metacyclic. It follows thatexp.V 0/ > p. Since V 0 is abelian in view of jV 0j � p3, U 0V 0 is also abelian (recallthat U 0 � Z.G0/), V 0 is cyclic of order > p (Theorem 65.7(d)) so V is a metacyclicA2-group and jV 0j D p2 (Corollary 65.3). It follows that all members of the set �1

are nonabelian.If E G G, where E is of order p3 and exponent p, then there is only one maximal

subgroup of G containing E, by the previous paragraph, so G=E is cyclic. ThenG0 < E, a contradiction since jG0j D p4 > jEj. Thus, E does not exist. Assume thatp > 2. Then, by Theorem 13.7 (see also Theorem 69.4), G is a 3-group of maximalclass and jGj D 32jG0j D 36. In that case, G has a nonabelian subgroup of index 33

(Theorem 9.6), a contradiction. Thus, p D 2.

(ii) Assume that G0 is nonabelian; then jG W G0j D p2 and G0 D ˆ.G/ is anA1-group. By Theorem 44.12, G0 must be metacyclic so p D 2, by (i); then G is of


maximal class (Taussky’s theorem) so Z.G0/ is cyclic, a contradiction (Lemma 1.4).Thus, G0 is abelian.

(iii) Let exp.G0/ D p. Then, by (ii), G0 Š Ep4 . If jG W G0j D p2, then p > 2

(Lemma 64.1(s)).Now assume that jG W G0j > p2 and assume, in addition, thatG=G0 has two distinct

noncyclic maximal subgroups U=G0 and V=G0. Then U is either abelian or an A2-group with the derived subgroup of order p, the same is true for V (Lemma 65.4(b)).Then jG0j � p3 (Lemma 64.1(u)), a contradiction. Thus, G=G0 has exactly p cyclicsubgroups of index p, so it is abelian of type .pn; p/ with n > 1. Let H=G0 be thenoncyclic maximal subgroup of G=G0. ThenH is either abelian or an A2-group withjH 0j D p. If F 2 �1 � fH g, then jF 0j � p2 (Lemmas 64.1(u) and 65.2(d)) and soF is an A2-group. If jF 0j D p3, then F has no maximal abelian subgroups (Lemma65.2(d)). But, if F1 is a maximal subgroup of F containingG0, thenF1must be abelian(Lemma 65.1). This contradiction shows that jF 0j D p2 and then jH 0j D p. Lookingat the determination of A2-groups (see �71), we see that F must be isomorphic to agroup (ii) of Proposition 71.3. In particular, p > 2. (H must be isomorphic with agroup of Proposition 71.1(i); then H must be a nonmetacyclicA1-group).Assume that there is x 2 G � G0 of order p. Then D D hx;G0i Š Ep5 since

jG W Dj � p2 and �1.D/ D D. If D � M 2 �1, then M is abelian (Lemmas65.1 and 65.4(a)). Take N 2 �1 � fM g. By Lemma 64.1(u) applied toM;N.2 �1/,jN 0j � p3. Considering N \M and using Lemma 65.4(b), we conclude that N=G0

is cyclic and jN 0j � p2, contrary to the result of the previous sentence. Thus, x doesnot exist so �1.G/ D G0.

(iv) By (i), p D 2 and the set �1 has only one nonmetacyclic member H and Ghas no normal elementary abelian subgroups of order 8. Take V 2 �1 � fH g; then Vis metacyclic so jV 0j � 4 (Corollary 65.3 and Lemma 65.1). Then H is nonabelian(apply Lemma 64.1(u) to H;V 2 �1). If H is an A1-group, then �1.H/ Š E8 G G

(Lemma 65.1), contrary to (i). If H is an A2-group with jH 0j D 2, then �1.H/ Š

E8 or E16 (Proposition 72.1), contrary to (i). Hence, H must be an A2-group ofProposition 71.4. If H is minimal nonmetacyclic of order 25, then jGj D 2jH j D 26

so jG W G0j D 4 and G is of maximal class (Lemma 64.1(s)), a contradiction. In othercases of Proposition 71.4, �1.H/ Š E8, a final contradiction.

3o. It follows from Proposition 72.2 and Lemma 64.1(u) that if a p-group G is anA4-group, then jG0j � p9. In this subsection we improve essentially that estimate.Since every p-group of order < p7 is an Ai -group with i � 4, in what follows we

consider only A4-groups of order � p7; then our A4-group is not of maximal class(Theorems 9.5 and 9.6).

Lemma 72.4. Suppose that a p-group G of order > p6 is an A4-group with non-abelian G0. Let jG W G0j D pi ; then i 2 f2; 3g.

�72 An-groups, n > 2 253

(a) If i D 2, then p > 2 and G0 is either metacyclic or an A2-subgroup withd.G/ D 3.

(b) If i D 3, then G0 is a metacyclic A1-subgroup.

Proof. We have jG W G0j D pi , i 2 f2; 3g.If i D 3, thenG0 is anA1-group so d.G0/ D 2 (Lemma 65.1); thenG0 is metacyclic

(Theorem 44.12).Now suppose that i D 2. If p D 2, then G is of maximal class (Lemma 64.1(s)),

a contradiction. Thus, p > 2. Assume that G0 is nonmetacyclic; then d.G0/ > 2

(Theorem 44.12) and G0 is anA2-subgroup with d.G/ D 3.

Proposition 72.5. If a p-group G is an A4-group, then jG0j � p7.

Proof. Assume that jG0j D p9. For U 2 �1, we have jU 0j D p4 so U 0 is nonmeta-cyclic, abelian (Lemma 64.1(u), Propositions 72.2, 72.3 and 72.1); in particular, allmembers of the set �1 are A3-groups (Lemmas 65.1 and 65.2(c)). If V 2 �1 � fU g,then U 0 \ V 0 D f1g (Lemma 64.1(u)) so U 0V 0 D U 0 � V 0 is abelian of order p8. SetL D �1.U

0V 0/; then j�1.L/j � p6 (Proposition 72.3) and L < U . If A < U isnonabelian subgroup, then jA\Lj � p5, contrary to Lemmas 65.1 and 65.4(b). Thus,jG0j � p8.Next we assume that jG0j D p8. If U;V;W 2 �1 are distinct with jU 0j � jV 0j �

jW 0j, then using Lemma 64.1(u) twice, we get jU 0j D jV 0j D p4 and jU 0 \ V 0j � p

since jU 0V 0j � 1p jG0j D p7. Then U and V are both A3-groups and U 0 and V 0

are nonmetacyclic abelian (Proposition 72.3(a,c)) so cl.U 0V 0/ � 2 (Fitting’s lemma),hence, if p > 2, thenU 0V 0 is regular (Lemma 64.1(a)). Now let p D 2. If U 0V 0 D G0,then G0 is abelian since then G0 D U 0 � V 0 (here we use Proposition 72.3(a) again).Suppose that U 0V 0 < G0. Then jG W U 0V 0j > jG W G0j � 8, by Lemma 64.1(s), soU 0V 0 is abelian again.Set L D �1.U

0V 0/ so that exp.L/ D p. By the previous paragraph, jLj � p5.If U1 is a maximal subgroup of U containing L, then U1 must be abelian (Lemma65.4(b)). Since jU 0j > p, U=L is cyclic (indeed, U1 is the unique abelian maximalsubgroup of U ), and so U 0 < L. Hence exp.U 0/ D p and Proposition 72.3(c) impliesthat U 0 D �1.U /. This is a contradiction since U 0 < L < U and exp.L/ D p.

Now we improve the estimate of Proposition 72.5.

Theorem 72.6. If a p-group G is an A4-group, then jG0j � p6.

Proof. Assume, by the way of contradiction, that a p-group G is an A4-group withjG0j D p7. By Propositions 72.2, 72.3(a) and Lemma 64.1(u), p2 � jU 0j � p4 andU 0 is abelian for all U 2 �1.

(i) We claim that the subgroup G0 is nonmetacyclic and then d.G0/ > 2 (Theorem44.12). Suppose that this is false. By Proposition 72.1, jU 0j D p3 for all U 2 �1

and so U 0V 0 D U 0 � V 0 for distinct U;V 2 �1 (Lemma 64.1(u)). It follows that each


U 0 is cyclic (otherwise, U 0 � V 0 is nonmetacyclic) and so U is metacyclic (Proposi-tion 72.1(b)). But then G is minimal nonmetacyclic (Proposition 72.1(a)), contrary toLemma 64.1(l).

(ii) If jG=G0j > p2, then G0 is abelian and p3 � j�1.G0/j � p4. Indeed, since

d.G0/ > 2, by (i) and Theorem 44.12, G0 is not an A1-group and so G0 must beabelian. Since G=G0 is noncyclic, there is U 2 �1 such that U=G0 is noncyclic. LetV1 and V2 be two distinct maximal subgroups of U containing G0. If j�1.G

0/j > p4,then V1 and V2 cannot be A1-groups or A2-groups (Lemmas 65.1 and 65.4(a,b)) andso they are abelian. It follows that jU 0j � p (Lemma 64.1(q)), a contradiction. Thusj�1.G

0/j � p4, and j�1.G0/j � p3 follows from d.G0/ > 2 and commutativity ofG0.

(iii) If G0 is nonabelian, then G=G0 Š Ep2 , by (ii), so p > 2 since G is not ofmaximal class (Taussky’s theorem), and G0 is an A2-group with d.G0/ D 3, by (i).Hence G0 is isomorphic to a group of Proposition 71.1 or Proposition 71.4(b). In anycase,�1.G

0/ Š Ep3 or Ep4 , which follows from �71.

(iv) If U 2 �1 and jU 0j D p4, then U 0 is abelian of type .p2; p; p/. If V 2 �1 andjV 0j D p3, then V 0 is abelian noncyclic since V is nonmetacyclic, in view of G0 < V .Indeed, U 0 is abelian nonmetacyclic (Proposition 72.3). Suppose that U 0 Š Ep4 . ThenProposition 72.3(c) (noting that jU j D 1

pjGj > p6) implies that �1.U / D U 0. Take

W 2 �1 � fU g. Then U 0 � ˆ.U / � ˆ.G/ < W so W is nonmetacyclic. Next,jU 0W 0j � 1

pjG0j D p6 (Lemma 64.1(u)) and we must have W 0 � U , �1.W

0/ � U 0

(see (iii)) so jW 0j � p3. By Proposition 72.1(b), W 0 is noncyclic so jU 0 \W 0j � p2.But then jU 0 \W 0j D p2 and jW 0j D p4 (Lemma 64.1(u)). Since W 0 is abelian andnonmetacyclic (Proposition 72.3(c)), we have j�1.W

0/j � p3 and so �1.W0/ — U 0,

contrary to what has just been proved.

(v) We claim that j�1.G0/j � p4. Indeed, assume that j�1.G

0/j � p5. ThenG0 is not an Ai -group, i D 1; 2 (Lemmas 65.1 and 65.4(a,b)) so G0 is abelian andE D �1.G

0/ is elementary abelian of order � p5. By (ii), G=G0 Š Ep2 so p > 2

(Taussky’s theorem). For each U 2 �1, p2 � jU 0j � p4 (Lemma 64.1(u)) so U isanA3-group (indeed, the nonabelian U cannot be anAi -group, i � 2, since E < U ;see Lemma 65.4(a,b)). If U=E is not cyclic, U has a maximal subgroup U0 ¤ G0

containingE (note thatG0 is maximal inU ). ThenU0 must be abelian and so jU 0j � p

(Lemma 65.2(c)), a contradiction. Hence U=E is cyclic; then U 0 < E and so U 0 iselementary abelian for each U 2 �1. Then (iv) forces that U 0 Š Ep3 for each U 2 �1

and U 0V 0 D U 0 � V 0 for any two distinct U;V 2 �1. In particular, E D �1.G0/ is

elementary abelian of order � p6. Let an A2-subgroup A be maximal in U . But inthat case jA \Ej � p5 since G0 < U , contrary to Lemma 65.4(a,b).

(vi) We claim that, for eachU 2 �1, U 0 is abelian of exponent p2. Also,�1.G0/ Š

Ep4 and, if G0 is nonabelian, then G0 D H � Cp , where H is a nonmetacyclic A1-group of order p6. Indeed, assume that there is U 2 �1 with U 0 Š Ep3 , and takeV 2 �1 � fU g. Since jU 0V 0j � p6 (Lemma 64.1(u)), we have jV 0 W .U 0 \V 0/j � p3.If jV 0j D p3, then U 0 \ V 0 D f1g and V 0 is noncyclic, by Proposition 72.1(b) and (i),

�72 An-groups, n > 2 255

sinceG0 < V . If jV 0j D p4, then jU 0 \V 0j � p (Lemma 64.1(u)) and�1.V0/ Š Ep3 ,

by (iv). In any case, j�1.U0V 0/j � p5, contrary to (v). Suppose that W 2 �1 with

W 0 Š Ep2 and take V0 2 �1 � fW g. Then jV 00j D p4, W 0 \ V 0

0 D f1g (Lemma64.1(u)), so again j�1.W

0V 00/j � p5, by (iv), which is a contradiction. We have

proved that if U 2 �1 and jU 0j D p2, then U 0 Š Cp2 and if jU 0j D p3, thenU 0 Š Cp2 � Cp. This together with (iv) proves our first assertion on exp.U 0/.Suppose that U 2 �1 is such that jU 0j is as large as possible. Let jU 0j D p4;

then U 0 is abelian of type .p2; p; p/, by (iv). Let V 2 �1 � fU g be such that jV 0 W

.U 0 \ V 0/j � p2. If jV 0j D p2, then U 0 \ V 0 D f1g and we have j�1.U0V 0/j � p4.

If jV 0j D p3, then jU 0 \ V 0j � p and so there are elements of order p in V 0 � U 0

since V 0 is abelian of type .p2; p/, by the previous paragraph. If jV 0j D p4, thenjU 0 \ V 0j � p2 (Lemma 64.1(u)) and again there are elements of order p in V 0 � U 0

since V 0 is abelian of type .p2; p; p/, by (iv). In any case, we get j�1.U0V 0/j � p4.

Suppose now that jU 0j D p3. If W 2 �1 � fU g, then jW 0j D p3, U 0W 0 D U 0 �W 0

(Lemma 64.1(u)) and both U 0 andW 0 are noncyclic ((i) and Proposition 72.1). Hencewe get j�1.U

0W 0/j � p4 again. These results together with (iii) and (v) give that�1.G

0/ Š Ep4 . IfG0 is nonabelian, then using (iii), we see thatG0 must be isomorphicto a group of Proposition 71.1(i) and we are done.

(vii) The group G is an irregular 3-group of order 39. Each maximal subgroup of Gis two-generator and Ã1.G/ D K3.G/ is of index 33 in G. In particular, G=G0 Š E9.Indeed, let U 2 �1 and let A < U be maximal. Since �1.G

0/ Š Ep4 , we havejA \ �1.G

0/j � p3 since G0 < U , and so A is nonmetacyclic. This implies that A0

is elementary abelian (Lemma 65.1 and Theorem 65.7(d)) and so A0 � �1.U0/ < U 0

since exp.U 0/ D p2, by (vi), and we get d.U / D 2 (Lemma 64.1(y)). Thus, allmembers of the set �1 are two-generator.Assume that d.G/ > 2. Using Theorem 70.1 and 70.2, we get in case p > 2 that

jGj � p5, a contradiction. If p D 2, then it follows from jGj � 210 that cl.G/ > 2.The quotient group G=K4.G/ is of order 27 or 28. If jG=K4.G/j D 28, then we knowthat G=K4.G/ is anA5-group, a contradiction (see the text following Theorem 70.5).If jG=K4.G/j D 27, then Theorem 70.5 implies that K4.G/ D f1g, a contradiction.We have proved that d.G/ D 2. Since G is not metacyclic, we get p > 2 (Lemma

64.1(n)). By Lemma 64.1(p), Ã1.G/ D K3.G/ is of index p3 in G. On the otherhand, j�1.G

0/j D p4 > jG=Ã1.G/j so G is irregular (Lemma 64.1(a)). This impliesthat p D 3 (Theorem 9.8(a)), G=G0 Š E9 and jGj D 39.

(viii) Let U;V 2 �1 be distinct. Then jU 0V 0j � 13 jG0j D 36 (Lemma 64.1(u))

and we claim that jU 0V 0j D 36. Since H D K3.G/ D ŒG;G0� � U 0V 0, we haveH D U 0V 0 D Ã1.G/ in view of jG W H j D 33. Also, �1.G

0/ D �1.H/ and His abelian of exponent 32. Finally, G=H is nonabelian of order 33 and exponent 3.Suppose that there are distinct U;V 2 �1 such that U 0V 0 D G0. We may assume thatjU 0j D 34 and V 0 covers G0=U 0. If jV 0j D 33, then G0 D U 0 � V 0. But then, by (iv),j�1.G

0/j � 35, contrary to (v). If jV 0j D 34, then jU 0 \ V 0j D 3. But j�1.V0/j D 33

and again we get j�1.G0/j � 35, a contradiction. Hence, for each distinct U;V 2 �1,


we have jU 0V 0j D 36. Since U 0V 0 G G and G0=.U 0V 0/ � Z.G=.U 0V 0//, we haveH D ŒG;G0� � U 0V 0. By (vii), H D U 0V 0 D Ã1.G/ (recall that jG W G0j D 32).Since j�1.U

0V 0/j � 34, we have j�1.U0V 0/j D 34, by (v), and so E D �1.H/ D

�1.G0/ Š E34 . IfH would be nonabelian, thenH is anA1-group, contrary to Lemma

65.1. Thus, H is abelian and so exp.H/ D 32 since exp.U 0/ D exp.V 0/ D 32 andH D U 0V 0.

We are now able to get the final contradiction. Take U 2 �1. Since G=H is non-abelian of order 33 and exponent 3, we get U=H Š E9, whereH D K3.G/ D Ã1.G/.On the other hand, d.U / D 2 (by (vii)) and so ˆ.U / D H . Let X be any of the fourmaximal subgroups of U . Then jX W H j D 3 and E D �1.H/ Š E34 is a normalsubgroup of X . Suppose that jX 0j � 32 so that X is an A2-subgroup. By the resultsof �71, X must be isomorphic to a group of Proposition 71.3(ii). In particular, X (oforder 37) has the unique abelian maximal subgroup of type .33; 3; 3; 3/. This is a con-tradiction since H is a unique maximal abelian subgroup of X and exp.H/ D 32 (by(viii)). We have proved that jX 0j � 3 for any maximal subgroup X of U . This impliesthat jU 0j � 33 for any U 2 �1 (Lemma 64.1(u)).We have proved that for each U 2 �1, U 0 is abelian of type .32; 3/ and so H D

U 0 � V 0, where V 2 �1 � fU g. This implies that H is abelian of type .32; 32; 3; 3/

and each maximal subgroup X of U must be an A2-group with jX 0j D 3 (Lemmas65.1 and 64.1(u)). By the results of �71, X must be the following group of Proposition71.1(i):

X D ha; b j a33

D b32

D 1; Œa; b� D c; c3 D Œa; c� D Œb; c� D 1i � C3;

where�1.X/ D ha32

; b3; ci �C3 D E D �1.H/ and X 0 D hci is a maximal cyclicsubgroup in X (see Lemma 65.1).We have U 0 < H and set U 0 \ E D E0 so that E0 Š E9. Also set he0i D Ã1.U

0/

so that he0i is a subgroup of order 3 in E0 and he0i is not a maximal cyclic subgroupin H . Let X1, X2, X3, X4 be all maximal subgroups of U and note that each oneis isomorphic to the above group X . Then X 0

1, X02, X

03, X

04 must be four distinct

subgroups of order 3 in E0 (Lemma 64.1(u)). It follows that for some i 2 f1; 2; 3; 4g,X 0

i D he0i. This is a contradiction since he0i is not a maximal cyclic subgroup in H(andH is contained in Xi ); see the last sentence of the previous paragraph.

�73

Classification of modular p-groups

We recall that a p-group G is modular if and only if any two subgroups of G arepermutable (for general groups there is another definition). Sections of modular p-groups are modular. A nonabelian Dedekindian p-group G is a modular 2-group ofthe form G D Q �E, whereQ Š Q8 and exp.E/ � 2 (Theorem 1.20). Such groupsare called Hamiltonian. Nonmodular p-groups of order p3 are D8 and the nonabeliangroup S.p3/ of order p3 and exponent p > 2.If G is a minimal nonmodular p-group, then there is N GG such that G=N is either

S.p3/ (p > 2) or D8 (see Theorem 44.13).The aim of this section is to correct the original Iwasawa’s proof [Iwa] of the theo-

rem on the structure of modular p-groups. ((Iwasawa’s proof contains essential gaps.Two gaps were filled by Napolitani [Nap], remaining ones – by the second author,who wrote this corrected proof. There exists another proof of Iwasawa’s theorem,due to R. Schmidt [Sch, Chapter 2]; that proof is based on entirely other ideas and isessentially shorter.) Our proof is self-contained.The following proposition is obvious.

Proposition 73.1. Let G be a modular p-group. Then �1.G/ is elementary abelian.

In particular, modular p-groups, p > 2, are powerful (apply Lemma 73.1 toG=Ã1.G/; see [LM] and �26). It follows from Theorem A.24.4 that non-Hamiltonianmodular 2-groups G are also powerful (see [LM] and �26.2). Indeed, by that theorem,G is Q8-free. One may assume that exp.G/ D 4. Assume that G has a minimal non-abelian subgroup H . Then 8 < jH j < 32 since exp.H/ D 4. Using Lemma 65.1,we see thatH has a nonabelian epimorphic image of order 8, contrary to what has justbeen said. Thus, H does not exist so G is abelian.

Proposition 73.2. Let G be a modular p-group. Then exp.�n.G// � pn and thegroup �n.G/=�n�1.G/ is elementary abelian for all n.

Use induction on n and Proposition 73.1.By Theorem 44.13, a p-group G is modular if and only if it is D8-free if p D 2 and

S.p3/-free if p > 2.

Remark. A two-generator modular p-group G is metacyclic. Assume that G is notmetacyclic. In view of Lemma 64.1(m), one may assume that K3.G/ˆ.G

0/ D f1g;


then G is minimal nonabelian such as in Lemma 65.1(a) with m C n > 2, jGj D

pmCnC1. Then G D ha; bi D haihbi has order pmCn < pmCnC1 D jGj, a contra-diction.

Since metacyclic p-groups, p > 2, are modular, it follows from the remark thata p-group G, p > 2, is modular if and only if every its two-generator subgroup ismetacyclic. It is interesting to classify the 2-groups satisfying this property. Notethat dihedral 2-groups are metacyclic but nonmodular. The group M2n is modular. Acentral product of two modular p-group is not necessary modular (for example, thegroup Q23 � C4 is not modular.

Proposition 73.3. Nonabelian modular groups G of order p4 are: (a)Mp4 , (b)Mp3 �

Cp , p > 2, (c) G D ha; b j ap2

D bp2

D 1; ab D a1Cpi, p > 2, (d) Q8 � C2.

Proposition 73.4. Suppose that G is a modular 2-group containing a normal elemen-tary abelian subgroup E such that G=E Š Q8. Then G D Q � E withQ Š Q8 andso G is Hamiltonian.

Proposition 73.5. Let G be a modular 2-group which is not Hamiltonian. Then eachquotient group �i.G/=�i�2.G/ is abelian. In particular, �2.G/ is abelian.

Proof. Let G be a modular 2-group of exponent 4. If G does not contain a quaternionsubgroup, then Proposition 73.3 implies that any two elements of G commute and soG is abelian. Suppose that Q8 Š Q � G. In that case,G is Hamiltonian, by TheoremA.24.4, and we are done. The result follows since all sections of G are modular.

Proposition 73.6. Let G be a modular p-group of exponent p . � 2/ which is notHamiltonian. Then

.uv/p��1

D up��1

vp��1

(�)

for any u; v 2 G, i.e., G is p�1-abelian.

Proof. (i) Suppose that D 2. If p D 2, then G D �2.G/ is abelian (Proposition73.5). Suppose that p > 2 and let a; b 2 G. Then H D ha; bi D haihbi of order� p4 is metacyclic (Lemma 64.1(m)), jH 0j D p so .ab/p D apbpŒb; a�.

p2/ D apbp.

The general case follows by induction on . Let > 2. Since G=�1.G/ is ofexponent p�1, we have, for x; y 2 G, .xy/p

��2

D xp��2

yp��2

z, where z 2

�1.G/. Since xp��2

; yp��2

2 �2.G/ and exp.�2.G// D p2, we get

.xy/p��1

D ..xy/p��2

/p D .xp��2

yp��2

z/p

D xp��1

yp��1

zp D xp��1

yp��1

:

�73 Classification of modular p-groups 259

Proposition 73.7. Let G be a non-Hamiltonian modular group of exponent p . �

2/. Set j�˛.G/ W �˛�1.G/j D p!˛ for ˛ D 1; : : : ; . Take elements a1; : : : ; a!�(of

order p) of G so that they form .mod ��1.G// a basis for the elementary abeliangroup �.G/=��1.G/. Now, a

p1 ; : : : ; a

p!�

can be extended .mod ��2.G// to abasis ap

1 ; : : : ; ap!�; a!�C1; : : : ; a!��1

of ��1.G/=��2.G/ and so on. Every ele-ment of G can be written in a unique way as a product of powers of these elementsa1; a2; : : : . In other words, the elements a1; a2; : : : form a basis of G. Looking at theseries �1.G/; : : : ;�.G/, we see that the constants o.a1/; o.a2/; : : : are uniquelydetermined (up to the ordering). Next, if a1; : : : ; ad is so constructed basis, thenQd

iD1 o.ai / D jGj.

Proof. Working in G=��2.G/, we may assume for a moment that ��2.G/ D f1g

so that exp.G/ D p2. We have to show that (commuting) elements ap1 ; : : : ; a

p!�are

linearly independent in��1.G/ .D �1.G/ Š Ep!��1 /. Indeed, if an1p1 : : : a

n!� p!�

D

1 for some integers n1; : : : ; n!�, then, by Proposition 73.6, we get .an1

1 : : : an!�

!�/p D

1, and so an1

1 : : : an!�

!� 1 .mod ��1.G//, and we conclude that ni 0 .mod p/.

Hence ap1 ; : : : ; a

p!�are linearly independent in ��1.G/ and so they are extendible

to a basis ap1 ; : : : ; a

p!�; a!�C1; : : : ; a!��1

of��1.G/ .D �1.G//.Now, in general, (without assuming D 2 ) we consider the elements

ap2

1 ; : : : ; ap2

!�; a

p!�C1; : : : ; a

p!��1

2 ��2.G/

and working in ��1.G/=��3.G/ (of exponent p2) we again show that these el-ements are linearly independent modulo ��3.G/ and therefore they are extendiblewith elements a!��1C1; : : : ; a!��2

2 ��2.G/ in such a way that

ap2

1 ; : : : ; ap2

!�; a

p!�C1; : : : ; a

p!��1

; a!��1C1; : : : ; a!��2

considered modulo ��3.G/, form a basis of ��2.G/=��3.G/. Then in the ob-vious way we continue with the construction of a “basis” a1; a2; : : : of G.Since for any two “basis” elements ai and aj we have hai ihaj i D haj ihai i, every

element of G can be written as a product of powers of elements a1; a2; : : : . Moreover,each element of G can be written in a unique way as a product of powers of elementsa1; a2; : : : (in that order). Indeed, by the construction of these elements, we haveo.a1/ o.a2/ D jGj and so the elements a1; a2; : : : form a basis of G.

Proposition 73.8. Let G be a modular p-group which is not Hamiltonian. Let a 2

�1.G/# (this assumption is essential: it cannot be avoided even in the case where G

is the abelian group of type .p3; p/)). Then there is a basis a1; : : : ; ar of G such thata 2 ha1i.

Proof. We use induction on jGj. Let a1; : : : ; ar be a basis of G such that o.a1/ �

o.a2/ � � o.ar /. Let a 2 �1.G/#. Set ha1; : : : ; ar�1i D G1 so that G1 is a


complement of har i inG. If a 2 �1.har i/, we are done. If a 2 G1, then, by induction,there is a basis b1; : : : ; br�1 of G1 such that a 2 hb1i. But then b1; : : : ; br�1; ar is abasis of G and we are done.It remains to consider the case a D cd D dc, where c 2 �1.G1/

# and d 2

�1.har i/#. By induction, there is a basis c1; : : : ; cr�1 of G1 such that c 2 hcr�1i.Set o.ar / D ps . If s D 1, then we replace ar with a. Since hai \ G1 D f1g ando.a/ D o.ar / D p, c1; : : : ; cr�1; a is a basis of G and we are done. Suppose s � 2

and replace ar with a0r so that d D .a0

r /ps�1

. Then c1; : : : ; cr�1; a0r is also a basis of

G. Since o.a1/ � o.a2/ � � o.ar / (and these numbers are invariants of G), wehave also o.cr�1/ � o.a0

r / D ps . Thus, there is an element l 2 hcr�1i of order ps

such that lps�1

D c.Since hl; a0

r i is a modular subgroup of exponent ps (noting that exp.�s.G// � ps),

it follows from Proposition 73.6 that .la0r /

ps�1

D lps�1

.a0r /

ps�1

D cd D a, and soo.la0

r / D o.a0r / D ps . On the other hand, hc1; : : : ; cr�1; la

0r i D G and G1 \ hla0

r i D

f1g. Hence c1; : : : ; cr�1; la0r is a basis of G and a 2 hla0

r i.

Proposition 73.9. LetG be a modular p-group which is not Hamiltonian. If jG0j D p,

then G D G1 � G2, where G1 D ha1; a2 j apm

1 D apn

2 D 1; aa2

1 D a1Cpm�1

1 ; m >

1; n � 1i and G2 is abelian of exponent � pm�1 with m > 2 if p D 2. HenceA D ha1; G2i is an abelian normal subgroup of G D hA; a2i and aa2 D a1Cpm�1

forall a 2 A.

Proof. By Proposition 73.8, we may choose a basis a1; a2; : : : ; as of G so that G0 �

ha1i. Since hai ; aj i \ G0 � hai ; aj i \ ha1i D 1 .i; j > 1/, we see that ha2; : : : ; asi

is abelian. Let a2 be an element of the smallest order among those basis elementsa2; : : : ; as which do not commute with a1. By Lemma 65.2(a), ha1; a2i is minimal

nonabelian. So, in view of Lemma 65.1, we may put Œa1; a2� D z, where z D apm�1

1

and o.a1/ D pm, m � 2. (We get ha1i ¤ G0 since G is nonabelian.)We have Œa1; ai � D zki with ki 2 f0; 1; : : : ; p � 1g .i > 2/, where (by assumption)

o.ai / � o.a2/ whenever ki ¤ 0. Then we replace the basis elements a3; : : : ; as withthe elements a�ki

2 ai D bi .3 � i � s/. We have o.bi / D o.ai / and

Œa1; bi � D Œa1; a�ki

2 ai � D Œa1; a2��ki Œa1; ai � D z�ki zki D 1

and so the abelian subgroup hb3; : : : ; bsi centralizes ha1; a2i. It is clear that fa1; a2;

b3; : : : ; bsg is also a basis of G. Indeed, ha1; a2; b3; : : : ; bsi D G and

jGj D o.a1/o.a2/o.a3/ : : : o.as/ D o.a1/o.a2/o.b3/ : : : o.bs/:

Denoting b3; : : : ; bs again with a3; : : : ; as , we may assume from the start that G D

G1 �G2, where G1 D ha1; a2i and G2 D ha3; : : : ; asi.For i � 3 we have

.a1ai /a2 D a

a2

1 ai D .a�12 a1a2/ai D a1.a

�11 a�1

2 a1a2/ai D a1zai D z.a1ai /:


Now, .G0 �/ A D ha1; a3; : : : ; asi G G, G D Aha2i, and A \ ha2i D f1g. Considerthe subgroup H D ha1ai ; a2i .i > 2/ which is nonabelian. Since H D ha1ai iha2i

and ha1ai i � A, we have H \ A D ha1ai i and so a2 normalizes ha1ai i. Thus

.a1ai /a2 D z.a1ai / D .a1ai /

x .x integer/ and so z D apm�1

1 2 ha1ai i. This impliesthat the order of ai .i > 2/ is at most pm�1.For each a 2 A, we have a D ar

1b with an integer r and b 2 G2 so that

aa2 D .ar1b/

a2 D .a1Cpm�1

1 /rb D .ar1/

1Cpm�1

b1Cpm�1

D .ar1b/

1Cpm�1

D a1Cpm�1

:

If p D 2 and m D 2, then K D ha1; a2 j a41 D a2n

2 D 1; aa2

1 D a31 D a�1

1 i is not

modular, since K=ha2n�1

2 i Š D8. Hence, if p D 2, then m > 2. Our proposition isproved.

Proposition 73.10. Let G be a modular p-group which is not Hamiltonian. Then eachsubgroup and each factor group is modular and is not Hamiltonian.

Subgroups and epimorphic images of modular p-groups are modular so the asser-tion follows since, by Theorem A.24.4, G is Q8-free.

Proposition 73.11. Let t; x 2 G, where exp.G/ D p. If xt D x1Cps

z, wherez 2 �1.Z.G//# and 1 � s < � 1 is a natural number which is � 2 if p D 2, then

xtp��1�s

D x1Cp��1

. In particular, if o.x/ � p�1, then Œx; tp��1�s

� D 1.

Proof. By induction on i , we first prove that xti

D x.1Cps/i

zi . Indeed, we have

xtiC1

D .x.1Cps/i

zi /t D .x1Cps

z/.1Cps/i

zi D x.1Cps/iC1

ziC1:

This gives xtp��1�s

D x.1Cps/p��1�s

, since s < � 1 and so zp��1�s

D 1.It remains to prove the following congruence:

.1C ps/p��1�s

1C p�1 .mod p/:(1)

According to P. Roquette, everything depends on the following formula:

.a C bps/p ap C ap�1bpsC1 .mod psC2/;(2)

where a; b are integers, p prime, s � 1 and, if p D 2, then s � 2. Indeed, we get

.a C bps/p D ap C pap�1bps CX

2�i�p�1

p

i

!

ap�ibipsi C bppsp :(3)

Therefore each member of the above sum is 0 .modp1Csi / and so 0 .modpsC2/

since 1C si � s C 2 in view of i � 2. For the last term on the right-hand side of (3)


we have sp � s C 2 for p � 3 and this holds also for p D 2 because in that case wehave assumed s � 2. Thus, the last term is also 0 .mod psC2/, and (2) is proven.It follows from (2) that

z 1C ps .mod psC1/ implies zp 1C psC1 .mod psC2/:(4)

To prove (4) we set z D 1C ps C bpsC1 D a C bpsC1, where a D 1C ps and b isan integer. From (2) follows (with sC 1 instead of s and noting that always sC 1 � 2)zp D .aC bpsC1/p ap C ap�1bpsC2 .mod psC3/ and so zp ap .1Cps/p

.mod psC2/. Again using (2) (for a D b D 1), we get .1 C ps/p 1 C psC1

.mod psC2/, proving (4).With iteration we get from (4) (with k D 1; 2; 3; : : : )

z 1C ps .mod psC1/ implies zpk

1C psCk .mod psCkC1/:(5)

If we set in (5) z D 1C ps , k D � 1 � s, where s < � 1, we get (1).

Proposition 73.12. LetG be a non-Hamiltonian modular p-group. Then each elementof Ãi.G/ .i � 1/ can be written as api

for some a 2 G.

Proof. We prove first this result for i D 1, working by induction on jGj. Let x 2

Ã1.G/ so that x D ap1 a

p2 : : : a

ps with some ai 2 G. Now, G=Ã2.G/ is of exponent

p2 and so, using Proposition 73.6, we get x D .a1a2 : : : as/pl where l 2 Ã2.G/.

We set a1a2 : : : as D b, H D hb;Ã1.G/i, and l D cp2

1 cp2

2 : : : cp2

r with somec1; c2; : : : ; cr 2 G. Thus x D bp.c

p1 /

p : : : .cpr /

p 2 Ã1.H/. We have H ¤ G,unless G is cyclic, so x D ap for some a 2 H , by induction.

Assume that i > 1 and use induction on i . Let y 2 Ãi .G/ so that y D bpi

1 : : : bpi

t D

.bpi�1

1 /p : : : .bpi�1

t /p for some b1; : : : ; bt 2 G. Here bpi�1

j 2 Ãi�1.G/, j D

1; : : : ; t , so y 2 Ã1.Ãi�1.G//. By the above, y D cp with c 2 Ãi�1.G/. Byinduction, c D dpi�1

and so y D dpi

.

Proposition 73.13. Let G be a non-Hamiltonian modular p-group. If we have G D

ha1; : : : ; ar i, then Ãi.G/ D hapi

1 ; : : : ; api

r i.

Proof. One may assume that fai gr1 is a minimal basis. First let i D 1. We have

ha1i : : : har i D G so by the product formula applied r�1 times, jG W hap1 i : : : ha

pr ij D

pr , Since pr D jG W ˆ.G/j D jG W Ã1.G/j, we are done.

Suppose that i > 2. By induction on i , Ãi�1.G/ D hapi�1

1 ; : : : ; api�1

r i. If y 2

Ãi�1.G/, then y D api�1

; a 2 G (Proposition 73.12) and so yp D api

2 Ãi .G/.This gives Ã1.Ãi�1.G// � Ãi.G/. On the other hand, if g 2 Ãi .G/ , then byProposition 73.12, g D bpi

D .bpi�1

/p ; b 2 G, and so g 2 Ã1.Ãi�1.G// which

gives Ãi.G/ D Ã1.Ãi�1.G//. By the above, this gives Ãi.G/ D hapi

1 ; : : : ; api

r i.


Proposition 73.14 (compare with [Sch, Lemma 2.3.4]). Let G be a p-group and letA GG be abelian and such that G=A D htAi is cyclic, all subgroups of A are normalin G and, if p D 2, when �2.A/ � Z.G/. Then G is modular.

Proof. Suppose that G is a counterexample of minimal order. Then, by induction,H < G is modular since H=.H \ A/ Š AH=A.� G=A/ is cyclic and H satisfiesthe hypothesis with respect to H \ A. Hence, G is minimal nonmodular. Then, byTheorem 44.18, there existsN GG withG=N 2 fD8; S.p

3/g. SinceG=AN is a cyclicepimorphic image of G=N , we have jG W AN j � p. But every subgroup of AN=N isnormal inG=N , and this implies that p D 2 andAN=N is the cyclic subgroup of order4 in G=N Š D8. Hence A=.A \ N/ Š C4 so A D .A \ N/hui, and A is modular,by the above. Let G D Ahvi. Then G D hu; vi since hu; vi is nonmodular since ithas an epimorphic image Š D8. Since v2 2 A, � W u ! uv is an automorphism oforder 2. Let o.u/ D 2m, o.v/ D 2n. By Theorem 1.2, we have to consider one of thefollowing three possibilities:

(i) uv D u�1. Then u2m�2

D .u2m�2

/v D u�2m�2

so u2m�1

D 1, a contradic-tion.

(ii) uv D u�1C2m�1

, m > 2. Then u2m�2

D .u2m�2

/v D u2m�2.�1C2m�1/ sou2m�1.�1C2m�2/ D 1, a contradiction.

(iii) uv D u1C2m�1

,m > 2, i.e.,G Š M2mC1 . In that case,G is modular, contraryto the assumption.

Theorem 73.15 ([Iwa]). A non-Hamiltonian p-group G is modular if and only if itcontains an abelian normal subgroup N with cyclic quotient group G=N and thereexists an element t in G with G D hN; ti and a positive integer s which is at least 2 inthe case p D 2 such that at D a1Cps

for all a 2 N .

Proof (Iwasawa–Napolitani–Janko). IfG is as in the theorem, it is modular, by Propo-sition 73.14.Suppose, conversely, that G is a modular p-group which is not Hamiltonian. This

property is inherited by sections (Proposition 73.10), so we may assume that the theo-rem is true for groups of smaller order. By Proposition 73.6, the mapping g 7! gp��1

,where p D exp.G/, is a homomorphism of G. The image of this homomorphism isÃ�1.G/ ¤ 1. There is a G-invariant subgroup Z D hzi of order p in Ã�1.G/. Byinduction, G=Z contains an abelian normal subgroup N=Z with properties stated inthe theorem. There is an element t 2 G such that G D hN; ti and at D a1Cps

zk forall a 2 N , where k 2 N [ f0g depends on a.There are two major possibilities for the structure of N . The subgroup N is either

(I) abelian or (II) nonabelian.

Case (I). N is abelian. Here z 2 Ã�1.G/ is the p�1-th power of an element c in

G, z D cp��1

. There are two possibilities for the position of c in G.


(˛) It is possible to choose c 2 N .

Since o.c/ D p, c is an element of a basis of G (Proposition 73.7). In particular,hci has a complement T inG which is generated by remaining elements of those basisofG that contains c. We haveG D T hciwith T\hci D f1g and soN D .T\N/�hci,by the modular law, and T covers G=N since TN � T hci D G. The quotient groupT=.T \N/ Š G=N is cyclic, and we choose t1 2 T \ .Nt/ so that T D hT \N; t1i;then G D hN; t1i. Since t�1t1 2 N and N is abelian, t and t1 induce the sameautomorphism on N . In view of Z \ T D f1g, we have at D at1 D a1Cps

for alla 2 T \N .If also ct D c1Cps

, then the fact that N D .T \ N/ � hci is abelian impliesxt D x1Cps

for all x 2 N and the theorem is proved. Therefore assume thatct D c1Cps

zk D c1CpsCkp��1

, where k 6 0 .mod p/. We may assume thatk 2 f1; 2; : : : ; p � 1g and so p � k 2 f1; 2; : : : ; p � 1g.We claim thatÃ�1.T \N/ D f1g. Otherwise, there is a 2 T \N of order p. We

have hai \ hci D f1g and hcai \ hci D f1g. Both elements ca and c are of order p

and they, clearly, are “linearly independent” modulo ��1.G/. Hence, by the processof constructing a basis of G in Proposition 73.7, there is a basis of G containing bothc and ca. Let T1 be a complement of hci in G containing ca. By the precedingargument, .ca/t D .ca/1Cps

and so .ca/t D ctat D cta1Cps

D c1Cps

a1Cps

, whichgives ct D c1Cps

, contrary to what has been assumed in the previous paragraph.We have proved thatÃ�1.T\N/ D f1g or, what is the same, exp.T\N/ � p�1.

Hence, if s � �1, thenG=hzi is abelian. Assuming that G is nonabelian (otherwise,there is nothing to prove), we conclude that G0 D hzi is of order p and so Proposition73.9 implies the validity of the theorem.It remains to examine the case s < �1 (and s � 2 for p D 2). We use Proposition

73.11 and get xtp��1�s

D x1Cp��1

, whenever xt D x1Cpz

z0 with z0 2 Z. From thisrelation we get, iterating (p � k) times,

xt.p�k/p��1�s

D x.tp��1�s/p�k

D x.1Cp��1//p�k

D x1C.p�k/p��1C.p�k2 /p

2.��1/C�� D x1�kp��1

:

Thus, setting t� D t1C.p�k/p��1�s

, we get

ct�

D ct.p�k/p��1�s�t D .c1�kp��1

/t D .ct /.1�kp��1/

D .c1CpsCkp��1

/.1�kp��1/ D c1CpsCkp��1�kp��1�kp��1Cs�k2p2��2

D c1Cps

:

Finally, for each a 2 T \ N (noting that ap��1

D 1 holds) we obtain also at�

D

.at.p�k/p��1�s

/t D .a1�kp��1

/t D at D a1Cps

. Thus yt�

D y1Cps

for eachy 2 N . Since G D hN; t�i, we are done.


(ˇ) It is not possible to choose c in N .

We claim that hc;N i D G. Indeed, if hc;N i < G, then we can choose a gen-

erator t1 of hti such that tph

1 c .mod N/ with h > 0. Hence c D btph

1 , where

b 2 N . It follows b D ct�ph

1 and then applying Proposition 73.6 we get bp��1

D

.ct�ph

1 /p��1

D cp��1

t�phC��1

1 D cp��1

D z, where b 2 N and o.b/ D p, acontradiction.

It follows that a suitable generator c0 of hci operates on N in the same way as t andso we may assume that c itself operates on N as t . In what follows we shall identify tand c. This means that we shall assume that o.t/ D p, tp

��1

D z, and so for eacha 2 N we have at D a1Cps

tkp��1

with k 2 f0; 1; : : : ; p � 1g depending on a.

Let fa1; a2; : : : ; arg be a basis of N such that Z D hzi � ha1i (such a basis existsby Proposition 73.8). For each ai , i D 1; 2; : : : ; r , we have either at

i D a1Cps

i or

ati D a

1Cps

i tki p��1

, ki 6 0 .mod p/.

We shall prove that it is possible to assume that ati D a

1Cps

i for i � 3. For thispurpose, we order the basis elements a2; : : : ; ar so that:

(1ˇ) there exists � such that ati D a

1Cps

i if and only if i D � C 1; : : : ; r ;

(2ˇ) if � > 1, then o.a2/ � o.aj / for each j with 2 � j � �.

If � � 2, the above statement is obviously true. Supposing � > 2, we observe thatfor each i , 2 � i � �, fa2; a

q2ai g (q integer) is a basis of ha2; ai i. Since

.aq2ai /

t D .at2/

qati D .a

1Cps

2 tk2p��1

/q.a1Cps

i /ki p��1

D .aq2/

1Cps

tk2qp��1

a1Cps

i tki p��1

D .aq2ai /

1Cps

t .k2qCki /p��1

and k2 6 0 .mod p/, we can solve the congruence equation k2q C ki 0 .mod p/,in q D qi (depending on i ) so that .a

qi

2 ai /t D .a

qi

2 ai /1Cps

. Hence the set fa1; a2;

aq3

2 a3; : : : ; aq�

2 a ; aC1; : : : ; ar g is a basis with the desired property.

We consider first the following special possibility:

(ˇ1) at2 D a

1Cps

2 tk2p��1

, k2 6 0 .mod p/, ap��1

2 ¤ 1, and ap��1

i D 1 for alli � 3. We shall prove that in this case either the theorem holds or we canchoose our abelian normal subgroup N D ha1; a2; : : : ; ar i in such a way thatalso at

2 D a1Cps

2 . Here we distinguish two subcases (i) s � � 1 and (ii)s < � 1.

(i) s � � 1. If at1 D a

1Cps

1 tk1p��1

with k1 6 0 .mod p/, then we multiply a2

with a Nq1 , where Nq is a solution of the congruence equation k1 NqCk2 0 .mod p/ and

we replace the basis fa1; a2; : : : ; ar g of N with the basis fa1; Oa2 D aNq1a2; a3; : : : ; arg

noting that o.a1/ � p�1 (since in our case .ˇ/, z � ha1i is not a p�1-th power of


any element in N ). Then we compute

. Oa2/t D .a

Nq1a2/

t D .a1Cps

1 tk1p��1

/ Nqa1Cps

2 tk2p��1

D aNq.1Cps/1 a

1Cps

2 t .k1 NqCk2/p��1

D .aNq1a2/

1Cps

D . Oa2/1Cps

;

as required. If, however, at1 D a

1Cps

1 D a1 (noting that s � � 1 and ap��1

1 D

1), then G0 D htp��1

i or G0 D hap��1

2 tk2p��1

i depending whether s > � 1 or

s D � 1. Here we have also used our assumption that ap��1

i D 1 and the fact that

ati D a

1Cps

i for all i � 3. But then our theorem follows from Proposition 73.9.

(ii) s < � 1. We replace here the subgroup N with the subgroup N � D ha1; a�2 ;

a3; : : : ; ar i, where a�2 D a˛

2 tp��1�s

with ˛ being a solution of the congruence ˛k2

1 .mod p/. Note that ˛ 6 0 .mod p/ and t˛k2p��1

D tp��1

D z. We note thattk2p��1

D zk2 2 Z.G/ and aps

2 (s � 1) is of order � p�1 and therefore aps

2 and

tp��1�s

commute (Proposition 73.11). Then we compute

.a�2 /

t D .a˛2 t

p��1�s

/t D .at2/

˛tp��1�s

D .a1Cps

2 tk2p��1

/˛tp��1�s

D a˛.1Cps/2 t˛k2p��1

tp��1�s

D a˛2a

˛ps

2 tp��1

tp��1�s

D a˛2a

˛ps

2 tp��1�s

.tp��1�s

/ps

D .a˛2 t

p��1�s

/.a˛ps

2 .tp��1�s

/ps

/:

The subgroup H D ha2; tp��1�s

i is of class 2. Indeed, by Proposition 73.11,

atp��1�s

2 D a1Cp��1

2 and soH 0 D hap��1

2 i is of order p. Therefore,

.a�2 /

ps

D .a˛2 t

p��1�s

/ps

D a˛ps

2 .tp��1�s

/ps

Œtp��1�s

; a˛2 �.ps

2 /

D a˛ps

2 .tp��1�s

/ps

;

since (in any case) p divides�ps

2

�

. Substituting this last result in the above relation,we get .a�

2 /t D a�

2.a�2 /

ps

D .a�2 /

1Cps

.

The subgroup N � is abelian. Indeed, tp��1�s

commutes with each ai , i ¤ 2,because each such ai is of order � p�1 (Proposition 73.11). Also, G D hN �; ti,.N �/t D N �, and so N � is normal in G with G=N � cyclic. We observe that a2 is oforder p and so ha�

2 i\ha1; a3; : : : ; ar i D f1g, which implies that fa1; a�2 ; a3; : : : ; arg

is a basis of N �. Indeed, by Proposition 73.6, .a�2 /

p��1

D .a˛2 t

p��1�s

/p��1

D

.ap��1

2 /˛tp2��2�s

D a˛p��1

2 is an element of order p (˛ 6 0 .mod p/) and

ha˛p��1

2 i \ ha1; a3; : : : ; ar i D f1g, where we have used the fact that 2 � 2� s � .All the statements in case .ˇ1/ are proved. Indeed, a�

2 is of order p, all other basis

elements ai of N � are of order � p�1 and so Ã�1.N�/ D ha

p��1

2 i is of order p.Hence z 2 ha1i cannot be a p�1-th power of any element of N �.


In view of what was proved so far in the case .ˇ/, it remains to consider the follow-ing two possibilities:

.ˇ2/ There exists an ai , i � 2, such that ap��1

i ¤ 1 and ati D a

1Cps

i .

.ˇ3/ All ai are of order � p�1 and, in addition, atj D a

1Cps

j for j � 3. (We are

in case .ˇ/ and so also a1 is of order � p�1 since z 2 ha1i and z is not ap�1-th power of any element in N .)

Assume that the condition .ˇ2/ is satisfied. Hence there exists an ai , i � 2, such

that ati D a

1Cps

i and ap��1

i ¤ 1. Here tp��1

D z 2 ha1i and therefore hti \ hai i D

f1g. Since tp��1

¤ 1 and ap��1

i ¤ 1, we get .tai /p��1

D tp��1

ap��1

i ¤ 1 sohtai i \ hai i D f1g. It follows that elements ai ; tai are contained in a basis of G sinceai and tai are both of order p and they are “linearly independent” mod ��1.G/

(see the proof of Proposition 73.7). Therefore, there exists a complement T of hai i inG containing tai . We have N D .N \ T / � hai i and tai operates in the same wayas t on N \ T . If tai does not transform each element of N \ T into its .1C ps/-th

power, then there exists Na 2 N \ T such that Natai D Nat D Na1Cps

tNkp��1

with Nk 6 0

.mod p/, and so tp��1

2 T . Since .tai /p��1

D tp��1

ap��1

i 2 T , it follows that

ap��1

i 2 T , a contradiction. Hence at D a1Cps

for each a 2 T \N and ati D a

1Cps

i .

But N D .T \ N/ � hai i is abelian and so xt D x1Cps

for all x 2 N and we aredone.

Assume, finally, that the condition .ˇ3/ is satisfied. We distinguish here two sub-cases: (i) s < � 1 and (ii) s � � 1.

(i) Suppose that s < � 1. If at1 D a

1Cps

1 tk1p��1

, k1 6 0 .mod p/ and ati D

a1Cps

i for i � 2, then we replace a1 with a�1 D a˛

1 tp��1�s

, where ˛ is such that˛k1 1 .mod p/. Applying Proposition 73.11, we prove thatN � D ha�

1 ; a2; : : : ; ar i

is an abelian normal subgroup of G with G D hN �; ti, .a�1 /

t D .a�1 /

1Cps

, and ati D

a1Cps

i for i � 2 (where fa�1 ; a2; : : : ; ar g is not necessarily a basis of N �). Indeed,

tp��1�s

commutes with each ai , i D 1; : : : ; r , and so N � is abelian. It remains tocompute

.a�1 /

t D .a˛1 t

p��1�s

/t D .at1/

˛tp��1�s

D .a1Cps

1 tk1p��1

/˛tp��1�s

D .a˛1 /

1Cps

t˛k1p��1

tp��1�s

D .a˛1 /

1Cps

tp��1

tp��1�s

D .a˛1 /

1Cps

.tp��1�s

/1Cps

D .a˛1 t

p��1�s

/1Cps

D .a�1 /

1Cps

:

If at2 D a

1Cps

2 tk2p��1

with k2 6 0 .mod p/, then we replace a2 with a�2 D

a˛2 t

p��1�s

, where ˛ is such that ˛k2 1 .mod p/. Applying Proposition 73.11again, we see thatN � D ha1; a

�2 ; a3; : : : ; ar i is an abelian normal subgroup ofG with

G D hN �; ti (where fa1; a�2 ; a3; : : : ; arg is not necessarily a basis of N �). It remains


only to compute

.a�2 /

t D .a˛2 t

p��1�s

/t D a˛.1Cps/2 t˛k2p��1

tp��1�s

D a˛.1Cps/2 tp

��1

tp��1�s

D .a˛2 t

p��1�s

/1Cps

D .a�2 /

1Cps

:

Then our theorem either holds or we have the case considered already in the previousparagraph (noting that the order of a�

2 is � p�1).

(ii) Suppose that s � � 1. Since o.ai / � p�1 for all i , then G is either abelianor G0 D hzi D Z is of order p. Then the theorem follows from Proposition 73.9.

Case (II). N is nonabelian. The idea here is to reduce this case to Case (I). ByProposition 73.9, N has the following structure. There is a basis fa1; a2; : : : ; arg ofN such that

apm

1 D apn

2 D 1; m > 1; n � 1; aa2

1 D a1Cpm�1

1 ;

aiaj D aj ai for i; j � 3; apm�1

i D 1 for i � 3;

N D ha1; a2i � ha3; : : : ; ar i; and if p D 2; then m > 2:

HereZ D hapm�1

1 i D N 0 � Z.G/ and, for each a 2 N , we have at D a1Cps

akpm�1

1 ,where k .mod p/ depends on a.Replacing t by a suitable ta˛

1aˇ2 (˛ and ˇ are integers .mod p/), we may assume

at1 D a

1Cps

1 and at2 D a

1Cps

2 :(6)

Indeed, we have at1 D a

1Cps

1 ak1pm�1

1 and at2 D a

1Cps

2 ak2pm�1

1 , where k1; k2 2

f0; 1; : : : ; p � 1g. It follows from aa2

1 D a1apm�1

1 that aa1

2 D a2a�pm�1

1 , and these

two relations imply aa

ˇ2

1 D a1Cˇpm�1

1 and aa˛

1

2 D a2a�˛pm�1

1 . We shall choose ˛ andˇ so that ˛ k2 .mod p/ and ˇ �k1 .mod p/. Then we compute (noting thats � 1)

ata˛

1a

ˇ

2

1 D .a1Cps

1 ak1pm�1

1 /aˇ2 D .a

1Cˇpm�1

1 /1Cps

ak1pm�1

1

D a1CpsC.ˇCk1/pm�1

1 D a1Cps

1

and similarly ata˛

1 aˇ2

2 D .a1Cps

2 ak2pm�1

1 /a˛1 a

ˇ2 D a

1Cps

2 a.�˛Ck2/pm�1

1 D a1Cps

2 .

SinceZ � Ã�1.G/, where p D exp.G/, a generator z D apm�1

1 ofZ is a p�1-

th power of some element c. Suppose that apm�1

1 D cp��1

and c D ae1

1 ae2

2 : : : aerr t

f0 .Then

apm�1

1 D cp��1

D ae1p��1

1 ae2p��1

2 tf0p��1

:(7)


Suppose that ap��1

1 ¤ 1. Then m D . If s D � 1 and ap��1

2 D 1, then

G0 D hap��1

1 i is of order p and our theorem holds by Proposition 73.9.

(i) If ap��1

1 ¤ 1, s D � 1, and ap��1

2 ¤ 1, then we replace a2 by a�2 D a2t

�1.

From (6) we get at�1

1 D a1�p��1

1 , and so

aa�

2

1 D aa2t�1

1 D .a1ap��1

1 /t�1

D a1a�p��1

1 ap��1

1 D a1;

since ap��1

1 2 Z.G/. Assume that, for an i � 3, we have ati D a

1Cp��1

i aki p��1

1 D

aiaki p��1

1 with ki 6 0 .mod p/. This gives Œai ; t � D aki p��1

1 and so H D hai ; ti

containsZ D hap��1

1 i. ButZ is normal inG and soH 0 D Z is of order p. By Lemma65.2(a), H is minimal nonabelian and by Remark, following Proposition 73.2, H is

metacyclic. Therefore �1.H/ Š Ep2 which implies �1.H/ D h�1.hai i/; ap��1

1 i:

In particular, �1.hti/ � �1.H/. Also Œai ; t�1� D Œai ; t �

�1 D a�ki p��1

1 , and so

at�1

i D aia�ki p��1

1 .

Now consider the subgroup K D hai ; a�2 i. We have a

a�2

i D aa2t�1

i D at�1

i D

aia�ki p��1

1 and so Œai ; a�2 � D a

�ki p��1

1 , which is a generator of Z. Thus K containsZ and since Z is normal in G, we have K 0 D Z. It follows that K is also minimal

nonabelian and metacyclic and we have �1.K/ D h�1.hai i/; ap��1

1 i D �1.H/: In

particular, �1.ha�2 i/ 2 �1.H/. We compute, using Proposition 73.6, .a�

2 /p��1

D

.a2t�1/p

��1

D ap��1

2 t�p��1

and since tp��1

2 �1.H/, we get 1 ¤ ap��1

2 2

�1.H/ D h�1.hai i/; ap��1

1 i. This is a contradiction since fa1; : : : ; ar g is a basisof N .We have proved that t centralizes ha3; : : : ; ar i. This implies that the subgroup

N � D ha1; a�2 ; a3; : : : ; ar i is abelian. Since t normalizes ha2i, it follows that t cen-

tralizes ap��1

2 . We compute

.a�2 /

t D .a2t�1/t D at

2t�1 D a

1Cp��1

2 t�1 D .a2t�1/a

p��1

2

D a�2a

p��1

2 D .a�2 /

1Cp��1

tp��1

;

since .a�2 /

p��1

D .a2t�1/p

��1

D ap��1

2 t�p��1

. Also, G D hN �; ti. If tp��1

D 1,

then t normalizes the abelian subgroup N � and xt D x1Cp��1

for each x 2 N � andwe are done. Suppose that tp

��1

¤ 1. Note that t induces an automorphism of orderp onN and so tp centralizesN . In particular, tp

��1

2 Z.G/. HenceA D N �htp��1

i

is a normal abelian subgroup of G, G D Ahti, and for each a 2 A, at a1Cp��1

.mod htp��1

i/. Finally, tp��1

2 Ã�1.G/ and this is exactly the situation studied inCase (I).


(ii) Suppose that ap��1

1 ¤ 1 .m D / and s < � 1. In this case we replace

a2 by a�2 D a2t

�p��1�s

. Since o.ai / � p�1 for i > 2, Proposition 73.11 implies

that ha�2 ; a3; : : : ; ar i is abelian. Again, Proposition 73.11 gives atp��1�s

1 D a1ap��1

1 ,

and conjugating this relation with t�p��1�s

(noting that ap��1

1 2 Z.G/) we get a1 D

at�p��1�s

1 ap��1

1 and so at�p��1�s

1 D a1�p��1

1 . We compute

aa�

2

1 D aa2t�p��1�s

1 D .a1Cp��1

1 /t�p��1�s

D .a1�p��1

1 /1Cp��1

D a1�p2��2

1 D a1

since 2 � 2 � and so N � D ha1; a�2 ; a3; : : : ; ar i is an abelian subgroup of G. It

remains to compute .a�2 /

t .

Let us consider the subgroup H D ha2; tp��1�s

i. Since, by Proposition 73.11,

atp��1�s

2 D a1Cp��1

2 , it follows that H 0 D hap��1

2 i is of order � p. By Proposition

73.11, aps

2 (s � 1 and if p D 2 then s � 2) commutes with tp��1�s

and so workingin the subgroup H of class � 2, we get

.a�2 /

1Cps

D .a2t�p��1�s

/1Cps

D a2t�p��1�s

.a2t�p��1�s

/ps

D

D a2t�p��1�s

aps

2 t�p��1

D a1Cps

2 t�p��1�s

t�p��1

:

On the other hand, .a�2 /

t D .a2t�p��1�s

/t D a1Cps

2 t�p��1�s

, and so by the above

we get finally .a�2 /

t D .a�2 /

1Cps

tp��1

.

We have G D hN �; ti. If tp��1

2 hap��1

1 i, then the abelian subgroup N � is nor-

mal in G and we have reduced this case to Case (I). So suppose tp��1

62 hap��1

1 i. By

Proposition 73.11, tp��1�s

centralizes a3; : : : ; ar and tp��1�s

induces an automor-phism of order � p on ha1i and ha2i and so tp

��1

D .tp��1�s

/ps

2 Z.G/. HenceN �� D hN �; tp

��1

i is a normal abelian subgroup of G, G D hN ��; ti, and, for each

x 2 N ��, we get xt D x1Cps

tkxp��1

alxp��1

1 , where kx , lx are integers depend-ing on x. Also, t and a1 are both of order p and they are “linearly independent”modulo ��1.G/. Indeed, if tn1a

n2

1 2 ��1.G/ (n1, n2 integers), then (Proposi-

tion 73.6) .tp��1

/n1.ap��1

1 /n2 D 1 and so n1 n2 0 .mod p/. Hence there isa basis a1; t; : : : of G (Proposition 73.7). In particular, there is a complement T ofha1i in G containing t . Since ha1i � N ��, we get N �� D ha1i � .N �� \ T /, by the

modular law. Take a 2 N �� \ T and assume that at D a1Cps

tkap��1

alap��1

1 with

la 6 0 .mod p/. Then ap��1

1 2 T , a contradiction. Hence, for each a 2 N �� \ T ,

we get at D a1Cps

tkap��1

. But N �� is abelian and so xt D x1Cps

tkxp��1

for allx 2 N ��. Since htp

��1

i 2 Z.G/ and tp��1

2 Ã�1.G/, we have again reduced thiscase to Case (I).


(iii) Suppose, finally, that ap��1

1 D 1. Then from relation (7) follows

apm�1

1 D ae2p��1

2 tf0p��1

; tp��1

¤ 1; f0 6 0 .mod p/(8)

since fa1; a2g is a basis of the minimal abelian subgroup ha1; a2i. Also, tp��1

2

�1.ha1; a2i/ D hapm�1

1 ;�1.ha2i/i but tp��1

62 �1.ha2i/.Denote pl D jha1; a2; ti W ha1; a2ij and note that (6) implies that t normalizes ha2i

(and ha1i) and so tpl

2 Nha1;a2i.ha2i/ D hap1 i � ha2i. Replacing a1 with ai

1 .i 6 0

.mod p//, we may assume (writing again a1 instead of ai1) that

tpl

D a�pk

1 a�hpf

2 ; where k � 1; h 6 0 .mod p/:(9)

Since Œapk

1 ; apf

2 � D 1 and �1.hti/ 6� �1.ha2i/, we get o.a�hpf

2 / � o.a�pk

1 / and so

o.tpl

/ D o.a�pk

1 /. On the other hand, o.tpl

/ D p�l and o.a�pk

1 / D pm�k and so � l D m � k. By assumption, > m, where pm D o.a1/, and so l > k � 1.If f D 0, then we consider the normal abelian subgroup M D ha1; a3; : : : ; ar i of

G. Since a�h2 D a

pk

1 tpl

, h 6 0 .mod p/, we have G D hM; ti. For each x 2 M ,

xt D x1Cps

akxpm�1

1 , where (8) gives apm�1

1 D .ae2

2 tf0/p

��1

2 Z.G/. This case isreduced to Case (I). Therefore we assume in the sequel that f � 1.

The element t centralizes tpl

D a�pk

1 a�hpf

2 , where h 6 0 .mod p/, k � 1,f � 1. On the other hand, t normalizes ha1i and ha2i and so (since fa1; a2g is a

basis of the subgroup ha1; a2i) t centralizes hapk

1 i and hapf

2 i. This implies that tp

centralizes hapk�1

1 i and hapf �1

2 i. We have tpl�1

2 htpi since l > k � 1 and so

l � 2. Thus tpl�1

centralizes hapk�1

1 ; apf �1

2 i.We consider an element

g D tpl�1

axpk�1

1 aypf �1

2(10)

and show that there exist integers x 6 0 .mod p/ and y 6 0 .mod p/ such thatgp D 1. (Hence the coset tp

l�1

ha1; a2i contains elements of order p.) Note that

tpl�1

centralizes haxpk�1

1 ; aypf �1

2 i and from (9) follows

tpl

apk

1 ahpf

2 D 1:(11)

We get from (10) in any case gp D tpl

.axpk�1

1 aypf �1

2 /p, and so if Œapk�1

1 ; apf �1

2 � D

1, then we take x D 1 and y D h so that (with the help of (11)) follows gp D

tpl

apk

1 ahpf

2 D 1. If Œapk�1

1 ; apf �1

2 � ¤ 1, then k D f D 1 (since ha1; a2i is minimalnonabelian) and Œa1; a2�

p D 1. If p > 2, then ha1; a2i is p-abelian and so setting

again x D 1 and y D h we get gp D tpl

apk

1 ahpf

2 D 1. However, if p D 2, then

Œa1; a2� D a2m�1

1 is an involution in Z.G/ and here we set x D 1 � 2m�2 (noting that


in this case m > 2 according to Proposition 73.9) and y D h. We obtain in that casealso

g2 D t2l

.a1�2m�2

1 ah2/

2 D t2l

a2.1�2m�2/1 a2h

2 Œa2; a1�.1�2m�2/h

D t2l

a2�2m�1C2m�1

1 a2h2 D 1;

since Œa2; a1�.1�2m�2/h D Œa2; a1� D a2m�1

1 .

We consider the subgroup K D htpl�k

; a1i noting that l > k � 1. Using Proposi-

tion 73.13 and (9), we get Ãk�1.K/ D htpl�1

; apk�1

1 i and

Ãk.K/ D htpl

; apk

1 i D ha�pk

1 a�hpf

2 ; apk

1 i D hapk

1 ; apf

2 i � Z.ha1; a2i/:

Since K=Ãk.K/ is of exponent pk , Proposition 73.6 implies that K=Ãk.K/ is pk�1-abelian. Take the element

tpl�1

axpk�1

1 2 Ãk�1.K/(12)

where x 6 0 .mod p/ is the integer from (10). Using Proposition 73.12, we see thatthere exists an element g� D t˛pl�k

aˇ1 (˛, ˇ integers) in K such that

.g�/pk�1

D tpl�1

axpk�1

1 :(13)

On the other hand, using the fact that K=Ãk.K/ is pk�1-abelian, we get

.g�/pk�1

D .t˛pl�k

aˇ1 /

pk�1

D t˛pl�1

aˇpk�1

1 s;(14)

where

s D a�pk

1 aıpf

2 2 Ãk.K/ .�; ı integers/:(15)

Using (12), (14), and (15), we get

axpk�1

1 D t .˛�1/pl�1

apk�1.ˇC�p/1 a

ıpf

2 :(16)

It follows that t .˛�1/pl�1

2 ha1; a2i and so ˛�1 0 .mod p/. We may set ˛�1 D d

(d integer), where ˛ 6 0 .mod p/ and together with (14) we get

t .˛�1/pl�1

D tdpl

D a�dpk

1 a�hdpf

2 :(17)

From (16) and (17) (noting that x 6 0 .mod p/) follows also ˇ 6 0 .mod p/ sincefa1; a2g is a basis of ha1; a2i.

Substituting (13) in (15), we get g D .g�/pk�1

aypf �1

2 and so the element g oforder p (not being contained in ha1; a2i) is contained in the subgroup H D hg�; a2i.


On the other hand,H is modular with d.H/ D 2 and so j�1.H/j D p2 and therefore

�1.H/ D hg; apn�1

2 i, where �1.ha2i/ D hapn�1

2 i.

From aa2

1 D a1Cpm�1

1 follows aa1

2 D a2a�pm�1

1 and aa

ˇ

1

2 D a2a�ˇpm�1

1 . From

at2 D a

1Cps

2 we get at˛pl�k

2 D au2 with a certain integer u 6 0 .mod p/. Then we

compute

ag�

2 D at˛pl�k

aˇ

1

2 D .au2 /

aˇ1 D .a

aˇ

1

2 /u D .a2a�ˇpm�1

1 /u D au2a

�ˇ upm�1

1 ;

since apm�1

1 2 Z.G/. Hence a�ˇ upm�1

1 2 H and �ˇu 6 0 .mod p/ implies apm�1

1 2

H . Thus apm�1

1 2 �1.H/ D hg; apn�1

2 i. This is a contradiction since hg; apn�1

2 i \

ha1; a2i D hapn�1

2 i. The theorem is proved.

Minimal nonmodular p-groups are described in [Jan1].

Corollary 73.16 (v. d. Waall). Every non-Dedekindian modular 2-group has a charac-teristic subgroup of index 2.

Proof. By Theorem A.24.4, G is Q8-free. Now the result follows from Ward’s Theo-rem 56.1.

As follows from v. d. Waall’s result, if p > 2 and G is nonabelian modular, itcontains a characteristic subgroup of index p.

Exercise. Let H be a powerful 2-group. If a 2-group G is lattice isomorphic with Hvia �, then G is also powerful. (Hint. Use Proposition 73.5.)

�74

p-groups with a cyclic subgroup of index p2

The title groups have been determined in [Nin] (see also the old paper [HT1] by Huaand Tuan), where these groups are given in terms of generators and relations withoutany comments about its subgroup structure and therefore this result was not very usefulfor applications. We shall classify here the title groups in a structural form.If a groupG of order 24 has no cyclic subgroups of index 4, it is elementary abelian.

Therefore, it suffices to classify the 2-groups with cyclic subgroup of index 4, whichhave the order > 24.

Theorem 74.1. LetG be a nonabelian group of order pm, p > 2,m > 3, and exponentpm�2. Then one and only one of the following holds:

(a) G is metacyclic, jG=Ã2.G/j D p4.

(b) m > 4, G is an L3-group.

(c) m D 4, G is regular, �1.G/ is of order p3 and exponent p.

(d) m D 4, G is irregular, p D 3, G is of maximal class.

Proof (Berkovich). Let Z < G be a cyclic subgroup of index p2 in G.Suppose that G has a normal subgroup E of order p3 and exponent p; then G D

EZ with jE \ Zj D p. We know that exp.Aut.E//p D p and so jG W CG.E/j � p.Therefore, if m > 4, then �1.G/ D E and G is an L3-group.Now let m D 4. If G is regular, then j�1.G/j 2 fp2; p3g. In the first case, G is

metacyclic (Lemma 64.1(a,m). In the second case, �1.G/ is of exponent p (Lemma64.1(a)). Next suppose that G is irregular. Then p D 3 (Lemma 64.1(a)). (Note thata 3-group of maximal class and order 34 has a cyclic subgroup of index 32, since it isirregular.)In what follows we assume that m > 4 and G has no normal subgroups of order

p3 and exponent p. According to Theorem 13.7 (see also Theorem 69.4), G is eithermetacyclic or a 3-group of maximal class. Let us consider these possibilities.Let G be metacyclic. In that case, G is regular. Since G has no cyclic subgroups of

index p, the subgroup Ã1.G/ is noncyclic. Therefore, jG=Ã2.G/j D p4. Let us showthatÃ2.G/ is cyclic. If not,G=Ã3.G/ is of order p6 and exponent p3, which is not thecase since exp.G/ D pm�2. Conversely, ifG is metacyclic with jG=Ã2.G/j D p4 andÃ2.G/ is cyclic, it has a cyclic subgroup of index p2 since Ã2.G/ D fxp2

j x 2 Gg

(Lemma 64.1(a)).

�74 p-groups with a cyclic subgroup of index p2 275

Now let G be a 3-group of maximal class, m > 4. Then G1, the fundamentalsubgroup of G, is metacyclic and exp.G1/ D exp.G/ (see �9). It follows that G1 hasa cyclic subgroup of index 3. In that case, Ã1.G1/ is cyclic of index p2 in G1 sojÃ1.G/j � 32. This is a contradiction since a p-group of maximal class and order> p3, p > 2, has no normal cyclic subgroups of order > p.

Theorem 74.2. Let G be a nonabelian group of order 2m,m > 4, and exponent 2m�2.Then one of the following holds:

(a) G is an L3-group.

(b) G is the uniquely determined group of order 25 with �2.G/ Š D8 � C2. Thegroup G has a normal elementary abelian self centralizing subgroup E D hz; u; ei oforder 8 and an element a of order 8 such that G D Ehai, where a4 D z, ea D eu,and ua D z. Here ˆ.G/ D ha2; ui is abelian of type .4; 2/, G0 D �1.ˆ.G// Š E4,Z.G/ D Ã1.ˆ.G// is of order 2, and c3.G/ D 4.

(c) G is a U2-group (all such groups are completely determined in �67). If m > 5,then cm�2.G/ D 2.

(d) G D WZ with W \ Z Š C4, where W is an abelian normal subgroup of type.4; 4/ and Z Š C2m�2 . We have W D �2.G/ and G is metacyclic. Also, 2 � jZ W

CZ.W /j � 4, (since exp.Aut.W //2 � 4; see �33) and cm�2.G/ D 4.

(e) G D QZ, whereQ Š Q8 is a normal subgroup ofG,Q\Z D Z.Q/,Z D hbi Š

C2m�2 and b either centralizesQ or b induces onQ an involutory outer automorphismin which case m > 5. Also we have cm�2.G/ D 4.

(f) G is the uniquely determined group of order 25 with�2.G/ D ha; bi � hui, whereha; bi Š Q8 and u is an involution with CG.u/ D �2.G/. Let hzi D Z.Q/; thena2 D b2 D z. There is an element y of order 8 in G such that y2 D ua, uy D

uz, ay D a�1, by D bu. Here ˆ.G/ D hy2; ui is abelian of type .4; 2/, G0 D

�1.ˆ.G// Š E4, Z.G/ D Ã1.G/ is of order 2, and c3.G/ D 4.

Proof (Janko). Suppose that G has a normal subgroup E Š E8; then G D EZ withZ Š C2m�2 and E \ Z Š C2. If K=E is the subgroup of order 2 in G=E, then�1.G/ � K. Therefore, if K is abelian, we get �1.G/ D E, and so G is an L3-group. Now let CG.E/ D E. Since exp.Aut.E//2 D 4, we get jGj D 25 andZ D hai is of order 8. In that case, jZ.G/j D 2. Therefore, setting z D a4, one canchoose e; u 2 E so that E D he; u; zi; ea D eu and ua D uz. The structure of G isdetermined and we get the group (b) stated in the theorem.From now on we assume that G has no normal elementary abelian subgroups of

order 8. Since G is neither cyclic nor of maximal class, it has a normal four-subgroupW0. Suppose that G=W0 is cyclic (of order 2m�2), then G D W0S with S Š C2m�2

and W0 \ S D f1g since G has no cyclic subgroups of index 2. But then W0 �

�1.S/ Š E8 is normal in G, contrary to our assumption. Hence, G=W0 is noncyclic.Let Z < G be cyclic of order 2m�2. Then W0 \ Z Š C2 and so W0Z is maximal


in G. It follows that G=W0 has a cyclic subgroup .W0Z/=W0 of order 2m�3 � 4

and index 2. Set F=W0 D ˆ.G=W0/ so that F < W0Z and F=W0 is cyclic oforder � 2. Since CW0Z.W0/ � F (indeed, jW0Z W CW0Z.W0/j � 2), it follows that�1.W0Z/ D W0. Let M=W0 be any cyclic subgroup of index 2 in G=W0. SinceM > F and �1.M/ � �1.F / D W0, we get �1.M/ D W0. If G=W0 is of maximalclass, then G is an U2-group and all such groups have been determined in �67. Note,that ifm > 5, then all cyclic subgroups of order 2m�2 are contained in the L2-subgroupW0Z so c2m�2.G/ D c2m�2.W0Z/ D 2.We may assume that G=W0 is not of maximal class. It follows that G=W0 is either

abelian of type .2m�3; 2/ or G=W0 Š M2m�2 in which case m > 5 (Lemma 64.1(t)).In any case, G=W0 has exactly two cyclic subgroups M1=W0 and M2=W0 of index2 and the third maximal subgroup M3=W0 is abelian of type .2m�3; 2/. Hence M3

cannot contain cyclic subgroups of index 2. But �1.Mi / D W0 and so Mi is eitherabelian of type .2m�2; 2/ orMi Š M2m�1 , i D 1; 2. In any case, cm�2.Mi / D 2 andtwo cyclic subgroups of order 2m�2 in Mi generateMi , i D 1; 2. Since exp..M1 \

M2/=W0/ D 2m�4, it follows that M1 \ M2 does not contain any cyclic subgroupof order 2m�2. We get (see the proof of Theorem 1.10) cm�2.G/ D cm�2.M1/ C

cm�2.M2/ D 2C2 D 4 andG is generated by its four cyclic subgroups of order 2m�2.We are now in a position to use Theorem 54.2. The groups (d) and (a) of our theorem

correspond to possibilities (a) and (b) of Theorem 54.2, respectively. In case (c) ofTheorem 54.2, we have here only the possibility jGj D 25 with �2.G/ Š Q8 � C2

(since in case�2.G/ D D8 �C2 the groupG would have a normal elementary abeliansubgroup of order 8). This leads to the group (f) of our theorem, and we are done.

Exercise. Let p > 2 and let G be a nonmetacyclic group of order pm > p4 with non-normal cyclic subgroup L of order pm�2. Suppose that G is not minimal nonabelian.Prove that G has a maximal subgroup H Š Mpm�1 .

Solution. We have p > 2. The quotient group G=LG is isomorphic to a Sylow p-subgroup of the symmetric group Sp2 so LG is cyclic of order > p since m > 4. Itfollows from Theorem 9.6 thatG is not a p-group of maximal class. Then, by Theorem69.3, G has a normal subgroup R of order p3 and exponent p. Since G has a cyclicsubgroup of index p2, then �1.G/ D R, G=R is cyclic so exp.�k.G// � pk for allk 2 N. By Proposition 10.28, G is generated by minimal nonabelian subgroups. Itfollows that G has a minimal normal subgroup H of exponent pm�2. By hypothesis,H < G. Then H has a cyclic subgroup of index p, and the result follows fromTheorem 1.2.

�75

Elements of order � 4 in p-groups

All results of this section are due to the second author.If G is a finite p-group and n 2 N a fixed natural number, then we define ��

n.G/ D

hx 2 G j o.x/ D pni. It is a known fact that �2.G/ has a strong influence on thestructure of a finite 2-group G. For example, if�2.G/ is metacyclic, then the 2-groupG is also metacyclic (this follows from the classification of minimal nonmetacyclicp-groups; see Theorem 66.1). In �52, all 2-groups G with the property j�2.G/j D 24

are classified. If j�2.G/j � 23, then we get for a 2-group G four infinite classes ofgroups (see Lemma 42.1). In this section we consider for the first time the case, wherefor �2.G/ we could have an infinite class of 2-groups. More precisely, we assumethat G is a finite 2-group with�2.G/ Š C2 �D, whereD is any 2-group of maximalclass. Then either G D �2.G/ or jG W �2.G/j D 2 and the structure of G is uniquelydetermined (Theorem 75.1). In fact, the most difficult part of the proof is to showthat jG W �2.G/j D 2. The exact determination of the structure of G is obtained byapplying a classification of so called U2-groups given in �67. A 2-group H is saidto be a Us-group (s � 2) with respect to the kernel R if H has a normal elementaryabelian subgroup R of order 2s , H=R is of maximal class and whenever T=R is acyclic subgroup of index 2 inH=R, then �1.T / D R.We shall determine also the structure of a finite 2-groupG with��

2.G/ Š C2�Q2n ,where Q2n is a generalized quaternion group of order 2n, n � 4 (see ‘Research prob-lems and themes I’, #563). If �2.G/ > ��

2.G/, then we show that G D �2.G/ Š

C2 �SD2nC1 , where SD2nC1 is the semidihedral group of order 2nC1 (Theorem 75.2).The case n D 3 was treated in �55.Finally, we show that a finite p-group G, all of whose noncyclic subgroups H have

the property H D �1.H/, is either cyclic or of exponent p or p D 2 and G is adihedral group D2n , n � 3 (Theorem 75.3). Here the case p > 2 is almost trivial butsome proving must be done in case p D 2; see also Exercise 1.113.

Theorem 75.1. Let G be a 2-group with �2.G/ D hui �D, where u is an involution,D is any 2-group of maximal class, and jDj D 2n, n � 4. If G ¤ �2.G/, thenjG W �2.G/j D 2, D is dihedral or generalized quaternion, CG.u/ D �2.G/, G is anonmetacyclic U2-group with respect to the kernel R D hui � Z.D/, G=R Š SD2n ,Z.G/ D Z.D/, and d.G/ D 2. More precisely, we set D D ha0; b j .a0/2

n�1

D

1; .a0/2n�2

D z; .a0/2n�3

D v; b2 D z�; � D 0; 1; .a0/b D .a0/�1i, and then


there is an element a 2 G ��2.G/ such that a2 D a0, ua D uz, ab D a�1vu, andthis determines the structure of G uniquely. (For n < 4, such groups G have beendetermined in Lemma 42.1.)

Proof. Set H D �2.G/ so that H D hui � D with jDj D 2n, n � 4. Let Zbe the unique cyclic subgroup of index 2 in D, where jZj D 2n�1 � 8. Since Dhas no normal four-subgroups, it follows that G has no normal elementary abeliansubgroups of order 8. We know that D has no normal abelian subgroups of type.4; 2/ and so H (and also G) has no normal abelian subgroups of type .4; 4/. Sethzi D �1.Z/ so that hzi D Z.D/. Let hvi be the cyclic subgroup of order 4 in Z,where v2 D z. Since hvi � Ã1.H/ D Ã1.Z/, it follows that hvi is normal in Gand so z 2 Z.G/. Obviously, W0 D hui � hzi D Z.H/ and so W0 is normal in G.Therefore, W D W0hvi is normal in G, W0 D �1.W /, and W is a normal abeliansubgroup of G of type .4; 2/. We see that W0 is the unique normal four-subgroup inH and then W is the unique normal abelian subgroup of H of type .4; 2/. It followsthat W0 is the unique normal four-subgroup in G and W is the unique normal abeliansubgroup of G of type .4; 2/.We are now in a position to use Proposition 50.6(a). It follows that N D CG.W /

is abelian of type .2m; 2/, where m � n � 1, N \ H D W0Z D hui � Z, G=N isisomorphic to a subgroup of D8, and W D �2.N /. Let r be an element in D � Z

so that r2 2 hzi and vr D v�1. Hence r induces an involutory automorphism onthe abelian group N and r inverts W D �2.N /. This implies that CN .r/ D W0 D

�1.W / and so, using Theorems 67.1, 67.2 and 67.3, we see that r inverts N=W0.Suppose that N ¤ W0Z. Take an element x 2 N � .W0Z/. We have xr D x�1w0

with some w0 2 W0. But then .rx/2 D rxrx D r2xrx D r2x�1w0x D r2w0 2 W0,and so o.rx/ � 4. This is a contradiction since rx 62 H D �2.G/.We have proved that N D W0Z. Since hvi is normal in G, vr D v�1, and ur D u

for any r 2 D � Z, it follows that CG.v/ covers G=H . If G ¤ H , there is s 2

CG.v/�H with us D uz, vs D v, so thatG=.W0Z/ is a four-group of automorphismsof W induced with hs; ri.From now on we assume G ¤ H . In that case jG W H j D 2, CG.W / D W0Z,

G=.W0Z/ Š E4, and for each x 2 G�H , ux D uz so that CG.u/ D hui�D D H . Itremains to determine the structure of G. Since S D CG.v/ coversG=H and CH .v/ D

W0Z, it follows that S is a nonabelian maximal subgroup of G with S \H D W0Z.Hence �2.S/ D W and so we may apply Lemma 42.1 for p D 2. It follows thateither S Š M2nC1 or S D ha; b j a2n�1

D b8 D 1; ab D a�1; a2n�2

D b4i, n � 4.In the second case,

hb2i D hvi D Z.S/; ha2i D S 0; ˆ.S/ D Z.S/ha2i; �1.S/ D �1.W / D hu; zi;

z D a2n�2

and CS .�1.S// D hb2; ai is the unique abelian maximal subgroup of Sand so hb2; ai D W0Z D hui �Z. SinceW0Z contains exactly two cyclic subgroupsZ1, Z2 of index 2, one of them, say Z1 D hai, is inverted by the element b 2 S �

.W0Z/. But Z1 \Z2 � hvi and so b inverts hvi, contrary to S D CG.v/.

�75 Elements of order � 4 in p-groups 279

We have proved that S Š M2nC1 . In particular, S has a cyclic subgroup S1 of index2 containing hvi and so S=W0 is a cyclic subgroup of index 2 in G=W0. But G=W0

contains DW0=W0 D H=W0 Š D=hzi, where D=hzi is a group of maximal class.ThusG=W0 is of maximal class and order 2n (n � 4) and so S=W0 is the unique cyclicsubgroup of index 2 in G=W0. Since �1.S/ D W0, it follows that G is a U2-groupwith the kernel W0.We are now in the position to use the classification of U2-groups given in �67 (see

Theorems 67.1 and 67.3). It follows that only the group G given in Theorem 67.3(d)satisfies our assumptions:

G D ha; b j a2n

D 1; a2n�1

D z; a2n�2

D v; b2 D z�; � D 0; 1;

ab D a�1vu; u2 D Œu; b� D 1; ua D uz; n � 4i;

where W0 D hu; zi, G=W0 Š SD2n , Z.G/ D hzi Š C2, G is non-metacyclic, andd.G/ D 2. Also, CG.u/ D hui �D D �2.G/, where D D ha2; bi. For � D 0, D isdihedral and for � D 1, D is generalized quaternion. Our theorem is proved.

Of course, groups of Theorem 75.1 have no normal elementary abelian subgroups oforder 8. Therefore, to prove that theorem, we must find all groups of �50 satisfying thehypothesis of Theorem 75.1. However, such proof will be not easier than the presentedproof.

Theorem 75.2. Let G be a 2-group with ��2.G/ D Q2n � C2, where n � 4. If

�2.G/ ¤ ��2.G/, then �2.G/ D G Š SD2nC1 � C2. (If n D 3, then such groups G

have been determined in Theorem 55.1.)

Proof. Set H D ��2.G/ D hui �Q, where u is an involution and Q Š Q2n , n � 4.

Assume �2.G/ ¤ H . Let hai be the unique cyclic subgroup of index 2 inQ and weset

Q D ha; b j a2n�1

D b4 D 1; a2n�2

D z; a2n�3

D v; b2 D z; ab D a�1i;

where Q1 D ha2; bi and Q2 D ha2; abi are the other two maximal subgroups of Qand Q1 Š Q2 Š Q2n�1 . Also, Z.H/ D hu; zi is normal in G. Since hvi � ha2i D

Ã1.H/, hvi and ha2i are normal subgroups in G and hzi � Z.G/.Since �2.G/ ¤ H , there exists an involution t 2 G �H and we setK D H hti. If

ut D uz, then Z.H/hti Š D8 and so o.ut/ D 4, a contradiction since the element utof order 4 is not contained inH D ��

2.G/. Hence t centralizes Z.H/. Since t cannotcommute with any element of order 4 in H , we get CH .t/ D Z.H/ D hu; zi.We act with t on the abelian group hui � hai and apply �55. It follows that t inverts

the quotient group hu; ai=Z.H/ and so at D a�1s with s 2 Z.H/. We compute.ta/2 D .tat/a D a�1sa D s, and so s D 1 (since ta cannot be of order 4 inview ta 62 H ) and at D a�1. Hence t inverts the maximal subgroup hu; ai of Hcontaining Z.H/.


Suppose that .Q1Z.H//t D Q1Z.H/. In that case bt D ba2iuj , where i , jare suitable integers. Working in Q, there is ak 2 hai (k is an integer) such thatbak

D ba�2i and so btak

D .ba2iuj /ak

D ba�2ia2iuj D buj , where t 0 D tak

is an involution in K � H . We compute .t 0b/2 D .t 0bt 0/b D bujb D uj z, and soo.t 0b/ D 4, a contradiction. Hence .Q1Z.H//t D Q2Z.H/. It follows bt D aba2rus

(r; s are some integers) and so hb; bt i D Q� Š Q Š Q2n with K D hui � .Q�hti/,whereQ�hti Š SD2nC1 .Let N D NG.Q1Z.H// so that jG W N j D 2, H � N , and G D N hti. Suppose

N ¤ H . By the above argument, N �H cannot contain involutions and so�2.N / D

H . By Theorem 75.1, we get jN W H j D 2 and N is a uniquely determined group. Inparticular, by the structure of N , if y 2 N �H , .Q1Z.H//y D Q2Z.H/, contrary toN D NG.Q1Z.H//. Hence, N D H and so G D H hti D K and we are done.

Let G be a 2-group. Of course, ��2.�

�2.G// D ��

2.G/. Therefore, if ��2.G/ D

C2�D, whereD is of maximal class and order> 8, thenD is a generalized quaterniongroup.

Remark. Let, as in Theorem 75.1,�2.G/ D C2�D, whereD is a 2-group of maximalclass. If D is dihedral, then c2.G/ D 2, and such groups are classified (see �43). Itfollows from the results of �54 that the 2-groups G with�2.G/ D E4 �D, whereD isdihedral, also known. Similarly, if n > 3 and �n�1.G/ D C2 �M2n or ��

n�1.G/ D

C2 � M2n , then cn�1.G/ D 4, and the classification of such groups follows fromresults of �55.

Theorem 75.3. LetG be a p-group all of whose noncyclic subgroups are generated byits elements of order p. Then one of the following holds: G is cyclic,G is of exponentp, G is a dihedral 2-group D2n of order 2n.

Proof. Suppose that p > 2 and assume that G is neither cyclic nor of exponent p.Let Z be a maximal cyclic subgroup of order ps , s � 2. Since Z ¤ G, there is asubgroup Y > Z with jY W Zj D p. It follows that Y is either abelian of type .ps ; p/

or Y Š MpsC1 . In any case,�1.Y / Š Ep2 so �1.Y / < Y , a contradiction.From now on assume p D 2. We may also assume thatG is not cyclic, exp.G/ � 4,

and G is not of maximal class. Let R be a normal four-subgroup of G. If CG.R/ hasan element v of order 4, then Rhvi contains an abelian subgroup of type .4; 2/, acontradiction. Hence H D CG.R/ is elementary abelian and jG W H j D 2. We haveexp.G/ D 4 and let s 2 G � H be an element of order 4 so that s2 2 H . SinceG has no abelian subgroups of type .4; 2/, it follows that hsi is a maximal abeliansubgroup of G. Then, by Suzuki (Proposition 1.8), G is of maximal class, contrary tothe assumption.

Problem. Classify the 2-groups G satisfying one of the following conditions:

(i) �2.G/ D E �D, where E is elementary abelian andD is of maximal class.

(ii) ��2.G/ D E �D, where E is elementary abelian andD is of maximal class.

�75 Elements of order � 4 in p-groups 281

(iii) ��n�1.G/ D E �M2n , where n > 3 and E is elementary abelian.

(iv) �n.G/ D E � C2n , where n > 1 and E is elementary abelian.

(v) ��n.G/ D E � C2n , where n > 1 and E is elementary abelian.

�76

p-groups with fewA1-subgroups

1o. Recall that G is an An-group if it contains a nonabelian subgroup of index pn�1

but all its subgroups of index pn are abelian (see �72). For example, a p-group G ofmaximal class and order pm is anAm�2-group since it contains a nonabelian subgroupof order p3 (see Theorems 9.5 and 9.6).Let ˛n.G/ be the number of An-subgroups in a p-group G. If G is an An-group,

then ˛i .G/ > 0 for i D 1; : : : ; n � 1, ˛n.G/ D 1 and j .G/ D 0 for j > n. ForH < G, we set ˇ1.G;H/ D ˛1.G/�˛1.H/. For example, ifH < G is abelian, thenˇ1.G;H/ D ˛1.G/. IfH < G, then �H

i denotes the set of all members of the set �i

which contain H . Given H < G, let �i .H/ D fU < H j ˆ.H/ � U; jH W U j D

pi g. If G is a p-group of maximal class containing an abelian subgroup of index p,then ˛1.G/ D pm�3.It is fairly difficult to compute ˛1.G/ even for groups with not very complicated

structure. Below we compute ˛1.G/ for three families of groups.Let X be a group and '2.X/ the number of noncommuting ordered pairs .x; y/ 2

X �X such that hx; yi D X (it follows that '2.X/ > 0 if and only if X is nonabelianand two-generator). Let k.X/ be the class number of X . We claim that (see �2; thisidentity is due to Mann)

X

H�X

'2.H/ D jX j.jX j � k.X//:(1)

Indeed, let fK1; : : : ;Krg (here r D k.X/) be the set of conjugacy classes of X . Thenthe number of commuting ordered pairs of G equals

rX

iD1

.X

x2Ki

jCX .x/j/ D

rX

iD1

jKi jjX j

jKi jD r jX j D jX jk.X/(2)

so the number of noncommuting ordered pairs of elements of X is identical withjX j2 � jX jk.X/. On the other hand, that number also equals

P

H�X '2.H/.

Examples. 1. Let us find ˛1.G/, where G is extraspecial of order p2mC1. We havek.G/ D jZ.G/j C jG�Z.G/j

pD p2m C p � 1. Let E � G be nonabelian and two-

generator. Then E 0 D G0 and E=E 0 Š Ep2 since E=E 0 � G=E 0 D G=G0 Š Ep2m . Itfollows that jEj D p3 so E is anA1-subgroup. We have

'2.E/ D .jE �ˆ.E/j/.jEj �pjˆ.E/j/ D .p3 �p/.p3 �p2/ D p3.p2 � 1/.p� 1/:

�76 p-groups with few A1-subgroups 283

It follows from (1) that

˛1.G/'2.E/ D ˛1.G/p3.p2 � 1/.p � 1/ D p2mC1.p2mC1 � p2m � p C 1/

D p2mC1.p � 1/.p2m � 1/;

and we get ˛1.G/ D p2m�2 p2m�1

p2�1. In particular, if jGj D p5, i.e., m D 2, then

˛1.G/ D p2.p2 C 1/.

2. Let G be a nonabelian group of order pm all of whose nonabelian two-generator

subgroups have the same order p3. Then, using (1), we get ˛1.G/ D pm�3.pm�k.G//

.p�1/.p2�1/.

3. Let G D S � Epn , where S is nonabelian of order p3. Then jGj D pnC3, k.G/ D

pn.p2 C p � 1/. If A < G is an A1-subgroup, then jAj D p3. Therefore, by the

formula in Example 2, we get ˛1.G/ D pnŒpnC3�pn.p2Cp�1/�

.p2�1/.p�1/D p2n.

Epimorphic images of an An-group are Ak-groups with k � n. Indeed, if N G G,where G is anAn-group, then all subgroups of index pn in G=N are abelian.The main result of this section is the following

Theorem A ([Ber30]). Let G be a nonabelian p-group. If ˛1.G/ � p2 C p C 1, thenG is an An-group, n 2 f1; 2; 3g.

Exercise 1. Prove that if G is metacyclic, distinct F;H 2 �1 are Ar -, As-groupsrespectively, and r � s, then G is anAsC1-group.

Solution. Since G0 is cyclic, we get F 0 � H 0. It follows from the above, Lemma64.1(u) that jG0j D pjH 0j D psC1 so G is anAsC1-group, by Theorem 72.1.

Exercise 2. Let G be a nonabelian p-group. Suppose that its maximal subgroupsM1; : : : ;Mk contain all A1-subgroups of G. Is it true that k � p?

Exercise 3. Suppose that maximal subgroups M1; : : : ;Mp together contain all A1-subgroups of a nonabelian p-group G. Is it true that jG W .

TpiD1Mi /j D p2?

Exercise 4. Let G D M � C , whereM is nonabelian of order p3 and C is cyclic oforder p2. Prove that G has an A1-subgroup of order p4 and find the number of suchsubgroups in G.

2o. Proof of Theorem A.We begin with the following

Remark 1. Let G be neither abelian nor anA1-group and let F1 2 �1 be nonabelian.We claim that ˇ.G;F1/ � p � 1. By Exercise 1.6(a), the set �1 has at least pnonabelian members. Let F2 2 �1 � fF1g be nonabelian and set D D F1 \ F2.Since D — Z.G/, at least p members of the set �D

1 , say F1; : : : ; Fp , are nonabelian.Let Ui � Fi be an A1-subgroup such that Ui — D, i D 1; : : : ; p (such Ui ex-ists, by Proposition 10.28). Then U1; : : : ; Up are distinct A1-subgroups of G andU2; : : : ; Up — F1 so ˇ1.G;F1/ � p � 1. (For more detailed description of suchgroups, see Lemma 76.5.)


Lemma 76.1 (= Exercise 1.8a (L. Redei)). Let G be an A1-group. Then G D ha; bi,Z.G/ D ˆ.G/, jG0j D p and one of the following holds:

(a) apm

D bpn

D cp D 1, Œa; b� D c, Œa; c� D Œb; c� D 1, jGj D pmCnC1,G D hbi .hai � hci/ D hai .hbi � hci/ (semidirect products with kernels in brackets)is nonmetacyclic. HereG0 D hci, Z.G/ D ˆ.G/ D hapi � hbpi � hci. If mC n > 2,then �1.G/ D h�1.hai/;�1.hbi/; ci Š Ep3 .

(b) apm

D bpn

D 1, m > 1, ab D a1Cpm�1

, jGj D pmCn is metacyclic. HereG0 D hapm�1

i, Z.G/ D ˆ.G/ D hapi � hbpi, �1.G/ D hapm�1

; bpn�1

i Š Ep2 . If�1.G/ — Z.G/, then n D 1 so G Š MpmC1 .

(c) a4 D 1, a2 D b2, ab D a�1, G Š Q8.

If j�1.G/j � p2, then G is metacyclic. The group G is nonmetacyclic if and onlyif G0 is a maximal cyclic subgroup of G. Next, jG=Ã1.G/j � p3 with equality if andonly if G is from (a) and p > 2. If, in (a), u 2 G �ˆ.G/, then hui 6E G. All membersof the set �1 have ranks at most 3. If N is normal in G and G=N is not cyclic, thenN � Z.G/.

In what follows G is a nonabelian group of order pm.Let us prove that if, in Lemma 76.1, G0 is not a maximal cyclic subgroup of G, then

G is metacyclic. Let G0 < L < G, where L is cyclic of order p2. By Theorem 6.1,L=G0 � C=G0, where C=G0 is a cyclic direct factor of G=G0; in particular, G=C iscyclic. It remains to show that C is cyclic. We get G0 D ˆ.L/ � ˆ.C/. It followsthat C=ˆ.C / as an epimorphic image of a cyclic group C=G0, is cyclic. In that case.C is also cyclic, as was to be shown.In what follows G is a nonabelian group of order pm.

Lemma 76.2. Suppose that a p-groupG is neither abelian nor anA1-group and �1 D

fH1; : : : ;Hp ; Ag, where A is abelian. Then H 01 D D H 0

p and G=H 01 is an A1-

group so jG0j D pjH 01j. In particular, d.Hi / � 3 for all i . If, in addition, d.Hi / D 2

for i D 1; : : : ; p, and L is a G-invariant subgroup of index p in H 01, then G=L is an

A2-group.

Proof. By Remark 1, H1; : : : ;Hp are nonabelian. Let jH 01j � � jH 0

pj. Since.H1=H

01/ \ .A=H 0

1/ � Z.G=H 01/, we get G=H

01=Z.G=H

01/ Š Ep2 so all maximal

subgroups of G=H 01 must be abelian so H

01 D D H 0

p since d.G/ D 2. Assumethat G=H 0

1 is abelian; then G0 D H 0

1. Let L be a G-invariant subgroup of index pin G0. Then, by Lemma 65.2(a), G=L is an A1-group since d.G=L/ D 2, so H 0

1 �

L < G0 D H 0, a contradiction. Thus, G=H 01 is an A1-group so jG0j D pjH 0

1j. ByLemma 76.1, d.Hi / D d.Hi=H

01/ � 3. Next suppose that d.Hi / D 2 for i D 1; : : : ; p

and L is taken as above; then Hi=L is an A1-group (Lemma 65.2(a)) so G=L is anA2-group.

Lemma 76.3. If all members of the set �2 are abelian, then d.G/ � 3. If, in addition,d.G/ D 3, then ˆ.G/ � Z.G/.


Proof. Suppose that d.G/ > 3. Take F 2 �1; then F contains at least p2 C p C 1

(abelian) members of the set �2 so it is abelian (Exercise 1.6(a)). In that case, sinceG is not two-generator, it is abelian (Lemma 76.1), contrary to the hypothesis. Nowsuppose that d.G/ D 3. Then CG.ˆ.G// � hH j H 2 �2i D G, completingthe proof.

In what follows we make use of the following fact: If N is a normal subgroup ofa p-group G, jG=N j > p2 and G=N is generated by subgroups of index p2, whoseinverse images in G are abelian, then N � Z.G/. Indeed, CG.N / � hH < G j N <

H; jG W H j D p2i D G.

Lemma 76.4. Let G be a nonmetacyclic A2-group of order pm > p4. Then

(a) d.G/ � 3, jG0j � p3, exp.G0/ D p, G=Z.G/ is of order p3 and exponent p.

(b) ˛1.G/ 2 fp; p C 1; p2; p2 C p; p2 C p C 1g.

(c) If ˛1.G/ < p2, then p > 2, d.G/ D 2, and cl.G/ D 3.

(c1) If ˛1.G/ D p, then G=.G0 \ Z.G// is an A1-group, G0 Š Ep2 .

(c2) If ˛1.G/ D p C 1, then jGj D p5, G0 D �1.G/ Š Ep3 , Z.G/ D

K3.G/ D Ã1.G/ Š Ep2 .

(d) If ˛1.G/ D p2, then d.G/ D 3, G=Z.G/ Š Ep2 , jG0j D p.

(e) If ˛1.G/ D p2 C p, then d.G/ D 3, G0 Š Ep2 , G0 � Z.G/ D ˆ.G/.

(f) If ˛1.G/ D p2 CpC 1, then p D 2, jGj D 26, G is special with G0 D Z.G/ D

ˆ.G/ D �1.G/ Š E8.

Proof. (a) Two inequalities follow from Lemmas 76.1 and 64.1(u). Next, exp.G0/ D p

(Theorem 65.7(d)). If d.G/ D 3, then ˆ.G/ � Z.G/, by the paragraph preceding thelemma. If p > 2, then jG=Ã1.G/j > p2 (Theorem 9.11) so G=Ã1.G/ is generatedby subgroups of index p2, and we get Ã1.G/ � Z.G/. Hence, the last assertion of (a)holds for p > 2. Now suppose that p D 2 and d.G/ D 2. However, this case does notoccur according to Propositions 71.1 to 71.5. Indeed, the results of �71 imply that anA2-group of order 2m > 24 with d.G/ D 2 must be metacyclic (this is also true form � 4, by Lemma 64.1(i)).

Assertions (b) to (f) are direct consequences of �71. Indeed, if G is a group ofProposition 71.1, then ˛1.G/ D p2, d.G/ D 3, G=Z.G/ Š Ep2 , and jG0j D p. IfG is a group of Proposition 71.3, then ˛1.G/ D p, p > 2, d.G/ D 2, cl.G/ D 3,G0 Š Ep2 , and NG D G=.G0 \ Z.G// is an A1-group since d. NG/ D 2 and j NG0j D

p (Lemma 65.2(a)). If G is a group of Proposition 71.4, then ˛1.G/ D p2 C p,d.G/ D 3, G0 Š Ep2 , and G0 � Z.G/ D ˆ.G/. If G is a group of Proposition71.5 (a), then ˛1.G/ D p2 C p C 1, p D 2, d.G/ D 3, jGj D 26, G is specialwith G0 D Z.G/ D ˆ.G/ D �1.G/ Š E8; moreover, G is isomorphic to a Sylow2-subgroup of the Suzuki simple group Sz.8/. If G is a group of Proposition 71.5(b),


then p > 2, d.G/ D 2, cl.G/ D 3, ˛1.G/ D p C 1, jGj D p5, G0 D �1.G/ Š Ep3 ,Z.G/ D K3.G/ D Ã1.G/ Š Ep2 . Our proof is complete since the A2-groups ofProposition 71.2 are metacyclic.

Let us give a proof, independent of �71, that if G is a nonmetacyclicA2-group with˛1.G/ < p

2, then cl.G/ D 3. It follows from Exercise 1.6(a) that d.G/ D 2. Assumethat cl.G/ D 2. Then G0 � Z.G/. Since G is nonmetacyclic, we get exp.G0/ D p, byLemma 76.4(a). In that case, G is an A1-group (Lemma 65.2(a)), a contradiction. Itremains to prove that cl.G/ D 3. This is the case if jG0j D p2. Now let jG0j D p3 (seeLemma 76.4(a)); then ˛1.G/ D p C 1 and G is nonmetacyclic with all two-generatormembers of the set �1 since the latter has no abelian members (Lemma 64.1(u)). Itfollows thatG=Ã1.G/ is nonabelian of order p3 and exponent p andÃ1.G/ D K3.G/

so jG W G0j D p2 (Lemma 64.1(p)), and we get jGj D jG W G0jjG0j D p5. Sinceall subgroups of order p3 that contain K3.G/, are abelian (G is an A2-group!) andgenerate G, we get K3.G/ D Z.G/ since jG W Z.G/j > p2, and so cl.G/ D 3.

Remarks. Suppose that A;B 2 �1 are distinct.

2. If A is abelian and B an A1-group, then jG=Z.G/j � p3. If, in addition, Ais the unique abelian member of the set �1, then G=Z.G/ is either of order p3 andexponent p or G=Z.G/ Š D8. Indeed, jG0j � p2 (Lemmas 64.1(u) and 76.1) sojG W Z.G/j � p3 (Lemma 64.1(q)). If A is the unique abelian member of the set �1,then jG W Z.G/j D p3 so G=Z.G/ has at most one cyclic subgroup of index p, andthe last assertion follows.

3. If A;B areA1-subgroups, then jG=Z.G/j � p4. Indeed, put D D A \ B . ThenZ.A/ D ˆ.A/ � ˆ.G/ � A\B D D and similarly Z.B/ < D so, comparing orders,we conclude that Z.A/ and Z.B/ are maximal in D (Lemma 76.1). It follows thatU D Z.A/ \ Z.B/ has index at most p2 inD and U � Z.G/ since CG.U / � AB D

G, and we get jG=Z.G/j � jG=U j � jG=DjjD=U j � p4.

The following lemma is the key result.

Lemma 76.5. Suppose that H is a nonabelian maximal subgroup of a p-group G.Then ˇ1.G;H/ � p � 1. If ˇ1.G;H/ D p � 1, then the following holds:

(a) d.G/ D 2, �1 D fH1 D H;H2; : : : ;Hp ; A D HpC1g, where A is abelian andall members of the set �1 � fH;Ag are A1-groups.

(b) H 01 D D H 0

p is of order p.

(c) G=H 01 is an A1-group so jG0j D p2. We also have d.H1/ � 3. If G is nonmeta-

cyclic, then G=H 01 is also nonmetacyclic.

(d) G=Z.G/ is either nonabelian of order p3 and exponent p or G=Z.G/ Š D8,Z.Hi / D Z.G/ andHi=G

0 is cyclic, i D 2; : : : ; p. In particular, cl.G/ D 3.

(e) If G0 Š Cp2 , thenH2; : : : ;Hp ; A are metacyclic and G has no normal subgroupsof order p3 and exponent p. If p > 2, then G is metacyclic (in that case, G is anA2-group, by Corollary 65.3).


(f) If G0 Š Ep2 , then cl.G/ D 3.

In what follows we assume that G is not an A2-group: then jGj D pm > p4. SincejG0j D p2, G is nonmetacyclic.

(g) d.H/ D 3.

In what follows we assume, in addition, that p D 2. Then, sinceG is not of maximalclass, we get jG W G0j > 4, by Taussky’s theorem.

(h) d.A/ D 2 so A is metacyclic.

(i) G has no normal elementary abelian subgroups of order 8.

(j) H2 is metacyclic.

(k) U D �1.H2/ D �1.A/ < ˆ.G/ is the unique normal abelian four-subgroupof G.

(l) If Z.G/ is cyclic, then H2 Š M2m�1 and G0 6� Z.G/. In that case, if NG D G=H 0

and NT D �1. NH/, then NT D �1. NG/ Š E8 and T D Q � L, where L Š C4,Q Š Q8,QGG andG=Q is cyclic soG0 < Q is cyclic of order 4. Since�1.G/ D T , it followsthat G has exactly 7 involutions. Next, G=G0 is abelian with cyclic subgroup of index2 and G has no elementary abelian subgroups of order 8.

(m) Suppose that Z.G/ is noncyclic. Then G0 Š C4 is a maximal cyclic subgroup ofG. Let T be as in (l). Then jT j D 16 and T D Q � L. where Q is nonabelian oforder 8. Next, H2=G

0 is cyclic since G0 6� Z.G/ D Z.H/ D ˆ.H/ and H2 has nocyclic subgroups of index 2. The quotient group A=G0 is also cyclic.

(n) G0 is a maximal cyclic subgroup of G.

Proof. The set �1.H/ \ �2 contains a member N ; then N 6� Z.G/ since H is non-abelian. Set �N

1 D fH1 D H; : : : ;Hp ;HpC1g. Since at most one member of theset �N

1 is abelian, one may assume that H1; : : : ;Hp are nonabelian. By Remark 1,ˇ1.G;H/ � p � 1. Next we suppose that ˇ1.G;H/ D p � 1.

(a–c) If Ui � Hi is anA1-subgroup not contained in N (see Theorem 10.28), thenU2; : : : ; Up are pairwise distinct. It follows that UpC1 does not exist so HpC1 D A

is abelian. Then N D H \ A is also abelian, and we get p � 1 D ˇ1.G;H/ �Pp

iD2 ˛1.Hi /. It follows that ˛1.Hi / D 1 (i D 2; : : : ; p) so H2; : : : ;Hp are A1-groups and the subgroups H1; : : : ;Hp together contain all A1-subgroups of G. As-sume that F 2 �1 � �N

1 . The intersection F \ Hi is abelian since Hi is an A1-subgroup, i D 2; : : : ; p. Therefore, if F is nonabelian, then all A1-subgroups of Fare contained in H1. In that case, F � H1 (Theorem 10.28), a contradiction. Thus,all members of the set �1 � �N

1 are abelian. Assuming that d.G/ > 2, we see thatthe set �1 has at least .p2 C p C 1/ � p D p2 C 1 > p C 1 abelian members,contrary to Exercise 1.6(a). Thus, �1 D �N

1 D fH1; : : : ;Hp ; Ag so d.G/ D 2 andˆ.G/ D N , completing the proof of (a). Now (b) follows from (a) and Lemma 76.2.By Lemma 76.2 again, G=H 0

1 is an A1-group and hence jG0j D pjH 01j D p2. In that


case, d.H1/ D d.H1=H01/ � 3 (Lemma 76.1). The last assertion in (c) follows from

Theorem 36.1.

(d) It follows from d.G/ D 2 that jG W Z.G/j > p2. By Remark 2, G=Z.G/ iseither nonabelian of order p3 and exponent p or G=Z.G/ Š D8 since A is the uniqueabelian member of the set �1. Now let i 2 f2; : : : ; pg. In view of d.G/ D 2, we getZ.G/ � ˆ.G/ < Hi so Z.G/ � Z.Hi /, and the equality jZ.G/j D jZ.Hi /j impliesZ.G/ D Z.Hi / .D ˆ.Hi //. Therefore, since G0 6� Z.G/, thenHi=G

0 is cyclic.

(e) Suppose thatG0 Š Cp2 . ThenHi is metacyclic sinceH 0i is not a maximal cyclic

subgroup in Hi in view of H 0i < G

0, i D 2; : : : ; p (Lemma 76.1). Assume that G hasa normal subgroup R of order p3 and exponent p. For i D 2; : : : ; p, we getHiR D G

soG=.R\Hi / D .Hi=.R\Hi //�.R=.R\Hi //, and we conclude thatHi=.R\Hi /

is cyclic; then G=.R \Hi / is abelian. On the other hand, G=.R \Hi / is nonabeliansince Cp2 Š G0 6� R \Hi Š Ep2 , and this is a contradiction. Thus, R does not existso A is metacyclic. Let p > 2. Since G0 is cyclic, G is regular (Theorem 7.1(c)) sop2 D j�1.G/j D jG=Ã1.G/j whence G is metacyclic (Theorem 9.11).

(f) If G0 Š Ep2 , then G0 6� Z.G/ ((a) and Lemma 65.2(a)) so cl.G/ D 3.

(g) By hypothesis,H is not anA1-group (otherwise,G is anA2-group). By Lemma65.2(a), d.H/ > 2. Since G=H 0 is an A1-group, we get d.H/ � 3, and our claim isproved.

In what follows we assume that p D 2 so jG W G0j > 4 since G is not of maximalclass (Taussky).

(h) By Theorem 71.6, the number of two-generator members in the set �1 is even.Since j�1j D 3 and d.H2/ D 2 < 3 D d.H/, by (g), we get d.A/ D 2, so A ismetacyclic. Clearly, A is noncyclic.

(i) Assume that E8 Š EGG. ThenE\A D �1.A/, by (h). We have CG.E\A/ D

EA D G so E \ A � Z.G/. Since E 6� A 2 �1 and j�1j D 3, the number ofmaximal subgroups of G=E is < 3 so G=E is cyclic. In that case, A D L �Z, wherejLj D 2 and Z is cyclic so A=L is a cyclic subgroup of index 2 in G=L (we haveL < �1.A/ � Z.G/). Since L < Z.G/ and G=Z.G/ Š D8, G=L is of maximal class(Theorem 1.2). However, G=L has a normal subgroup E=L Š E4 and jG=Lj � 24,and this is a contradiction. Thus, E does not exist.

(j) Assume that H2 is nonmetacyclic. Then �1.H2/ Š E8 (Lemma 65.1) and�1.H2/ GG, contrary to (i).

(k) Since the two-generator 2-group G has no cyclic subgroups of index 2, the sub-group ˆ.G/ is noncyclic. We have H2 \ A D ˆ.G/ is noncyclic so U D �1.H2/ D

�1.A/ Š E4. Assume that V ¤ U is a G-invariant four-subgroup. Then V — H2; A

so G=V is cyclic. Since G has no cyclic subgroups of index 2 and jGj > 24, we get�1.G/ D UV Š E8, contrary to (i). Thus, V does not exist.Let NG D G=H 0 and NT D �1. NH/; then NT Š E8 is normal in NG since NH is abelian

of rank 3, by (g). Since NG is minimal nonabelian, by (c), we get �1. NG/ D NT so�1.G/ � T . Assume that jGj > 25. Then NG= NT Š G=T is cyclic since T 6� A;H2,


by (h) and (j). Since d.T / D 3, it follows that T is nonabelian, by (i). By (k), U < T ,where U is the unique normal abelian four-subgroup of G. Let Q < T be minimalnonabelian; then jQj D 8, T D QZ.T / (Lemma 64.1(i)) and Z.T / GG since T GG.

(l) Suppose that Z.G/ is cyclic. Then the noncyclic abelian subgroup ˆ.G/ has acyclic subgroup Z.G/ of index 2 (recall that 8jZ.G/j D jGj D 4jˆ.G/j and Z.G/ <ˆ.G/ since d.G/ D 2) so �1.ˆ.G// D U 6� Z.H2/ (otherwise, CG.U / � H2A D

G and E4 Š U � Z.G/, contrary to the assumption) so H2 Š M2m�1 (Lemma76.1). Let T be defined as in the previous paragraph. Then Z.T / is cyclic (otherwise,since Z.T / G G, we get Z.T / D U , by (k), and CG.U / � TA D G; then Z.G/ isnoncyclic, a contradiction), so, by the product formula,H D QZ.G/, whereQ Š Q8

(Appendix 16) sinceQ — A;H2. SinceQ is the unique subgroup isomorphic to Q8 inT (Appendix 16), we getQ GG, and then G=Q is cyclic since d.G/ D 2 andQ 6� A.In that case,G0 < Q is of order 4 soG0 Š C4 and G0 6� Z.G/ D Z.H2/. Then G=G0,as abelian group of type .jG=Qj; 2/, has exactly two cyclic subgroups H2=G

0 andA=G0 of index 2 (H=G0 is not cyclic since d.H/ D 3). Assume that E8 Š E < G.Then E \A D �1.A/ D U so CG.U / � EA D G, contrary to the cyclicity of Z.G/.Thus, G has no elementary abelian subgroups of order 8. By the above and Appendix16, T has exactly 7 involutions so G has also 7 involutions since �1.G/ D T (seeTheorem 43.10).

(m) Suppose that Z.G/ is noncyclic; then U D �1.Z.G//. We have G0 ¤ U

(otherwise, G is an A1-group, by (a) and Lemma 65.2(a), a contradiction). Since Uis the unique normal four-subgroup of G, we get G0 Š C4. Let T=H 0 D �1.H=H

0/

be as in (l); then T=H 0 Š E8 (see the proof of (l)), T is nonabelian and T D QZ.T /,whereQ is nonabelian of order 8 (Lemma 64.1(i)). As above, G=T is cyclic so G0 <

T . Since U < T \Z.G/, we get T D Q�L, where jLj D 2. We know that�1.G/ D

�1.T /. Since Z.T / D U ¤ G0, we get G0 6� Z.T / so G0 6� Z.G/ hence cl.G/ D 3.Since U < H2, H2 has no cyclic subgroups of index 2 (otherwise, �1.H2/ D U �

Z.G/ so H2 is abelian). Next, H2=G0 is cyclic since G0 6� Z.G/ D Z.H/ D ˆ.H/.

Since G=G0 is abelian with cyclic subgroup of index 2, it follows that A=G0 is cyclic(otherwise,H=G0 is cyclic soH is metacyclic, which is a contradiction).

(n) By (l) and (m), G0 Š C4. Assume that G0 < Z, where Z Š C8. ThenG0=H 0 < Z=H 0 so G0=H 0 is not a maximal cyclic subgroup of the A1-group G=H 0.It follows that G=H 0 is metacyclic (Lemma 76.1). In that case, G is metacyclic, byTheorem 36.1, which is a contradiction. Thus, G0 < G is maximal cyclic.

Remarks. 4. Let G be a p-group and let members H1; : : : ;Hk , k > 1, of the set�1 contain together all A1-subgroups of G. Suppose that there exists A 2 �1 �

fH1; : : : ;Hkg such that A \ Hi is abelian for all i D 2; : : : ; k. Then A is abelian.Assuming that this is false, we can take an A1-subgroup U � A such that U 6� H1

(Proposition 10.28). Since, for i D 2; : : : ; k, also U 6� Hi (recall that A \ Hi isabelian, by hypothesis), we get a contradiction.

5. For a nonabelian p-group G, the following two conditions are equivalent: (a) G


is of maximal class with abelian subgroup A of index p. (b) Whenever H � G isnonabelian of order pk , then ˛1.H/ D pk�3. In the proof we use induction on jGj.Let us prove that (a) ) (b). By Fitting’s lemma, all members of the set �1 � fAg areof maximal class so, by Hall’s enumeration principle and induction,

˛1.G/ DX

H2�1�fAg

˛1.H/ D p pm�1�3 D pm�3:

Next, let L < G be nonabelian. Then L � H 2 �1 and the subgroup H of maximalclass has an abelian subgroupH\A of index p so, by induction, L is of maximal classand ˛1.L/ D pl�3, where jLj D pl . Now we prove that (b) ) (a). In that case, allproper nonabelian subgroups of G are of maximal class, by induction. Take T � G,where T is an A1-subgroup of G. Setting jT j D pk , we get 1 D ˛1.T / D pk�3 sok D 3, i.e., all A1-subgroups of G have the same order p3. Assume that G is not ofmaximal class. Then we get CH .T / 6� T (Lemma 64.1(i)). Let F be a subgroup oforder p2 in CH .T / such that Z.T / < F . Then jF T j D p4 and ˛1.F T / D p2 ¤

p D p4�3, a contradiction. Thus, G is of maximal class. Next, let R G G be of orderp2 and set U D CG.R/; then jG W U j D p and, by induction, U is abelian since it isnot of maximal class.

6. Let d.G/ D 3 and ˆ.G/ 6� Z.G/; then the set �2 has a nonabelian member N(Lemma 76.3). We claim that the set �2 contains at least p2 nonabelian members.Let A1; A2 2 �2 be distinct abelian; then A D A1A2 2 �1 contains exactly p C 1

members of the set �2 and all of them are abelian since ˆ.G/ D A1 \ A2 � Z.A/.Assume that B 2 �2 � �1.A/ is abelian. Then CG.ˆ.G// � AB D G, contrary tothe hypothesis. Thus, �1.A/\�2 is the set of all abelian members of the set �2 so thelast set contains exactly j�2j � .pC 1/ D p2 nonabelian members. Thus, in our case,the set �2 has p2, p2 C p or p2 C p C 1 nonabelian members.

7. LetG be a nonabelian p-group with d D d.G/ > 3. Let T D fT1; : : : ; Trg � �2

and M D fM1; : : : ;Msg � �1 be the sets of nonabelian members of the sets �2,�1, respectively (T ¤ ¿, by Lemma 76.3, M ¤ ¿, by Exercise 1.6(a)). We claimthat r > .p � 1/s. Let i be the number of members of the set M, containing Ti ,i D 1; : : : ; r , and let �j be the number of members of the set T contained in Mj ,j D 1; : : : ; s. Then, by double counting,

1 C C r D �1 C C �s;

and, in view of i D p C 1 for all i , that formula can be rewritten as follows:

.p C 1/r D �1 C C �s:(3)

Note that Mj contains exactly pd�2 C pd�3 C C p C 1 members of the set �2.Since, by Exercise 1.6(a), the set �1.Mj / contains at most pC1 abelian members, we


get �j � pd�2 C C p2 for j D 1; : : : ; s, and we deduce from (3) the followinginequality: .p C 1/r � .pd�2 C C p2/s. We get

r �pd�2 C C p2

p C 1s �

p2

p C 1s D

.p2 � 1/C 1

p C 1s > .p � 1/s;

as was to be shown. By Exercise 1.6(a), we have s � j�1j�.pC1/ D pd�1C Cp2.

8. Given a set M of subgroups of a p-group G, let ˛1.M/ denote the number ofA1-subgroups that contain together members of the setM. Suppose that G is neitherabelian nor anA1-group. We claim that if, for eachK 2 �2, we have ˛1.�

K1 / � pC1,

then G is an A2-group. By hypothesis, there is K 2 �2 which is not contained inZ.G/. Indeed, this is the case if d.G/ D 2. If d.G/ > 2, our claim is obvious.Set �K

1 D fH1; : : : ;HpC1g and let ˛1.H1/ � � ˛1.HpC1/. Since at most onemember of the set �K

1 is abelian,H1; : : : ;Hp are nonabelian. Assume that ˛1.H1/ >

1; then ˛1.H1/ D p, by Remark 1 and hypothesis. We get

p C 1 � ˛1.�K1 / D ˛1.H1/C

pC1X

iD2

ˇ1.Hi ;K/ D p C

pC1X

iD2

ˇ1.Hi ;K/

so ˇ1.H2;K/ D 1 and we get p D 2 and ˇ1.H3;K/ D 0 (Remark 1) whence H3

is abelian, and we conclude that K is abelian; in that case,H2 is an A1-group. Thus,all members of the set �2 are abelian so d.G/ � 3 (Lemma 76.3). Assume that G isnot an A2-group. If d.G/ D 2, then �1 D �K

1 D fH1;H2; A/, where ˛1.H1/ D 2,˛1.H2/ D 1 and A is abelian. In this case, ˇ1.G;H1/ D 1 so jH 0

1j D 2 (Lemma76.5), and the equality ˛1.H1/ D 2 is impossible (Lemma 76.4(c)), a contradiction.Now assume that d.G/ D 3; then j�1.H1/ \ �2j D p C 1 > 1 so jH 0

1j D 2 (Lemma64.1(q)), and again we get a contradiction.

Lemma 76.6. Suppose that ˛1.G/ > 1. Then ˛1.G/ � p.

(a) If ˛1.G/ D p, then G is an A2-group with d.G/ D 2 and jG0j D p2. If G isnonmetacyclic, it is of class 3.

(b) If ˛1.G/ D pC1, thenG is an A2-group with d.G/ D 2. IfG is not metacyclic,then p > 2, G is of order p5 with G0 Š Ep3 and of class 3.

Proof. The inequality ˛1.G/ � p follows from Remark 1. By Remark 8, G is anA2-group. All remaining assertions follows from Lemma 76.4(c).

Lemma 76.7. Let N 2 �2 and �N1 D fH1; : : : ;HpC1g. Then:

(a) ˛1.G/ � ˛1.N /CPpC1

iD1 ˇ1.Hi ; N /.

(b) If N is nonabelian, then ˇ1.G;N / � p2 � 1.

(c) If ˛1.G/ < p2, then NG.K/=K has only one subgroup of order p for eachnonabelian K < G.


Proof. (a) is obvious, (b) follows from (a) and Remark 1. It remains to prove (c).Assume that K < G is nonabelian and NG.K/=K contains an abelian subgroup L=Kof type .p; p/; then 1Cˇ1.L;K/ � ˛1.K/Cˇ1.L;K/ D ˛1.L/ � ˛1.G/ � p2 � 1

so ˇ1.L;K/ < p2 � 1, contrary to (b). Thus, L=K does not exist so NG.K/=K hasonly one subgroup of order p so it is either cyclic or generalized quaternion.

Lemma 76.8. Suppose that G is an An-group, n > 2. (a) If d.G/ > 2 then ˛1.G/ �

p2. (b) If d.G/ D 3 and ˆ.G/ D Z.G/, then ˛1.G/ � 2p2 C p � 1. (c) If d.G/ D 3

and jG W Z.G/j D p2, then ˛1.G/ � 2p2 � 1.

Proof. (a) There is N 6� Z.G/ for some N 2 �2 since hM j M 2 �2i D G. If allsuch N are abelian, then ˛1.G/ D

P

H2�1˛1.H/ � p.j�1j � .p C 1// � p3 > p2

(Exercise 1.6(a) and Lemma 76.6). If N is nonabelian, then ˛1.G/ � p2 (Lemma76.7(b)).

(b) If L 2 �1 is nonabelian, then jL W Z.L/j D jL W Z.G/j D p2 so jL0j D p

(Lemma 64.1(q)). Suppose thatL is neither abelian nor anA1-group (such anL existssince n > 2). Then, by Lemma 76.4, ˛1.L/ � ˛1.H/ � p2, where H � L is anA2-subgroup. In view of jG W Z.G/j D p3, the set �1 has at most one abelian member(Lemma 64.(u)) so this set has at least p2 C p nonabelian members, and, by Hall’senumeration principle, ˛1.G/ D ˛1.L/C

P

M 2�1�fLg ˛1.M/ � p2 Cp2 Cp� 1 D

2p2 C p � 1 (we have here equality if and only if ˛1.L/ D p2 and all nonabelianmembers of the set �1 � fLg areA1-groups).

(c) In our case, jG0j D p and all members of the set �2 are abelian, the set �1

has exactly p2 nonabelian members (Remark 6). If L 2 �1 is neither abelian noran A1-group, then ˛1.L/ � p2 (see the proof of (b)) so we get ˛1.G/ D ˛1.L/ CP

M 2�1�fLg ˛1.M/ � p2 C .p2 � 1/ D 2p2 � 1.

Theorem 76.9. A p-group G with 1 < ˛1.G/ < p2 is a two-generator A2-group. Inparticular, ˛1.G/ 2 fp;p C 1g.

Proof. Suppose that G is a counterexample of minimal order; then p2 > ˛1.G/ >

p C 1 (Lemma 76.6(a,b)) so p > 2. By Lemma 76.8(a), d.G/ D 2 so ˆ.G/ > Z.G/.By Lemma 76.7(c), ˆ.G/ is abelian. By Proposition 10.28 and induction, G is anA3-group. Then there is H 2 �1, which is an A2-group, and so, by Proposition 10.28,Lemmas 76.4 and 76.6, d.H/ D 2, jH 0j > p and ˛1.H/ 2 fp;p C 1g.

(i) Assume that A;U 2 �1, where A is abelian and U is an A1-group. ThenG=Z.G/ is of order p3 and exponent p (Remark 2), and Z.G/ < ˆ.G/ < H . In thatcase, jH W Z.G/j D p2 so jH 0j D p (Lemma 64.1(q)), contrary to what has been saidin the previous paragraph. Thus, a pair .A;U / does not exist.

Let F1; : : : ; Fs; V1; : : : ; Vu be all nonabelian members of the set �1, where˛1.Fi / D p and ˛1.Vk/ D 1 (since ˆ.G/ is abelian, the set �1 has no membersH with ˛1.H/ D p C 1, by Lemma 76.4). Since the set �1 has at most one abelianmember (Lemma 64.1(u)), we get s C u � j�1j � 1 D p.


(ii) Assume that A 2 �1 is abelian. Then u D 0, by (i), so s D p and, since ˆ.G/is abelian, p2 > ˛1.G/ D

PpiD1 ˛1.Fi/ D p p D p2, a contradiction.

Thus, all members of the set �1 are nonabelian, i.e., s C u D p C 1. As in theprevious paragraph, s < p so u � 2. Then G and all members of the set �1 aretwo-generator so G is either metacyclic or G=K3.G/ is nonabelian of order p3 andexponent p and then jG W G0j D p2 (Lemma 64.1(p)).

(ii1) Assume that G is metacyclic. Then V1; V2 2 �1 are distinct A1-subgroups soG is anA2-group (Exercise 1), a contradiction. Thus, G is nonmetacyclic.

(ii2) Since u > 1, then jG0j � pjV 01V

02j D p3 so jGj D jG W G0jjG0j � p5.

Since G is not an A2-group, we get jGj D p5. In that case, G has a nonabeliansubgroup of order p3. Since all nonabelian groups of order p3 areA1-groups, we getby Proposition 2.3, ˛1.G/ � p2, a final contradiction.

Let G be a p-group of maximal class and order p5 with abelian subgroup of indexp. Then ˛1.G/ D p2 and G is anA3-group.

Remark 9. Let a p-group G be neither abelian nor anA1-group. If, for eachK 2 �2,we have ˛1.�

K1 / � p2, then G is either anA2- or A3-group. Indeed, letH1 2 �1 be

nonabelian and take K 2 �1.H1/ \ �2. Set �K1 D fH1; : : : ;HpC1g. Since at most

one member of the set �K1 is abelian, we get ˛1.H1/ < p2, i.e., H1 is either A1- or

A2-group (Theorem 76.9) so G is anAn-group, n 2 f2; 3g.

Theorem 76.10. If ˛1.G/ D p2, then G is an An-group, n 2 f2; 3g. Next supposethat G is an A3-group. Then jG0j D p3 and ˛1.H/ 2 f1; pg for every nonabelianH < G.

(a) If the set �2 has a nonabelian memberN , thenN is an A1-group and one of thefollowing holds:

(a1) G is metacyclic with ˛1.H/ D p for all H 2 �1, i.e., all members ofthe set �1 are A2-groups.

(a2) d.G/ D 3, �1 D fF1; : : : ; Fp2Cp; Ag, where ˛1.Fi / D p for all i andA is abelian, Z.G/ D Z.N / is of index p4 and the set �2 has exactly p2

nonabelian members which are A1-groups.

(b) If all members of the set �2 are abelian, then d.G/ D 2 and one of the followingholds:

(b1) G is metacyclic, �1 D fH1; : : : ;Hp ; Ag, where ˛1.Hi / D p for all i ,H 0

1 D D H 0p, A is abelian.

(b2) �1 D fH1; : : : ;Hp ; Ag, where ˛1.Hi / D p,H 01 D D H 0

p is of orderp2, A is abelian, G0 is noncyclic, G=H 0

1 is an A1-group.

(b3) p D 2, �1 D fH1;H2;H3g, ˛.H1/ D 2, ˛.H2/ D ˛.H3/ D 1. Wehave jG=Z.G/j D 16.


Proof. Suppose that G is not anA2-group; then G is anA3-group (Proposition 10.28and Theorem 76.9).

(a) Suppose that the set �2 has a nonabelian member N . Let �N1 D fH1;H2; : : : ;

HpC1g; then all members of the set �N1 are neither abelian nor A1-groups. Using

Remark 1, we get

p2 D ˛1.G/ � ˛1.N /C

pC1X

iD1

ˇ1.Hi ; N / � 1C .p � 1/.p C 1/ D p2;

and hence ˛1.N / D 1, ˇ1.Hi ; N / D p � 1, ˛1.Hi / D p, d.Hi / D 2 and jH 0i j D p2

for all i , H1; : : : ;HpC1 are A2-groups (Remark 1, Lemmas 74.4 and 76.6(a)) andthese p C 1 subgroups contain together all A1-subgroups of G. If G is metacyclic,it is as in (a1). Next we assume that G is nonmetacyclic. We also have d.G/ �

d.H1/C 1 D 3.Suppose, in addition, that d.G/ D 2 so N D ˆ.G/. Then G and all members

of the set �1 are two-generator (Lemma 76.6(a)). Since G is nonmetacyclic, we getp > 2 and jG W G0j D p2 (Lemma 64.1(p)). By the above and Lemma 76.6(a), Hi

has the unique abelian maximal subgroup Ai , all i . Then the subgroups A1 ¤ A2

are normal in G; therefore, A D A1A2 is of class at most 2 (Fitting’s lemma). ByLemma 64.1(l), G is not minimal nonmetacyclic so we may assume that H1 is notmetacyclic. Since A > A1 and cl.G/ � cl.H1/ > 2 (Lemma 76.4(c)), we get A 2 �1.Then A1 D H1 \ A D ˆ.G/ D N , a contradiction since N is nonabelian.Thus, d.G/ D 3. Then, by Remark 6, the set�2 has at leastp2 nonabelian members,

say L1; : : : ; Lp2 , and all Li are A1-groups since G is an A3-group and jG W Li j D

p2. Since ˛1.G/ D p2, all A1-subgroups of G are members of the set �2 so eachnonabelian H 2 �1 is an A2-group. If U1; : : : ; UpC1 are all abelian members of theset �2, then A D U1U2 2 �1, by the product formula, and jA0j � p (Lemma 64.1(q))so A is not anA2-group since ˛1.A/ < ˛1.G/ D p2 (Lemma 76.4). It follows that Ais abelian; moreover, A is the unique abelian member of the set �1 (Lemma 76.8(b)).Thus, �1 D fF1; : : : ; Fp2Cp; Ag, where ˛1.Fi/ D p and jF 0

i j D p2 (Lemma 76.6).Next, in view of d.Fi / D 2, we get Z.Fi/ < ˆ.Fi/ D ˆ.G/ < A; then CG.Z.Fi // �

FiA D G so Z.Fi / � Z.G/ and, by Lemma 64.1(q), jFi W Z.Fi/j D pjF 0i j D p3.

Thus, jG W Z.G/j � p4. By Lemma 64.1(q), jG W Z.G/j D pjG0j � pjF 01j � p3.

SinceG=Z.G/ is noncyclic andA is the unique abelian member of the set�1, it followsthat Z.G/ � Fi for some i . By the above, Z.Fi / � Z.G/ so Z.G/ D Z.Fi /, and weget jG W Z.G/j D jG W Fi jjFi W Z.G/j D p4 so jG0j D 1

pjG W Z.G/j D p3

and G is as stated in (a2) since, if N < Fi , then Z.N / D ˆ.N/ < ˆ.G/ < A soCG.Z.N // � AN D G hence Z.N / � Z.G/ and jG W Z.N /j D p4 D jG W Z.G/j.

(b) Suppose that all members of the set �2 are abelian. Then p2 D ˛1.G/ DP

M 2�1˛1.M/ since the intersection of any two distinct members of the set �1 is

abelian (here we use Hall’s enumeration principle). It follows from Remark 6 and


Lemma 76.8(b,c) that d.G/ D 2 (indeed, if d.G/ D 3, then ˆ.G/ � Z.G/, byLemma 76.3, and, by Lemma 76.8(b,c), ˛1.G/ � 2p2 � 1 > p2).

Let fH1; : : : ;Hs ; V1; : : : ; Vug be the set of nonabelian members of the set �1, where˛1.Hi / D p, ˛1.Vj / D 1 (since ˆ.G/ 2 �2 is abelian, the set �1 has no member Hwith ˛1.H/ D p C 1; see Lemma 76.4). By assumption, s > 0. Next, jG0j � jH 0

i j D

p2 so the set �1 has at most one abelian member (Lemma 64.1(q)).We have p2 D ˛1.G/ D spCu � spC.pC1�s/ D s.p�1/C.pC1/. If s < p,

then p2 � .p�1/2 CpC1 D p2 �pC2 so p D 2, s D 1, u D 2. Then ˛1.H1/ D 2

so cl.H1/ D 3 (Lemma 76.4). By Remark 3, jG=Z.G/j � p4. Since Z.G/ < H1 andcl.H1/ D 3, we get jH1 W Z.G/j > p2 so jG=Z.G/j D p4. Assume that jG0j D p2.Then G=V 0

1 is an A1-group (Lemma 65.2(a)) hence H1=V0

1 is abelian and H01 D V 0

1

is of order p, a contradiction. Since jG0j � 2jV 01V

02j � 8, we get jG0j D 8 so G is as

in (b3). Thus, s � p.Since s < p C 1, we get s D p. It follows from p2 D ˛1.G/ D sp C u D p2 C u

that u D 0. Thus, �1 D fH1; : : : ;Hp ; Ag, where A is abelian and ˛1.Hi / D p fori D 1; : : : ; p.Suppose that G is metacyclic. Then jG W Z.G/j D pjG0j D p4 (Lemma 64.1(q)

and Theorem 72.1) and G is as stated in (b1).Suppose that G is not metacyclic. Then G=H 0

i is an A1-group for all i (Lemma76.2) whence jG0j D pjH 0

1j D p3. By Proposition 72.1(b), G0 is noncyclic so G is asstated in (b2).

Theorem 76.11. Suppose that G is a p-group with p2 < ˛1.G/ � p2 C p � 1. ThenG is a two-generator A3-group with jG0j > p and one of the following holds:

(a) �1 D fH1;H2; : : : ;Hp ; Ag, where A is abelian, H1 is an A2-group withd.H1/ D 3, ˛1.H1/ D p2, ˛1.Hi / D 1, i D 2; : : : ; p, H 0

1 D D H 0p is

of order p, G=H 01 is an A1-group so jG0j D p2, ˛1.G/ D p2 C p � 1.

(b) �1 D fH1; : : : ;Hp ; Bg, where ˛1.Hi / D p, i D 1; : : : ; p, ˛1.B/ D 1,˛1.G/ D p2 C 1. If G is nonmetacyclic, then p > 2, jG W G0j D p2 andjGj 2 fp5; p6g.

(c) G is metacyclic, �1 D fF1; : : : ; Fs ;H1; : : : ;Hug, where s C u D p C 1,˛1.Fi / D p, s � 2, ˛1.Hj / D p C 1, ˛1.ˆ.G// D 1 and jG0j D p3.

Proof. By Lemma 76.4(b), G is not anA2-group.Assume that G is not an A3-group. Then there exists a nonabelian H 2 �1 which

is anAn-group, n � 3. By Theorem 76.9, ˛1.H/ � p2. Then, by Lemma 76.5,

p2 C p � 1 � ˛1.G/ D ˛1.H/C ˇ1.G;H/ � p2 C p � 1

so ˛1.G/ D p2Cp�1 and ˛1.H/ D p2, ˇ1.G;H/ D p�1 so jH 0j D p. LetL < Hbe anA2-subgroup. Since jL0j D p, we get, by Lemma 76.4, ˛1.L/ � p2 D ˛1.H/,contrary to Proposition 10.28. Thus, H is an A2-group so G is an A3-group. Thisargument also shows that the set �1 has no member U with ˛1.U / > p

2.


If U 2 �1, ˛1.U / D p2, then ˇ1.G;U / D p�1 so, by Lemma 76.5, G is as in (a).Next we assume that ˛1.U / < p2 for all U 2 �1; then d.G/ � d.U / C 1 D 3

(Lemma 76.1 and Theorem 76.9). Let

�1 D fF1; : : : ; Fs;H1; : : : ;Ht ; B1; : : : ; Bu; A1; : : : ; Avg; s C t > 0;

where ˛1.Fi / D p, ˛1.Hj / D p C 1 (see Lemma 76.6), ˛1.Bk/ D 1 and Al areabelian. Since sC t > 0 and jF 0

i j; jH 0j j > p, we get v � 1 (Lemma 64.1(u)). We have

(�) If t > 0, then u D v D 0 since the set �1.H1/ has no abelian members.

(��) If d.G/ D 2, then uv D 0. Indeed, if uv > 0, then all maximal subgroupsof G=B 0

1 are abelian, by Exercise 1.6(a), a contradiction since jT 0j � p2 forT 2 fF1; : : : ; Fs ;H1; : : : ;Ht g (Lemma 76.4).

(i) Suppose that d.G/ D 2. Then uv D 0, by (��). Put N D ˆ.G/.

(i1) Suppose, in addition, thatN is nonabelian so it is anA1-group since jG W N j D

p2. Then s C t D j�1j D p C 1,

p2 C p � 2 � ˇ1.G;N / DX

M 2�1

ˇ1.M;N/ D s.p � 1/C tp

D s.p � 1/C .p C 1 � s/p D p2 C p � s;

and we get s � 2 and t � p � 1. Let Ui be (the unique so normal in G) abelianmaximal subgroup of Fi , i D 1; 2, and set A D U1U2. Since F1 \ F2 D ˆ.G/ isnonabelian, we get U1 ¤ U2. Then A .> U1/ is of class� 2 (Fitting). If A 2 �1, thenˆ.G/ D A\F1 D A1 is abelian, a contradiction. Thus, A D G, so cl.G/ D 2. In thatcase, all members of the set �1 are metacyclic (Lemma 76.4) so G is also metacyclic(Lemma 64.1(e,f,h)), and G is as in (c).

(i2) In what follows we assume thatN D ˆ.G/ is abelian; then t D 0. If s D pC1,then ˛1.G/ D sp D .pC 1/p > p2 Cp� 1, a contradiction. Thus, uC v > 0. SincejG0j � jF 0

1j D p2, we get v � 1 (Lemma 64.1(u)).If v D 1, then u D 0, by (��); then s D p and we get p2 C1 � ˛1.G/ D sp D p2,

a contradiction. Thus, v D 0 so s C u D p C 1 and u D p C 1� s > 0. We have

p2 C 1 � ˛1.G/ DX

M 2�1

˛1.M/ D ps C u D .p � 1/s C p C 1 � p2 C p � 1:

It follows that p � s � p C 1 � 1p�1

, and so s D p and u D 1. In that case,

˛1.G/ D ps C u D p2 C 1, �1 D fF1; : : : ; Fp ; Bg, where ˛1.Fi / D p, ˛1.B/ D 1.Thus, G and all members of the set �1 are two-generator so G is either metacyclic orp > 2 and G=K3.G/ is nonabelian of order p3 and exponent p (Lemma 64.1(n,p)).Suppose that G is not metacyclic; then jG W G0j D p2. In that case, jG0j �

pjF 01B

0j � p4 so jGj D jG W G0jjG0j � p6. Since G is an A3-group, we getjGj 2 fp5; p6g so G is as in (b).


(ii) Now let d.G/ D 3. By Lemma 76.8(b,c), ˆ.G/ — Z.G/. Then, by Lemma76.3 and the last sentence of Remark 6, the set �2 has exactly p2 nonabelian memberswhich are A1-groups so this set has two distinct abelian members U1 and U2. ThenM D U1U2 2 �1, by the product formula and j�1.M/\�2j D pC1. It follows fromU1 \ U2 � Z.M/ that all members of the set �1.M/ \ �2 are abelian so the set �2

has exactly p C 1 abelian members, say U1; : : : ; UpC1. We have jM 0j � p (Lemma64.1(u)) and �2 D fL1; : : : ; Lp2 ; U1; : : : ; UpC1g, where all Li are A1-groups, allUi < M . Since Li \M D ˆ.G/ is abelian for all i , we get ˛1.M/Cp2 � ˛1.G/ �

p2 C p � 1 so ˛1.M/ � p � 1 andM is either abelian or A1-group (Remark 1).

(ii1) Suppose thatM is abelian. Then v D 1 (Lemma 64.1(u)) and so t D 0. Also,u D 0 since all pC 1 abelian members of the set �2 lie inM . Hence, s D j�1j � v D

p2 C p. By the hypothesis and Hall’s enumeration principle (Theorem 5.2),

p2 < ˛1.G/ DX

H2�1

˛1.H/�pX

H2�2

˛1.H/ D ps�p p2 D p.p2 Cp/�p3 D p2;

a contradiction. Thus, v D 0.

(ii2) Now letM be an A1-group; then u D 1 since all Ui < M . We get t D 0, by(�), so s D j�1j � u D p2 C p. Therefore, by Hall’s enumeration principle,

p2 C 1 � ˛1.G/ DX

H2�1

˛1.H/ � pX

H2�2

˛1.H/

D 1C sp � p p2 D 1C .p2 C p/p � p3 D p2 C 1:

Note that d.G/ D 3 but all members of the set �1 are two-generator. We have cl.G/ �

cl.F1/ D 3. Then p D 2 (Theorem 70.1) and G=K4.G/ is of order 27 or 28 (Theorem70.4). However, by remarks following Theorem 70.5, the above groups of order 27

and 28 areA4-groups andA5-groups, respectively, a contradiction.

Lemma 76.12. LetG be anA4-group and letR 2 �2 be an A2-group. Then ˛1.G/ >

p2 C p C 1, unless p D 2 and G satisfies the following conditions:

(a) d.G/ D 3, ˛1.G/ D 22 C 2C 1 D 7.

(b) �2 D fR;R1; R2; R3; A1; A2; A3g, where ˛1.R/ D 4, jR0j D 2, d.R/ D 3,R1; R2; R3 are A1-groups and A1; A2; A3 are abelian.

(c) �1 D fF1; F2; F3;H1;H2;H3; Ag, where A is abelian and

(c1) �1.Fi / D fR;Ri ; Ai g, ˛1.Fi/ D 22C1 D 5, jF 0i j D 4,R0 D R0

i , Fi=R0

is an A1-group, i.e., Fi is a group of Theorem 76.11(a), i D 1; 2; 3.

(c2) �1.H1/ D fA1; R2; R3g, �1.H2/ D fA2; R3; R1g, �1.H3/ D fA3; R1;

R2g so ˛1.Hi / D 2 and jH 0i j D 4 henceHi is an A2-group, i D 1; 2; 3.

(d) Z.G/ < ˆ.G/, jG0j D 8, jG=Z.G/j D 16, Z.G/ D Z.Ri /, i D 1; 2; 3.

All nonabelian members of the set �1 are two-generator.


Proof. Let �R1 D fF1; : : : ; FpC1g; then F1; : : : ; FpC1 are A3-groups. By Theorem

76.9, ˛1.Fi / � p2 for all i . By Lemma 76.6, ˛1.R/ � p.If ˛1.R/ D p, then ˇ1.Fi ; R/ � p2 � p for all i , and we are done since

˛1.G/ � ˛1.R/C

pC1X

iD1

ˇ1.Fi ; R/ � p C .p2 � p/.p C 1/ D p3 > p2 C p C 1:

Now suppose that ˛1.R/ D p C 1. Then ˛1.Fi/ � p2 C p, by Theorems 76.10and 76.11, so ˇ1.Fi ; R/ � p2 � 1 for all i , and we are done since

˛1.G/ � ˛1.R/C

pC1X

iD1

ˇ1.Fi ; R/ � pC1C.p2�1/.pC1/ D p3Cp2 > p2CpC1:

If ˛1.R/ > p C 1, then ˛1.R/ � p2 (Theorem 72.1 and 76.4). Since Fi is anA3-group with R 2 �1.Fi /, we get ˇ1.Fi ; R/ � p � 1 (Lemma 76.5), and so

˛1.G/ � ˛1.R/C

pC1X

iD1

ˇ1.Fi ; R/ � p2 C .p � 1/.p C 1/

D 2p2 � 1 � p2 C p C 1:

(4)

If p > 2, then ˛1.G/ � 2p2 � 1 > p2 C p C 1, and we are done in this case.Now let 2p2 � 1 D p2 C p C 1; then p D 2 and ˛1.G/ D 7. In that case, as it

follows from (4), ˛1.R/ D 4 and ˇ1.Fi ; R/ D 1 so ˛1.Fi/ D 5 for i D 1; 2; 3. ByLemma 76.5 applied to the pair R < Fi , we get jR0j D 2, d.Fi / D 2 and �1.Fi/ D

fR;Ri ; Ai /, where Ri is an A1-group with R0 D R0i and Ai is abelian, jF 0

i j D 4,i.e., Fi is a group of Theorem 76.11(a), i D 1; 2; 3. By Lemma 76.5, d.G/ � 1 C

d.F1/ D 3.Assume that d.G/ D 2. Since all members of the set �1 are also two-generator, G

is metacyclic since p D 2 (Lemma 64.1(n)). In that case, jG0j D 24 so jF 0i j D 23

(Theorem 72.1), contrary to what has been said in the previous paragraph.Thus, d.G/ D 3. Then ˆ.Fi / D ˆ.G/ and so fR;R1; R2; R3; A1; A2; A3g D

S3iD1 �1.Fi / D �2, where A1; A2; A3 are abelian so ˆ.G/ is abelian, ˛1.Ri / D 1

(i D 1; 2; 3).Set A D A1A2; thenA 2 �1 andA3 < A (if A3 — A then CG.ˆ.G// � AA3 D G

so ˆ.G/ � Z.G/, a contradiction since jR W ˆ.G/j D 2 and R is nonabelian). SinceF1; F2; F3 together contain allA1-subgroups of G and A\Fi is abelian for i D 2; 3,then A is abelian (Remark 4).Set �1 � �R

1 D fH1;H2;H3; Ag. One may assume that Ai < Hi , i D 1; 2; 3.SinceHi \ A > ˆ.G/ is an abelian maximal subgroup ofHi andHi \ A 2 �2, thenRi — Hi since RiAi D Fi ¤ Hi , i D 1; 2; 3. Since R — Hi and j�1.Hi /\�2j D 3,i D 1; 2; 3, we get

H1 D fA1; R2; R3g; H2 D fA2; R1; R3g; H3 D fA3; R1; R2g:


Now, R, R1, R2, R3 together contain all A1-subgroups of G. Each of the four A1-subgroups S1, S2, S3, S4 of R satisfies Sjˆ.G/ D R, j D 1; 2; 3; 4, sinceˆ.G/ is anabelian maximal subgroup of R, and so, in view of R — Hi , no one of S1; S2; S3; S4

is contained in Hi , i D 1; 2; 3. Thus, ˛1.Hi / D 2, i D 1; 2; 3. By Lemma 76.6(a),Hi is an A2-group with d.Hi / D 2 and jH 0

i j D 4, i D 1; 2; 3. Hence, all nonabelianmembers of the set �1 are two-generator.By Lemma 64.1(u), jG0j � 2jH 0

1A0j D 8 and so Lemma 64.1(q) implies that jG W

Z.G/j D 2jG0j � 24. On the other hand, G D RiA with A \ Ri D ˆ.G/. Wehave Z.Ri / D ˆ.Ri / < ˆ.G/ < A, jˆ.G/ W Z.Ri /j D 2, by the product formula,and CG.Z.Ri // � RiA D G. Thus, Z.Ri / � Z.G/, i D 1; 2; 3. Assume thatjG W Z.G/j < 24; then jG W Z.G/j D 23. In that case, all members of the set�1, containing Z.G/, must be abelian or A1-subgroups since nonabelian members ofthe set �1 are two-generator. Since A is the unique abelian member of the set �1,this set also contains an A1-group, contrary to what has been said about �1. Thus,Z.G/ D Z.Ri / (i D 1; 2; 3) so jG W Z.G/j D 24 and jG0j D 8 (Lemma 64.1(q)).

Definition 1. A 2-group G of Lemma 76.12 is called a 2-group of type G7;1.

We shall see (see Appendix, below) that groups of type G7;1 do not exist.

Lemma 76.13. Let a p-group G be an An-group, n > 2, let H1 2 �1 andˇ1.G;H1/ D p; then H1 is nonabelian (Lemma 76.6). In that case, d.G/ D 2 so�1 D fH1; : : : ;HpC1g, and one of the following holds:

(a) If H2 is neither abelian nor an A1-group, then p D 2, ˛1.H2/ D 2, H3

is abelian, H 01 D H 0

2 is of order 4, G=H 02 is an A1-group so jG0j D 8 and

jG=Z.G/j D 16.

(b) All p members of the set �1 � fH1g are A1-groups, H 01 D H 0

2 D D H 0pC1

and G=H 01 is an A1-group so jG0j D p2.

(c) All p members of the set �1 � fH1g are A1-groups and H 02; : : : ;H

0pC1 are

pairwise distinct. Set Q D H 02 : : :HpC1; then Q Š Ep2 , G=Q is an A1-group

soH 01 � Q and jG0j D p3.

Proof. Let R 2 �1.H1/ \ �2 and set �R1 D fH1;H2; : : : ;HpC1g. Since H1 is

nonabelian (Lemma 76.6(a)), we get R — Z.G/ so the set �R1 has at most one abelian

member.

(i) Suppose that R is nonabelian; then all subgroups Hi are nonabelian and wehave ˇ1.Hi ; R/ � p � 1 for i > 1 (Remark 1), and we get p D ˇ1.G;H1/ �PpC1

iD2 ˇ1.Hi ; R/ � p.p � 1/ so that p D 2 and ˇ1.Hi ; R/ D 1 for i D 2; 3.Therefore, by Lemma 76.5, applied to the pair R < Hi , i D 2; 3, we have

jR0j D 2; d.Hi / D 2; jH 0i j D 4; �1.H2/ D fR;L2; A2g;

�1.H3/ D fR;L3; A3g;


where ˛1.Li / D 1 and A is abelian. Since, in view of ˇ.G;H1/ D 2, H1, H2, H3

contain together allA1-subgroups of G, the members of the set �1 are notA1-groups.Next, A2; A3 GG. Set A D A2A3; then cl.A/ � 2 (Fitting). Since cl.G/ � cl.H2/ D

3 > 2, we get A 2 �1. Then A is abelian since A \Hi D Ai is abelian for i D 2; 3

(Remark 4). Therefore, in view of jG0j � jH 02j D 4, A is the unique abelian member

of the set �1 (Lemma 64.1(u)).

Since R 6� A, we get d.G/ > 2 so d.G/ D 3 since d.H2/ D 2; then j�1.Hi / \

�2j D 3, i D 1; 2; 3, hence �1.H2/ [ �1.H3/ � �2. We have Hi \ A D Ai 2

�2 and Ai is the unique abelian member of the set �1.Hi /, i D 2; 3. Also putA1 D H1 \ A.2 �2/ and �

Ai

1 D fHi ; Fi ; Ag, i D 1; 2; 3. It follows that �1 D

fH1;H2;H3; F1; F2; F3; Ag.

Since F1 \H1 D A1 and ˛1.F1/ D 2 D ˇ1.G;H1/, F1 is an A2-group (Lemma76.6). For i D 2; 3, set Si D F1 \Hi ; then Si 2 �2 and, since Si ¤ A1, the uniqueabelian member of the set �1.F1/, Si is anA1-subgroup (Lemma 76.5). We also haveSi ¤ R, i D 2; 3, and S2 ¤ S3 so �1.F1/ \ �2 D fS2; S3; A1g. One may assume,without loss of generality, that �S2

1 D fF1;H2; F2g; then �S3

1 D fF1;H3; F3g. ByLemma 76.5, d.Hi / D d.Fj / D 2, i D 2; 3 and j D 1; 2; 3.

Let �1.H1/ \ �2 D fR;U;A1g. Then H1 \ Fi D U for i D 2; 3 so �U1 D

fH1; F2; F3g. By the above, �2 D fR;S2; S3; U;A1; A2; A3g and

A1 D H1 \ F1 \ A; A2 D H2 \ F2 \ A; A3 D H3 \ F3 \ A;

R D H1 \H2 \H3; U D H1 \ F2 \ F3; S2 D H2 \ F1 \ F2;

S3 D H3 \ F1 \ F3:

By Lemma 64.1(u), jG0j � 2jH 02A

0j D 8. Assume that jG0j D 4. Then all non-abelian maximal subgroups of G have the same derived subgroup which is equal toG0. By Lemma 64.1(q), jG W Z.G/j D 8. If K=Z.G/ is a maximal subgroupof G=Z.G/ such that K ¤ A, then jK 0j D 2 (Lemma 64.1(q)), a contradiction.Thus, jG0j D 8 so jG=Z.G/j D 16 (Lemma 64.1(q)). If Z.G/ — H 2 �1, thenG D HZ.G/ so jG0j D jH 0j D 4, which is a contradiction. Thus, Z.G/ < ˆ.G/.Since R \ U D ˆ.G/, we get jU 0j D 2 (Lemma 64.1(q)).

Let us consider the quotient group NG D G=Z.G/ which is of rank 3 and order16. The subgroups NR, NS2, NS3 and NU are four-groups so NG has at least 9 involutions.Since exp. NG/ D 4, it follows that NG is nonabelian. Let ND be a minimal nonabeliansubgroup of NG. Then, by Lemma 64.1(i), NG D NDZ. NG/, and now it is easy to see thatNG Š D8 � C2 (see Appendix 16). Then NG contains exactly two distinct elementaryabelian subgroups of order 8. Let NK Š E8 be such that NK ¤ NA. Then K 2 �1 isnonabelian of rank � 3, contrary to what has been proved already.

Thus, all members of the set �1.H1/\ �2 are abelian.


(ii) Since R is abelian, we have

p D ˇ1.G;H1/ �

pC1X

iD2

ˇ1.Hi ; R/ D

pC1X

iD2

˛1.Hi /:

Thus, for nonabelianH 2 �R1 �fH1gwe have ˛1.H/ 2 f1; pg so d.H/ D 2 (Lemmas

76.1 and 76.6), and so d.G/ � 1C d.H/ D 3.

(ii1) Assume that H2 is neither abelian nor an A1-group. Then ˛1.H2/ D p D

ˇ1.G;H1/ (Lemma 76.6) so Hi is abelian for i > 2. Next, H2 is an A2-subgroupand jG0j � jH 0

2j D p2 (Lemma 76.6) so the set �1 has at most one abelian member(Lemma 64.1(u)). In that case, p D 2 and �R

1 D fH1;H2;H3 D Ag; here A isabelian, ˛1.H2/ D 2 and jH 0

2j D 4 (Lemmas 76.6 and 76.4).Assume that d.G/ D 2. Then H 0

1 D H 02 has order 4 and G=H

02 is an A1-group

(Lemmas 76.6 and 76.2) so jG0j D 8, G=Z.G/j D 2jG=G0j D 16, and G is as in (a).Now suppose that d.G/ D 3. Then j�2 \ �1.H1/j D 2 C 1 D 3. Take S 2

.�2 \ �1.H1// � fRg; then S is abelian, by (i). Set �S1 D fH1; F2; F3g. Since H1

and H2 together contain all A1-subgroups of G, F2 and F3 are not A1-subgroups(they are nonabelian since A is the unique abelian maximal subgroup of G). SinceFi \ H1 D S .2 �1.H1// is abelian, we get 1 < ˛1.Fi/ � ˇ1.G;H1/ D 2 so˛1.Fi / D 2 and Fi is an A2-group, i D 2; 3 (Lemma 76.6(a)). Then F2 and F3

together contain 4 distinct A1-subgroups and all of them are not contained in H1

(indeed, F2 \H1 D S D F3 \H1 is abelian), i.e., ˇ1.G;H1/ � 4 > 2, contrary tothe hypothesis.

Thus, all nonabelian members of the set �R1 � fH1g are A1-groups. It follows that

all nonabelian members of the set �1 � fH1g areA1-groups.

(ii2) Assume that d.G/ D 3. Then p D ˇ1.G;H1/ DP

F 2�1�fH1g ˛1.F / �

p2 � 1 since the set �1 � fH1g has at least p2 � 1 nonabelian members (Lemma64.1(q)) which areA1-subgroups. However, p2 � 1 > p, and we get a contradiction.

Thus, d.G/ D 2. ThenH2; : : : ;HpC1 areA1-groups, by (ii1) and hypothesis.

(ii3) Let H 02 D H 0

3 .Š Cp). In that case, CG.H02/ � H2H3 D G so H 0

2 � Z.G/.Then all maximal subgroups of the quotient group G=H 0

2 are abelian (Exercise 1.6(a))so G=H 0

2 is nonabelian hence anA1-group (Lemma 65.2(a)). We get

H 01 D D H 0

pC1; jG0j D j.G=H 02/

0jjH 02j D p p D p2;

and G is as in (b).

(ii4) Now let H 02; : : : ;H

0pC1 be all distinct. Set Q D H 0

2H03. Then, as above, all

maximal subgroups of the quotient group G=Q are abelian so H 02 : : :H

0pC1 D Q Š

Ep2 andH 01 < Q. To fix ideas, assume thatH

02 ¤ H 0

1. ThenH1=H02 is nonabelian and

ˇ1.G=H02;H1=H

02/ D p�1 so, by Lemma 76.5, jG0j D j.G=H 0

2/0jjH 0

2j D p2p D p3

and G=Q is anA1-group. Thus, G is as in (c).


Now we are ready to complete the proof of Theorem A for odd p.

Theorem 76.14. Let G be a p-group, p > 2. Suppose that ˛1.G/ 2 fp2 C p;p2 C

p C 1g. Then G is an An-group, n 2 f2; 3g.

Proof. Assume that the theorem is false. Then there is an Ak-group M 2 �1 withk > 2, and so ˛1.M/ � p2 (Theorem 76.9), hence ˇ1.G;M/ � p C 1.Assume that jM 0j D p; then d.M/ > 2 (Lemma 65.2(a)). Let S < M be an

A2-subgroup. Then, by Lemma 76.4, ˛1.S/ � p2. If S 62 �1.M/, then, by Lemma76.5,

˛1.M/ � ˛1.S/C 2.p � 1/ � p2 C p C .p � 2/ � p2 C p C 1

so, by Theorem 10.28, ˛1.G/ > ˛1.M/ � p2 C p C 1, a contradiction. Thus,S 2 �1.M/. By Lemma 76.5, ˇ1.M;S/ � p and ˇ1.G;M/ � p � 1, so

˛1.G/ � ˛1.S/C ˇ1.G;M/C ˇ1.M;S/ � p2 C p C p � 1 > p2 C p C 1

since p > 2, a contradiction. It follows that

(A�) jM 0j � p2 so ˇ1.G;M/ � p (Lemma 76.5). Therefore,

p2 C p C 1 � ˛1.G/ D ˛1.M/C ˇ1.G;M/ � ˛1.M/C p;

hence p2 � ˛1.M/ � p2 C 1 and soM is anA3-group of Theorem 76.10 orTheorem 76.11; then G is anA4-group.

By Lemma 76.12, since p > 2, we have

(B�) Any member of the set �2 is either abelian or an A1-group.

(i) Let ˛1.M/ D p2; then M is a group of Theorem 76.10 so jM 0j D p3, andˇ1.G;M/ 2 fp;p C 1g. By (B�), M is not a group of Theorem 76.10(a1) since theset �1.M/ \ �2 has no members which areA2-groups.

(i1) LetM be a group of Theorem 76.10(a), i.e., d.M/ D 3, jM 0j D p3, �1.M/ D

fF1; : : : ; Fp22Cp; Ag, where ˛1.Fi/ D p for all i and A is abelian. By (B�), Fi 62 �2,i D 1; : : : ; p2, so j�1.M/ \ �2j D 1 hence A D ˆ.G/ and d.G/ D 2. Let �1 D

fM1 D M;M2; : : : ;MpC1g; then at most one member of the set �1 is abelian andˇ1.G;M/ D

PpC1iD2 ˛1.Mi / since ˆ.G/ D A is abelian.

Let ˛1.G/ D p2 Cp; then ˇ1.G;M/ D p so jM 0j � p2 (Lemma 76.13), contraryto the first paragraph of (i).Thus, we have ˛1.G/ D p2 C p C 1. In view of ˇ1.G;M/ D p C 1 > p D

jfM2; : : : ;MpC1gj, one may assume that M2 is neither abelian nor an A1-group(Dirichlet’s principle); then ˛1.M2/ D p. It follows from

p C 1 D ˇ1.G;M/ D ˛1.M2/C

pC1X

iD3

˛1.Mi / D p C

pC1X

iD3

˛1.Mi /


that j�1j � 3 members of the set �1 are abelian and exactly one its member, sayM3,is an A1-group. Since jG0j � jM 0j D p3 > p, the set �1 has at most one abelianmember (Lemma 64.1(u)) so we get p D 3. Then �1 D fM1 D M;M2;M3;M4 D

Ag, where A is abelian. However, by Lemma 76.2, we getM 03 D M 0, a contradiction

since jM 0j D p3 > p D jM 03j.

(i2) Let M be a group of Theorem 76.10(b1,b2), i.e., d.M/ D 2, �1.M/ D

fF1; : : : ; Fp; Ag, where ˛1.Fi / D p, F 01 D D F 0

p has order p2, A is abelian.

It follows from (B�) that Fi 62 �1.M/ \ �2 for all i so A D ˆ.G/ and d.G/ D 2.Let �1 D fM1 D M; : : : ;MpC1g. As in (i1), one has to consider two possibilities:˛1.G/ 2 fp2 C p;p2 C p C 1g.Let ˛1.G/ D p2 Cp; then ˇ1.G;M/ D p so jM 0j � p2 (Lemma 76.13), contrary

to the first paragraph of (i).Now let ˛1.G/ D p2CpC1. In that case, ˇ1.G;M/ D pC1. Therefore, as in (i1),

one may assume thatM2 is neither abelian nor anA1-group. In that case, as in (i1), weget p D 3 and �1 D fM1 D M;M2;M3;M4 D Ag, whereA is abelian, ˛1.M3/ D 1.By Lemma 76.2, we getM 0

3 D M 0, a contradiction since jM 0j � p2 > p D jM 03j.

All the above, together with (A�) and (B�), yields

(C�) ˛1.M/ D p2 C 1, i.e.,M is a group of Theorem 76.11(b).

(ii) In view of (C�), Theorem 76.11(b) and Lemma 76.5, we must have ˇ1.G;M/ �

p so that ˛1.G/ D p2 C p C 1; then ˇ1.G;M/ D p.By Theorem 76.11(b), d.M/ D 2, �1.M/ D fF1; : : : ; Fp; Bg, where ˛1.Fi / D p,

i D 1; : : : ; p, ˛1.B/ D 1. By (B�), ˆ.G/ D B so d.G/ D 2. Set �1 D fM D

M1; : : : ;MpC1g; then ˇ1.Mi ; B/ � p � 1, i > 1 (Lemma 76.5). We have

˛1.G/ D ˛1.M/C

pC1X

iD2

ˇ1.Mi ; B/ � p2 C 1C .p � 1/p > p2 C p C 1;

a final contradiction.

Theorem 76.15. A 2-group G satisfying ˛1.G/ D 6 is an An-group, n 2 f2; 3g.

Proof. Assume that this is false. Then, by Theorems 76.9–76.11, G is anA4-group sothere is an A3-group H 2 �1. Since ˛1.H/ � 4 (Theorem 76.9) and ˇ1.G;H/ > 0

(Proposition 10.28), we get ˛1.H/ 2 f4; 5g so ˇ1.G;H/ 2 f2; 1g.

(i) Let ˛1.H/ D 4; then ˇ1.G;H/ D 2, jH 0j � 4 (Lemma 76.13). On the otherhand, H is a group of Theorem 76.10, and so jH 0j D 8, a contradiction.

(ii) Let ˛1.H/ D 5. Then H is a group of Theorem 76.11 so jH 0j � 4. However,ˇ1.G;H/ D 1 so jH 0j D 2 (Lemma 76.5), a final contradiction.

Appendix by Z. Janko: Nonexistence of groups of typesG7;1 andG7;2

Theorem A.1. 2-groups of type G7;1 do not exist.


Proof. Let G be a 2-group of type G7;1, i.e., G is a 2-group of Lemma 76.12 with˛1.G/ D 7. We shall use freely the notation and the results of Lemma 76.12.By Lemma 76.4(c), Hi is metacyclic for i D 1; 2; 3 since p D 2 and ˛1.Hi / D 2.

Here we note that jGj � 26 and so jHi j � 25 since R 2 �2 and R is anA2-group andso jRj � 24. In particular, R1, R2, R3, A1, A2, A3 are also metacyclic. Next, R isnonmetacyclic since d.R/ D 3.Since A1 is a maximal subgroup of A and d.A1/ D 2, we have d.A/ � 3. Assume

that d.A/ D 2. In that case, all members of the set �1 are two-generator so we canuse the results of �70. By Theorem 70.1, cl.G/ > 2. Then Theorem 70.2 implies that.G=K3.G//

0 Š E8. But K3.G/ ¤ f1g and so jG0j > 8, contrary to Lemma 76.12(d).Hence we have d.A/ D 3 and so �1.A/ Š E8.Since Hi is metacyclic and jHi j � 25 (i D 1; 2; 3), we can use Proposition 71.2.

It follows that H 0i Š C4 (Exercise 1) and Hi is isomorphic to one of the following

groups:

Hi D ha; b j a2m

D 1; m � 4; b2n

D a�2m�1

; n � 2; � D 0; 1;(5)

ab D a1C2m�2

i;

Hi D ha; b j a8 D 1; b2n

D a4�; n � 2; � D 0; 1; ab D a�1C4�; � D 0; 1i:(6)

However, in case (5) we get Ã1.Hi / D ha2; b2i (this subgroup is abelian) and wesee that ha; b2i, ha2; bi, ha2; b2ihabi are three maximal subgroups of Hi and theyare all A1-groups, contrary to ˛1.Hi / D 2 (Lemma 76.12). (Indeed, to check that anonabelian metacyclic p-group is anA1-group, is suffices, in view of Lemma 65.2(a),to show that its derived subgroup has order p. For example, if, in (5), K D ha; b2i,then ab2

D a.1C2m�2/2

D a1C2m�1

so K 0 D ha2m�1

i has order 2 and K is an A1-group.) We have proved that Hi must be isomorphic to a group given in (6). Herewe distinguish two essentially different cases � D 0 (splitting case) and � D 1 (non-splitting case).

(i) We consider first the splitting case � D 0 so that

H1 D ha; b j a8 D b2n

D 1; n � 2; ab D a�1z� ; z D a4; � D 0; 1i;

and we also set u D b2n�1

and v D a2 so that v2 D z and u are involutions. SincejH1j D 2nC3, we have jGj D 2nC4, n � 2. But A1 D hb2i � hai is the uniqueabelian maximal subgroup of H1 and so A \ H1 D A1. The fact that d.A/ D 3

implies that there is an involution t 2 A � A1 so that �1.A/ D hu; z; ti because�1.H1/ D hu; zi < �1.A/. We have Z.H1/ D hb2; zi D Z.G/ (indeed, Z.G/ � H1

and jZ.G/j D jZ.H1/j) and ˆ.H1/ D hb2i � hvi D ˆ.G/ since ˆ.H1/ � ˆ.G/ andd.G/ D 3; in particular, u D b2n�1

2 Z.G/. Also,H 01 D hvi Š C4 is normal inG and

jG0j D 8 (Lemma 76.12). Sinceˆ.G/ D hb2i�hvi andH1 > G0 > hvi, it follows, by

the modular law, thatG0 D hui�hvi is abelian of type .2; 4/. Note thatR1,R2,R3 are


metacyclic of order 2nC2 because �1.H1/ D fA1; R2; R3g, �1.H2/ D fA2; R1; R3g

andH1 andH2 are metacyclic of order 2nC3.We have G D hH1; ti D ha; b; ti, hŒa; b�i D hvi, Œa; t � D 1 (recall that a; t 2 A)

and, since G=hvi is nonabelian, we get Œb; t � 2 G0 � hvi. On the other hand, �1.A/ D

hu; z; ti is normal in G and so Œb; t � D tbt 2 �1.A/. It follows that Œb; t � D uzı

with ı 2 f0; 1g since Œb; t � D b�1bt 2 H1 \�1.A/ D �1.H1/ and Œb; t � 2 G0 � hvi

(recall that z D v2). Also note thatR 2 �2 is anA2-group of order 2nC2 and thereforefour A1-subgroups S1, S2, S3, S4 which are contained in R are of order 2nC1. Since˛1.G/ D 7, fS1; S2; S3; S4; R1; R2; R3g is the set of all A1-subgroups in G. Inparticular, allA1-subgroups of G of order < 2nC2 are contained in R.Let ı D 1 so that Œb; t � D uz. Since uz 2 hb2; zi D Z.G/, huzi G G and so

hb; ti=huzi is abelian which implies that hb; ti0 D huzi Š C2. It follows that hb; ti

is an A1-subgroup (Lemma 65.2(a)) containing hb2n�1

D u; z; ti Š E8 and so hb; ti

is nonmetacyclic. Also, hb; ti is of order � 2nC2 (since hu; z; ti \ hbi D hui) and sohb; ti is of order 2nC2 and therefore hb; ti is one of R1, R2, R3 (see the last sentenceof the previous paragraph). This is a contradiction since R1, R2, R3 are metacyclic.Suppose that ı D 0; then Œb; t � D u D b2n�1

is an involution, and therefore wehave either hb; ti Š M2nC1 .n > 2/ or hb; ti Š D8 .n D 2/. It follows that hb; ti is anA1-subgroup of order 2nC1 and so hb; ti < R. On the other hand, R > ˆ.G/ and soR � ˆ.G/hb; ti. But ˆ.G/hb; ti 2 �1. This is a contradiction since R 2 �2.

(ii) We consider now the non-splitting case � D 1 so that

H1 D ha; b j a8 D 1; b2n

D a4 D z; n � 2; ab D a�1z�; � D 0; 1i

and we set v D a2, w D b2n�1

and u D vw. We see again that A1 D hb2; ai D

A\H1 is the unique abelian maximal subgroup inH1. Since d.A/ D 3 and�1.A1/ D

�1.H1/ D hu; zi, there is an involution t 2 A�A1 such that�1.A/ D hu; z; ti Š E8.Also, jH1j D 2nC3, jGj D 2nC4 and the metacyclicA1-subgroups R1, R2, R3 are oforder 2nC2. Since R 2 �2 and R is an A2-group with ˛1.R/ D 4, four A1-subgroupsS1, S2, S3, S4, which are contained in R, are of order 2nC1.We have Z.H1/ D hb2i Š C2n , Z.H1/ D Z.G/, ˆ.G/ D ˆ.H1/ D hb2ihvi since

ˆ.H1/ � ˆ.G/ and d.G/ D 3. Also, H 01 D hvi Š C4 is normal in G, jG0j D 8

(Lemma 76.12) and so G0 D hv;wi D hui � hvi is abelian of type .4; 2/.We have G D hH1; ti D ha; b; ti, hŒa; b�i D hvi, where t 2 �1.A/ � �1.H1/,

Œa; t � D 1 and, since G=hvi is nonabelian (recall that o.v/ D 4 < 8 D jG0j), we getŒb; t � 2 G0 � hvi. On the other hand, �1.A/ D hu; z; ti is normal in G and so

b�1bt D Œb; t � D tbt 2 �1.A/ \H1 D �1.H1/ D �1.G0/ D hui � hzi:

Since z D v2 2 hvi, it follows that Œb; t � D u or uz. Note that o.b/ D 2nC1 � 8 andso, taking into account that ub D .vw/b D v�1w D vwz D uz, we have ˆ.G/ D

hb2; vi < hb; ui Š M2nC2 noting that b2n�1

D w, v D uw�1 and so v 2 hb; ui.Therefore hb; ui is an A1-subgroup of order 2nC2 contained in H1 and consequently


hb; ui D R2 or R3. Since t normalizes hb; ui, we have jhb; ti W hb; uij D 2 andso hb; ti is of order 2nC3 which implies that hb; ti is a maximal subgroup of G withhb; ti0 � hz; ui and so hb; ti0 D hz; ui Š E4. It follows that hb; ti is a nonmetacyclicmember of the set �1. Hence hb; ti D F2 or F3 and so hb; ti is an A3-group. On theother hand, QA D hb2i � hui � hti is an abelian nonmetacyclic subgroup of order 2nC2

in hb; ti. Hence QA is the unique abelian maximal subgroup of hb; ti and so QA D A2 orA3 (see, in Lemma 76.12, lists of the members of the sets �1.F2/ and �1.F3/). Thisis a contradiction since A1, A2, A3 are metacyclic.

Definition 2. A 2-groupG is said to be a group of typeG7;2, if it satisfies the followingconditions:

(1G7;2) d.G/ D 2, �1 D fH1;H2;H3g, ˛1.H1/ D 4, ˛1.H2/ D 2, ˛1.H3/ D 1.

(2G7;2) �1.H1/ D fF1; : : : ; F6; Ag, where ˛1.Fi / D 2, i � 6, A D ˆ.G/ isabelian, jH 0

1j D 8. We have ˇ1.G;H1/ D 3 so ˛1.G/ D 7.

(3G7;2) jG W Z.G/j D 25, Z.G/ D Z.H1/ < ˆ.H1/.

(4G7;2) H 01 D H 0

2 �H 03, G=H

01 is anA1-group so jG0j D 24, ˇ1.G=H

02;H1=H

02/ D

1, ˇ1.G=H03;H1=H

03/ D 2.

It follows from Definition 2 that a 2-group G of type G7;2, if it exists, is an A4-group. It will be proved in the Appendix that groups of type G7;2 do not exist.

Theorem A.2. Groups of type G7;2 do not exist.

Proof. We prove first that a two-generator 2-group X which is an A2-group is meta-cyclic. Indeed, if jX j > 24, then the result follows from Lemma 76.4(c). So letjX j D 24 and let Y be a nonabelian subgroup of order 23. Since d.X/ D 2, we haveCX .Y / � Y . It follows from Lemma 64.1(i) that X is of maximal class and so X ismetacyclic and we are done.Let G be a 2-group of type G7;2, where we use the notation from Definition 2.

By Lemma 76.6(a), H2; F1; : : : ; F6 are all two-generator A2-groups. Therefore, thegroups H2; F1; : : : ; F6 are all metacyclic. But A D ˆ.G/ < H2 and so A is alsometacyclic. Hence H1 is minimal nonmetacyclic and so, by Theorem 66.1, H1 is anA2-group of order 25. This is a contradiction sinceH1 contains a proper subgroup F1

which is anA2-group.

Now we are ready to prove

Theorem 76.16. If a 2-group G satisfies ˛1.G/ D 7, then G is an An-group, n 2

f2; 3g.

Proof. Let G be a counterexample of minimal order. If H 2 �1, then ˛1.H/ <

˛1.G/ D 7, so, by Lemma 76.6 and Theorems 76.9–76.11, 76.15, H is either abelianor an An-group, n � 3. Therefore, G is an A4-group so one can choose H 2 �1 sothat H is an A3-group. In that case, 4 � ˛1.H/ � 6 (Theorem 76.9) so H is one of


theA3-groups of Theorems 76.10, 76.11 or 76.15. Since G7;1-groups do not exist, byAppendix, it follows from Lemma 76.12 that

(i) The set �2 has no members which are A2-groups. It follows that H is not agroup of Theorems 76.10(a1) and 76.11(c).

(ii) Let H 2 �1 be a group of Theorem 76.10(a2,b1,b2). In that case, ˛1.H/ D 4,jH 0j D 8, the set �1.H/ has exactly one abelian member A and all other its membersare A2-groups. It follows from (i) that A D ˆ.G/ so, in view of jG W Aj D 22, weget d.G/ D 2, and then Z.G/ < ˆ.G/ D A. Set �1 D fH1 D H;H2;H3g; thenA 2 �1.Hi /, i D 1; 2; 3, so

˛1.H2/C ˛1.H3/ D ˇ1.G;H/ D ˛1.G/ � ˛1.H/ D 7 � 4 D 3:

One may assume that ˛1.H2/ D 2 and ˛1.H3/ D 1 (˛1.H2/ ¤ 3 since the set�1.H2/ has one abelian member). Thus,H2 is anA2-group (Lemma 76.6(a)) andH3

is anA1-group. We have d.H2/ D 2 D d.H3/ (Lemmas 76.1 and 76.6(a)).Assume thatH is a group from Theorem 76.10(b1,b2); then d.H/ D 2. In that case,

by the above, G and all members of the set �1 are two-generator so G is metacyclic(Lemma 64.1(n)). SinceH2 is anA2-group andH3 is anA1-group, Exercise 1 showsthat G is anA3-group, a contradiction.Thus,H is a group from Theorem 76.10(a2). Then d.H/ D 3, jH 0j D 8, �1.H/ D

fF1; : : : ; F6; Ag, where ˛1.Fi / D 2 and d.Fi / D 2 for i � 6 and A is abelian,jH W Z.H/j D 2jH 0j D 16 (Lemma 64.1(q)). By �65, H2=Z.H2/ 2 fD8;C4 � C4g

since d.H2/ D 2.Suppose that H2=Z.H2/ Š D8. Then Z.H2/ � ˆ.H2/ � ˆ.G/ D A since

d.H2/ D 2, and so jA W Z.H2/j D 4. Since H3 is an A1-group, we get Z.H3/ D

ˆ.H3/ < ˆ.G/ D A and jA W ˆ.H3/j D 2. Then Z.H2/ \ Z.H3/ � Z.G/ and jA W

Z.G/j � 8 so jG W Z.G/j D jG W AjjA W Z.G/j � 25. By the above, jH W Z.G/j D 24

so Z.G/ D Z.H/ has index 25 in G.Now suppose thatH2=Z.H2/ is abelian of type .4; 4/. We have Z.H2/ < ˆ.H2/ <

ˆ.G/ < H3. Assume that Z.H2/ 6� Z.H3/. ThenH3=Z.H2/ is cyclic since Z.H3/ D

ˆ.H3/ (Lemma 76.1). Thus, G=Z.H2/ has the cyclic subgroupH3=Z.H2/ of index 2and the abelian subgroup H2=Z.H2/ of type .4; 4/ and index 2, which is impossible.It follows that Z.H2/ � Z.H3/ so Z.H2/ � Z.G/, and we get jG W Z.G/j D jG W

H2jjH2 W Z.H2/j � 2 24 D 25. Since Z.G/ < ˆ.G/ < H2, we get Z.H2/ D Z.G/and so jG W Z.G/j D 25. Comparing the orders, we get Z.G/ D Z.H/.Since jH 0j D 8 and jF 0

i j D 4, the equality FiZ.G/ D H is impossible for i � 6.It follows that Z.G/ < Fi for all i . In particular, Z.G/ < ˆ.H/. We see that G is agroup of type G7;2. By the Appendix, groups of type G7;2 do not exist, and this is acontradiction.

(iii) Now let ˛1.H/ D 22 C 1, i.e., H is a group of Theorem 76.11. As we havenoticed in (i), H is not a group of Theorem 76.11(c). We have jH 0j > 2 (Theorem76.11) and ˇ1.G;H/ D 2.


(iii1) Let H 2 �1 be a group of Theorem 76.11(a), i.e., d.H/ D 2, jH 0j D 4,�1.H/ D fF1; F2; Ag, where d.F1/ D 3, ˛1.F1/ D 4, jF 0

1j D 2, ˛1.F2/ D 1 and Ais abelian. If d.G/ D 3, then F1 2 �1.H/ � �2, contrary to (i). Thus, d.G/ D 2 so�1 D fH1 D H;H2;H3g. We have ˆ.G/ 2 fF2; Ag, by (i).Assume that F2 D ˆ.G/. Then

7 D ˛1.G/ D ˛1.H1/C

3X

iD2

ˇ1.Hi ; F2/ D 5C

3X

iD2

ˇ1.Hi ; F2/

so ˇ1.H2; F2/ C ˇ1.H3; F2/ D 2. By Remark 1, we get ˇ1.Hi ; F2/ D 1, i D 2; 3.In that case, ˛1.Hi / D 2 so jH 0

i j D 4, d.Hi / D 2, i D 2; 3 (Lemma 76.6(a)).Thus, G and all members of the set �1 are two-generator so G is metacyclic (Lemma64.1(n)). Then, by Exercise 1, applied to H2 and H3, G is anA3-group so G is not acounterexample.Therefore, we must have ˆ.G/ D A. Then we have ˛1.H2/ C ˛1.H3/ D 2. If

˛1.Hi / D 1, i D 2; 3, then G and all members of the set �1 are two-generator so Gis metacyclic (Lemma 64.1(n)). In that case, by Exercise 1, G must beA2-group so itis not a counterexample.Thus, ˛1.H2/ D 2 so jH 0

2j D 4 andH3 is abelian. By Lemma 76.2, H 0 D H 02 and

G=H 0 is anA1-group. It follows that jG0j D 2jH 0j D 8 so jG W Z.G/j D 16 (Lemma64.1(q)). Since d.G/ D 2, we get Z.G/ < ˆ.G/ < Hi so jHi W Z.G/j D 8, i D

1; 2; 3. SinceH1=Z.G/ is noncyclic and two-generator. we get Z.G/ < ˆ.H1/ < Fi ,i D 1; 2. Then jFi W Z.G/j D 4, i D 1; 2. Since d.F1/ D 3 and Z.F1/ > ˆ.F1/, weget, by Lemma 76.8(c), ˛1.F1/ � 2 22 � 1 D ˛1.G/, contrary to Proposition 10.28.

(iii2) Let H 2 �1 be a group from Theorem 76.11(b). Retaining the notation ofTheorem 76.11, we get, in view of (i), ˆ.G/ D B so d.G/ D 2. Set �1 D fH1 D

H;H2;H3g. Then, as above, ˇ1.Hi ; R/ D 1, and we conclude that ˛1.Hi / D 2 andd.Hi / D 2 for i D 2; 3 (Lemma 76.6(a)). In that case, G and all members of theset �1 are two-generator so G is metacyclic (Lemma 64.1(n)). Then, by Exercise 1,applied toH2 andH3, G is anA3-group so G is not a counterexample.

(iv) It remains to consider the case where H 2 �1 is a group of Theorem 76.15,i.e., ˛1.H/ D 6. In that case, ˇ1.G;H/ D 1 so jH 0j D 2 and jG0j D 4, d.G/ D 2,�1 D fH1 D H;H2; Ag, where H2 is an A1-group and A is abelian (Lemma 76.5).By Lemma 64.1(q), jG W Z.G/j D 8 and, since Z.G/ < ˆ.G/ < H , we get jH W

Z.G/j D 4 so jH W Z.H/j D 4. SinceH is not anA1-group, we get d.H/ D 3. Then,by Lemma 76.8(c), ˛1.H/ � 2 22 � 1 > 6, a final contradiction.

Theorem A follows immediately from Lemma 76.6, Theorems 76.9–76.11, 76.14–76.16 and the Appendix.If G is a 2-group of maximal class and order 26, then ˛1.G/ D 8 and G is an

A4-group. I do not know if, for odd p, there exists an A4-group G with ˛1.G/ D

.p2 C p C 1/C 1.


If G is a nonabelian p-group such that for each K 2 �2, we have ˛1.�K1 / < p

2 C

2pC1, thenG is anAn-group, n � 4. Assume that this is false. Then there isH 2 �1

such that H is an Ak-group with k � 4. By Theorem A, ˛1.H/ � p2 C p C 2. IfK 2 �1.H/ is abelian, then, since the set �K

1 has at most one abelianmember, we have˛1.�

K1 / D ˛1.H/C

P

M 2�K1

�fHg ˛1.M/ � .p2 CpC2/C .p�1/ D p2 C2pC1

(Remark 1), a contradiction. If K is nonabelian, then, by Remark 1,

˛1.�K1 / D ˛1.H/C

X

M 2�K1

�fHg

ˇ1.M;K/

� .p2 C p C 2/C p.p � 1/ D 2p2 C 2 > p2 C 2p C 1;

a final contradiction.

3o. A lower estimate of ˛1.G/ in terms of d.G/. To facilitate the proof of Theo-rem B, we begin with the following remarks.

Remarks. 10. Let G be a nonabelian p-group, d.G/ D 3 and H 2 �1. Let us provethat ˇ1.G;H/ � p.p � 1/. This is the case if H is abelian since then ˇ1.G;H/ D

˛1.G/ � p2 > p.p � 1/ (Theorem 76.9). Next we assume that H is nonabelian.If ˛1.G/ D p2, then ˛1.H/ � p (Theorem 76.10) so ˇ1.G;H/ � p2 � p D

p.p � 1/. Now let ˛1.G/ � p2 C 1. The result is true if ˛1.H/ � p C 1. So,in view of Lemma 76.6 and Theorem 76.9, one may assume that ˛1.H/ � p2.Take R 2 �1.H/ \ �2. Assuming that R is nonabelian, we get, using Remark 1,ˇ1.G;H/ �

P

F 2�R1

�fHg ˇ1.F;R/ � p.p � 1/ since all p C 1 members of the set

�R1 are nonabelian. If R is abelian, then the set �R

1 has at least p nonabelian membersso, by Lemma 76.6, ˇ1.G;H/ �

P

F 2�R1 �fHg ˛1.F / � p2 � .p � 1/p.

11. Suppose that G is a p-group such d D d.G/ � 3. We claim that if H 2 �1

is nonabelian, then ˇ1.G;H/ � pd�2.p � 1/. We use induction on d . Remark 10is the basis of induction so assume that d > 3. If X 2 �1, then d.X/ � d � 1 �

3. Since j�1.H/ \ �2j D pd�2 C C p C 1 > p C 1, the set �1.H/ \ �2

has a nonabelian member R (Exercise 1.6(a)). Then all members of the set �R1 D

fH1 D H; : : : ;HpC1g are nonabelian. Let d.Hi / D di , where di � d � 1. Byinduction, ˇ1.Hi ; R/ � pdi �2.p�1/ � pd�3.p�1/, i > 1. Therefore, ˇ1.G;H/ �PpC1

iD2 ˇ1.Hi ; R/ � pd�3.p � 1/ p D pd�2.p � 1/, as required.

Theorem B. Suppose that a p-group G is neither abelian nor an A1-group and d D

d.G/. Then ˛1.G/ � pd�1.

Proof. By Lemma 76.6 and Theorem 76.9, the theorem is true for d � 3 so wemay assume in what follows that d > 3. If X 2 �1, then d.X/ � d � 1. Weproceed by induction on d . If H 2 �1 is nonabelian and such that d.H/ � d ,then, by induction, ˛1.H/ � pd�1, and we are done. Therefore, one may as-sume that, for all nonabelian H 2 �1, we have d.H/ < d ; then, for such H we


have d.H/ D d � 1.� 3/. It follows from Lemma 76.3 that there is a nonabelianR 2 �2. Set �R

1 D fH1; : : : ;HpC1g. By induction, ˛1.Hi / � pd�2 and, byRemark 11, ˇ1.Hi ; R/ � pd�3.p � 1/ for all i . Therefore, ˛1.G/ � ˛1.H1/ CPpC1

iD2 ˇ1.Hi ; R/ � pd�2 C pd�3.p � 1/ p D pd�1.

4o. Groups of exponent p. We assume that all groups, considered in this subsection,have exponent p. If G is anA1-group of exponent p, then jGj D p3.Now let G be an A2-group; then jGj D p4. If the set �1 has exactly one abelian

member, then cl.G/ D 3 and ˛1.G/ D p. If the set �1 has two distinct abelianmembers, then G D S � C and ˛1.G/ D p2, where jC j D p.

Remarks. 12. Suppose that G is a nonabelian group of exponent p and order � p5

andH 2 �1. By Proposition 2.3, p2 j ˛1.H/, p2 j ˛1.G/ so p2 j ˇ1.G;H/.

13.We claim that if G is nonabelian of order p6 and exponent p, then ˛1.G/ � p3.Let R 2 �2 be such that R ¤ Z.G/. Set �R

1 D fH1; : : : ;HpC1g. If R is nonabelian,

then, by Remark 12 and Proposition 10.28, ˛1.G/ � ˛1.R/ CPpC1

iD1 ˇ1.Hi ; R/ >

p2.p C 1/ > p3. If R is abelian, one may assume that H1; : : : ;Hp are nonabelianso ˛1.Hi / � p2 .i D 1; : : : ; p/, by Proposition 2.3. In that case we have ˛1.G/ �Pp

iD1 ˛1.Hi / � p2 p D p3. Now let R D Z.G/. Then G D A �E, where A is anA1-group and E Š Ep3 so, by Remark 3, ˛1.G/ D p6.

Theorem C. If G is nonabelian of order pm and exponent p, m > 3, then:

(a) IfH 2 �1, then ˇ1.G;H/ � pm�4.p � 1/.

(b) ˛1.G/ � pm�3.

(c) If ˛1.G/ D pm�3, then G is of maximal class with abelian subgroup of index pand m � p.

Proof. In all three cases we proceed by induction on m.

(a) One may assume that m > 4, in view of Remark 1. Let H 2 �1 be nonabelianand R 2 �1.H/ \ �2. Then the set �R

1 D fH1 D H; : : : ;HpC1g has at most oneabelian member.Suppose that HpC1 is abelian; then R is also abelian. In that case, by induction,

ˇ1.G;H/ �Pp

iD2 ˛1.Hi / � p.m�1/�3 .p � 1/ D pm�4.p � 1/.Now suppose that the set �R

1 has no abelian members. Then, by induction, we get

ˇ1.G;H/ �PpC1

iD2 ˇ1.Hi ; R/ � p.m�1/�4.p � 1/ p D pm�4.p � 1/, and (a) isproven.

(b) In view of Remarks 12 and 13, one may assume that m > 6. There is R 2 �2

such that R ¤ Z.G/ since G is neither abelian nor A1-group. Set

�R1 D fH1; : : : ;HpC1g:


Suppose that R is abelian. One may assume in that case that H1; : : : ;Hp are non-abelian. By induction, we have

˛1.G/ D

pX

iD1

˛1.Hi / � p.m�1/�3 p D pm�3:

Now let R be nonabelian. Then, by induction and (a), we have

˛1.G/ D ˛1.R/C

pC1X

iD1

ˇ1.Hi ; R/ � p.m�2/�3 Cp.m�1/�4.p� 1/.pC 1/ D pm�3;

completing the proof of (b).

(c) Let ˛1.G/ D pm�3. As in (b), we prove that there is an abelian A 2 �1.

(i) Let jZ.G/j D p. Then G is of maximal class (Remark 5) and, by Theorem9.5, since exp.G/ D p, we must have m � p. It is easy to check that then indeed˛1.G/ D pm�3 (see Remark 5). Next we assume that jZ.G/j > p.

(ii) As in Remark 13, we have jG W Z.G/j > p2. Therefore, there is a nonabelianH1 2 �1 containing Z.G/ as a subgroup of index > p. Take Z.G/ < R 2 �1.H1/ \

�2 and set �R1 D fH1; : : : ;HpC1g. One may assume thatH1; : : : ;Hp are nonabelian.

It follows from Z.G/ < Hi and jZ.G/j > p that Hi is not of maximal class so, byinduction, ˛1.Hi / > pm�4. If HpC1 is abelian, then R is also abelian, and we have˛1.G/ �

PpiD1 ˛1.Hi / > p

m�4 p D pm�3, contrary to the hypothesis. If HpC1 isnonabelian, then, by (a), we get

˛1.G/ � ˛1.H1/C

pC1X

iD2

ˇ1.Hi ; R/ > pm�4 C pm�5.p � 1/ p D pm�3;

a contradiction.

5o. M3-groups. A p-group G is said to be an M3-group, if all its A1-subgroupshave the same order p3. The groups of maximal class with abelian subgroup of indexp areM3-groups (Remark 5). If anA2-group G is also anM3-group, then jGj D p4

and ˛1.G/ 2 fp;p2g.

Theorem C1. Let G be a nonabelian M3-group of order pm and exponent p, m > 3.Then:

(a) IfH 2 �1, then ˇ1.G;H/ � pm�4.p � 1/.

(b) ˛1.G/ � pm�3.

(c) If, in addition, ˛1.G/ D pm�3, then G is of maximal class with abelian sub-group of index p.


Theorem C2. Let G be a nonabelian group of order pm. Suppose that G has an A1-subgroup of order pa but has no A1-subgroups of order > pa. Then ˛1.G/ � pm�a.If, in addition, ˛1.G/ D pm�a, then all A1-subgroups of G have the same order pa.

The proofs are omitted since they are repetitions of the proof of Theorem C. Forp D 2, see �90. See also Problem 155.

6o. p-groups with few conjugate classes of A1-subgroups. Given a p-group G, let�1.G/ denote the number of conjugate classes of A1-subgroups in G.GivenH 2 �1, we define (i) N

1.G;H/ is the number of conjugateG-classes ofA1-subgroups not contained inH , (ii) N�1.H/ is the number ofG-classes ofA1-subgroupscontained in H . We have N�1.H/ � �1.H/ and, as a rule, the strong inequality holds.Obviously, N�1.G/ D �1.G/.

Theorem 76.17 (compare with Lemma 76.5). Let G be a p-group and letH 2 �1 benonabelian. Then N

1.G;H/ � p � 1. If N1.G;H/ D p � 1, then:

(a) d.G/ D 2 and �1 D fH1 D H; : : : ;Hp ; Ag, where A is the unique abelianmember of the set �1.

(b) H 01 D D H 0

p ,G=H01 is anA1-group so jG0j D pjH 0

1j and d.Hi / � 3, i � p.

(c) If jH 0i j D p for some i > 1, then H2; : : : ;Hp are A1-groups and ˇ1.G;H/ D

p � 1.

Lemma 76.18. Let G be a p-group and let a nonabelian H 2 �1 be not an A1-subgroup. Suppose that N�1.H/ D 1. Then (a) H has no proper nonabelian G-invariant subgroups and the set �1.H/ has exactly one abelian member, (b) d.G/ D 2,(c) jH 0j > p.

Proof. Assume that a nonabelian A 2 �1.H/ is G-invariant and let U � A be anA1-subgroup. By Proposition 10.28, H has an A1-subgroup V that is not containedin A. Since UG � A, U and V are not conjugate in G, a contradiction. This provesthe first assertion of (a).Assume that jH 0j D p; then d.H/ > 2 since H is not an A1-group (Lemma

65.2(a)). Let U < H be an A1-subgroup. Then U 0 D H 0 so U G H . It follows thatH D NG.U / so jG W NG.U /j D p. In that case, H contains exactly p subgroupsconjugate with U in G so, by hypothesis, ˛1.H/ D p. However, by Theorem B,˛1.H/ � p2, contrary to what has just been said. Thus, jH 0j > p, proving (c).The set �1.H/ has exactly one abelian member (otherwise, jH 0j D p), and this

completes the proof of (a). Then, since all members of the set �1.H/\�2 are abelian,we get j�1.H/ \ �2j D 1 (Lemma 64.1(q)), hence d.G/ D 2, completing the proofof (b).

Proof of Theorem 76.17. Let N 2 �2 \ �1.H/; then N 6� Z.G/ since H is non-abelian. Set �N

1 D fH1 D H; : : : ;HpC1g. Since the set �N1 has at most one abelian

member, one may assume that H1; : : : ;Hp are nonabelian. By Proposition 10.28,


there exists an A1-subgroup Bi � Hi such that Bi 6� N , i D 2; : : : ; p. For i ¤ j ,we have BG

i N D Hi ¤ Hj D BGj N so BG

i ¤ BGj , and we conclude that Bi and Bj

are not conjugate in G. Thus, N1.G;H/ � jfB2; : : : ; Bpgj D p � 1.

Now let N1.G;H/ D p � 1. For i D 2; : : : ; p, all A1-subgroups of Hi that are

not contained in N , are G-conjugate with Bi so HpC1 must be abelian (otherwise,N1.G;H/ > p� 1); thenN is also abelian. In that case, N�1.Hi / D 1 for i D 2; : : : ; p

so, by Lemma 76.18(b), d.G/ D 2; then �1 D �N1 and N D ˆ.G/. By Lemma

76.2, H 01 D D H 0

p and G=H01 is an A1-subgroup so jG0j D pjH 0

1j and, sinceH 0

i � ˆ.Hi /, we get d.Hi / D d.Hi=H0i / � 3 for i D 2; : : : ; p (Lemma 76.1). Now,

(c) follows easily from Lemmas 65.2(a) and 76.18.

IfG is a group of maximal class and order pn > p3 with abelian subgroup of indexp, then �1.G/ D p and, ifH 2 �1 is nonabelian, then N.G;H/ D p � 1.

Corollary 76.19 (compare with Lemma 76.6(b)). Let G be a p-group. If �1.G/ D

p C 1, then d.G/ D 2 and ˆ.G/ is abelian.

Proof. Since G is neither abelian nor anA1-group, there existsR 2 �2 such that R 6�

Z.G/ (Lemma 76.3). Set �R1 D fH1; : : : ;HpC1g; then at most one member of the set

�R1 is abelian. Suppose thatH1 is nonabelian; then N

1.G;H1/ � p, by hypothesis. IfN1.G;H1/ D p� 1, then d.G/ D 2 and R D ˆ.G/ is abelian (Theorem 76.17). Nowwe assume that N

1.G;H1/ D p; then N�1.H1/ D 1 (otherwise, �1.G/ > p C 1). Inthat case, d.G/ D 2 andH1 has no proper nonabelian G-invariant subgroups (Lemma76.18(a,b)) so ˆ.G/, as a proper G-invariant subgroup of H1, is abelian.

Proposition 76.20. Suppose that a p-group G is neither abelian nor an A1-group.Let all proper nonabelian normal subgroups of G be members of the set �1. Thend.G/ � 3 and one of the following holds:

(a) d.G/ D 2 and either cl.G/ D 2 or G=G0 has a cyclic subgroup of index p.

(b) d.G/ D 3. If R is a G-invariant subgroup of index p in G0, then G=R is anA2-group.

Proof. Suppose that d.G/ > 2. By hypothesis, all members of the set �2 are abelianso ˆ.G/ � Z.G/ and d.G/ D 3 (Lemma 76.3). Let R be a G-invariant subgroup ofindex p in G0. We claim that G=R is an A2-group. Without loss of generality, onemay assume that R D f1g; then jG0j D p. It follows from d.G/ D 3 that G is not anA1-group (Lemma 76.1). Let B < G be anA1-subgroup. In view of B 0 D G0 we getB GG so B 2 �1, by hypothesis, and G is anA2-group.Now suppose that d.G/ D 2. Let M=G0 < G=G0 be of index p2. Then, by

hypothesis, M is abelian. It follows that M � CG.G0/. If G is not of class 2,

then G=G0 is not generated by subgroups of index p2 so it has a cyclic subgroup ofindex p.

Proposition 76.21. Let G be a nonabelian p-group.


(a) (Compare with Theorem 76.9) If d.G/ > 2. then �1.G/ � p2.

(b) If d.G/ > 3, then �1.G/ > p2 C p C 1.

Proof. (a) Suppose that all members of the set �2 are abelian. Then d.G/ D 3 andˆ.G/ � Z.G/ (Lemma 76.3). Let fM1; : : : ;Msg be the set of all nonabelian membersof the set �1; then s � p2 (Exercise 1.6(a)). It remains to show that �1.G/ � s. LetAi � Mi be an A1-subgroup; then Mi D Aiˆ.G/ is the unique member of the set�1, containing Ai , i D 1; : : : ; s. It follows that A1; : : : ; As are pairwise not conjugatein G, and we are done in this case.Now suppose that the set �2 has a nonabelian member N . Take T 2 �1.N / \ �3;

then T — Z.G/. Since G=T is generated by any p C 2 subgroups of order p, itfollows that T is contained in at most p C 1 abelian members of the set �2. In thatcase, there are p2 distinct subgroups L1=T; : : : ; Lp2=T of order p in G=T such thatL1; : : : ; Lp2 are nonabelian; obviously, Li 2 �2 for all i . Let Ai � Li be an A1-subgroup not contained in T (Proposition 10.28), i D 1; : : : ; p2; then Li D AiT .Since A1; : : : ; Ap2 are pairwise not conjugate in G, we get �1.G/ � p2.

(b) Let d.G/ > 3. Suppose that all members of the set �3 are abelian. Then allmembers of the set �4 are contained in Z.G/ so d.G/ D 4 (otherwise,G D hH j H 2

�4i is abelian). If jG W Z.G/j D p2, then G D BZ.G/ for eachA1-subgroup B < G

so allA1-subgroups are normal in G. Therefore, �1.G/ D ˛1.G/ � p3 (Theorem B).Since, p3 > p2 C p C 1, we are done in this case. Thus, jG W Z.G/j 2 fp3; p4g. Itfollows that at least j�3j�1 D p3 Cp2 Cp members of the set �3 are not contained inZ.G/. Let T1; T2 and T3 be those distinct members of the set �3 that are not containedin Z.G/. Since any p C 2 distinct subgroups of order p generate G=Ti , it followsthat at least p2 members of the set �Ti

2 are nonabelian, i D 1; 2; 3. Let i ¤ j ,

i; j � 3 and K 2 �Ti

2 \ �Tj

2 . Then K D TiTj is determined uniquely. It follows

that the set M DS3

iD1 �Ti

2 contains at least 3p2 � 3 distinct nonabelian members.Let M0 D fH1; : : : ;Hkg be the set of all nonabelian members in the set M, wherek � 3p2 � 3, and let Ls � Hs be an A1-subgroup, s � k. Since the intersection ofany two distinct members of the setM0 is abelian,Hs in the unique member of the setM

0 containing Ls , s � k. It follows that L1; : : : ; Lk are not pairwise G-conjugate so�1.G/ � k � 3p2 � 3 > p2 C p C 1.Now suppose that the set �3 has a nonabelian member T . Set �T

2 D fH1; : : : ;Hkg,k D p2 C p C 1. Let Li � Hi be an A1-subgroup not contained in T , i D 1; : : : ; k

(Li exists, by Proposition 10.28), and let L � T be anA1-subgroup. We have LiT D

Hi for i � k. Since L;L1; : : : ; Lk are pairwise not conjugate in G, we get �1.G/ �

k C 1 > k D p2 C p C 1.

Let G be a nonabelian p-group. GivenM � �1, let �1.M/ be the number of con-jugate G-classes of A1-subgroups contained in the members of the setM (obviously,�1.�1/ D �1.G/, unless G is anA1-group).


Remarks. 14. Suppose that every A1-subgroup is contained in the unique maximalsubgroup of a nonabelian p-group G. Then ˆ.G/ is abelian and one and only one ofthe following holds: (i) d.G/ D 2, (ii) d.G/ D 3 and ˆ.G/ � Z.G/. Indeed, thegroups from (i) and (ii) satisfy the hypothesis (if H < G is an A1-subgroup, then, inboth cases, Hˆ.G/ is the unique maximal subgroup of G containing H ). Now let Gsatisfy the hypothesis. Then all members of the set �2 are abelian. By Lemma 76.3,d.G/ � 3 and, if d.G/ D 3, then ˆ.G/ � Z.G/.

15. Suppose that G is a nonabelian p-group with d.G/ > 2. We claim that thefollowing assertions are equivalent: (a) �1.�

K1 / � p for all K 2 �2, (b) G is an

A2-group with ˛1.G/ D p2. Let (b) holds. Then all nonabelian members of theset �1 are A1-subgroups and jG W Z.G/j D p2 (Lemma 76.4(d)). If K 2 �2 �

fZ.G/g, then KZ.G/ 2 �1 so �1.�K1 / D p and (b) ) (a). It remains to prove the

reverse implication. Assume that H1 2 �1 is neither abelian nor minimal nonabelian.Then there is a nonabelian K1 2 �1.H1/. In that case, ˛1.�

K1 // � p C 2 > p, a

contradiction. Thus H1 does not exist so G is A2-group.

Exercise 5. LetG be a p-group and let the set �1 have exactly p nonabelian members.Then: (a) �1 D fH1; : : : ;Hp ; Ag, where A is abelian. (b) H 0

1 D D H 0p has index

p in G0.

Exercise 6. Let G be a nonabelian two-generator p-group and jG W G0j D pn > p2.Prove that if G has only one normal subgroup of index pn, then G=G0 is abelian oftype .pn�1; p/ and exp.G/ � pn. (Hint. Consider the quotient group G=R, where Ris a G-invariant subgroup of index p in G0. Use Lemma 65.2(a).)

�77

2-groups with a self-centralizingabelian subgroup of type .4; 2/

The results of this section are taken from [Jan9]. Here we classify finite 2-groups Gwhich possess an abelian subgroup A Š C4 � C2 such that CG.A/ D A. If A isnormal in G, then G=A is isomorphic to a subgroup of Aut.A/ Š D8. Therefore wemay assume in the sequel that such a subgroup A is not normal in G. We investigatefirst the case where A � ˆ.G/ and prove the following basic result.

Theorem 77.1. Let G be a 2-group which possesses a self-centralizing abelian sub-group A of type .4; 2/ but G does not possess any self-centralizing abelian normalsubgroup of type .4; 2/. If A � ˆ.G/, then G has no normal elementary abeliansubgroups of order 8.

Proof. We assume thatA � ˆ.G/. LetW0 D �1.A/ so thatW0 Š E4 and Z.ˆ.G// <A (the inclusion is strong sinceA is not normal inG). It follows from jG W CG.W0/j �

2 that W0 � Z.ˆ.G//. Since Z.ˆ.G// < A, it follows that W0 D Z.ˆ.G// so W0 isa normal four-subgroup in G.We claim that W0 is the unique normal four-subgroup of G. Indeed, if W1 were

another normal four-subgroup ofG, thenA does not centralizeW1 and so CG.W1/ is amaximal subgroup of G not containing A, contrary to our assumption that A � ˆ.G/.Since A < ˆ.G/ and jG=ˆ.G/j � 4, we get jGj � 26.Assume that W0 � Z.G/. Let B be a normal subgroup of G such that W0 <

B � ˆ.G/ and jB W W0j D 2. Then G stabilizes the chain B > W0 > f1g and soG=CG.B/ is elementary abelian. In particular, ˆ.G/ � CG.B/ which contradicts thefact that Z.ˆ.G// D W0. Hence jG W CG.W0/j D 2 and so Z.G/ D hzi < W0.Set W0 D hz; ui and T D CG.W0/. For each x 2 G � T , ux D uz and for eacha 2 A �W0, CT .a/ D CT .A/ D A since A D ha;W0i and W0 � Z.T /.Suppose that A1 is a normal elementary abelian subgroup of order 16 of T D

CG.W0/. Take an element a 2 A � W0. Since a2 2 W0 � Z.T /, the element ainduces on A1 an automorphism of order � 2 and so jCA1

.a/j � 4 since ha;A1i isnot of maximal class (Proposition 1.8). Since CT .a/ D A, we get CA1

.a/ D W0 andA \ A1 D W0. Set C D AA1 so that jC j D 25, Z.C / D W0 and jC W A1j D 2.Set hsi D Ã1.A/ < W0 and we see that all four elements in A�W0 are square roots

of s in C . Let aa1 (a 2 A�W0, a1 2 A1) be any square root of s in C . Then we have

�77 2-groups with a self-centralizing abelian subgroup of type .4; 2/ 317

s D .aa1/2 D aa1aa1 D a2.a�1a1a/a1 D saa

1a1, and so aa1 D a1 which implies

that a1 2 W0 hence aa1 2 A � W0. It follows that four elements in A � W0 are theonly square roots of s in C and they form a single conjugate class in C (it follows fromCT .a/ D A that CC .a/ has index 4 in C ). Since T centralizes s, it follows that A�W0

is a conjugacy class in NT .C / so hA �W0i D A is normal in NT .C / which impliesC D T (in view of W0 � Z.T / we have jT W Aj D jT W CT .A/j � 4 D jC W Aj)and jGj D 2jT j D 2jC j D 26. Since W0Z.T / < A, we conclude that Z.T / D W0. Itfollows that A1 is the unique abelian subgroup of index 2 in T so A is characteristicin T ; then A1 is normal in G. It follows from the previous sentence that T=W0 Š E23

(otherwise, T contains an abelian subgroup of index 2 and exponent 4, which is notthe case). By Lemma 1.1, jT 0j D 4 so T 0 D W0 D ˆ.G/ D Z.T /, and T is special.Let b be any element in C � A1 D T � A1 so that b D a1a for an a 2 A � W0

and a1 2 A1 (recall that A \ A1 D W0). Hence CA1.b/ D CA1

.a/ D W0. Inparticular, b2 2 W0 and so four elements in hW0; bi � W0 are either involutions (incase b2 D 1) or square roots of the involution b2. Conversely, let bx (x 2 A1) be anysquare root of b2 or an involution if b2 D 1 in C �A1. Then b2 D .bx/2 D bxbx D

b2.b�1xb/x D b2xbx, which implies xb D x and so x 2 CT .b/ D W0. Hence fourelements in hW0; bi � W0 are all possible square roots of b2 if hW0; bi Š C4 � C2

(or involutions if hW0; bi Š E8) in T � A1. By Exercises 17–19 in �10 and Theorem1.17(a), each group of order 25 contains at most 19 involutions, unless it is elementaryabelian or isomorphic to D8 �E4. It follows that T �A1 contains at least 12 elementsof order 4 and so, by what has been proved already, the set of squares of these elementsis W #

0 . Thus, each involution in W0 has exactly four square roots in T � A1 and, inaddition, there are exactly four involutions in T � A1.Let z 2 W0 be the central involution in G. By the previous paragraph, there are

exactly four square roots of z in C and if c is one of them, then hW0; ci Š C4 � C2

and all square roots of z in C D T form the set hW0; ci � W0. Hence hW0; ci �W0

is a normal subset in G and so hW0; ci is normal in G in view of z 2 Z.G/. SinceZ.T / D W0, A1 is the unique abelian subgroup of index 2 in T so CT .c/ D hW0; ci.Thus hW0; ci is a self-centralizing (inG) normal abelian subgroup of G of type .4; 2/,contrary to the hypothesis.We have proved that T D CG.W0/ has no normal elementary abelian subgroups of

order 16.Let U be a noncyclic abelian normal subgroup of G. ThenW0 � U (because of the

uniqueness of W0) so U � CG.W0/ D T . By the previous paragraph, U 6Š E24 so Ghas no normal elementary abelian subgroups of order 16.Suppose that E8 Š E G G. By the previous paragraph, E � T , and E > W0, and

so A \E D W0.Let B be a maximal G-invariant abelian subgroup containing E. Since B is non-

cyclic, we have B � T . We know that CG.B/ D B .Suppose that B D E. ThenG=E, as a subgroup of order jG W Ej D 8 of Aut.E/ Š

GL.3; 2/ Š PSL.2; 7/, is isomorphic to D8. Set V0 D AE so that V0=E D ˆ.G=E/


and therefore V0 is normal inG. Indeed, ifM=E is a maximal subgroup of G=E, thenA � M since A � ˆ.G/, and so V0=E D AE=E < M=E, proving our claim.Since T stabilizes the chain E > W0 > f1g and CT .E/ D E, it follows that

T=E Š E4. Take an element a 2 A � W0 so that four elements in A � W0 aresquare roots of a2 ¤ 1. Let ae0 .e0 2 E/ be any square root of a2 in V0. Then wehave a2 D .ae0/2 D ae0ae0 D a2.e0/ae0, which gives .e0/a D e0 and so e0 2 W0

since CT .a/ D A. It follows that A � W0 is the set of all square roots of a2 inV0 and so A D hA � W0i is normal in T . Since A is not normal in G, we havefor every x 2 G � T , Ax ¤ A. Since Ax � V0 and A \ Ax D W0, we haveV0 � E D Ea D .A � W0/ [ .Ax � W0/ and so x sends four elements in A � W0

onto four elements in Ax � W0. But G=E Š D8 and so (since T=E Š E4 andV0=E D ˆ.G=E/) there is y 2 G � T with y2 2 V0 � E. Since V0 is nonabelian,A, Ax and E are all abelian subgroups of index 2 in V0. Since hy2;W0i equals eitherA or Ax , one of these subgroups is normal in G since it is normalized by hy; T i D G;then both of them are normal in G, and this is a contradiction.We have proved that B > E and so jBj � 24. Since �1.B/ D E (recall that G

has no normal elementary abelian subgroups of order 24), B is an abelian group ofrank 3. Take an element a 2 A�W0. Since a2 2 W0 � Z.T /, a induces an involutoryautomorphism on B with CB.a/ D W0 � �1.B/ (since a 62 B in view of CT .a/ D

A). Applying Proposition 51.2, we see that a inverts Ã1.B/ and B=W0. Suppose thatsome e 2 E�W0 .D E�A/ is a square inB . Then a inverts (centralizes) e .2 Ã1.B//

so e centralizes ha;W0i D A, a contradiction. Suppose that each involution inW0 is asquare in B . Then �2.B/ D hb1i � hb2i � hb3i, where o.b1/ D o.b2/ D 4, b3 is aninvolution in E �W0, and hb2

1; b22i D W0. Since a inverts B=W0, we get ba

1 D b�11 w

with w 2 W0 and so ba1 D b�1

1 w D b1.b21w/ D b1w1, where w1 2 W0. Similarly,

ba2 D b2w2 and ba

3 D b3w3 with w2; w3 2 W0. Since C�2.B/.a/ D W0, w1, w2, w3

must be three distinct involutions in W0 (otherwise, if, for example, w1 D w2, then.b1b2/

a D b1b2, which is not the case). But then .b1b2b3/a D .b1b2b3/.w1w2w3/ D

b1b2b3 2 �2.B/�W0, a contradiction. Hence we have B D hbi � hwi � hei, where

o.b/ D 2m; m > 1; b2m�1

D z; hzi D Ãm�1.B/ D Z.G/; W0 D hz;wi;

and e 2 E � W0. Also we set v D b2m�2

so that o.v/ D 4, v2 D z, and �2.B/ D

hE; vi is abelian of type .4; 2; 2/.Set V D AB and assume that V < T (V � T since A;B < T ). Set QV D NT .V /

so that j QV W V j � 2. Since B is normal in G, V � B D aB (a 2 A �W0) is a normalsubset in V � D NG.V / � QV . Set a2 D w0 2 W0 so that four elements in A �W0 aresquare roots of w0 in V � B . An element ab0 2 V � B (b0 2 B) is a square root ofw0 if and only if w0 D .ab0/2 D ab0ab0 D a2.b0/ab0 D w0.b0/ab0 or .b0/a D .b0/�1.If .b0/a D .b0/�1 and .b00/a D .b00/�1 (b0; b00 2 B), then .b0b00/a D .b0/a.b00/a D

.b0/�1.b00/�1 D .b0b00/�1, and so the set B0 of elements of B which are inverted by ais a subgroup of B . Therefore, the number of square roots of w0 in V �B equals jB0j.It follows from CT .a/ D A that B0 \E D W0. We have jBj D 2mC2 .m > 1/ (recall


that B is abelian of type .2m; 2; 2/) so jV � Bj D 2mC2 and therefore jB0j � 2mC1.All conjugates of a in QV lie in V � B and all these conjugates are square roots of w0

since QV � T and T centralizes w0 2 W0. Since C QV.a/ D A, we get j QV W Aj � 2mC1.

But jV j D 2mC3 and so j QV j D 2mC4, j QV W V j D 2, and jB0j D 2mC1. It followsthat B0 covers B=E and so we may choose b 2 B0 such that B0 D hbi � hwi, wherew 2 W0 � hzi (since b2m�1

D z).

Suppose for a moment that QV D NT .V / D NG.V /. Then looking at G=B , we get,by Proposition 1.8, that G=B is of maximal class and V=B is a noncentral subgroupof order 2 in G=B . Let R=B be a cyclic subgroup of index 2 in G=B . Then R isa maximal subgroup of G and R \ V D B . In particular, A 6� R, contrary to ourassumption A � ˆ.G/. We have proved that V � D NG.V / > QV D NT .V / so thatjV � W QV j D 2, G D T V �, and T \ V � D T \ NG.V / D NT .V / D QV .

Assume that CV �.a/ D A D C QV.a/ and take an element y 2 V � � QV so that

y 2 G � T . Then all 2mC2 elements in V � B form a single conjugate class in V �.Since w0 has exactly 2mC1 square roots in V � B , it follows that w0 D a2 ¤ z (sincehzi D Z.G/) and so .w0/y D w0z. Hence y sends 2mC1 square roots of w0 in V � B

onto 2mC1 square roots of w0z in V � B . Since e 62 B0 (e 2 E � W0), we have.ae/2 D w0z D aeae D a2eae D w0eae, and so ea D ez. On the other hand, ainverts B0 D hb;wi and so setting v D b2m�2

, we get v2 D z and va D v�1 D vz.But then .ev/a D ezvz D ev, contrary to CB.a/ D W0. We have proved thatQA D CV �.a/ > A, where j QA W Aj D 2 and QA \ T D A.

Take an element y 2 QA � A so that y acts non-trivially on W0. But y centralizesa and so y centralizes a2 D w0 which gives w0 D z D b2m�1

, where hzi D Z.G/.Suppose y2 2 W0 so that hW0; yi Š D8. In that case there are involutions in hW0; yi�

W0 and so we may assume that y is an involution. Act with the involution y on E.Since y acts non-trivially onW0 and jCE .y/j D 4, there is an involution e0 2 E �W0

with .e0/y D e0. We have 1 ¤ Œa; e0� 2 W0 and so Œa; e0�y D Œay ; .e0/y � D Œa; e0�,which gives Œa; e0� D z or .e0/a D e0z. But a inverts B0 and so va D v�1 D vz

(where v D b2m�2

) which gives .ve0/a D vze0z D ve0 62 W0. This is a contradictionsince CB.a/ D W0.

We have proved that for each y 2 QA�A, y2 2 A�W0 and so we may assume thaty2 D a. Since a2 D z centralizesE, it follows that y induces on E an automorphismof order 4 and so CE .y/ D hzi. Hence for an e 2 E � W0, we get ey D ew0 withw0 2 W0 �hzi andwy

0 D w0z. This gives ea D ey2

D .ew0/y D .ew0/.w0z/ D ez.

On the other hand, a inverts B0 D hb;w0i and so va D v�1 D vz. This gives.ve/a D vzez D ve 62 W0, contrary to CB.a/ D W0.

The contradiction in the previous paragraph shows that we must have V D AB D

T . We set again a2 D w0 2 W0 .a 2 A � W0/ and v D b2m�2

.m > 1/ so thatv2 D z, where hzi D Z.G/ and B D hE; bi. We have jG=Bj D 4. If G=B Š E4,then ˆ.G/ � B and this contradicts our assumption A � ˆ.G/. Hence G=B Š C4

and G=B acts faithfully on E (since T D Bhai and a acts non-trivially on E). If


y 2 G � T , then for an element e 2 E �W0, we have ey D ew with w 2 W0 � hzi

and wy D wz. Then we compute ey2

D .ew/y D .ew/.wz/ D ez. But y2 2 T � B

and T D Bhai and so a acts in the same way on E as the element y2 which impliesea D ez. By Proposition 51.2, a inverts Ã1.B/. If m > 2, then v 2 Ã1.B/ andso va D v�1 D vz. We get in that case .ev/a D ezvz D ev which contradicts toCB.a/ D W0.We have proved that we must have m D 2 and so b D v, jBj D 24, and jGj D 26.

The argument of the previous paragraph shows that a does not invert any element (oforder 4) in B � E and so the subgroup B0 of all elements of B inverted by a is equalB0 D W0. Indeed, if a inverts an element s 2 B � E, then sa D s�1 D sz and so.es/a D ezsz D es, contrary to CB.a/ D W0. Hence a2 D w0 has exactly jB0j D 4

square roots in T � B D V � B D .AB/ � B and they all lie in A � W0. SincehA � W0i D A, A is normal in T D AB . For each x 2 T � B , x2 2 W0 sinceCB.x/ D CB.a/ D W0, and x (acting in the same way on B as the element a) invertson B exactly the elements of W0. It follows that x2 has exactly four square roots inT �B . Hence each involution inW0 has exactly four square roots in T �B and (sincejT � Bj D 16) T � B contains exactly four involutions.Let a0 2 T � B so that .a0/2 D z. Set A� D W0ha0i Š C4 � C2 and A� is normal

in G. Indeed, hzi D Z.G/ and so G normalizes the subset f.W0ha0i/ � W0g of allsquare roots of z in T � B . We have CG.W0/ D T and CT .a

0/ D ha0iCB.a0/. But

CB.a0/ D CB.a/ D W0 and so A� is a self-centralizing abelian normal subgroup of

type .4; 2/ in G. This is a final contradiction and our theorem is proved.

In our next result we shall determine the structure of the groups appearing in Theo-rem 77.1.

Theorem 77.2. Let G be a 2-group which possesses a self-centralizing abelian sub-group A of type .4; 2/ but G does not possess any self-centralizing abelian normalsubgroup of type .4; 2/. If A � ˆ.G/, then G has the following properties:

(i) G has no normal elementary abelian subgroups of order 8.

(ii) G has the unique normal four-subgroup W0 D �1.A/.

(iii) G has a normal metacyclic subgroup N such that �2.N / D W is abelian oftype .4; 4/, CG.W / � N , �1.W / D W0, CG.W0/ � N , jG W CG.W0/j D 2,and G=N Š C4 or D8.

(iv) N is either abelian of type .2k ; 2kC1/ or .2k ; 2k/, k � 2, orN is minimal non-abelian and more precisely N D ha; b j a2m

D b2n

D 1; ab D a1C2m�1

i,where either m D n with n � 3 or m D nC 1 with n � 2.

Proof. Suppose that our 2-group G satisfies all the assumptions of our theorem to-gether with A � ˆ.G/. By Theorem 77.1, G has no normal elementary abeliansubgroups of order 8. By the first three paragraphs of the proof of Theorem 77.1,we know that W0 D Z.ˆ.G// D �1.A/ is the unique normal four-subgroup of G,


jG W CG.W0/j D 2, Z.G/ is of order 2, and jGj � 26. SetW0 D hz; ui, T D CG.W0/,where hzi D Z.G/. For each t 2 A �W0, CT .t/ D A.We apply now the results of �50, since G is neither abelian nor of maximal class.

It follows that G possesses a normal metacyclic subgroup N such that CG.�2.N // �

N , G=N is isomorphic to a subgroup of D8 andW D �2.N / is abelian of type .4; 2/or .4; 4/. In any case, W0 D �1.W / D �1.N / is the unique normal four-subgroupof G and W0 6� Z.G/. If A � N , then A � W D �2.N /. Since CG.A/ D A,we have A D W and then A is normal in G, a contradiction. Hence A 6� N and soA \ N D W0. Since A � ˆ.G/, G=N is not elementary abelian. Hence G=N isisomorphic to C4 or D8. If T D CG.W0/ does not contain N , then T covers G=Nand G=N acts faithfully on W (since CG.W / � N ). In that case CG.W0/=CN .W0/

cannot contain elements of order 4 (since that group centralizesW0 D �1.W /) and soG=N is elementary abelian, a contradiction. Hence T D CG.W0/ � N .Assume that G=N Š C4. If N is abelian of type .2j ; 2/, j � 2, then G=N acts

faithfully on �2.N / Š C4 � C2. If N > �2.N /, then there is a characteristiccyclic subgroup Z Š C4 of N so that Z is normal in G. But then jG W CG.Z/j �

2 and therefore A (being contained in ˆ.G/) centralizes Z, a contradiction. ThusN D �2.N / is a normal abelian self-centralizing subgroup of type .4; 2/ in G, acontradiction. We have proved that �2.N / D W is abelian of type .4; 4/. Supposethat Z� is a G-invariant cyclic subgroup of order 4 contained in N . But then againA centralizes Z�, a contradiction. Hence there is no such Z� and so we may applyProposition 50.4. It follows thatN is either abelian of type .2n; 2n/ or .2nC1; 2n/withn � 2 or N is minimal nonabelian and more precisely N D ha; b j a2m

D b2n

D

1; ab D a1C2m�1

i; where either m D n with n � 3 or m D nC 1 with n � 2.Assume that G=N Š D8. The structure of N in that case is already determined

by Theorem 50.1. The minimal case N Š C4 � C2 cannot occur because in thatcase N would be a self-centralizing normal abelian subgroup of type .4; 2/ of G, acontradiction.

Finally, we consider the case where A 6� ˆ.G/.

Theorem 77.3. Suppose that G is a 2-group that possesses a self-centralizing abeliansubgroup A of type .4; 2/. If �1.A/ 6� ˆ.G/, then G possesses an involution t suchthat CG.t/ D hti �D, whereD is isomorphic to one of the following groups: C4, D8,Q2n , n � 3, or SD2m , m � 4. Such groups G have been classified in �48, �49, and�51.

Proof. Suppose that�1.A/ 6� ˆ.G/. There is a maximal subgroupM of G such thatA � M contains an involution t . It follows that A0 D A \ M Š C4 and CG.t/ D

hti �D, where D D CM .t/ � A0. We have CD.A0/ D A0 and so (by a well-knownresult of M. Suzuki) either D D A0 Š C4 or D is a 2-group of maximal class. In thesecond caseD is isomorphic to D8, Q2n , n � 3 or SD2m , m � 4.


Theorem 77.4. Suppose that G is a 2-group that possesses a self-centralizing abeliansubgroup A of type .4; 2/. If W0 D �1.A/ � ˆ.G/ but A 6� ˆ.G/, then for anya 2 A�W0, CG.a/ D hai�M0, jhai\M0j D 2, whereM0 D W0 orM0 is isomorphicto one of the following groups: D2n , n � 3 or SD2m , m � 4 and G > CG.a/.

Proof. Suppose thatM is a maximal subgroup of G which does not contain A. ThenA \ M D W0 D �1.A/. Let a 2 A � W0 so that G D M hai and CG.a/ D

haiCM .a/ with CM0.W0/ D W0, where M0 D CM .a/. By a result of M. Suzuki,

either M0 D W0 or M0 is of maximal class. In the second case, M0 is isomorphicto one of the following groups: D2n , n � 3, or SD2m , m � 4. If G D CG.a/, thenˆ.G/ D ˆ.CG.a// D ˆ.M0/ is cyclic, contrary to our assumption.

Exercise. Describe the subgroup structure of the holomorph of the abelian group oftype .4; 2/.

�78

Minimal nonmodular p-groups

The main results of this section are taken from [Jan10].

1o. We know that a 2-group is modular if and only if it is D8-free (Theorem 44.13).Here we classify minimal nonmodular 2-groups G, i.e., nonmodular 2-groups all ofwhose proper subgroups are modular. Hence there is N G G such that G=N Š D8-free but each proper subgroup of G is D8-free. We shall use freely all results onmodular p-groups from �73 and Appendix 24. In this section a 2-group is said to beHamiltonian if it is nonabelian and all its subgroups are normal. A Dedekindian 2-group is Hamiltonian if it is nonabelian. A metacyclic 2-group H is called ordinarymetacyclic with respect to A if H possesses a cyclic normal subgroup A such thatH=A is cyclic andH centralizes A=Ã2.A/, or what is the same,H=Ã2.A/ is abelian.In other words, a metacyclic 2-group is ordinary if and only if it is powerful (see �26).Below we use freely the following facts:

(�) If minimal nonmodular 2-group G is of maximal class and order � 24, thenG Š Q24 . Moreover, a nonabelian 2-group G with cyclic subgroup of index 2 ismodular if and only if G 2 fQ8;M2ng. It follows that any two involutions of G arepermutable so �1.G/ is elementary abelian.

(��) [Wil] Let G be a Q8-free modular 2-group. Then d.�1.G// D d.G/. If, inaddition, d.G/ D 2, then G is ordinary metacyclic.

(��) LetN be a normal subgroup of a 2-groupG such thatG=N Š D8. IfL=N isthe unique cyclic subgroup of index 2 in G=N and x 2 G � L, then x2 2 N . Indeed,all elements if .G=N/ � .L=N/ are involutions.

Until the end of this subsection, G is a minimal nonmodular 2-group. In that case,G contains two non-permutable cyclic subgroups A and B . Since the subgroup hA;Bi

is nonmodular, we get G D hA;Bi so d.G/ D 2. By hypothesis, G has a normalsubgroup N such that G=N Š D8. It follows from d.G/ D 2 D d.G=N/ thatN � ˆ.G/. The following proposition clears up the structure of N .

Remark 1. Suppose that G is a modular 2-group of order 16 containing a subgroupH Š E8. Then, G is Q8-free in view of the existence ofH . SinceG has no subgroupsŠ D8, it is either abelian or minimal nonabelian. Assume that G is nonabelian. ThenG D ha; b j a4 D b2 D c2 D 1; c D Œa; b�; Œa; c� D Œb; c� D 1i and G=ha2i Š D8, acontradiction. Thus, G is abelian.


Proposition 78.1. We have d.N / � 2, so N is metacyclic (Theorem 44.13).

Proof. Suppose that d.N / � 3. ThenN possesses aG-invariant subgroup R such thatN=R Š E8. The quotient group G=R is also minimal nonmodular. We want to obtaina contradiction. To this end, we may assume that R D f1g; then N Š E8. Let S=Nbe any subgroup of order 2 in G=N . Then S (being modular) is abelian, by Remark 1,so S centralizes N . On the other hand, G=N Š D8 is generated by its subgroups oforder 2, and so N � Z.G/.Assume that Z.G/ > N ; then Z.G/=N D Z.G=N/ D ˆ.G=N/ so Z.G/ D ˆ.G/.

In that case,G is minimal nonabelian. Then�1.G/ D N and jG0j D 2 (Lemma 65.1)so G0 < N , a contradiction since G=N Š D8 is nonabelian. Thus, N D Z.G/.Let L=N be the unique cyclic subgroup of index 2 in G=N . Then L is abelian

since N D Z.G/. If L D N � L1 with L1 Š C4, then Ã1.L/ D Ã1.L1/ is oforder 2 and so Ã1.L/ � Z.G/, a contradiction since Ã1.L/ 6� N D Z.G/. HenceL does not split over N and so L D NC , where C Š C8 and C \ N D C0 is oforder 2. We have ˆ.L/ D ˆ.C/ D C1 Š C4, where C0 < C1, and ˆ.G/ D C1N

is abelian of type .4; 2; 2/. For each x 2 G � L, x2 2 N , by (��), and so there isb 2 G�Lwith b2 2 N�C0 (otherwise,ˆ.G/ D Ã1.G/ D C1). Since CG.C1/ D L,Aut.C1/ Š C2 and C1 is normal in G, it follows that b inverts C1. In that case,D D hC1; bi is (nonabelian) metacyclic of order 24 and exponent 22 soD=hb2i Š D8,a contradiction.

Proposition 78.1 is a variant of Theorem 44.13. It follows from Theorem 44.13 thatthe subgroup N of Proposition 78.1 is metacyclic.

Remark 2. Let, as in the following four propositions, the subgroup N be cyclic. Weclaim that then Ã2.G/ D ˆ.N/. One may assume that N > f1g. In view of Ã2.G/ �

Ã2.N /, it suffices to show that exp.G=Ã2.N // D 8. Without loss of generality, onemay assume that Ã2.N / D f1g, i.e., N is cyclic of order 4. Then ˆ.G/ is abelianof order 8. One may assume that ˆ.G/ is abelian of type .4; 2/ (otherwise, G has acyclic subgroup of index 2, contrary to (�)). Since ˆ.G/ D Ã1.G/ is not generatedby involutions, there exists x 2 G such that o.x2/ D 4; then o.x/ D 8, and we aredone. It follows from the obtained result and Lemma 64.1(m) that G is metacyclic ifand only if G=ˆ.N/ is metacyclic.

Proposition 78.2. Suppose that N is cyclic and some proper subgroup of G is not Q8-free. Then G is isomorphic to Q24 or to the uniquely determined group X of order 25

with �2.X/ Š Q8 � C2 given in �52.

Proof. Suppose that N is cyclic and G has a maximal subgroup M which is not Q8-free. Since M is modular, it follows that M is Hamiltonian (�73 or Appendix 24),i.e., M D Q � E with Q Š Q8 and exp.E/ � 2. In particular, exp.M/ D 4 andÃ1.M/ D ˆ.M/ D Ã1.Q/. We have N < M since N � ˆ.G/ so jN j � 4; in thatcase, jGj D jG=N jjN j � 8 4 D 32. If jG0j D 2, it follows from d.G/ D 2 that G

�78 Minimal nonmodular p-groups 325

is minimal nonabelian (Lemma 65.2(a)), a contradiction sinceM < G is nonabelian.Hence jG0j � 4 so G has at most one abelian maximal subgroup (Lemma 65.2(c)).

(i) Suppose that jN j D 4; then jGj D 32 and jˆ.G/j D 8. Let L=N Š C4 be theunique cyclic subgroup of index 2 inG=N .Š D8/. By (�), G is not of maximal class.In that case, the metacyclic subgroup L is noncyclic (Lemma 64.1(t)) so jÃ1.L/j D14 jLj D 4 and therefore L is not Hamiltonian. Suppose that L is abelian. We haveÃ1.L/ > Ã1.N / D Ã1.M/. Let K be the maximal subgroup of G distinct fromMand L. We know that K must be nonabelian and, since G=N Š D8, we get K=N Š

E4. By (�), K is not of maximal class. Since N Š C4 does not lie in Z.M/, we haveCG.N / D L. Then jK W CK.N /j D 2 soK 6Š M16 (otherwise,N D ˆ.K/ D Z.K/),and we conclude (Theorem 1.2) that exp.K/ D 4. Take k 2 K � CK.N /; thenk2 2 N , in view of K=N Š E4, and therefore hN;ki Š Q8. It follows that Kis Hamiltonian (�73 or Appendix 24) and so Ã1.K/ D Ã1.N / < Ã1.L/. Henceˆ.G/ D Ã1.G/ D Ã1.L/ is of order 4, a contradiction since jˆ.G/j D 8.We have proved that L is nonabelian. In particular, N D hni 6� Z.L/ since

L=N is cyclic, and if we set L D hN; li, then nl D n�1 and l4 2 hn2i. Ifo.l/ D 4, then L=hl2i Š D8, a contradiction. Hence o.l/ D 8 and so L Š M16

with hl4i D hn2i D L0 and ˆ.L/ D Z.L/ D hl2i Š C4. Note that �2.L/ D Nˆ.L/

is abelian of type .4; 2/. Set K D CG.N / so that K is the maximal subgroup of Gdistinct fromM and L and (noting that hl2i > hn2i) we get (see also Exercise 1.133)ˆ.G/ D ˆ.M/ˆ.L/ˆ.K/ D hn2ihl2iˆ.K/ D hl2iˆ.K/. Since K=N Š E4, wehave ˆ.K/ � N . But ˆ.G/ > N and so we must have ˆ.K/ D N (otherwise,ˆ.G/ D hl2i is of order 4). It follows that the modular subgroup K has a cyclic sub-group of index 2 andK is Q8-free since K cannot be Hamiltonian (because jKj D 16

and jÃ1.K/j D 4; see �73 and Appendix 24). HenceK is either abelian of type .8; 2/or K Š M16. In any case, �2.K/ D ˆ.L/N D ˆ.G/ is abelian of type .4; 2/. Itfollows that�2.G/ D M Š Q8 � C2 and consequently G is the uniquely determinedgroup of order 25 described in �52.

(ii) Now let jN j D 2. In that case, jGj D 16 and jG0j D 4 so jG W G0j D 4. ByTaussky’s theorem, G is of maximal class so, by (�), G Š Q16.

It is easy to check that the group G of Proposition 78.2 of order 25 is the group F

from Appendix 24. Indeed, take u 2 G � M and put U D hui; then jU j D 8 sinceM D �2.G/. Assume that UQ D QU . Then UQ is of maximal class (Theorem 1.2),contrary to (�). It follows that U \ Q is normal in G and G=.U \ Q/ Š D8. ByAppendix 24, G Š F .

Proposition 78.3. Suppose that N > f1g is cyclic and all proper subgroups of G areQ8-free. Then Ã2.G/ D ˆ.N/ and G=ˆ.N/ is minimal nonabelian of order 24 andexponent 4. Thus G=ˆ.N/ is isomorphic to one of the following groups:

(a) hx; y jx4 D y2 D 1; Œx; y� D z; z2 D Œx; z� D Œy; z� D 1i is nonmetacyclic,

(b) hx; y jx4 D y4 D 1; xy D x�1i is metacyclic.


Proof. By Remark 2, Ã2.G/ D ˆ.N/ so G=Ã2.G/ is 2-generator nonabelian ofexponent 4 and order 24. Therefore, by hypothesis, G=Ã2.G/ is minimal nonabelian.By Lemma 65.1 that G=Ã2.G/ D G=ˆ.N/ is isomorphic to the group (a) or (b).

In the next two propositions we determine the groups G of Proposition 78.3.

Proposition 78.4. Suppose that N > f1g is cyclic and all proper subgroups of G areQ8-free. If G=ˆ.N/ is as in Proposition 78.3(a), then G has a normal elementaryabelian subgroup E D hn; z; ti of order 8 such that G=E is cyclic. We set G D

hE; xi, where o.x/ D 2sC1, s � 1, E \ hxi D hni, Œt; x� D z, Œz; x� D n�, � D 0; 1,and G D hx; ti. We have G=hx2i Š D8, ˆ.G/ D hx2i � hzi, �1.G/ D E, and G isQ8-free. If � D 0, then G is minimal nonabelian nonmetacyclic. If � D 1, then s � 2,G0 D hz; ni Š E4 and Z.G/ D hx4i.

Proof. LetM=ˆ.N/ D �1.G=ˆ.N//.Š E8/; then N is a maximal cyclic subgroupof M . Set jN j D 2s , s � 1. Let S=N be any subgroup of order 2 in M=N . SinceM is D8-free and Q8-free, S cannot be of maximal class. It follows that S is eitherabelian of type .2s; 2/ or S Š M2sC1 (s > 2). In any case, there exists an involution inS � N . Hence �1.M/ coversM=N Š E4 and, sinceM is modular, E D �1.M/ iselementary abelian of order 8, by the product formula. It follows thatE is normal inGand E \ N D �1.N / D hni � Z.G/. In particular, ˆ.G/ D N�1.ˆ.G// is abelianof type .2s ; 2/, s � 1 since �1.ˆ.G// Š E4 centralizes ˆ.G/. Take an involutiont 2 E �ˆ.G/.Let K ¤ M be a maximal subgroup of G such that K=N Š E4; then K \ M D

ˆ.G/ since d.G/ D 2. Suppose that N is a maximal cyclic subgroup of K. Then,by the argument of the previous paragraph, �1.K/ covers K=N and so there is aninvolution r 2 K �M since E8 Š �1.K/ ¤ �1.M/ .D E/; then r 62 E. Since Ghas no elementary abelian subgroups of order 24, there is in E an involution u suchthat ru ¤ ur . Then hr; ui is dihedral, a contradiction. We have proved that there is anelement x 2 K�M such that hx2i D N and so o.x/ D 2sC1, s � 1. Since hx; ti D G

(indeed, involutions xN and tN lie in different maximal subgroups of G=N Š D8)and t 2 E � ˆ.G/, we get G D Ehxi with E \ hxi D �1.N / D hni Š C2 and soG=E is cyclic of order 2s and jGj D 2sC3. In particular,G0 < E and so jG0j 2 f2; 4g.Since d.G/ D 2 and G is not minimal nonabelian, we get jG0j D 4 (Lemma 65.2(a)).We have Œx; t � ¤ 1 since hx; ti D G is nonabelian, so z D Œx; t � is an involution

in .E \ ˆ.G// � hni since hxi is not normal in G. Indeed, if hxi were normal in G,then G=ˆ.N/ D htˆ.N/i hxi=ˆ.N/ is metacyclic, which is not the case. It followsthat E D hn; z; ti, where z 2 ˆ.G/, and hxi is not normal in G. Thus NG.hxi/ isa maximal subgroup of G and so z 2 ˆ.G/ � NG.hxi/ and therefore z normalizeshxi. Since hx; zi cannot be of maximal class, by (�), we have either Œx; z� D 1 orŒx; z� D n (in which case hx; zi Š M2sC1 , s > 1). Let us consider the first possibility.We have CG.z/ � hE;xi D G so z 2 Z.G/ \G0. Then G=hzi has a cyclic subgrouphx; zi=hzi of index 2. It follows that then G=hzi is neither abelian nor isomorphic to


M2sC1 (otherwise, G=hzi has two distinct cyclic subgroups of index 2 so G has twodistinct abelian maximal subgroups, and we get jG0j < 4, a contradiction). It follows(Lemma 64.1(t)) that G=hzi is of maximal class so it is isomorphic to Q16, by (�),contrary to the existence of E. Now let hx; zi Š M2sC1 , s > 1. Then hxi induces anautomorphism of order 4 on E.Let u be an involution in G � E. Then, by Remark 1, F D Ehui Š E16. But

G=E is cyclic and so F=E D .Ehx2s�1

i/=E and so x2s�1

is an element of order 4contained in F �E, a contradiction. We have proved that �1.G/ D E.

Proposition 78.5. Suppose that N > f1g is cyclic and all proper subgroups of G areQ8-free. If G=ˆ.N/ is metacyclic, then G is also metacyclic and we have one of thefollowing possibilities:

(i) G D hx; y jx4 D y2sC1

D 1; s � 1; xy D x�1i, where G is minimal non-abelian and N D hy2i.

(ii) G D hx; a jx2sC1

D a8 D 1; s � 2; x2s

D a4; ax D a�1i, where G is anA2-group with N D hx2i, Z.G/ D N (see �65), G0 D ha2i Š C4 and G is ofclass 3.

In both cases (i) and (ii), G is a minimal non-Q8-free 2-group.

Proof. By Remark 2, ˆ.N/ D Ã2.G/ andG is also metacyclic. Set jN j D 2s , s � 1.If s D 1, then G is isomorphic to the group (b) of Proposition 78.3 and we are done.Now we assume that s � 2.Let S=N be a subgroup of order 2 in G=N . By (�), S is not of maximal class so S

is either cyclic of order 2sC1 or S is abelian of type .2s ; 2/ or s > 2 and S Š M2sC1 .If ˆ.G/ is cyclic, then G has a cyclic subgroup of index 2, contrary to (�). Also,M2sC1 , s > 2, having cyclic center, cannot be the Frattini subgroup (Burnside). TakingS D ˆ.G/, we see that ˆ.G/ is abelian of type .2s ; 2/, s � 2.Let �1.N / D hni so that hni � ˆ.N/ \ Z.G/. For any subgroup S=N of order

2 in G=N , by the previous paragraph, S=hni is abelian so centralizes N=hni. SinceG=N is generated by its subgroups of order 2, we get N=hni � Z.G=hni/.Suppose for a moment thatG is minimal nonabelian. Since ˆ.G/ is abelian of type

.2s; 2/, we get at once (Lemma 65.1): G D hx; y jx4 D y2sC1

D 1; s � 1; xy D

x�1i, where N D hy2i; this is a group of part (i). In what follows we assume that Gis not minimal nonabelian. In particular, jG0j � 4 (Lemma 65.2(a)).We will determine the structure of all three maximal subgroups of G. Let M be a

maximal subgroup of G such that M=N Š E4. If N is a maximal cyclic subgroupof M , then for each subgroup S=N of order 2 of M=N , there is an involution inS � N . Hence �1.M/ covers M=N and, since M is D8-free and Q8-free, �1.M/

is elementary abelian and �1.M/ \ N D hni so that �1.M/ Š E8, by the productformula. This is a contradiction since G is metacyclic. It follows that N is not a max-imal cyclic subgroup of M . Let M0 be a maximal cyclic subgroup of M containingN so that M0 Š C2sC1 is of index 2 in M . By (�), M is not of maximal class and


so M is either abelian of type .2sC1; 2/ or M Š M2sC2 , s � 2 (Lemma 64.1(t)).In any case, N � Z.M/ and M=hni is abelian since M 0 � hni. Let K (¤ M ) beanother maximal subgroup of G with K=N Š E4. Then K is either abelian of type.2sC1; 2/ or K Š M2sC2 , s � 2, and again N � Z.K/ and K=hni is abelian. Weget CG.N / � MK D G so N � Z.G/. Since G is not minimal nonabelian, we getˆ.G/ ¤ Z.G/. We have proved that N D Z.G/.Let L=N be the unique cyclic subgroup of index 2 in G. Then L is abelian and

using Lemma 64.1(q), we get jG0j D 4. By Lemma 64.1(u), L is the unique abelianmaximal subgroup ofG and soM Š K Š M2sC2 withM 0 D K 0 D hni. In particular,G is anA2-group (since jG0j D 4, this also follows from Corollary 65.3).We have G0 > hni and, since G0 6� N D Z.G/ and jˆ.G/=N j D 2, we get

ˆ.G/ D NG0, N \G0 D hni D �1.N / and cl.G/ D 3.Since G is metacyclic, there exists a cyclic normal subgroup Z of order 8 such that

Z > G0. ButN \Z D N \G0 D hni and soNZ D L which determines the structureof the maximal subgroup L and shows that L does not split over N . It follows that Lis abelian of type .2s ; 2/.Set Z D hai. By (��), we get hx2i D N for each x 2 G � L. Hence G D Zhxi

with Z \ hxi D hni for a fixed x 2 G � L (since �1.G/ D �1.L/, x is not aninvolution). In view of jG0j D 4 and G0 < Z, we get either ax D a�1 or ax D a�1n,where n D a4. However, if ax D a�1n, then we replace Z D hai with Z� D haui,where u 2 N is such that u2 D n. Then we have .au/x D a�1nu D a�1u�1 D

.au/�1. Since h.au/2i D ha2i D G0, we may assume from the start that ax D a�1

and so the structure of G is completely determined.

Remark 3. Let G be a 2-group and let N be a G-invariant metacyclic subgroup ofˆ.G/. We claim thatN is ordinary metacyclic. One may assume thatN is nonabelian;thenN has no cyclic subgroups of index 2 since Z.N / must be noncyclic. There existsa maximal cyclic subgroup A of N such that N 0 < A and N=A is cyclic. We have toprove that N=Ã2.A/ is abelian. This is the case if jA W N 0j > 2. Now we assume thatjA W N 0j D 2 and obtain a contradiction. Under our assumption, N=N 0 is abelian withcyclic subgroup of index 2 (indeed, by the choice of A and Frobenius–Stickelbergertheorem on abelian groups, A=N is a direct factor of N=N 0). In particular, N has noepimorphic images which is abelian of type .4; 4/. Now consider the quotient groupNN D N=Ã2.N /. Since N has no cyclic subgroups of index 2, we get j NN j D 16. Byassumption, NN is nonabelian so NN D hx; y j x4 D y4 D 1; xy D x�1i. In that case,NN=hy2i Š D8 has a cyclic center, a contradiction since N=Ã2.N / � ˆ.G=Ã2.N //.

In the rest of this subsection we consider the case d.N / D 2.

Proposition 78.6. Suppose that d.N / D 2. Then G=ˆ.N/ is the minimal nonabeliannonmetacyclic group of order 25 and exponent 4. In particular, G=ˆ.N/ has theunique epimorphic image isomorphic to Q8. Each maximal subgroup of G is Q8-freeand N is ordinary metacyclic.


Proof. We want to determine the structure of G=ˆ.N/. Since G=ˆ.N/ is also mini-mal nonmodular, we may assume for a moment that ˆ.N/ D f1g so that N Š E4 andjGj D 25. Let S=N be any subgroup of order 2 in G=N . Since S 6Š D8, S is abelianand so N � Z.G/ since such subgroups S generate G.Suppose that Z.G/ D N . Let L=N be the unique cyclic subgroup of index 2 in

G=N ; then L is abelian. If L D N � R with R Š C4, then Ã1.L/ D Ã1.R/ 6� N

and Ã1.L/ � Z.G/, contrary to our assumption. Hence L D NL1 with L1 Š C8 andL0 D L1 \N Š C2; thus, L is abelian of type .8; 2/. We haveˆ.L/ D ˆ.L1/ Š C4,where ˆ.L/ > L0. For each x 2 G � L, x2 2 N , by (��), and ˆ.G/ D Ã1.G/ D

ˆ.L/N . This implies that there exists b 2 G � L such that b2 2 N � L0. Assumethat this is false. Then all elements in .G=L0/ � .L=L0/ are involutions so G=L0 isgenerated by involutions and, since G=L0 is not dihedral, it is not generated by twoinvolutions. SinceG=L0 is minimal nonmodular, all its involutions commute soG=L0

is elementary abelian; then d.G/ � 4, a contradiction. Since ˆ.L/ 6� Z.G/, we getCG.ˆ.L// D L so b inverts ˆ.L/. But thenD D hˆ.L/; bi is nonabelian metacyclicof order 24 and exponent 4; in that case,D=hb2i Š D8, a contradiction.We have proved that Z.G/ > N and so Z.G/ D ˆ.G/. It follows that each maximal

subgroup of G is abelian and so G is minimal nonabelian. In particular, jG0j D 2

and since G0 covers Z.G/=N D .G=N/0, we have Z.G/ D N � G0 is elementaryabelian of order 8. It follows that G is the uniquely determined minimal nonabeliannonmetacyclic group of order 25 and exponent 4: G D ha; b j a4 D b4 D 1; Œa; b� D

c; c2 D Œa; c� D Œb; c� D 1i, where Z.G/ D ha2; b2; ci, G0 D hci, and G=ha2c; b2ci

is the unique quotient group of G which is isomorphic to Q8. In particular, G is notQ8-free.We return now to the remaining case ˆ.N/ > f1g. Assume that N is not Q8-free.

Then N (being modular) is Hamiltonian (see �73 and Appendix 24). But d.N / D

2 and so N Š Q8. This is a contradiction since Z.N / is cyclic and N < ˆ.G/

(Burnside). We have proved that N is Q8-free and so N=Ã2.N / must be abelian.By Remark 3, N is ordinary metacyclic.Suppose that a maximal subgroupM of G is not Q8-free. ThenM (being modular)

is Hamiltonian and so M D Q � E, Q Š Q8, exp.E/ � 2. In particular, ˆ.M/

is of order 2 and exp.M/ D 4. In view of N < ˆ.G/ < M , we get exp.N / D 4

and N is abelian of type .4; 2/ since N is metacyclic. In that case, jGj D 26. Weget ˆ.M/ D Ã1.M/ D Ã1.N / D ˆ.N/ and so M=ˆ.N/ is an elementary abeliansubgroup of order 16 in the minimal nonabelian group G=ˆ.N/, contrary to Lemma65.1. Thus, all maximal subgroups of G are Q8-free.

Proposition 78.7. Suppose that d.N / D 2. Then for each maximal subgroup M ofG we have d.M/ D 3. Also, ˆ.N/ D Ã2.G/, E D �1.G/ D �1.ˆ.G// Š E8,E � Z.ˆ.G//, and either G=E Š Q8 with �2.G/ D ˆ.G/ being abelian of type.4; 2; 2/ or G=E is noncyclic and ordinary metacyclic.

Proof. By (�), �1.G/ is elementary abelian.


Set F D ˆ.G/ so we have F=ˆ.N/ D ˆ.G=ˆ.N// D �1.G=ˆ.N// Š E8

(Proposition 78.6). Since ˆ.N/ � ˆ.F /, we get ˆ.N/ D ˆ.F /. Thus d.F / D 3

and F is D8-free and, by Proposition 78.6, Q8-free, E D �1.F / Š E8 (by (��)) isnormal in G. Let M be any maximal subgroup of G so that M=ˆ.N/ is abelian oftype .4; 2; 2/, ˆ.N/ � ˆ.M/ and so d.M/ D 3. ButM is also D8-free and Q8-freeand therefore, by (��), �1.M/ Š E8 which implies �1.M/ D �1.F / D �1.G/

since �1.G/ � �1.M/ in view of �1.G=N/ D �1.M=N/ (Lemma 65.3).We have ˆ.N/ � Ã2.G/ (Proposition 78.6). On the other hand, exp.G=Ã2.G// D

4 and so each maximal subgroup of G=Ã2.G/, being modular and Q8-free, is abelian.Thus G=Ã2.G/ is minimal nonabelian of exponent 4 and so jG=Ã2.G/j � 25. Itfollows Ã2.G/ D ˆ.N/.If G=E is not D8-free, then there is a normal subgroup N � of G such that E � N �

and G=N � Š D8. By Proposition 78.1, N � must be metacyclic, a contradiction.Hence G=E is D8-free so it is modular.Suppose that G=E is not Q8-free. Then G=E is Hamiltonian (�73 or Appendix 24;

by the previous paragraph, G=E is modular). Since d.G=E/ D 2, we get G=E Š Q8

so �2.G/ D ˆ.G/ since E D �1.G/. On the other hand, G=E cannot act faithfullyon E since Aut.E/ has no subgroups Š Q8, and so CG.E/ � ˆ.G/. In particular,ˆ.G/ is abelian of type .4; 2; 2/.We assume that G=E is Q8-free. In that case, by (��), G=E is ordinary metacyclic

but noncyclic since ˆ.G/ � E. There is a cyclic normal subgroup S=E of G=E withthe cyclic factor-group G=S . Let s 2 S be such that S D hE; si and let r 2 G � S besuch that G D hS; ri. Since E � ˆ.G/, we have G D hr; si.Since S D hE; si is a proper subgroup of G, it follows that S is D8-free and,

by Proposition 78.6, Q8-free. Since S=hsiS is a subgroup of the symmetric groupSjS jWhsij D S4, whose Sylow 2-subgroup is isomorphic to D8, that quotient group isabelian and contains a subgroup isomorphic S=.Shsi/ Š E4 so hsi is normal in S ands induces an automorphism of order � 2 on E which implies that s2 centralizes E.Since hE; ri < G, we get (as in the previous paragraph) that r2 centralizes E. On

the other hand, ˆ.G/ D hE; r2; s2i and so we get again E � Z.ˆ.G//.

We summarize our results in a somewhat different form.

Theorem 78.8 (Janko). Let G be a minimal nonmodular 2-group of order > 25. Theneach proper subgroup of G is Q8-free and G=Ã2.G/ is minimal nonabelian of order24 or 25.

(a) Suppose that jG=Ã2.G/j D 24. If N is any normal subgroup of G such thatG=N Š D8, then N is cyclic. If G=Ã2.G/ is nonmetacyclic, then G is Q8-freeand �1.G/ Š E8 with G=�1.G/ cyclic. If G=Ã2.G/ is metacyclic, then G isalso metacyclic and G is either minimal nonabelian or an A2-group.

(b) Suppose that jG=Ã2.G/j D 25. Then G=Ã2.G/ is nonmetacyclic, G is notQ8-free and�1.G/ Š E8 withG=�1.G/ Š Q8 or G=�1.G/ is ordinary meta-


cyclic (but not cyclic). Moreover, if N is any normal subgroup of G such thatG=N Š D8, then N is ordinary metacyclic but noncyclic.

Remark 4. Let us change the last paragraph of the proof of Proposition 78.1 by thefollowing argument. Let A be a minimal nonabelian subgroup of G. Let jAj > 8.If Z.A/ is cyclic, setting N D .N \ A/ � N1, we get G D A � N1, d.G/ D 4, acontradiction. Let Z.A/ Š E4. Then A is metacyclic without cyclic subgroup of index2. Since A is D8-free, jAj D 25 andG D A�N1, whereN < N1 is of order 2. In thatcase,G is modular, a contradiction. If Z.A/ Š E8, thenA is D8-free soA D G. In thatcase, G0 < N , which is a contradiction. Now let jAj D 8; then A Š Q8. By Lemma1.1, jG0j D 4. If L is as in the last paragraph of the proof of Proposition 78.1, thenG0 < L and G0 6� N D �1.L/ so G0 Š C4. Then A0 D �1.G

0/ and AN=A0 Š E16,a contradiction since G=A0 is minimal nonabelian in view of j.G=A/0 D G0=A0 Š C2

(Lemma 65.2(a)) so must be j�1.G=A0/j � 8 (Lemma 65.1).

2o. We recall that a p-group G is modular if and only if any subgroups X and Y ofG are permutable, i.e., XY D YX . We turn now to the case p > 2.

Proposition 78.9. Let G be a modular p-group with p > 2 and d.G/ D 2. Then G ismetacyclic.

Proposition 78.10. Let G be a minimal nonmodular p-group, p > 2, which is gener-ated by two subgroups A and B of order p. Then G Š S.p3/ (the nonabelian groupof order p3 and exponent p).

Proof. Since G is a p-group, G1 D hAGi and G2 D hBGi are proper normal sub-groups of G and so G1 and G2 are modular. It follows that G1 and G2 are ele-mentary abelian. But hG1; Bi D hA;Bi D hG2; Ai D G, and so G1 and G2 aretwo distinct maximal subgroups of G. By Lemma 64.1(u), we have jG0j D p andG0 � G1 \ G2. Thus G=G0 is abelian and G=G0 is generated by elementary abeliansubgroups G1=G

0 and G2=G0. Hence G=G0 is elementary abelian and d.G/ D 2 im-

plies that G=G0 Š Ep2 . Thus, G Š S.p3/ since the metacyclic nonabelian group oforder p3 is modular.

Proposition 78.11 (see Theorem 44.13). Let G be a minimal nonmodular p-group.Then G possesses a normal subgroup N such that d.N / � 2, N � Ã1.G/, and G=Nis a nonmodular group of order p3. If p D 2, then G=N Š D8 and if p > 2, thenG=N Š S.p3/, N D Ã1.G/, and N is metacyclic. The subgroup N is metacyclic. Inparticular, if p > 2, then G is nonmetacyclic.

Proof. There are non-permutable cyclic subgroups hai and hbi. It follows G D ha; bi

and so d.G/ D 2. Since hap; bpi � ˆ.G/, the subgroups E D hap ; bi and F D

ha; bpi are proper subgroups of G. Hence E and F are modular and so E D hapihbi,F D haihbpi, and G D hE;F i. Set N D hapihbpi so that jE W N j D jF W N j D p.It follows that N is normal in G, N � Ã1.G/, and d.N / � 2. It remains to determine


the structure of NG D G=N D hNa; Nbi, where NG is a minimal nonmodular p-groupgenerated by elements Na and Nb of order p. If p D 2, then NG is dihedral, and, by (�),NG Š D8. If p > 2, then Proposition 78.10 implies that NG Š S.p3/. In that case wehave N D Ã1.G/ and Proposition 78.9 implies that N is metacyclic.

Proposition 78.12. Let G be a minimal nonmodular p-group, p > 2, with jGj > p4.Then �1.G/ is elementary abelian of order � p3.

Proof. Let A and B be subgroups of order p in G such that AB ¤ BA. Then G D

hA;Bi and so Proposition 78.10 implies that G Š S.p3/, a contradiction. We haveproved that AB D BA and so hA;Bi is abelian. Hence�1.G/ is elementary abelian.Assume that each proper subgroup of G is metacyclic. By Proposition 78.11, G is

nonmetacyclic and so jGj � p4 (Theorem 69.1 for p > 2), a contradiction.LetM be a nonmetacyclic maximal subgroup of G. SinceM is modular, Proposi-

tion 78.9 implies that d.M/ � 3. By Proposition 78.11, G is not a 3-group of maximalclass (since it has a maximal subgroupH which satisfiesH=Ã1.H/ Š S.33/, so non-modular. Therefore, if j�1.G/j < p3, then, by Theorem 13.7, G must be metacyclicso modular (Proposition 78.11), a contradiction. Thus, �1.G/j � p3.

Remark 5.Minimal nonmodular p-group of exponent p is isomorphic to S.p3/. In-deed, p > 2. All proper subgroups of G are elementary abelian so G Š S.p3/, byLemma 65.3.

Theorem 78.13. Let G be a minimal nonmodular p-group, p > 2, with jGj > p4. IfÃ1.G/ is cyclic, then �1.G/ Š Ep3 and G=�1.G/ is cyclic of order � p2 (i.e. G isan L3-group; see ��17, 18).

Proof. By Proposition 78.11, G=Ã1.G/ Š S.p3/. By assumption, N D Ã1.G/ iscyclic. By Proposition 78.12, E D �1.G/ is elementary abelian of order � p3. ButjE \ N j D p and E does not cover G=N , and so E Š Ep3 , by the product formula.On the other hand, there is a 2 G � N with hapi D N . Since jG W haij D p2 andjhai \Ej D p, we get G D hE; ai, by the product formula, and we are done.

Proposition 78.14. Let G be a minimal nonmodular p-group, p > 2, with jGj >

p4. Suppose that d.Ã1.G// D 2 and let M be any maximal subgroup of G. Thend.M/ � 3.

Proof. Suppose that this is false. LetM be a maximal subgroup of G with d.M/ � 4.Set N D Ã1.G/ so that M=N Š Ep2 and N=ˆ.N/ Š Ep2 . Since ˆ.N/ � ˆ.M/,we must have ˆ.M/ D ˆ.N/ so thatM=ˆ.N/ Š Ep4 . We shall study the structureof G=ˆ.N/ (which is also minimal nonmodular of order > p4 and exponent p2) andso we may assume ˆ.N/ D f1g which impliesM Š Ep4 . Since�1.G/ is elementaryabelian, we haveM D �1.G/. If x 2 G�M , then xp 2 Z.G/#. There is y 2 G�M

such that yp 2 N �hxpi sinceN D Ã1.G/; then again yp 2 Z.G/#. SinceN Š Ep2 ,we get N � Z.G/. If Z.G/ > N , then Z.G/=N D Z.G=N/ D ˆ.G=N/ and


so Z.G/ D ˆ.G/. But then G is minimal nonabelian. By Lemma 65.1, we getj�1.G/j � p3, a contradiction.We have proved that Z.G/ D N . By Lemma 1.1, jGj D p5 D pjZ.G/jjG0j and

so jG0j D p2. Since G=N Š S.p3/, G0 ¤ N.D Z.G//, G0 \ Z.G/ Š Cp andG0 < M so that CG.G

0/ D M . Since N D Ã1.G/, where is v 2 G � M suchthat vp 2 N � G0. It follows that the subgroup H D hG0; vi is nonabelian. Weclaim that H=hvi Š S.p3/. Assume that this is false. Then that quotient group iselementary abelian. Set NG D G=hvi; then C NG.

NG0/ � h NM; Nvi D NG so NG D Z. NG/.Since NG=Z. NG/ Š Ep2 , the group NG is minimal nonabelian so j NG0j D p (Lemma65.1), a contradiction. We have proved that each maximal subgroup of G is generatedby three elements.

Theorem 78.15 (Janko). Let G be a minimal nonmodular p-group, p > 2, with jGj >

p4. Then Ã1.G/ is metacyclic and G=Ã1.G/ Š S.p3/ (nonabelian group of orderp3 and exponent p). If Ã1.G/ is noncyclic, then ˆ.G/ D Ã1.G/�Cp, �1.ˆ.G// D

�1.G/ Š Ep3 , G=�1.G/ is metacyclic and for each maximal subgroup M of G wehave d.M/ D 3.

Proof. Set N D Ã1.G/ .< ˆ.G// and suppose that d.N / D 2. By Proposition78.12, �1.G/ is elementary abelian of order � p3 and �1.G/ \ N Š Ep2 . Since�1.G/ does not cover G=N Š S.p3/, N�1.G/ is contained in a maximal subgroupM of G. By Proposition 78.14, d.M/ � 3 and the modularity ofM implies d.M/ D

d.�1.M// [Suz1]. This implies�1.G/ Š Ep3 and .N�1.G//=N D ˆ.G=N/. Thusˆ.G/ D N�1.G/ and so for each maximal subgroup X of G, we have d.X/ D 3

since X � ˆ.G/ and d.X/ D d.�1.X// D d.�1.G//.We know that Aut.�1.G// does not possess an automorphism of order p2 (see, for

example, �33). There are elements a; b 2 G such that N D hapihbpi and ap and bp

centralize�1.G/. Hence ˆ.G/ D N � Z with jZj D p.If G=�1.G/ is nonmodular, then (Proposition 78.11) there is a normal subgroup K

of G with K � �1.G/, G=K Š S.p3/, and d.K/ � 2. This is a contradiction sinced.K/ D d.�1.K// D 3. Hence G=�1.G/ is modular and since d.G=�1.G// � 2,G=�1.G/ is metacyclic and our theorem is proved.

�79

Nonmodular quaternion-free 2-groups

Modular Q8-free 2-groups are classified in [Iwa] (see �73). Here we classify nonmod-ular Q8-free 2-groups. The original proof of the corresponding classification theorem,given in [Wil2], depends on the structure theory of powerful 2-groups. In addition,in the proof of Lemma 10 and 13 in [Wil2] there are some gaps. Our new proof ofthe classification theorem is completely elementary and does not involve powerful 2-groups. Nevertheless, the proof is very involved and reaches probably a deepest resultever proved in the finite 2-group theory.We first prove some easy preliminary results. Then we state the Main Theorem 79.7

and afterwards we describe in great detail the groups appearing in the Main Theorem.Propositions 79.8 to 79.11, describing these groups, are also of independent interestsince they are needed by applying the Main Theorem in future investigations. Afterthat the proof of the Main Theorem follows.

Lemma 79.1. In a Q8-free 2-group X there are no elements x; y with o.x/ D 2k > 2

and o.y/ D 4 so that xy D x�1. If D � X and D Š D8, then CX .D/ is elementaryabelian.

Proof. If y2 D x2k�1

, then hx; yi Š Q2kC1 . If hxi \ hyi D f1g, then we have

hx; yi=hx2k�1

y2i Š Q2kC1 . Suppose that D � X , where D D ha; t j a4 D t2 D

1; at D a�1i Š D8. if v is an element of order 4 in CX .D/, then o.tv/ D 4 and tvinverts a, a contradiction. Hence CX .D/ must be elementary abelian.

Lemma 79.2. Let X be a Q8-free 2-group with elements a and b of order 4 such thatŒa; b2� D Œa2; b� D 1. If Œa; b� ¤ 1, then ha; bi is minimal nonabelian nonmetacyclicof order 24 and therefore Œa; b� D a2b2 and ab is an involution.

Proof. Without loss of generality, one may assume that X D ha; bi holds. We haveha2; b2i � Z.X/. Set Œa; b� D c; then c ¤ 1. We compute

1 D Œa2; b� D Œa; b�aŒa; b� D cac and 1 D Œa; b2� D Œa; b�Œa; b�b D ccb:

By Lemma 79.1, c must be an involution and, by the displayed equalities, Œc; a� D

Œc; b� D 1. Hence hci is normal in X and X=hci is abelian. It follows that X 0 D hci

and so X , by Lemma 65(a), is minimal nonabelian (and so of class 2) and thereforeexp.X/ D 4. By assumption, X 6Š Q8, and X 6Š D8 since D8 has only one cyclic

�79 Nonmodular quaternion-free 2-groups 335

subgroup of order 4. We have proved that jX j � 24. Considering X=hci, we concludethat jha; bij � 25.On the other hand, c D Œa; b�, a2, and b2 are central involutions in X . Set V D

ha2c; b2ci so that V � Z.ha; bi/ and jV j D 4. We consider X=V and compute

ab D aŒa; b� D ac D a�1.a2c/; ba D bŒb; a� D bc D b�1.b2c/:

Since X=V is Q8-free, Lemma 79.1 implies that at least one of a2 or b2 is contained

in V . Hence a2 D b2c or b2 D a2c and so in any case c D a2b2 2 V sincea2c b2c D a2b2. It follows from the displayed equalities that X=V Š E4 so jX j D

jV jjE4j D 24.

Lemma 79.3 (see �78 and especially Proposition 78.4). Let G be a Q8-free minimalnonmodular 2-group of order> 8. ThenG has a normal elementary abelian subgroupE D �1.G/ D hn; z; ti of order 8 with G=E cyclic. There is an element x 2 G � E

of order 2sC1, s � 1, such that G D hE;xi, E \ hxi= hni, and tx D tz, zx D

zn�, � D 0; 1, where in case � D 1 we must have s > 1 and we have in that caseG0 D hn; zi Š E4 and Z.G/ D hx4i. If � D 0, then G is a minimal nonabeliannonmetacyclic group so E Š E8. In any case, hx2i is normal in G, G=hx2i Š D8,and ˆ.G/ D hx2; zi is abelian of type .2s ; 2/.

Lemma 79.4. Let V be a minimal non-quaternion-free 2-group. Then there is a normalsubgroup U of V such that V=U Š Q8 and U < ˆ.V / so that d.V / D 2. We haveˆ.V /=U D Z.V=U / so that for each x 2 V �ˆ.V /, x2 2 ˆ.V /� U . In particular,there are no involutions in V �ˆ.V /.

We use very often the following

Lemma 79.5 (= Lemma 64.1(u)). If A and B are two distinct maximal subgroups of ap-group G, then jG0 W .A0B 0/j � p.

For the sake completeness we also state the Iwasawa’s result in a suitable form.

Proposition 79.6 ([Iwa] and �73). A 2-group G is modular if and only if G is D8-free. A 2-group G is modular and Q8-free if and only if G possesses a normal abeliansubgroup A with cyclicG=A and there is an element g 2 G and an integer s � 2 suchthat G D hA;gi and ag D a1C2s

for all a 2 A (and so, if exp.G/ � 4, then G isabelian).

Main Theorem 79.7 (B. Wilkens). A finite 2-group G is nonmodular and quaternion-free if and only if G is one of the following groups:

(a) (Wilkens group of type (a) with respect to N ) G is a semidirect product hxi N ,where N is a maximal abelian normal subgroup of G with exp.N / > 2 and, if tis the involution in hxi, then every element in N is inverted by t .

(b) (Wilkens group of type (b) with respect toN ) G D N hxi, where N is a maximalelementary abelian normal subgroup of G and hxi is not normal in G.


(c) (Wilkens group of type (c) with respect to N , x, t ) G D hN;x; ti, where N is anelementary abelian normal subgroup of G and t is an involution with ŒN; t � D 1.If o.xN/ D 2k , then G=N Š M2kC1 , k � 3, and x2k

¤ 1; furthermore

Œx2k�1

; N � D 1 and ht; x2k�1

i Š D8.

We analyze now in great detail the above Wilkens groups of types (a), (b), and (c).In what follows we call these groups Wx-groups, where x 2 fa; b; cg so, for example,Wb-group is a Wilkens group of type (b).

Remark 1. A nonabelian 2-group G D hti N is said to be generalized dihedral withbase N , if t is an involution and N is an (abelian) subgroup of index 2 in G suchthat t inverts N . We claim that if also G D ht1i N1 is generalized dihedral withbase N1, then N1 D N . Assume that this is false. We have N \ N1 D Z.G/ andjG W .N \N1/j D 4, G D N [ Nt is a partition and all elements of the coset Nt areinvolutions and invert N . Therefore, if x 2 N1 \ Nt , then x centralizes and invertsN \ N ˛ so exp.N \ N1/ D 2. It follows that N1 D hx;N \ N1i is elementaryabelian, which is a contradiction since exp.N1/ > 2. In particular, the base of G ischaracteristic in G.

Proposition 79.8. Let G be a Wa-group with respect to N . Then �1.G/ D N hti,where t is an involution in G �N inverting N and N is characteristic in �1.G/ andso in G. If G is a Wa-group with respect to N1, then N D N1. Also, G is not D8-free but G is Q8-free. If z 2 �1.Z.G//, then G=hzi is either abelian or a Wa- orWb-group.

Proof. By hypothesis,G is a semidirect product hxiN , whereN is a maximal abeliannormal subgroup of G with exp.N / > 2 and, if t is the involution in hxi, then everyelement in N is inverted by t . Since G=N is cyclic, we have �1.G/ � N hti. Thecoset Nt consist of involutions and so �1.G/ D N hti. It follows that �1.G/ is ageneralized dihedral group with respect N . By Remark 1, N is the unique base of�1.G/ so it is characteristic in �1.G/ and in G. Thus, if G is also a Wa-group withrespect to N1, then N1 D N .Since t invertsN and exp.N / > 2, G is not D8-free. Suppose that G is not Q8-free.

Let V be a minimal non-Q8-free subgroup of G so that V has a normal subgroup Uwith V=U Š Q8 and ˆ.V / > U ; it follows that d.V / D 2. Since V 6� N and G=Nis cyclic, we see that V=.V \N/ > f1g is cyclic. Let t 0 be an element in V �N suchthat .t 0/2 2 N . Then Nt 0 is the involution in G=N and so all elements in Nt 0 areinvolutions. In particular, all elements in the set S D .V \ N/t 0 are involutions. ByLemma 79.4, S � ˆ.V / and so also hSi D .V \N/ht 0i � ˆ.V /. But then V=ˆ.V /,as an epimorphic image of V=.V \N/, is cyclic, a contradiction.Let z 2 Z.G/ be an involution. We want to determine the structure of G=hzi. We

know that G D hxi N is a semidirect product. Let t be the involution in hxi. Wehave z 2 N and t inverts N and so t inverts N=hzi. If exp.N=hzi/ > 2, then G=hzi isa Wa-group.


Let exp.N=hzi/ D 2. Set E D htiN . Then E=hzi D �1.G=hzi/ is a maximalelementary abelian normal subgroup of G=hzi. If hx; zi D hxi � hzi is not normalin G, then G=hzi is a Wb-group. Suppose that hx; zi is normal in G. Then G0 �

hx; zi \N D hzi so G0 D hzi. In that case, G=hzi is abelian, and we are done.

Remark 2. Let a 2-group G of exponent > 2 have two distinct elementary abeliansubgroups E and E1 of index 2. Then G D D � L, where D Š D8 and exp.L/divides 2. Since exp.G/ > 2, G is nonabelian. It follows that E \ E1 D Z.G/ hasindex 4 in G. Let A be a minimal abelian subgroup of G; then jA W .A \ Z.G//j > 2

so G D AZ.G/, by the product formula. In that case, A \ Z.G/ D Z.A/. If Z.G/ D

Z.A/�L, thenG D A�L and exp.L/ divides 2. Since A has two distinct elementaryabelian subgroup of index 2, then A Š D8 (Lemma 65.1).

Proposition 79.9. (i) A 2-group G is aWb-group with respect to E if and only if E isa maximal normal elementary abelian subgroup E such that G=E is cyclic and G isnot D8-free.

(ii) Let G be a Wb-group with respect to E. Then G is Q8-free. We have j�1.G/ W

Ej � 2. If j�1.G/ W Ej D 2, then G has exactly two maximal normal elementaryabelian subgroups E and E1 and we have �1.G/ D EE1. In that case, if G is alsoa Wb-group with respect to E1, then j�1.G/ W E1j D 2 and �1.G/ Š D8 � E2s .Let z 2 �1.Z.G//. Then G=hzi is either abelian or aWb-group or G=hzi Š D � F ,where exp.F / � 2 and either D Š D8 or D Š M2n , n � 4 (in which case G=hziis modular and nonabelian). Finally, if G is any 2-group with an elementary abeliannormal subgroup E0 such that G D hE0; yi (and so G=E0 is cyclic) and hyi is notnormal in G, then G is aWb–group with respect to any maximal elementary abeliannormal subgroup E of G containing E0.

Proof. (i) Let G be a nonmodular 2-group possessing a maximal elementary abeliannormal subgroup E such that G=E is cyclic. We have G D hE;xi for some x 2 G

and if hxi 6E G, then G is a Wb-group.Assume that hxi is normal in G. In that case, G0 � hxi \ E and, since G is

nonmodular (and so nonabelian), G0 D hxi\E D hzi Š C2. We have jG W CG.x/j D

2 since jG W CG.x/j � jG0j for each x 2 G. We setE1 D CE .x/ so that jE W E1j D 2

and E1 � Z.G/. Let t be an involution in E � E1 and let V be a complement ofhzi in E1 so that G D V � hx; ti, by the product formula. If jG=Ej D 2s > 2, thenhx; ti Š M2sC2 , s � 2 since a 2-group of maximal class has no cyclic epimorphicimages of order 2s > 2, and so for each a 2 A D hxi � V , at D a1C2s

sinceexp.V / divides 2. But Proposition 79.6 implies that G is modular, a contradiction.Hence jG=Ej D 2 and hx; ti Š D8. In that case Qx D xt is an involution in G � E,G D Eh Qxi, h Qxi is not normal in G, and so G is a Wb-group.

(ii) Conversely, let G be a Wb-group with respect to E so that G D hE;gi, whereE is a maximal normal elementary abelian subgroup of G and hgi is not normal in G.


Set Z D hgi \ E so that jZj � 2 and Z � Z.G/. Set S D NG.hgi/ so that S ¤ G

and S \ E < E . Since G D hgiS , we get NG.S/ D hgiNE .S/, by the modularlaw, so, in view of NE .S/ > E \ S , there is an involution n 2 E � S normalizingS . Since Œn; g� 2 Œn; S� � and Œn; g� 2 ŒE; g� � E, we get Œn; g� 2 S \ E and 1 ¤

u D Œn; g� 62 hgi since n 62 S D NG.hgi/. We have Œn; g� D ng�1ng D nng D u.On the other hand, ˆ.S/ D hg2i and so hg2i is normal in hS; ni so that ng2

D nz

with z 2 Z. Hence hgi normalizes hn; ng ; Zi and acts nontrivially on the four-grouphn; ng ; Zi=Z, where hg2i � Z. It follows that hn; gi=hg2i Š D8 and so G is notD8-free.We claim that G is Q8-free. Indeed, if V is a minimal non-Q8-free subgroup of G,

then, by Lemma 79.4, there are no involutions in V � ˆ.V / so that ˆ.V / � V \ E.But then V=ˆ.V / is cyclic since V=.V \E/ is, a contradiction.Set W=E D �1.G=E/ so that �1.G/ � W and j�1.G/ W Ej � jW W Ej D 2.

Suppose that j�1.G/ W Ej D 2 so that �1.G/ D W and there is an involutiont 2 W �E. Since E is a maximal normal elementary abelian subgroup ofG,W is notelementary abelian so hti is not normal inW , and we conclude that W is a Wb-groupwith respect to E. Let t1 2 W � E be an involution, t1 ¤ t . Then tE D t1E sinceW W Ej D 2. In that case t t1 2 E is involution so all involutions in W � E commutewith t . SetE1 D CW .t/ so that E1 �E is the set of all involutions inW �E, by whathas just been proved. Since hE1 � Ei D E1, E1 is normal in G and E and E1 arethe only maximal normal elementary abelian subgroups of G. If G is also a Wb-groupwith respect to E1, then G=E1 must be cyclic and so jW W E1j D 2 and in that caseW D �1.G/ Š D8 � E2s , by Remark 2.Let z be a central involution in G, where G D hE;xi and hxi is not normal in G;

then z 2 E since E is a maximal elementary abelian subgroup of G. If z 2 hxi,then hxi=hzi is not normal in G=hzi so G=hzi is a Wb-group with respect to E=hzi.Suppose that z 62 hxi. If hx; zi is not normal in G, then again G=hzi is a Wb-group.Assume that hx; zi D hxi�hzi is normal inG. If hxi\E D f1g, then hx; zi\E D hzi

and G0 � hx; zi \ E D hzi and so G=hzi is abelian. Assume that hxi \ E ¤ f1g sothat hx; zi \ E D �1.hxi/ � hzi Š E4 and suppose that G=hzi D NG is nonabelian.Then NG D h Nxi NE with h Nxi \ NE Š C2 and both h Nxi and NE are normal in NG so thatNG0 D h Nxi \ NE. Let Nt be an involution in NE which does not centralize h Nxi. If o. Nx/ D 4,then h Nx; Nti Š D8 and if o. Nx/ > 4, then h Nx; Nti Š M2n , n � 4 since h Nx; Nti \ NE Š E4

(see Theorem 1.2). We have NE D .h Nxi\ NE/�hNti� NV , where .h Nxi\ NE/� NV D C NE . Nx/

so that NG D NV � h Nx; Nti and we are done.

Proposition 79.10. Let G be a Wc-group with respect to N , x, t . Then G is Q8-free but is not D8-free. We have �1.G/ D hx2k�1

; tiN Š D8 � E2s (k � 3) andG D �1.G/hxi so that G=�1.G/ is cyclic of order � 4. Also, N is the uniquemaximal normal elementary abelian subgroup of G. No subgroup of order � 8 in hxi

is normal in G. The involution z D x2k

lies in G0 \ Z.G/ and G=hzi is aWb-group.If z0 2 �1.Z.G// and z0 ¤ z, then z0 2 N and G=hz0i is aWc-group.


Proof. By hypothesis, G D hN;x; ti, where N is an elementary abelian normal sub-group of G and t is an involution with ŒN; t � D 1. If o.xN/ D 2k , then G=N Š

M2kC1 , k � 3, and x2k

¤ 1; furthermore Œx2k�1

; N � D 1 and ht; x2k�1

i Š D8 so that

G is not D8-free. We set a D x2k�1

and z D a2. We have

.G=N/0 D �1.Z.G=N// D .haiN/=N and �1.G=N/ D .ht; aiN/=N D W=N;

where W D ht; aiN and Z.W / D N . Each involution in G is contained in W andW D �1.W / and so �1.G/ D W Š D8 � E2s (Remark 2) and G D �1.G/hxi

so that G=�1.G/ is cyclic of order � 4. By the structure of G=N , if X is a normalsubgroup of G with N � X � W , then X 2 fN;W; haiN g, where haiN is abelianof type .4; 2; : : : ; 2/ so that N is a maximal normal elementary abelian subgroup ofG. Let N1 be any maximal normal elementary abelian subgroup of G; then N1 <

�1.G/ D W . Assume that N1 ¤ N . Since N1 does not cover W=N (since allelements in .haiN/ � N are of order 4), we get j.NN1/ W N j D 2, NN1 G G and so(by the above) NN1 D haiN , a contradiction. Thus, N is the unique maximal normalelementary abelian subgroup of G so it is characteristic in G.Let Y � hxi with jY j � 8 and assume that t normalizes Y . Since t inverts hai

and hai < Y , it follows that Y hti is of maximal class and order 2m, m � 4, and so.Y hti/=hzi Š D2m�1 is isomorphic to a proper subgroup of G=N Š M2n , n � 4,which is not the case. Thus, Y is not normal in G.Let G be not Q8-free and let V � G be minimal non-Q8-free. By Lemma 79.4,

there are no involutions in V � ˆ.V / and so ˆ.V / � V \ N . On the other hand,ˆ.V / is contained in the maximal subgroup hxiN of G and note that�1.hxiN/ D N

(sinceN \ hxi D hzi and a centralizesN ). It follows that�1.ˆ.V // D V \N . Notethat G=N has exactly three involutions: Na, Nt , andN.at/, where all elements in thecoset Na are of order 4 and all elements in cosets Nt and N.at/ are involutions. Let.V \N/s (s 2 V ) be an involution in V=.V \N/. Then Ns is an involution in G=N .If all elements in the coset Ns are involutions, then s 2 ˆ.V /, contrary to the abovefact that�1.ˆ.V // D V \N . It follows thatNs D Na and so .V \N/s D .Na/\V

is the unique involution in V=.V \N/. SinceG=N is Q8-free, we get that V=.V \N/

is cyclic. But then V=ˆ.V / is also cyclic, a contradiction.We shall determine the structure of G=hzi D NG. Since �1.G/ D �1.G/=hzi is an

elementary abelian normal subgroup of NG, NG D �1.G/h Nxi, and h Nxi is not normal inNG (noting that z 2 hxi and hxi is not normal in G), NG is a Wb-group. Let z0 be aninvolution in Z.G/ and z0 ¤ z. Then z0 2 N and hz0i \ ha; ti D f1g so that G=hz0i isnot D8-free. Obviously, G=hz0i is a Wc-group.

Proposition 79.11. Let G be one of the Wilkens groups. Suppose that there is aninvolution z 2 Z.G/ such that G=hzi is modular (i.e., D8-free).

(i) If G=hzi is abelian, then G is a Wb-group and more precisely: G D D � E

where exp.E/ � 2 and D D hx; t jx2n

D t2 D 1, n � 1, Œx; t � D z, z2 D


Œx; z� D Œt; z� D 1i. (If n D 1, then D Š D8, and if n > 1, then D is minimalnonabelian nonmetacyclic with �1.D/ 6� Z.D/.)

(ii) If G=hzi is nonabelian, then there is another involution z0 2 Z.G/ such that hz0i

is a characteristic subgroup of G and G=hz0i is aWb-group.

Proof. (i) Suppose that G=hzi is abelian. Then G0 D hzi and by Propositions 79.8,79.9, and 79.10, G is either a Wa- or Wb-group.Suppose that G is aWa-group. Then G D hxi N (a semidirect product), where N

is a maximal normal abelian subgroup of G with exp.N / > 2 and if t is the involutionin hxi, then t inverts each element of N . Suppose n 2 N with o.n/ > 2. Thenhn; ti is dihedral and so n2 2 hn; ti0. It follows that n2 2 hzi and therefore N=hzi iselementary abelian with Ã1.N / D hzi and jN W �1.N /j D 2. Suppose hxi > hti

and let v 2 hxi with v2 D t . Let n 2 N with o.n/ D 4. We have Œn; v� ¤ 1 and soŒn; v� D z which gives nv D nz. But then nt D nv2

D .nz/v D nvzv D nzz D n

since z 2 Z.G/. This is a contradiction and so hxi D hti. If n 2 N with o.n/ D 4,then D D hn; ti Š D8. Let E be a complement of hzi in �1.N /, where t centralizes�1.N /. We get G D hti N D D � E, whereD Š D8 and exp.E/ � 2.Suppose that G is a Wb-group. Then G D hN;xi, where N is a maximal normal

elementary abelian subgroup of G and hxi is not normal in G. We have z 2 N andG0 D hzi. If z 2 hxi\N , then hxi is normal inG, a contradiction. Hence z 62 hxi andhx; zi D hxi � hzi is normal in G. Since hx; zi contains exactly two cyclic subgroupshxi and hxzi of index 2 not containing hzi, we have jG W NG.hxi/j D 2. There ist 2 N � NG.hxi/ such that xt D xz. Also note that NG.hxi/ centralizes hxi (sinceG0 D hzi). Let E be a complement of hz; hxi \ N i in NN .hxi/. Then G D D � E,whereD D hx; t jx2n

D t2 D 1; n � 1; Œx; t � D z; z2 D Œx; z� D Œt; z� D 1i.

(ii) Suppose that G=hzi is nonabelian. By Propositions 79.8, 79.9, and 79.10, G isa Wb-group. Then G D hN;xi, where N is a maximal normal elementary abeliansubgroup of G and hxi is not normal in G. If z 2 hxi \N , then the fact that hxi is notnormal in G gives that hxi=hzi is not normal in G=hzi and so G=hzi is a Wb-group,a contradiction. Hence z 62 hxi. If hx; zi is not normal in G, then again G=hzi is aWb-group, a contradiction. Hence hx; zi D hxi � hzi is normal in G. Assume firstthat hxi \N D f1g. Then hx; zi \N D hzi andG0 � hx; zi \N D hzi and soG=hziis abelian, a contradiction. Hence hxi \N > f1g and so hx; zi \N D �1.hxi/� hzi.SinceG0 � hx; zi\N andG=hzi is nonabelian (by assumption), we get 1 ¤ jG0j � 4

and G0 6� hzi. We have �1.hxi/ � Z.G/. Set �1.hxi/ D hx0i and S D NG.hxi/ sothat hx0; zi � Z.G/, jG W S j D 2 and jN W .S \ N/j D 2 because hxi is not normalinG and the abelian normal subgroup hx; zi has exactly two cyclic subgroups hxi andhxzi of index 2. Therefore, we have Œx; s� D z or Œx; s� D x0z for an s 2 N � S andso ˆ.G/ D hx2; Œhxi; N �i D hx2i � hzi.Suppose o.x/ � 8 so that ˆ.ˆ.G// D hx4i � hx0i and hx0i is a characteristic

subgroup of G. But hxi=hx0i is not normal in G=hx0i (since hxi is not normal in G)and so G=hx0i is a Wb-group, and we are done in this case.


Suppose o.x/ D 4 so that x2 D x0. If hxi 6� Z.S/, then there is an involutiont in S \ N which inverts hxi and so hx; ti Š D8. But then G=hzi is not D8-free, acontradiction: G=hzi is modular. Hence hxi is central in S and so G0 D hŒx; s�i D

hx0zi (since in case G0 D hŒx; s�i D hzi, G=hzi would be abelian). But then hx; si

is the minimal nonabelian nonmetacyclic group of order 24 with hx; si0 D hx0zi andhx; si=hzi Š D8, contrary to our assumption that G=hzi is modular.

Lemma 79.12. Let G be a Wb-group with respect to N . Suppose in addition that�1.G/ D N . Then for each element g 2 G such that G D hN;gi, N \ hgi D hg0i isof order 2 and G=hg0i is also aWb-group (and so G=hg0i is nonmodular).

Proof. It is enough to show that hgi is not normal in G (because then hgi=hg0i is alsonot normal inG=hg0i). Suppose false. ThenG0 � N \ hgi D hg0i and soG0 D hg0i.We have jG W CG.g/j D 2 and let t be an involution in N �CN .g/. If o.g/ D 4, thenhg; ti Š D8 and so gt is an involution inG �N , a contradiction. Hence o.g/ > 4 andhg; ti Š M2n , n � 4. If V is a complement of hg0i in CN .g/, then G D V � hg; ti.But then G is modular (see Proposition 79.6), a contradiction.

Proof of the Main Theorem

Let G be a nonmodular quaternion-free 2-group of a smallest possible order which isnot isomorphic to any Wilkens group. Hence any proper nonmodular subgroup andany proper nonmodular factor group is isomorphic to a Wilkens group. We shall studysuch a minimal counter-example G and our purpose is to show that such a group Gdoes not exist.

(i) There is a central involution z ofG such thatG=hzi is nonmodular (and soG=hziis isomorphic to a Wilkens group).

Suppose that this is false. Then for each z 2 �1.Z.G//, G=hzi is modular. Supposethat z0 2 �1.Z.G// is such that G=hz0i is modular. Since G is nonmodular, there isa minimal nonmodular subgroup K of G which is isomorphic to a group of Lemma79.3. Obviously, K is a Wb-group and so K ¤ G. Since G=hz0i is modular, we havez0 2 K. If K Š D8, then hz0i D �1.Z.G// D K 0. Suppose that K has a normalelementary abelian subgroup E D hn; z; ti of order 8 such that K D hE;xi, o.x/ D

2sC1, s � 1, E \ hxi D hni, tx D tz, zx D zn�, � D 0; 1, �1.K/ D E, and in case� D 1 we have s > 1 and Z.K/ D hx4i. If � D 1, then we must have z0 D n. Butthen K=hz0i is nonmodular since K=hx2i Š D8. Hence we have � D 0 in which caseK is minimal nonabelian nonmetacyclic with Z.K/ D hx2; zi D ˆ.K/ andK 0 D hzi.SinceK=hz0i is modular (andK=hni is nonmodular), we have either z0 D z (and thenK=hz0i is abelian) or z0 D zn (in which case s > 1 and K=hzni Š M2sC2).Let H be a maximal subgroup of G containing K. Since H is nonmodular and

H ¤ G, H is a Wilkens group. Since H=hz0i is modular, we may use Proposition79.11. If H=hz0i is nonabelian, then there is another involution z0

0 in Z.H/ such thathz0

0i is a characteristic subgroup inH andH=hz00i is nonmodular. But then z0

0 2 Z.G/


andG=hz00i is nonmodular, contrary to our assumption. HenceH=hz0imust be abelian

and so K=hz0i is also abelian. In particular, z0 D z, where hzi D K 0. In any case�1.Z.G// D hzi is of order 2. By Proposition 79.11(a), we haveH D D�E0, whereexp.E0/ � 2 and

D D hy; t jy2m

D t2 D 1; m � 1; Œy; t � D z; z2 D Œz; y� D Œz; t � D 1i:

If m D 1, then D Š D8 and if m > 1, then D is minimal nonabelian nonmetacyclicwith E8 Š �1.D/ 6� Z.D/ and z D z0, where hzi D D0 D H 0 D �1.Z.G//.Suppose m D 1. Then Z.H/ D hzi � E0 is elementary abelian. If jE0j � 4,

then acting with an element x 2 G � H on Z.H/, we see that jCZ.H/.x/j � 4 andCZ.H/.x/ � Z.G/, contrary to the fact that �1.Z.G// is of order 2. Hence jE0j � 2.If D D H Š D8, then CG.D/ � D would imply that G is of maximal class andthen G=hzi Š D8, a contradiction. If D D H Š D8 and CG.D/ 6� D, then Lemma79.1 implies that G Š D8 � C2, contrary to j�1.Z.G//j D 2. Hence we must haveH D D � hti, where t is an involution with CG.t/ D H . Since H=hzi is elementaryabelian, exp.G=hzi/ � 4 and therefore G=hzi is abelian since G=hzi is modular (andQ8-free) of exponent � 4. In particular,D is normal in G. We have CH .D/ D hz; ti.If CG.D/ > hz; ti, then CG.t/ D G, a contradiction. Hence CG.D/ D hz; ti andAut.D8/ Š D8 implies that G=hz; ti Š D8, a contradiction: G=hzi is modular.Suppose m > 1. Here Z.D/ D ˆ.D/ D hy2; zi is abelian of type .2m�1; 2/ and

Z.H/ D hy2; zi �E0 so that Ã1.Z.H// D hy4i. If m > 2, then�1.hy4i/ is of order

2 and �1.hy4i/ � Z.G/, contrary to the fact that �1.Z.G// D hzi. Thus we have

m D 2, jDj D 24, and Z.H/ D hy2; zi � E0 is elementary abelian. Suppose thatE0 ¤ f1g. Then acting with an element x 2 G �H on Z.H/, we get jCZ.H/.x/j � 4

and CZ.H/.x/ � Z.G/, a contradiction. It follows D D H and so jGj D 25.If x 2 G �H is of order 8, then x4 2 Z.H/ � hzi (with hzi D H 0) since ˆ.H/ D

Z.H/ and z is not a square in H . But then Z.H/ � Z.G/, a contradiction. Henceexp.G/ D 4 and the fact that G=hzi is modular gives that G=hzi is abelian. It followsG0 D H 0 D hzi. SinceH D D D hy; ti and jG W CG.y/j D jG W CG.t/j D 2, we getjG W CG.H/j � 4. But jH W CH .H/j D jH W Z.H/j D 4 and so CG.H/ must coverG=H . But then E4 Š Z.H/ � Z.G/, a final contradiction.

(ii) The factor group G=hzi (z 2 �1.Z.G//) is not isomorphic to a Wa-group.

Suppose that this is false. Then G=hzi has a maximal normal abelian subgroupN=hzi of exponent> 2 such thatG=N is cyclic of order� 2 and ifL=N D �1.G=N/,then for each element x 2 L �N , x2 2 hzi, and x inverts each element of N=hzi.If all elements in L�N are involutions, then each y 2 L�N inverts each element

inN which implies thatN is abelian of exponent > 2, N is a maximal normal abeliansubgroup of G and if G D hN;gi, then G is a semidirect product of N and hgi andthe involution in hgi inverts N . Thus, G is a Wa-group, a contradiction. Hence, thereis v 2 L �N with v2 D z.Let n 2 N be of order 8. Then nv D n�1z� (� D 0; 1) and therefore .n2/v D

.n�1z�/2 D n�2, contrary to Lemma 79.1. Thus, exp.N / D exp.N=hzi/ D 4.


Suppose that N is abelian. Let s and l be elements of order 4 in N such thathsi \ hli D f1g. In that case o.sl/ D 4 and using Lemma 79.1, we get sv D s�1z,lv D l�1z and consequently .sl/v D s�1zl�1z D .sl/�1, a contradiction. HenceN is abelian of type .4; 2; : : : ; 2/ and since exp.N=hzi/ D 4, z 62 Ã1.N /. We setÃ1.N / D htiwith t ¤ z and so jN W �1.N /j D 2, where each element inN��1.N /

is of order 4 and has square equal to t . Also, hti D Ã1.N / is central inG. Suppose thata 2 N ��1.N / is such that a2 D t and (by Lemma 79.1) av D a�1z D a.zt/. ByLemma 79.2, hv; ai is minimal nonabelian nonmetacyclic of order 24 with Œv; a� D zt

and va is an involution. Suppose that v does not commute with an involution u 2

�1.N /. Then uv D uz, o.au/ D 4, and .au/v D a�1zuz D .au/�1, contraryto Lemma 79.1. It follows that CN .v/ D �1.N /, Œhvi; N � D htzi, L0 D htzi,�1.N / D Z.L/, ˆ.L/ D hz; ti, where hti D Ã1.N /. We have L �N D v�1.N / [

.va/�1.N /, where all elements in v�1.N / are of order 4 and their squares are equalz and all elements in .va/�1.N / are involutions. Thus E D hvai�1.N / D �1.L/

is elementary abelian of index 2 in L and also E D �1.G/ (since G=N is cyclic) isthe unique maximal normal elementary abelian subgroup ofG. Now, L=E is a normalsubgroup of order 2 in G=E with cyclic factor group G=L. Thus G=E is abelian. IfG=E were cyclic, then the fact that G is not D8-free gives that G is a Wb-group, acontradiction. Thus G=E is abelian of type .2s ; 2/, s � 1. If s > 1, then we setK=E D �1.G=E/ Š E4. Since K � L, K is not D8-free and so K < G implies thatK must be a Wilkens group. But E D �1.K/ and K=E is noncyclic, a contradiction(see Propositions 79.8 to 79.10). Hence s D 1, G=E Š E4, and so exp.G/ D 4. Wehave G=N Š C4 and so for each x 2 G �L, x2 2 L�N and so x2 2 E �N . Sincethe square of each element inG is contained in hzi or hti or inE�N , it follows that tzis not a square in G. Hence�1.G=htzi/ D E=htzi. If G=htzi were nonmodular, thenG=htzimust be a Wilkens group and then .G=htzi/=.E=htzi/ Š G=E must be cyclic,a contradiction. It follows that G=htzi is modular and since exp.G/ D 4, G=htziis abelian and so G0 D htzi. Suppose that x 2 G � L is such that x2 D E � N .Then Œv; x� 2 htzi and so Œv; x2� D Œv; x�2 D 1. But then v centralizes E and sinceL D hE; vi, we get that L is abelian, a contradiction. Thus, N is nonabelian and soN 0 D hzi.The subgroup N is nonmodular because a Q8-free modular 2-group of exponent 4

is abelian. Let S be a minimal nonabelian subgroup of N . Then S 0 D hzi and S isnormal inN . Since v invertsN=hzi, v normalizes S and so S is normal inL D hN; vi.Since exp.S/ D 4 and S is Q8-free, it follows that either S Š D8 or S is minimalnonabelian nonmetacyclic of order 24. If S Š D8, then Ã1.S/ D hzi and N 0 D hzi

implies that CN .S/ covers N=S . In that case, Lemma 79.1 implies that CN .S/ iselementary abelian and so Ã1.N / D hzi, contrary to exp.N=hzi/ > 2. It follows thatwe have the second possibility:

S D ha; t j a4 D t2 D 1; Œa; t � D z; z2 D Œa; z� D Œt; z� D 1i:

We put b D at and compute b2 D a2t2Œt; a� D a2z, so that S 0 D hzi, o.b/ D 4,


hai\hbi D f1g, and S D ha; biwith Œa; b� D a2b2 D z. Again, sinceN 0 D S 0 D hzi,we get that CN .S/ coversN=S and CN .S/\ S D Z.S/ D ˆ.S/ D hz; a2i. Supposethat y 2 CN .S/ is of order 4. Since hai \ hbi D f1g, there is an s 2 fa; bg so thathsi \ hyi D f1g and o.sy/ D 4. We get .sy/v D s�1zy�1z D s�1y�1 D .sy/�1,contrary to Lemma 79.1. Thus, CN .S/ is elementary abelian and N D SCN .S/,S \ CN .S/ D Z.S/ and so the structure of N is completely determined.We act with hvi on S D ha; bi and get (using Lemma 79.1) av D a�1z D a.a2z/,

bv D b�1z D ba2, which together with v2 D z determines uniquely the structure ofT D Shvi.Suppose that v does not centralize CN .S/ D Z.N /. Then there is an involution

s 2 CN .S/ � S such that sv D sz. Then o.as/ D 4 and .as/v D a�1zsz D

.as/�1, contrary to Lemma 79.1. Hence CN .S/ D CN .T / and L D TCL.T / withT \ CL.T / D Z.T / D hz; a2i.We have �1.S/ D hz; a2; abi Š E8, hz; a2i D Z.T /,

.av/2 D avav D av2v�1av D aza�1z D 1;

.ab/av D .abz/v D a�1zb�1zz D a�1b�1z D aa2bb2z

D aba2.a2z/z D ab:

Hence F D hz; a2; ab; avi Š E24 is an elementary abelian maximal subgroup of Tand so from T 0 � hz; a2i and jT j D 25 D 2jT 0jjZ.T /j follows that T 0 D Z.T / D

hz; a2i. Finally, T=hz; a2i is elementary abelian and therefore Z.T / D T 0 D ˆ.T / Š

E4 and so T is a special group of order 25. For each x 2 T � F , CF .x/ D Z.T / andso the set T0 D T �F has exactly four square roots of z, four square roots of a2, foursquare roots of za2 and so T0 must contain exactly four involutions in T � S . If t0 isone of them, then CF .t0/ D Z.T /, and F and hz; a2; t0i Š E8 are the only maximalnormal elementary abelian subgroups of T (containing all involutions of T ).We have CN .S/ D CN .T / D Z.L/ and so if we set U D FZ.L/ and V D

hz; a2; t0iZ.L/, then L D UV , U \ V D Z.L/, U and V are the only maximalnormal elementary abelian subgroup of L and they are of distinct orders and so U andV are normal in G and jU W Z.L/j D 4, jV W Z.L/j D 2, L0 D ˆ.L/ D hz; a2i.For each t0 2 V � Z.L/, CU .t0/ D Z.L/ and so both U an V are self-centralizingin L. Also, U is the unique abelian maximal subgroup of L (otherwise, by a result ofA. Mann, jL0j � 2). Now, G=N is cyclic, L=N D �1.G=N/, and so �1.G/ D L,which is a Wb-group with respect to U . In particular, L < G.If G=U is cyclic, then (since G is not D8-free) G is a Wb-group, a contradiction.

Hence G=U is noncyclic. But L=U � Z.G=U / and G=L is cyclic (since L � N andG=N is cyclic) and so G=U is abelian of type .2n; 2/. Set K=U D �1.G=U / Š E4.If G ¤ K, then K is a Wilkens group with �1.K/ D L. By the structure of L, L hasno abelian maximal subgroups of exponent > 2 and so K is not a Wa-group. Also,jK W �1.K/j D 2 and so K is not a Wc-group. Hence K must be a Wb-group withrespect to U . But K=U Š E4, a contradiction. Hence G D K and so G=U Š E4


implies exp.G/ D 4. Since �1.G/ D L, all elements in G � L are of order 4 and ifx 2 G �L, then x2 2 U �N , where N D hZ.L/; ab; ai and U \N D Z.L/� habi.If CG.V / > V , then CL.V / D V implies that there is y 2 G � L with y2 2 V ,

contrary to y2 2 U �N . We have proved that CG.V / D V and soG=V acts faithfullyon V . We have G=V Š .U hxi/=Z.L/, where U=Z.L/ Š E4 and hxiZ.L/=.Z.L/ Š

C4/ with x 2 G � L. Thus G=V Š D8 or G=V Š C4 � C2. But L=V Š U=Z.L/ Š

E4 is a four-subgroup inG=V and for each x 2 G�L, x2 2 U �N so that hxi\V D

f1g. Hence all elements in .G=V / � .L=V / are of order 4 and so G=V Š C4 � C2.If L0 D hz; a2i D Z.L/, then V Š E8. But G=V Š C4 � C2 cannot act faithfully

on V Š E8. We have proved that Z.L/ > L0 and so jZ.L/j � 8.We act with hxi on Z.L/, where x 2 G � L. Since x2 2 L, hxi induces an

automorphism of order� 2 on Z.L/. Since jZ.L/j � 8, it follows that jCZ.L/.x/j � 4.Suppose that x centralizes an involution u 2 Z.L/ � L0 so that u 2 Z.G/. SinceG=hui is nonabelian and of exponent 4, G=hui must be nonmodular. Thus G=hui

is a Wilkens group with �1.G=hui/ D L=hui because �1.G/ D L and u is not asquare in G. Then U=hui and V=hui are the only maximal normal elementary abeliansubgroups of G=hui and both G=U and G=V are noncyclic and so G=hui cannot bea Wb-group. Also, G=hui cannot be a Wa-group since �1.G=hui/ D L=hui andjG=Lj D 2. If G=hui is a Wa-group, then (by the first part of the proof of (ii))L must possess a nonabelian maximal subgroup N0 containing hui such that N 0

0 D

hui. This is a contradiction since u 2 Z.L/ � L0. We have proved that for eachx 2 G � L, CZ.L/.x/ � L0. Since jCZ.L/.x/j � 4, we must have CZ.L/.x/ D L0 D

Z.G/ D hz; a2i.Suppose that x 2 G�L is such that x2 2 U�N . We haveU D hz; a2; ab; aviZ.L/

and so .ab/x2

D ab. On the other hand, x2 2 L � N and x2 inverts N=hzi. Thusax2

D a�1z� , bx2

D b�1z� .�; � D 0; 1/, and so, noting that b2 D a2z we get

ab D .ab/x2

D a�1z�b�1z� D a�1b�1z�C� D aa2bb2z�C� D abz1C�C�;

which gives � C � 1 .mod 2/ and so � D 1 or � D 1.Suppose � D 1. Then ax2

D a�1z D a.a2z/ and we apply Lemma 79.2 inthe group G=ha2zi D NG. We have (using bar convention) o. Nx/ D o. Na/ D 4 andŒ Nx2; Na� D 1 D Œ Nx; Na2�. If Œ Nx; Na� D 1, then Œx; a� 2 ha2zi (and hx; ai is of class � 2

since hx; ai0 � ha2zi) and so Œx2; a� D Œx; a�2 D 1, a contradiction. Thus Œ Nx; Na� ¤ 1

and so (by Lemma 79.2) o.xa/ D 2. Hence .xa/2 2 ha2zi, contrary to the fact thatxa 2 G � L and .xa/2 2 U �N .Let � D 1. Then bx2

D b�1z D b.a2/ and we apply Lemma 79.2 to G=ha2i D NG.We have o. Nx/ D o. Nb/ D 4 and Œ Nx2; Nb� D 1 D Œ Nx; Nb2�. If Œ Nx; Nb� D 1, then Œx; b� 2 ha2i

(and hx; bi is of class � 2 since hx; bi0 � ha2i) and so Œx2; b� D Œx; b�2 D 1, acontradiction. Thus Œ Nx; Nb� ¤ 1 and so (by Lemma 79.2) o.xb/ D 2. Hence .xb/2 2

ha2i, contrary to the fact that xb 2 G � L and .xb/2 2 U � N . Our claim (ii) isproved.


(iii) G=hzi (z 2 �1.Z.G//) is not isomorphic to a Wb-group.

Suppose this is false. Then G=hzi has a maximal normal elementary abelian sub-group N=hzi so that G=N is cyclic and G=hzi is not D8-free . If N is elementaryabelian, then G is a Wb-group with respect to N , a contradiction. Hence N is notelementary abelian and so Ã1.N / D hzi and G ¤ N . Set L=N D �1.G=N/.

(˛) Let N be abelian. Then jN W �1.N /j D 2 and for each a 2 N � �1.N /,a2 D z. Also, hN � �1.N /i D N , �1.G/ � L, and �1.G=hzi/ � L=hzi. Letv 2 L �N with v2 D z and let x 2 N ��1.N /. By Lemma 79.2, Œv; x� D 1 and sov centralizes N . But then L is abelian with Ã1.L/ D hzi and L G G, a contradiction:N=hzi is a maximal normal elementary abelian subgroup of G=hzi. Thus, for eachv 2 L �N , v2 ¤ z.Suppose that for each x 2 L � N , x2 2 N � �1.N /. If G D hN;gi, then hgi

covers G=�1.N /. But then G=�1.N / is cyclic and (since G is not D8-free) G is aWb-group, a contradiction. We have proved that there is x 2 L�N with x2 2 �1.N /.Suppose that there are no involutions in L � N . There is x 2 L � N such that

x2 2 �1.N / � hzi. Suppose that x does not commute with an a 2 N � �1.N /.By Lemma 79.2, xa is an involution, a contradiction. Hence x commutes with allelements in N ��1.N / and therefore L is abelian of type .4; 4; 2; : : : ; 2/. We have�1.L/ D �1.N / D �1.G/ and L=�1.L/ Š E4. Since G is nonabelian, we haveL < G. Because N=�1.N / is a normal subgroup of order 2 in G=�1.N / and G=Nis cyclic, G=�1.N / is abelian of type .2s; 2/, s � 2. For each y 2 L � N , y2 D x2

or y2 D x2z. Hence, if g 2 G is such that G D hN;gi, then �1.hgi/ D hx2i or�1.hgi/ D hx2zi and so Ã1.L/ D hz; x2i � Z.G/. We may assume (by a suitablenotation) that hgi � hxi. SetM D hgi�1.N / so thatM is a maximal subgroup of G.Suppose that G=hx2i is nonmodular so that it is a Wilkens group. But�1.G=hx

2i/ D

S0=hx2i, where S0 D �1.N /hxi. On the other hand, G=S0 is noncyclic (since L=S0

andM=S0 are two nontrivial cyclic subgroups ofG=S0 with .L=S0/\.M=S0/ D f1g).This is a contradiction and soG=hx2i is modular. Since�1.G=hzi/ D N=hzi, we mayuse Lemma 79.12 and we see that G=hx2; zi is nonmodular. This is not possible sinceG=hx2i is modular. Thus, there are involutions in L �N .Let t be an involution in L � N . If t centralizes an element a 2 N � �1.N /,

then ta 2 L � N and .ta/2 D z, a contradiction. Thus, CN .t/ � �1.N /. For anyx 2 L � N , CN .x/ D CN .t/ and so x2 2 �1.N /. It follows that exp.L/ D 4.Suppose that t does not centralize�1.N /. Let w be a fixed involution in �1.N / �

CN .t/ and let a be a fixed element in N ��1.N /. We have wt D wu with 1 ¤ u 2

CN .t/ and .tw/2 D .twt/w D wuw D u and so u 2 CN .t/�hzi (since tw 2 L�N ).Since CN .tw/ D CN .t/ < �1.N /, we have Œtw; a� ¤ 1. By Lemma 79.2, .tw/ais an involution. We get 1 D .twa/2 D twatwa D wtatwa D wuatwa D uaat ,and so at D a�1u D aa2u D a.uz/. We consider the factor group L=huzi D NL andwe see that o. Na/ D 4, o.Nt / D 2, Œ Na; Nt � D 1, Na2 D Nz. Also, wt D wu D wz.uz/

gives Nw Nt D Nw Nz so that h Nw; Nti Š D8 with Z.h Nw; Nti/ D hNzi. Hence h Nw; Ntih Nai is thecentral product D8 � C4, contrary to Lemma 79.1 (applied in NL). We have proved


that t centralizes�1.N / so that S D hti ��1.N / is an elementary abelian maximalsubgroup of L. Since L is nonabelian and exp.L/ D 4, L is nonmodular.Let a 2 N ��1.N /. Then by Proposition 51.2, at D a�1s, s 2 �1.N /. If s D 1,

then t invertsN ,G=N is cyclic and soG would be a Wa-group, a contradiction. Thus,s ¤ 1 and .ta/2 D .tat/a D a�1sa D s and so s 2 �1.N / � hzi. Set ta D y sothat y2 D s and since CN .y/ D CN .t/, we have for each n 2 CN .t/ D �1.N /,.ny/2 D n2y2 D s and Œt; N � D hzsi D L0. The group L has exactly three abelianmaximal subgroups: S D �1.G/, N , and �1.N /hyi, where only S is elementaryabelian and all three are normal in G with Ã1.�1.N /hyi/ D hsi, s 2 �1.N / � hzi.We have �1.N / D Z.L/ and z; s and zs are central involutions in G.If G D L, then G would be a Wb-group, a contradiction. Thus L < G. Since

G=N is cyclic of order 2s , s � 2, we have G D hN;gi with g2s�1

2 L � N . Ifg2s�1

2 L � N � S , then hgi covers G=S , G=S is cyclic and so G would be a Wb-group with respect to S , a contradiction. It follows that g2s�1

2 S � �1.N /. If Lis not maximal in G, then there is a subgroup K > L with jK W Lj D 2 and so Kis a Wilkens group. Since S D �1.K/ and K=S Š E4, we have a contradiction.Indeed, .Shg2s�2

i/=S and L=S are two distinct subgroups of order 2 in K=S . HencejG W Lj D 2 and for each g 2 G � L, g2 2 S ��1.N /.We apply now Lemma 79.2 in the group G=hzsi D NG, where hzsi D L0. We have

for some elements g 2 G � L and a 2 N � �1.N /, o. Ng/ D o. Na/ D 4 and sinceag2

D a.zs/, we get Œ Na; Ng2� D 1 D Œ Na2; Ng�. If Œ Ng; Na� D 1, then Œg; a� 2 hzsi and soŒg2; a� D Œg; a�2 D 1, a contradiction. Hence Œ Ng; Na� ¤ 1 and so (by Lemma 79.2)o. Ng Na/ D 2 which gives .ga/2 2 hzsi, contrary to ga 2 G � L and (by the above).ga/2 2 S ��1.N /. We have proved that N must be nonabelian.

(ˇ) Let N be nonabelian.

This case is very difficult. We have Ã1.N / D N 0 D hzi. Let D be a minimalnonabelian subgroup ofN so thatD0 D hzi. Since d.D/ D 2 andD=hzi is elementaryabelian, we have D=hzi Š E4 and therefore D Š D8. The subgroup D is normal inN and since D0 D N 0, CN .D/ covers N=D. By Lemma 79.1, CN .D/ is elementaryabelian. We have CN .D/ D Z.N /, N D DZ.N / with D \ Z.N / D Z.D/ D hzi sothat Z.N / is normal in G. Set D D ha; u j a4 D u2 D 1; a2 D z; au D a�1i sothat A D ha;Z.N /i, E1 D hu;Z.N /i, and E2 D hau;Z.N /i are all abelian maximalsubgroups of N , where A is of exponent 4 (all elements in A � Z.N / are of order 4,Ã1.A/ D hzi), and E1 and E2 are both elementary abelian. Thus A is normal in G.All elements inN �A are involutions, Z.N / D �1.A/, N is a Wa-group with respectto A and also a Wb-group with respect to E1 and E2.Since N=A is a normal subgroup of order 2 in G=A and G=N is cyclic, it follows

that G=A is abelian. If G=A were cyclic, then we have G D hA;gi with some g 2 G

and since �1.hgi/ is of order 2 and inverts A (of exponent > 2), G is a Wa-group, acontradiction. It follows that G=A is abelian of type .2m; 2/, m � 1, and L=A Š E4,where L=N D �1.G=N/. Also note that N=Z.N / Š E4.


Suppose that for each l 2 L � N , l2 2 A � Z.N /. This is equivalent to assumingthat L=Z.N / Š C4 � C2 which also implies that both E1 and E2 are normal in L.If G D hN;gi, then either Eg

1 D E2 (in which case G > L and G=Z.N / Š M2n ,n � 4, and G is a Wc-group) or E

g1 D E1, E1 is normal in G, hgi covers G=E1 and

since G is not D8-free, G is a Wb-group. In both cases we have a contradiction.Thus, L=Z.N / is not isomorphic to C4 � C2 and so L=Z.N / is either elementary

abelian or L=Z.N / Š D8. In any case, there is l 2 L � N with l2 2 Z.N /.

(ˇ1) Let L=Z.N / Š D8. Then Ex1 D E2 for x 2 L � N , and so G D L. Indeed,

G=N acts on the set fE1; E2g and so, if G > L, then L would normalize E1.Suppose in addition that there are no involutions in L � N . Let l 2 L � N with

l2 2 Z.N / so that l2 ¤ 1 and o.l/ D 4. Let a0 be any element in A � Z.N / so thata2

0 D z. If Œl; a0� ¤ 1, then la0 is an involution in L � N , a contradiction. Thus lcentralizes each element in A � Z.N / and since hA � Z.N /i D A, Ahli is an abelianmaximal subgroup of G D L. Also, Z.N / D Z.G/ (since Z.L=Z.N // D A=Z.N /)and so we may use the relation jGj D 2jG0jjZ.G/j which gives jG0j D 4. But G0

covers A=Z.N / D .G=Z.N //0 and so G0 Š C4 is a cyclic subgroup of order 4 inA inverted by u (since u inverts A) and so we have G0hui Š D8. We may assumeG0 D hai < D Š D8 so thatD is normal inG. By Lemma 79.1, CG.D/ is elementaryabelian and so CG.D/ cannot cover G=N (since there are no involutions in G � N ).Hence CG.D/ D Z.N / and so l 2 G � N induces an outer automorphism on D.Hence al D a�1, contrary to Lemma 79.1.We have proved that there are involutions in G �N and let t be one of them so that

�1.G/ D G. Since G=Z.N / Š D8, there is k 2 G � N such that k2 2 A � Z.N /.Our ultimate goal is to show that Z.N / D Z.G/. Suppose Z.N / ¤ Z.G/.Assume there is s 2 G�N with s2 D z and let a0 2 A�Z.N /. If Œs; a0� ¤ 1, then

(Lemma 79.2) Œs; a0� D s2a20 D zz D 1, a contradiction. Thus s centralizesA�Z.N /

and hA � Z.N /i D A and so Z.N / � Z.G/. Since Z.G=Z.N // D A=Z.N /, we haveZ.N / D Z.G/. This is a contradiction and so there is no s 2 G � N with s2 D z. Inparticular, the involution t does not centralize any element in A � Z.N /.Now, Ahti < G maximal. If x 2 At is of order 8, then o.x2/ D 4 and x2 2

A � Z.N /. But then t centralizes x2 (noting that CA.t/ D CA.x/), a contradiction.Hence exp.Ahti/ D 4 and since Ahti is nonabelian, Ahti is nonmodular and thereforeAhti must be a Wilkens group. If X is an abelian maximal subgroup of Ahti distinctfrom A, then X \ A � Z.N / (recalling that t does not commute with any element inA � Z.N /) and since jA W .X \ A/j D 2, we get X � Z.N / and so Z.N / D Z.G/, acontradiction. Hence A is the unique abelian maximal subgroup of Ahti.If all elements in .Ahti/�A are involutions, then t inverts each element of A and so

t centralizes Z.N / and then Z.N / D Z.G/, a contradiction. It follows that there is anelement c of order 4 in At such that c2 2 Z.N /� hzi. But Œc; a� ¤ 1 (since t does notcentralize a) and so (by Lemma 79.2) ac is an involution for each a 2 A�Z.N /. Sincet does not centralize Z.N / (otherwise Z.N / D Z.G/), Z.N /hti contains less than2jZ.N /j � 1 involutions. All jZ.N /j elements ca (a 2 A� Z.N /) are involutions and


so Ahti contains at least 2jZ.N /j � 1 involutions. This shows that �1.Ahti/ D Ahti.Since A (of exponent > 2) is the unique abelian maximal subgroup of Ahti, Ahtimustbe a Wa-group. In that case t inverts A and so t centralizes Z.N / and Z.N / D Z.G/,a contradiction.We have proved that Z.N / D Z.G/ and so Z.N /hki D Ahki is an abelian maximal

subgroup of G (noting that k2 2 A � Z.N / and so Z.N /hki=Z.N / is cyclic). Usingthe relation jGj D 2jG0jjZ.G/j we get jG0j D 4 and so G0 Š C4 (since G0 coversA=Z.N /). We may assume (as before) that G0 � D and so D is normal in G withCG.D/ D Z.N / (since CG.D/ is elementary abelian and jG W Z.G/j D 8). Theinvolution t induces an outer automorphism on D and so Dhti Š D24 . It follows thatG D .Dhti/ � E2m for some m � 1, which is a Wa-group, a contradiction.

(ˇ2) We have proved that we must have L=Z.N / Š E8 and so exp.L/ D 4.In that case we prove first that there are involutions in L�N . Suppose false. If v is

any element in L � N and a0 2 A � Z.N /, then 1 ¤ v2 2 Z.N / and a20 D z so that

we may apply Lemma 79.2. If Œv; a0� ¤ 1, then va0 is an involution, a contradiction.Hence L D hL � N i centralizes A D hA � Z.N /i. In particular, A � Z.N / whichcontradicts the fact that Z.N / < A. We have proved that there are involutions inL�N

and so L D �1.L/ D �1.G/ (noting that G=N is cyclic and L=N D �1.G=N/).Assume CG.D/ > Z.N / so that CG.D/ covers L=N and (Lemma 79.1) CG.D/

is elementary abelian. In that case, L Š D8 � E2m and Ã1.L/ D hzi which con-tradicts our assumption that N=hzi is a maximal normal elementary abelian subgroupof G=hzi. Hence CG.D/ D Z.N /. If D were normal in L, then L=Z.N / Š D8

since Aut.D/ Š D8. This is a contradiction and so NG.D/ D N . In particular,Z.L/ � Z.N / and Z.N / > hzi.

(ˇ2a) We assume that G > L. Then L D �1.L/ (being nonmodular) is a Wa- orWb-group. In particular, L has an abelian maximal subgroup B . Since B \ N is anabelian maximal subgroup ofN , we get B\N 2 fA;E1; E2g. HenceB\N � Z.N /and so Z.N / � Z.L/. By the result in the previous paragraph, we get Z.N / D Z.L/.Since jL W Z.L/j D 8, B is the unique abelian maximal subgroup of L and so B isnormal inG. Using the relation jLj D 2jL0jjZ.L/j, we get jL0j D 4. SinceL0 � Z.L/,L0 Š E4 and L0 > hzi.It is easy to see that jG W Lj D 2. Suppose that this is false. Let K < G be

such that jK W Lj D 2. Since L D �1.K/, K must be a Wa- or Wb-group. Bythe uniqueness of B in L, follows that K=B is a cyclic group (of order 4). SinceN < L and G=N is cyclic, G=L is cyclic. But L=B is a normal subgroup of order2 in G=B and so G=B is abelian. We have L=B � �1.G=B/. On the other hand,G=L is cyclic and K=L D �1.G=L/ so that �1.G=B/ � K=B . Since K=B Š

C4, we get �1.G=B/ � L=B and so �1.G=B/ D L=B . Hence G=B is cyclic. IfL is of type (a) (in the case exp.B/ > 2), then all elements in L � B are involutionsand so G is also a Wa-group, a contradiction. If L is of type (b), then B must beelementary abelian and (sinceG is not D8-free)G is also a Wb-group, a contradiction.We have proved that jG W Lj D 2 and G=B Š E4 (because in case G=B Š C4,


G would be a Wa- or Wb-group). But G=N Š C4 and so for each x 2 G � L,x2 2 B �N .

We assume first that exp.B/ > 2. Then L is a Wa-group. In that case B=Z.N / Š

E4, all elements in L � B are involutions and �1.B/ D Z.N /. Indeed, if jB W

�1.B/j D 2, then acting by conjugation with an involution in L � B on B , we seethat jL0j D 2, a contradiction. Hence we must have �1.B/ D Z.N / so that B isabelian of type .4; 4; 2; : : : ; 2/, B \ N D A, u inverts B , each element in B � N

is of order 4, and if k 2 B � N , then k2 D z0 2 L0 � hzi (noting that ku D k�1

and so k2 2 L0) and L0ha; ki D ha; ki Š C4 � C4 (if k2 D z, then ak wouldbe an involution in B � N , contrary to �1.B/ D Z.N /). Let x 2 G � L so thatx2 2 B � N and we may assume that x2 D k, where k2 D z0 2 L0 � hzi. Inparticular, CG.z0/ � hL; xi D G and so L0 � Z.G/. Let x0 be an arbitrary elementin G � L so that .x0/2 D k0 2 B �N and .k0/2 D z0 2 L0 � hzi. Consider the factorgroup NG D G=hz0i. We have o. Na/ D o.x0/ D 4 and Œ Na2; x0� D 1 D Œ Na; .x0/2� and sowe may use Lemma 79.2. If Œ Na; x0� ¤ 1, then o. Nax0/ D 2 and so .ax0/2 2 hz0i � N ,a contradiction. Hence we must have Œ Na; x0� D 1 and so Œa; x0� 2 h.x0/4i. It followsthat Œa; x� D z�

0 .� D 0; 1/. Consider the element y D xu 2 G � L; then Œa; y� D

Œa; xu� D Œa; u�Œa; x�u D zz�0 ¤ 1. On the other hand, Œa; y� 2 hy4i, y4 2 L0 � hzi.

Thus � D 1 and y4 D zz0. Finally, consider the factor group QG D G=hzz0i so thato. Qy/ D o. Qk/ D 4 and Œ Qy; Qk2� D 1 D Œ Qy2; Qk� and apply again Lemma 79.2. SinceŒk; x� D 1, we get Œk; y� D Œk; xu� D Œk; u�Œk; x�u D k2 D z0, and so Œ Qk; Qy� ¤ 1.Thus o. Qk Qy/ D 2 and so .ky/2 2 hzz0i � N . This is a contradiction since ky 2 G�L.

We study now the case exp.B/ D 2 and L is a Wb-group. We may assume thatB \ N D E1. Since G=B Š E4, exp.G/ D 4. Let t be an involution in B � N .Since B � E1 D Z.N /hui, t centralizes u 2 D. But NL.D/ D N and so Œa; t � 2

L0 � hzi. Suppose that L0 � Z.G/. Let x 2 G � L so that x2 2 B � N and(by the above) Œa; x2� D z0 2 L0 � hzi. Consider the factor group G=hz0i D NG

so that o. Na/ D o. Nx/ D 4. If Œ Nx; Na� D 1, then Œx; a� 2 hz0i. But hx; ai0 D hz0i

and so hx; ai is of class 2. Thus Œx2; a� D Œx; a�2 D 1, a contradiction. HenceŒ Nx; Na� ¤ 1 and so (by Lemma 79.2) o. Nx Na/ D 2, which gives .xa/2 2 hz0i, contraryto xa 2 G � L and .xa/2 2 B � N . We have proved that L0 6� Z.G/. Again,let x 2 G � L so that hxi induces an involutory automorphism on Z.L/ D Z.N /.Suppose Z.N / > L0. Then there is an involution z0 2 Z.L/ � L0 centralized by xand so z0 2 Z.G/. Since G=hz0i is nonmodular (noting that D \ hz0i D f1g), G=hz0i

is a Wilkens group. All squares of elements of G lie either in B-N or in L0. Indeed,let as .s 2 B/ be any element in L � B . Then .as/2 D a2.a�1sas/ D zŒa; s� 2 L0.Therefore z0 is not a square of any element inG which implies L=hz0i D �1.G=hz

0i/.Let y 2 L be such that Œy;L� � hz0i. But hz0i \ L0 D f1g and so Œy;L� D f1g

and therefore y 2 Z.L/. Hence Z.L=hz0i/ D Z.L/=hz0i. Since jL W Z.L/j D 8, theelementary abelian group B=hz0i is the unique abelian maximal subgroup of L=hz0i.It follows thatG=hz0imust be a Wb-group with respect to B=hz

0i. But thenG=B mustbe cyclic, a contradiction. We have proved that L0 D Z.L/ and so jGj D 26. Now,


A D hai � hz0i D haiL0 Š C4 � C2 is normal in G and is self-centralizing in L(since B is the unique abelian maximal subgroup of L). If CG.A/ 6� L, then thereis g 2 CG.A/ � L such that g2 2 A � N , contrary to g2 2 B � N . Thus A isself-centralizing inG and soG=A Š D8 since Aut.A/ Š D8. On the other hand, N=Ais a normal subgroup of order 2 in G=A and G=N Š C4 so that G=A is abelian. Thisis a contradiction. We have proved that the case G > L is not possible.

(ˇ2b) It remains to study the case G D L D �1.G/. Since G=Z.N / Š E8,exp.G/ D 4. If for each x 2 G � N , x2 2 hzi, then G=hzi is elementary abelian,contrary to our assumption that N=hzi is a maximal normal elementary abelian sub-group of G=hzi. Hence there is k 2 G �N such that k2 2 Z.N /� hzi. It follows thatk2 2 Z.G/ and so hzi < Z.G/ � Z.N /.Let z0 2 Z.G/ � hzi. Since G=hz0i is nonmodular (noting that D \ hz0i D f1g

with D Š D8), it follows that G=hz0i is a Wilkens group with �1.G=hz0i/ D G=hz0i

and so G=hz0i cannot be a Wc-group. Hence G=hz0i must be a Wb-group by ourprevious result (ii). Thus, G has a maximal subgroup N1 containing z0 such thatN1=hz

0i is a maximal normal elementary abelian subgroup of G=hz0i. By our previousresult (iii)(˛), hz0i D Ã1.N1/ D N 0

1 (since N1 must be nonabelian). In particular,z0 is a square of an element in G � N and z0 is a commutator in G. Conversely, ifk 2 G � N , then k2 2 Z.G/ so that ˆ.G/ � Z.G/. Hence Z.G/ D G0 D ˆ.G/

and so G is a special group. Now, N \ N1 is a maximal subgroup of N and sinceÃ1.N \N1/ � hzi \ hz0i D f1g, N \N1 is elementary abelian and soN \N1 D E1

(or E2) containing Z.N / and (by the structure of N1 Š N ) Z.N1/ is a subgroup ofindex 2 in N \N1. But Z.N / \ Z.N1/ � Z.G/ and so jZ.N / W Z.G/j � 2.Let jZ.N / W Z.G/j D 2 and let s be an involution in Z.N / � Z.G/. Let t be

an involution in G � N . We have Œt; s� D s0 with s0 2 Z.G/ and s0 ¤ 1 sinceZ.N / ¤ Z.G/. If n 2 N , then Œtn; s� D Œt; s�Œn; s� D Œt; s� D s0 2 Z.G/. Hence, foreach x 2 G�N , Œx; s� D s0, where s0 is a fixed involution in Z.G/. Take an involutionz0 2 Z.G/�hzi. LetN1 be a maximal subgroup ofG such that hz0i D Ã1.N1/ D N 0

1.Then N \ N1 is an elementary abelian maximal subgroup of N containing Z.N /. Iff1 2 N1 � N , then Œf1; s� D z0 D s0. Let N2 (¤ N1) be a maximal subgroup of Gsuch that hzz0i D Ã1.N2/ D N 0

2. Then again, N \N2 � Z.N / and if f2 2 N2 �N ,then Œf2; s� D zz0 D s0. Hence zz0 D z0 and so z D 1, a contradiction.We have proved that Z.N / D Z.G/. Suppose that Z.G/ possesses a four-subgroup

hz1; z2i such that hz1; z2i \ hzi D f1g (which is equivalent with the assumptionjZ.G/j � 8). Let k1; k2 2 G � N be such that k2

1 D z1 and k22 D z2. Sup-

pose that k1 centralizes all elements (of order 4) in A � Z.N /. Then k1 centralizesA D Z.N /hai D hA � Z.N /i and so Ahk1i is an abelian maximal subgroup of G.Using a result of A. Mann (Lemma 79.5 with respect to maximal subgroups Ahk1i

and N ), we get jG0j � 4. But G0 D Z.G/ is of order � 8, a contradiction. We mayassume that Œa; k1� ¤ 1 and so (Lemma 79.2) Œa; k1� D a2k2

1 D zz1. We have eitherŒa; k2� D 1 or Œa; k2� ¤ 1 in which case (Lemma 79.2) Œa; k2� D a2k2

2 D zz2.We have Œa; k1k2� D Œa; k1�Œa; k2� D zz1Œa; k2�, and so either Œa; k1k2� D zz1


or Œa; k1k2� D zz1zz2 D z1z2. But the element k1k2 is contained in N and soŒa; k1k2� 2 hzi, a contradiction.We have proved that we must have Z.N / D Z.G/ Š E4 and so jGj D 25. Let

z0 2 Z.G/� hzi and let k1; k2 2 G �N be such that k21 D z0 and k2

2 D zz0. SupposeŒk1; k2� D 1 so that hk1; k2i Š C4 � C4, k1k2 2 N and .k1k2/

2 D k21k

22 D z

and therefore we may assume that k1k2 D a. Hence hk1; k2i is an abelian maximalsubgroup of G normalized by u, where u inverts each element in

hk1; k2i \N D hai � hz0i Š C4 � C2 and G D hk1; k2ihui:

We know that there are involutions inG�N and so there is an element k 2 hk1; k2i�N

such that uk is an involution. This gives .uk/2 D ukuk D 1, ku D k�1 and so uinverts each element of the abelian group hk1; k2i. In this case G is a Wa-group, acontradiction. Hence Œk1; k2� ¤ 1 and using Lemma 79.2 we get Œk1; k2� D k2

1k22 D

z0.zz0/ D z, o.k1k2/ D 2, k1k2 2 N . Because k1k2 62 Z.hk1; k2i/, we may assumek1k2 D u (an involution inN �A). Since Œk1; u� D Œk1; k1k2� D Œk1; k2� D z (whichgives uk1 D uz) and D D ha; ui Š D8 is not normal in G, we have Œa; k1� ¤ 1.By Lemma 79.2, Œa; k1� D a2k2

1 D zz0 and k1a is an involution in G � N . We haveuk1a D .uz/a D .uz/z D u. Hence hu; k1ai Š E4 with hu; k1ai \ Z.G/ D f1g andso hu; k1a;Z.G/i Š E16. Since G is not D8-free, it follows that G is a Wb-group.This is our final contradiction and so our statement (iii) is completely proved.

(iv) The factor group G=hzi (z 2 �1.Z.G//) is not isomorphic to a Wc-group.

Suppose that the assertion (iv) is is false. ThenG=hzi is a Wc-group. Hence we mayset�1.G=hzi/ D H=hzi (implying that �1.G/ � H ) so thatH D H1H2, whereH1

and H2 are normal subgroups in H with H1 \H2 D hzi, H1=hzi Š D8, H2=hzi iselementary abelian and G=H is cyclic of order � 4. Let Z=hzi be the unique cyclicsubgroup of index 2 inH1=hzi and set Z.H1=hzi/ D Z0=hzi so that jZ W Z0j D 2 andjZ0 W hzij D 2. Let Z0H2 D N , ZH2 D A, so that Z.H=hzi/ D N=hzi and N=hziis the unique maximal normal elementary abelian subgroup of G=hzi, G=N Š M2n ,n � 4, A=N D .G=N/0, H=N D �1.G=N/ Š E4. If E1=N and E2=N are othertwo subgroups of order 2 in H=N (distinct from A=N ), then EG

1 D E2, E1=hzi

and E2=hzi are elementary abelian and A=hzi is abelian of type .4; 2; : : : ; 2/. Also,NG.E1/ D NG.E2/ > H and jG W NG.E1/j D 2. Finally, G possesses a subgroup Ssuch that G D HS , H \ S D Z, and S=hzi is cyclic so that S is abelian. In fact, Sis either cyclic or abelian of type .2n; 2/, n � 4. Also, S is cyclic if and only if Z0 iscyclic (since Z0=hzi D �1.S=hzi/). We have H 0 � Z0 andH 0 covers Z0=hzi.

(˛) Suppose Z0 (and so also S ) is cyclic.

We have H1=hzi Š D8 and .H1=hzi/0 D Z0=hzi. Since H 0

1 covers Z0=hzi andZ0 is cyclic, we get that H 0

1 D Z0 is of index 4 in H1. By a very well-known resultof O. Taussky, H1 is of maximal class. Since H1 is Q8-free, we get H1 Š D24 . Aninvolution t 2 H1 � Z inverts Z and so H3 D hZ0; ti Š D8. The subgroup H3

is normal in H1 and ŒH3;H2� � H1 \ H2 D hzi implies that H3 is normal in H .


Since CH1.H3/ D hzi and H1=hzi Š D8 Š Aut.H3/, we get that CH .H3/ covers

H=H1. By Lemma 79.1, CH .H3/ is elementary abelian. In particular, �1.H/ D

H D �1.G/. Since H is nonmodular, H is a Wilkens group with �1.H/ D H

and so H is either a Wa- or Wb-group. It follows that H has an abelian maximalsubgroup QA which is unique (by a result of A. Mann) since H 0 � H 0

1 and jH 01j D 4

(see Lemma 79.5). In particular, QA is normal in G. Also, QA \H1 D Z Š C8 sinceZ is the unique abelian maximal subgroup of H1 Š D24 and so exp. QA/ > 2. Ift 2 H1 � Z, then H must be a Wa-group and the involution t inverts on QA. LetQN D �2. QA/ D Z0�1. QA/ (since QA=Z is elementary abelian) so that QN is normal inG, and QN=hzi is a normal elementary abelian subgroup of G=hzi. By the uniquenessof N=hzi, QN � N and so QA D Z QN � ZN D A and therefore QA D A is abelianof type .8; 2; : : : ; 2/. Let G1 > H be such that jG1 W H j D 2. Hence G1 ¤ G andG D HS1, where S1 D S \ G1. It follows that G1 is also a Wilkens group with�1.G1/ D H . Since jG1 W H j < 4, G1 is of type (a) or (b). But A D QA is the uniqueabelian maximal subgroup of H and exp.A/ > 2. Thus G1 must be a Wa-group withrespect to A and so G1=A Š C4. On the other hand, we know that S1 \ H D Z.Hence, if x 2 S1 �H , then x2 2 Z � A. This is a contradiction since G1=A Š C4

and therefore x2 2 H � A.

(ˇ) We have Z0 Š E4 and so S splits over hzi.

We setZ D hai�hzi so thatZ0 D ha2i�hzi and replacing awith az (if necessary),we may put S D hsi � hzi with �2.hsi/ D hai D S \ H . If jH 0

1j D 4, thenjH1 W H 0

1j D 4, H 01 D Z0, and (by a result of O. Taussky) H1 is of maximal class. In

that case Z0 would be cyclic, a contradiction.We have proved that jH 0

1j D 2 and so Z0 D H 01 � hzi. We set H 0

1 D hz0i so thatz0 is a central element in H . But S is abelian, G D HS , and S \H D Z > Z0 D

hz; z0i. It follows CG.hz; z0i/ � hH;Si D G and so hz; z0i � Z.G/.We show that there are exactly two possibilities for the structure of H1. Since

H1=Z0 Š E4, we have exp.H1/ D 4. Suppose thatH1 is minimal nonabelian. IfH1

is metacyclic (of exponent 4), then we know that H1 is not Q8-free. Thus H1 mustbe nonmetacyclic and we know that there is only one such minimal nonabelian groupof order 24 and exponent 4. In particular, there is an element b 2 H1 � Z such thatb2 D z, Œa; b� D a2b2 D a2z D z0, where H 0

1 D hz0i, z0 is not a square in H1,and ab is an involution so that �1.H1/ D hz; z0; abi. Suppose now that H1 is notminimal nonabelian. Then there is a subgroup D Š D8 in H1 which covers H1=hzi.ThusH1 D D � hzi andH 0

1 D D0 D hz0i. We haveD \Z D hai orD \Z D hazi

and all elements in H1 � Z are involutions inverting ha; zi. Replacing a with az (ifnecessary), we may assume that D \ Z D hai. If t is an involution in D � Z, thenD D ha; ti, where z0 D a2 is a square inH1.Suppose thatH2 is nonabelian. ThenH 0

2 D hzi sinceH2=hzi is elementary abelian.Let H4 be a minimal nonabelian subgroup of H2 so that H 0

4 D hzi, H4=hzi is ele-mentary abelian and d.H4/ D 2. Thus H4=hzi Š E4 and soH4 Š D8. The subgroupH4 is normal inH2 andH2 centralizesH4=hzi. We have ŒH1;H4� � H1 \H2 D hzi


and so H1 also centralizes H4=hzi and H4 is normal in H . Thus, H centralizesH4=hzi and therefore there is no h 2 H inducing an outer automorphism on H4

(because otherwise such an element h would act nontrivially on H4=hzi). It fol-lows that CH .H4/ covers H=H4. But H=H4 is nonabelian since H4 � H2 andH=H2 Š H1=hzi Š D8. This contradicts Lemma 79.1. We have proved thatH2 mustbe abelian and so H2 is either abelian of type .4; 2; : : : ; 2/ or elementary abelian. Inany case, N D Z0H2 D hz0i �H2 since z0 2 Z.G/.

(ˇ1) Suppose that H2 is abelian of type .4; 2; : : : ; 2/, where Ã1.N / D Ã1.H2/ D

hzi. Set E D �1.H2/ so that �1.N / D hz0i � E and all elements in H2 � E are oforder 4. Let h be an arbitrary element inH2 �E so that h2 D z.We consider first the possibility that H1 is minimal nonabelian nonmetacyclic. Let

x be any element of order 4 in H1 so that x2 D z or x2 D zz0, where hz0i D H 01.

Suppose that Œx; h� ¤ 1. By Lemma 79.2, Œx; h� D x2h2 D x2z. On the other hand,Œx; h� � H1 \ H2 D hzi and so Œx; h� D z which gives x2 D 1, a contradiction.Hence Œx; h� D 1 and since H1 is generated by its elements of order 4 (noting that�1.H1/ Š E8) andH2 D hH2 �Ei, we get ŒH1;H2� D f1g. We apply Lemma 79.1in the factor group NH D H=hzz0i. We have NH1 Š D8 and hH1; hi D NH1 � h Nhi is thecentral product of NH1 with h Nhi Š C4, where NH1 \ h Nhi D Z. NH1/, a contradiction.We consider now the possibility, whereH1 D D � hzi with

D D ha; t j a4 D t2 D 1; at D a�1i; a2 D z0; and hz0i D D0:

If Œa; h� ¤ 1, then Œa; h� 2 H1 \ H2 D hzi and so Œa; h� D z. On the other hand,Lemma 79.2 implies Œa; h� D a2h2 D z0z, a contradiction. Hence Œa; h� D 1 andso a centralizes H2 D hH2 � Ei. It follows that A D ZH2 D hai � H2 is anabelian maximal subgroup of H with exp.A/ D 4 and A is normal in G. We have�1.H1/ D H1 and so �1.H/ contains the maximal subgroup H1E of H , whereE D �1.H2/ with jH2 W Ej D 2. If Œt; h� D 1, then D D ha; ti Š D8 centralizeshhi Š C4, contrary to Lemma 79.1. Hence Œt; h� ¤ 1 and since Œt; h� 2 H1 \H2 D

hzi, we get Œt; h� D z with z D h2. Thus, ht; hi Š D8 and so th is an involutionin H � .H1E/. We have proved that �1.H/ D H and since �1.G/ � H , we getalso �1.G/ D H . Also, H 0 D hz; z0i Š E4 and therefore, by a result of A. Mann(Lemma 79.5), A is the unique abelian maximal subgroup of H . Take a subgroup G1

of G with H < G1 < G and jG1 W H j D 2. It follows that G1 must be a Wa-groupwith�1.G1/ D H since jG1 W H j D 2 and A is the unique abelian maximal subgroupof H (with exp.A/ > 2). In that case G1=A is cyclic. On the other hand, settingS1 D S \ G1, we have G1 D HS1 and S1 \ H D Z. Thus, if g 2 S1 � H , theng2 2 Z � A. This contradicts the fact that G1=A Š C4.

(ˇ2) We have proved that H2 (and so also N D hz0i �H2) is elementary abelian.Suppose first that H1 is minimal nonabelian nonmetacyclic. In this case z0 is not asquare in H1, where hz0i D H 0

1. There is b 2 H1 � Z with b2 D z, t D ab is aninvolution, and a2 D zz0. Now, A D ZN D haiN is normal in G and .htiN/s D


hbiN (recalling that S D hsi � hzi) since G=N Š M2n , n � 4, and so G actsnontrivially on the four-group H=N . The subgroup �1.H/ contains the maximalsubgroup htiN of H , where �1.H1/ D hz; z0; ti Š E8 and N \ H1 D hz; z0i. Ift centralizes H2, then htiN is elementary abelian. But .htiN/s D hbiN and hbiN

is not elementary abelian, a contradiction. Hence t does not centralize H2 and soŒH2; t � D hzi since ŒH2; t � � H1 \ H2 D hzi. It follows that htiN is nonabelianand so hbiN is also nonabelian. In particular, H 0 D hz; z0i Š E4, ŒH2; b� D hzi,and so hbi is normal in H2hbi. Let u be an involution in H2 with Œb; u� D z sothat hb; ui Š D8 and therefore bu is an involution. But bu 2 H � .htiN/ and so�1.H/ D H . It follows that H must be a Wa- or Wb-group. In that case H mustpossess an abelian maximal subgroupM which is also unique (by a result of A. Mannsince jH 0j D 4). We have M � H 0 D hz; z0i. If M does not contain H2, then Mcovers H=H2 which is nonabelian (since H=H2 Š H1=hzi Š D8), a contradiction.HenceM � H2 and soM � hH 0;H2i D N . Since htiN and hbiN are nonabelian,we get that M D A D haiN is abelian (of exponent 4). Thus H is a Wa-groupwith respect to A and so the involution t must invert each element in A. In particular,at D a�1 D aa2 D a.zz0/. This is a contradiction sinceH 0

1 D hz0i.It remains to investigate the case, whereH1 D D � hzi with

D D ha; t j a4 D t2 D 1; at D a�1i; a2 D z0; hz0i D D0

andS D hsi � hzi; hsi \H D hai; .htiN/s D hatiN

since G=N Š M2n , n � 4. Obviously, in this case �1.H/ D H D �1.G/. Supposethat t does not centralize H2. Then Œt;H2� � H1 \ H2 D hzi and so Œt;H2� D hzi

andH 0 D hz; z0i Š E4. Then htiN and hatiN D .htiN/s are nonabelian. ButH is aWilkens group with �1.H/ D H and so H must have an abelian maximal subgroupU which is unique (by a result of A. Mann). If U does not contain H2, U coversH=H2 and this is a contradiction sinceH=H2 Š D8. Hence U � hH 0;H2i D N andso U D A D haiN is abelian of exponent 4. Thus, H is a Wa-group with respectto A. In particular, the involution t inverts each element in A and so t centralizesH2,a contradiction. Hence t centralizes H2 and so htiN is elementary abelian. In thatcase hatiN D .htiN/s is also elementary abelian. In particular, D D ht; ati Š D8

centralizesH2 and soH D D �H2. It follows that G is a Wc-group, a contradiction.Our statement (iv) is proved.We have proved that the nonmodular factor group G=hzi (z 2 �1.Z.G//) (accord-

ing to our statement (i)) is not isomorphic to any Wilkens group (according to (ii), (iii),and (iv)). This is a final contradiction and so the Main Theorem is proved.

�80

Minimal non-quaternion-free 2-groups

Here we classify minimal non-quaternion-free 2-groups. A 2-group G is minimalnon-quaternion-free if G is not quaternion-free but each proper subgroup of G isquaternion-free. Recall that a 2-group G is modular if and only if G is D8-free.The main theorem is a consequence of results of the previous section.

Theorem 80.1. Let G be a minimal non-quaternion-free 2-group. Then G possessesa unique normal subgroup N such that G=N Š Q8. We have N < ˆ.G/ and soG=ˆ.G/ Š E4 and �1.G/ � ˆ.G/. If R is any G-invariant subgroup of index 2in N , then H2 D G=R is the minimal nonabelian metacyclic group of order 24 andexponent 4: H2 D ha; b j a4 D b4 D 1; ab D a�1i, where H2=ha

2b2i Š Q8,X=hb2i Š D8, and X=ha2i Š C4 �C2. In particular, if jGj > 8, thenG has a normalsubgroup S such that G=S Š D8 and so G is nonmodular.

Proof. The group G possesses a normal subgroup N such that G=N Š Q8. SupposethatA is a maximal subgroup ofG not containingN . ThenAN D G soA=.A\N/ Š

G=N Š Q8, a contradiction. Thus, N < ˆ.G/ so d.G/ D 2. It follows from�1.G=N/ D ˆ.G=N/ that �1.G/ � ˆ.G/. We may assume jGj > 8.Let R be any G-invariant subgroup of index 2 in N .We shall determine the structure of G=R. For that purpose we may assume R D f1g

so that jN j D 2, N < ˆ.G/, N � Z.G/, jˆ.G/j D 4, jGj D 24 and ˆ.G/=N D

Z.G=N/ D .G=N/0, where G=N Š Q8. By Theorem 1.2, G has no cyclic subgroupsof index 2 so ˆ.G/ is a four-subgroup. In that case, N < Z.G/ since G is not ofmaximal class soˆ.G/ D Z.G/ andG is minimal nonabelian. Since�1.G/ D ˆ.G/

is a four-subgroup, G is metacyclic (of exponent 4). Since H2 D ha; b j a4 D b4 D

1; ab D a�1i is the unique nonabelian metacyclic group of order 16 and exponent 4,we get G Š H2.Next we do not assume that R D f1g. Let M ¤ N be a G-invariant subgroup of

index 2 in ˆ.G/; then L D M \ N is of index 2 in N , by the product formula. Bythe previous paragraph, G=L Š H2 so G=M 2 fC4 � C2;D8g. In particular, G isnonmodular. This argument shows thatN is the unique G-invariant subgroup of index8 in G such that G=N Š Q8.

Let G be a minimal non-Q8-free 2-group of order > 8 such that each proper sub-group of G is modular. Then G is a minimal nonmodular 2-group and such groups

�80 Minimal non-quaternion-free 2-groups 357

have been classified in �78. Therefore, we may assume in the sequel thatG has a max-imal subgroup H which is nonmodular. Since H is Q8-free, we are in a position toapply Theorem 79.7 classifying nonmodular Q8-free 2-groups. Also, we assume thatthe reader is familiar with Propositions 79.8, 79.9 and 79.10 describing the Wilkensgroups of types (a), (b), and (c), which appear in Theorem 79.7.

Theorem 80.2. LetG be a minimal non-quaternion-free 2-group which has a nonmod-ular proper subgroup. Then G has one of the following properties:

(i) �1.G/ D ˆ.G/ D Ahti, where A is a maximal normal abelian subgroupof G with exp.A/ > 2, t is an involution inverting each element in A, andG=A Š Q8, D8, or C4 � C2. If G=A Š D8 or C4 � C2, then A is abelian oftype .4; 2; : : : ; 2/.

(ii) �1.G/ D EE1, where E ¤ E1 are the only maximal normal elementaryabelian subgroups of �1.G/ and j�1.G/ W Ej D 2, j�1.G/ W E1j � 2. Wehave either �1.G/ D ˆ.G/ (and then G=�1.G/ Š E4) or G=�1.G/ Š Q8.If G=�1.G/ Š Q8, then also j�1.G/ W E1j D 2 which implies �1.G/ Š

D8 � E2s .

(iii) �1.G/ D E is elementary abelian andG=E is isomorphic toQ8,M2n , n � 4,or C2m � C2, m � 1. If G=E Š M2n or C2m � C2 with m � 2, then �2.G/

is abelian of type .4; 4; 2; : : : ; 2/.

Proof. Let G be a minimal non-quaternion-free 2-group possessing a maximal sub-group H which is nonmodular. It follows that H is a Wilkens group of type (a), (b),or (c). In particular, H=�1.H/ is cyclic, where �1.H/ D �1.G/ since �1.G/ �

ˆ.G/. If �1.G/ is not elementary abelian, then we know (by the structure of theWilkens group H ) that �1.G/ is nonmodular and so in that case each maximalsubgroup M of G is a Wilkens group. It follows that M=�1.G/ is cyclic, where�1.M/ D �1.G/. In that case, X D G=�1.G/ Š E4 or Q8. Here we have used atrivial fact that noncyclic 2-group X , all of whose maximal subgroups are cyclic, mustbe isomorphic to E4 or Q8.

(i) Suppose that H is a Wilkens group of type (a). Then H is a semidirect productH D hxi A, where A is a maximal normal abelian subgroup of H with exp.A/ > 2

and if t is the involution in hxi, then t inverts each element of A. We have �1.H/ D

�1.G/ D Ahti and A is a characteristic subgroup in H (Proposition 79.8) and soA is normal in G. By the previous paragraph, G=�1.G/ Š E4 or Q8. However, ifG=�1.G/ Š Q8, then G=A (having a cyclic subgroupH=A of index 2) is of maximalclass. But such a group does not have a proper factor group isomorphic to Q8. HenceG=�1.G/ Š E4 and so �1.G/ D ˆ.G/. Since d.G=A/ D 2, we get G=A Š Q8, D8

or C4 � C2.Suppose thatA is not a maximal normal abelian subgroup ofG. LetB be a maximal

normal abelian subgroup of G containing A so that B \ H D A, jB W Aj D 2, andG D hxi B with hxi \ B D f1g. Let b 2 B � A and we compute .bbt /t D btb D


bbt 2 A, since bt 2 B � A so that bbt D s 2 �1.A/. If s D 1, bt D b�1 and so tinverts each element of the abelian group B . It follows that G D hxi B is a Wilkensgroup of type (a). In that caseG would be Q8-free (Proposition 79.8), a contradiction.Hence s ¤ 1 and .bt/2 D btbt D bbt D s shows that o.bt/ D 4. Let a be anelement of order 4 in A. We have abt D a�1 and ha; bti ¤ G (since a 2 ˆ.G/) andso ha; bti is Q8-free, contrary to Lemma 79.1. We have proved that A is a maximalnormal abelian subgroup of G.Suppose that G=A Š D8 or C4 � C2. In that case there is a maximal subgroup K

of G such that K=A Š E4 and �1.K/ D �1.G/. Then K is a Wilkens group oftype (a) or (b) since jK W �1.K/j D 2 (and so K cannot be a Wilkens group of type(c)). Suppose that K is a Wilkens group of type (a) with respect to a maximal normalabelian subgroup A1 of K with exp.A1/ > 2. We know that A1 � �1.K/ D �1.G/,j�1.G/ W A1j D 2 and K=A1 is cyclic. Since K=A is noncyclic, we have A1 ¤

A. All elements in A1 � A are involutions and if t0 2 A1 � A, then t0 inverts andcentralizes each element in A \ A1 and so A \ A1 is elementary abelian. It followsthat exp.A1/ D 2, a contradiction. We have proved that K is a Wilkens group oftype (b) with respect to a maximal normal elementary abelian subgroup E of K. Weknow that j�1.K/ W Ej D 2 (Proposition 79.9). Since �1.K/ D �1.G/, we havejA W A \Ej D 2 and so A is abelian of type .4; 2; : : : ; 2/.

(ii) Assume that H is a Wilkens group of type (b) with respect to E and j�1.H/ W

Ej D 2, whereH=E is cyclic. We have�1.H/ D �1.G/ and�1.G/ � ˆ.G/ so that�1.G/ is nonmodular. Indeed, �1.G/ has exactly two maximal normal elementaryabelian subgroups E and E1, where�1.G/ D EE1 and so�1.G/ is a Wilkens groupof type (b) (and so nonmodular). By the first paragraph of the proof, G=�1.G/ Š E4

or Q8.Suppose that G=�1.G/ Š Q8. It is easy to see that E is not normal in G. Suppose

false. Since G=E has a cyclic subgroup H=E of index 2 and G=�1.G/ Š Q8, we getthat G=E must be of maximal class. But there is no 2-group of maximal class havingQ8 as a proper homomorphic image. Hence E is not normal in G and so for each x 2

G �H , Ex D E1. In particular, j�1.G/ W E1j D 2 and F D E \ E1 D Z.�1.G//.Take e 2 E �F , e1 2 E1 �F so thatD D he; e1i Š D8 and if V is a complement ofhŒe; e1�i in F , then �1.G/ D D � V Š D8 � E2s .

(iii) Suppose that H is a Wilkens group of type (b) with respect to E and E D

�1.H/ so that E D �1.G/ (since E � ˆ.G/), E is normal in G and G=E isnoncyclic with the cyclic subgroup H=E of index 2.Let F=E be a proper subgroup of G=E such that F=E Š E4. Then F is abelian.

Indeed, since F ¤ G, F is Q8-free. If F is not D8-free, then F must be a Wilkensgroup. But then F=�1.F / must be cyclic. This is a contradiction since �1.F / D E.It follows that F is D8-free. Since exp.F / � 4, F must be abelian (Proposition 79.6).Since each proper subgroup of G=E is Q8-free, G=E cannot be semidihedral or

Q2m withm � 4. Suppose that G=E Š D2n , n � 3. In that caseG=E is generated byits four-subgroups and so E � Z.G/. This is a contradiction because in that case H

�80 Minimal non-quaternion-free 2-groups 359

would be abelian (noting thatH=E is cyclic). We have proved thatG=E is isomorphicto Q8, M2n , n � 4, or C2m � C2, m � 1.Suppose that G=E is not isomorphic to Q8 or C2 � C2. Set F=E D �1.G=E/ so

that F=E Š E4. By the above, F is abelian. Obviously, F D �2.G/ is abelian oftype .4; 4; 2; : : : ; 2/.

(iv) Finally, assume that H is a Wilkens group of type (c). We have �1.H/ D

�1.G/ Š D8�E2s andH=�1.G/ is cyclic of order� 4 (Proposition 79.10). The sub-group �1.G/ has exactly three abelian maximal subgroups E1, E2, A, where E1 andE2 are elementary abelian andA is abelian of type .4; 2; : : : ; 2/. By the structure ofH ,E1 andE2 are not normal inH andH=E1 \E2 Š M2n , n � 4. By the first paragraphof the proof, G=�1.G/ Š Q8 since jH=�1.G/j � 4. Thus H=E1 \E2 Š M24 .On the other hand, �1.G/ is normal in G and so NG.E1/ D NG.E2/ D K is a

maximal subgroup of G distinct from H . Since K � �1.G/, K is nonmodular andthereforeK is a Wilkens group. ButK cannot be a Wilkens group of type (c) since E1

and E2 are normal in K. Hence K is either a Wilkens group of type (a) with respectto A or K is a Wilkens group of type (b) with respect to E1 or E2. The group G withsuch a maximal subgroup K has been considered in (i) and (ii) and so we do not gethere new possibilities for the structure of G. Our theorem is proved.

Let

H2 D ha; b j a4 D b4 D 1; ab D a�1i;

H1 D ha; b j a4 D b4 D 1; c D Œa; b�; c2 D 1; Œa; c� D Œb; c� D 1i:

Exercise 1. Let G D H1. Then 1, a2, b2 are only squares in G. In particular, a2b2 isnot a square.

Solution. Let g D aibj ck 2 G. Then g2 D .aibj /2. If i is even, then g2 D

b2j 2 f1; b2g. If j is even, then g2 D a2i 2 f1; a2g. If i and j are odd, theng2 D .ab/2 D ab2ab D ab2ac D a2b2c.

Exercise 2. If G D H1, then G=ha2b2i Š H2.

Solution. Since G=ha2b2i is nonabelian of exponent 4, it suffices to show that it ismetacyclic. Note that a2b2 2 ˆ.G/. Let Ai , i D 1; 2; 3, be all maximal subgroups ofG; then these subgroups are abelian of type .4; 2; 2/. It follows that ha2b2i is a directfactor of Ai , i D 1; 2; 3, so all maximal subgroups of G=ha2b2i are abelian of type.4; 2/. Then, by Lemma 65.1, G=ha2b2i is metacyclic, as was to be shown.

Exercise 3. Classify Q8-free minimal nonabelian 2-groups. (Hint. G is Q8-free if andonly if G=Ã2.G/ is metacyclic.)

Exercise 4. (a) A direct product of two Q8-free 2-groups is not necessarily Q8-free.(Hint. (a) The group D8 � C4 has an epimorphic image isomorphic to Q8 � C4.)


(b) A direct product of two D8-free 2-groups is not necessarily D8-free. (Hint. Thegroup Q8 � C4 has an epimorphic image isomorphic to D8 � C4.)

(c) Suppose that G is Q8-free. If D8 Š D � G, then CG.D/ is elementary abelian.

(d) Suppose that G is D8-free. If Q8 Š Q � G, then CG.Q/ is elementary abelian.

Problem 1. Study the structure of nonabelianH2-free 2-groups.

Problem 2. Study the structure of minimal non-H2-free 2-groups.

Let S.p3/ be a nonabelian group of order p3 and exponent p; then p > 2. Obvi-ously, S.p3/-free p-groups are metacyclic (see Theorem 9.11). It is easy to check thatif G is a minimal non-S.p3/-free p-group, then p D 3 and G is a minimal nonmeta-cyclic group of order 34 (see Theorem 69.1).

�81

Maximal abelian subgroups in 2-groups

Abelian subgroups in 2-groups G play an important role. Therefore, it is not verysurprising that our assumption that every two distinct maximal abelian subgroups havecyclic intersection determines completely the structure of G. We obtain five classes of2-groups with this property. More precisely, we prove here the following result.

Theorem 81.1. Let G be a nonabelian 2-group in which any two distinct maximalabelian subgroups have cyclic intersection. Then Z.G/ is cyclic, each abelian sub-group of G is of rank at most 2, the intersection of any two distinct maximal abeliansubgroups is equal Z.G/, and G has (at least) one abelian subgroup of index 2. More-over, G is isomorphic to one of the following groups.

(a) Group of maximal class (dihedral, semidihedral or generalized quaternion).

(b) M2n D ha; t j a2n�1

D t2 D 1; n � 4; at D a1C2n�2

i.

(c) G D D � C (central product), where D Š D2n , n � 3, is dihedral of order 2n,C Š C2m , m � 2, is cyclic of order 2m and D \ C D Z.D/.

(d) G D hx; t j .xt/2 D a; a2m

D t2 D 1; m � 2; x2 D ab; b2n�1

D 1; n �

3; bt D b�1; Œa; x� D Œa; t � D 1; tx D tb; a2m�1

D b2n�2

i, where jGj D 2mCn,m � 2, n � 3, Z.G/ D hai Š C2m , G0 D hbi Š C2n�1 , and M D hx; ai is aunique abelian maximal subgroup of G. We have CG.t/ D hti � hai Š C2 �C2m andhb; ti Š D2n .

(e) G D hg; h j g2n

D h2m

D 1; m � 3; n � 3; g2n�1

D h2m�1

; hg D h�1i,where G is metacyclic, jGj D 2mCn�1 since hgi \ hhi D hg2n�1

i Š C2. Also,Z.G/ D hg2i Š C2n�1 , G0 D hh2i Š C2m�1 and M D hh; g2i is a unique abelianmaximal subgroup of G.

The more general problem to determine the structure of a nonabelian p-group Gsuch that A \ B D Z.G/ for any two distinct maximal abelian subgroups A and B isvery difficult. First we show that a p-group G has this property if and only if CG.x/

is abelian for each x 2 G � Z.G/ (Theorem 81.2). Then we show that such a 2-groupG has either an abelian subgroup of index 2 or G is of class 2 and G0 is elementaryabelian (Theorem 81.3). In Corollary 81.5 we get a new result for an arbitrary finite2-group.We also classify 2-groups G such that A=Z.G/ is cyclic for each maximal abelian

subgroup A of G (a problem of Heineken–Mann). It is surprising that such groups


have the property that CG.x/ is abelian for each element x 2 G � Z.G/. Then wemay use our Theorems 81.2 and 81.3 to classify such groups (Theorem 81.4). In thisclassification we distinguish the cases, where G has an abelian subgroup of index 2and the case where jG W Aj > 2 for each maximal abelian subgroup A of G.

Proof of Theorem 81.1. LetG be a nonabelian 2-group in which any two distinct max-imal abelian subgroups have cyclic intersection. Since Z.G/ is contained in each max-imal abelian subgroup of G, it follows that Z.G/ is cyclic.Suppose that G possesses an elementary abelian subgroup E of order 8. Let A be

a maximal abelian subgroup containing E and set F D �1.A/ so that E � F . LetB � G be such that A < B and jB W Aj D 2 and let x 2 B � A. Then x2 2 A andtherefore x induces on F an automorphism of order � 2. It follows that jCF .x/j � 4

and the abelian subgroup CF .x/hxi is contained in a maximal abelian subgroup Cwhich is distinct from A since x 62 A. But A\C � CF .x/ and so A\C is noncyclic,a contradiction. We have proved that each abelian subgroup of G is of rank at most 2.We may assume that G is not of maximal class (case (a) of Theorem 81.1) and so

G possesses a normal four-subgroup U . SetM D CG.U / so that jG W M j D 2 sinceZ.G/ is cyclic. Let A be a maximal abelian subgroup of G which contains U so thatA � M . Suppose that A ¤ M and let y 2 M � A be such that y2 2 A. Let B be amaximal abelian subgroup of G containing the abelian subgroup U hyi. Then B ¤ A

(since y 62 A) and A \ B � U , a contradiction. We have proved that whenever U is anormal four-subgroup of G, thenM D CG.U / is an abelian maximal subgroup of G.If x is any element in G � M , then CM .x/ D Z.G/ and Z.G/hxi is a maximal

abelian subgroup of G. Thus, the intersection of any two distinct maximal abeliansubgroups of G is equal to Z.G/ and this statement also holds for 2-groups of maxi-mal class.Suppose that G has more than one normal four-subgroup. By Theorem 50.2, G D

D � C with D Š D8, D \ C D Z.D/ and C is either cyclic of order � 4 or Cis of maximal class (distinct from D8). Let U be a four-subgroup in D so that U isnormal in G. By the above, CG.U / is abelian and so C must be cyclic. We haveobtained a group stated in part (c) of Theorem 81.1. In the sequel we assume that Ghas a unique normal four-subgroup U and set M D CG.U / so that M is an abelianmaximal subgroup of rank 2 with �1.M/ D U .

(i) First assume�2.G/ 6� M . Then there is an element y 2 G �M of order � 4 sothat y2 2 U . We have U hyi Š D8 and so there is an involution t 2 G �M . Since tdoes not centralize U andM is abelian of rank 2, we get CG.t/ D hti�CM .t/, whereCM .t/ is cyclic of order 2m, m � 2. Indeed, if m D 1, then G is of maximal class.Also, we have t 62 ˆ.G/, G has no elementary abelian subgroups of order 8 and Gis not isomorphic to M2s , s � 4, since M2s has only three involutions. We are nowin a position to use Theorem 48.1. It follows that G has a subgroup S of index � 2,where S D AL, L is normal in G, L D hb; t j b2n�1

D t2 D 1; bt D b�1i Š D2n ,n � 3, A D hai Š C2m , m � 2, A\L D Z.L/ D hzi, Œa; t � D 1, CG.t/ D hti � hai,

�81 Maximal abelian subgroups in 2-groups 363

�1.G/ D �1.S/ D �2.A/ �L, �2.A/\L D Z.L/ and if jG W S j D 2, then there isan element x 2 G � S such that tx D tb.Since hbi is a unique cyclic subgroup of index 2 in L, hbi is normal in G. Set

B D �2.A/ �L D �1.G/, �2.A/ D hli, l2 D z, and hvi D �2.hbi/ so that hvi andhli D Z.B/ are normal in G. Hence hl; vi Š C4 � C2 is normal in G. Set u D lv

so that U D hz; ui D �1.hl; vi/ Š E4 is a unique normal four-subgroup in G. Weknow that M D CG.U / is abelian and jG W M j D 2. Note that b centralizes U andut D .lv/t D lv�1 D lvz D uz. If va D v�1 D vz, then A D hai > �2.A/ D hli

and we replace a with a0 D at . In that case o.a0/ D o.a/, �2.ha0i/ D hli and

ua0

D .lv/at D .avz/t D av�1z D lv D u;

so that ha0i centralizes U and S D ha0iL. Writing again a instead of a0, we mayassume from the start that A D hai centralizes U . Hence CS .U / D ha; bi is of index2 in S and therefore M D CG.U / covers G=S and G D M hti. But M is abelianand t centralizes hai and so hai � Z.G/. On the other hand, CG.t/ D hti � hai andCM .t/ D hai so that A D hai D Z.G/.If G D S , then G D L � A, where L Š D2n , n � 3, A Š C2m , m � 2, and

L \ A D Z.L/. We have obtained groups stated in part (c) of Theorem 81.1. In whatfollows we assume that jG W S j D 2 and we know that in that case there is an elementx 2 G � S such that tx D tb. We may assume that x 2 M � S . Indeed, if x D tx0

with x0 2 M � S , then tb D tx D t tx0

D tx0

.SinceM is abelian and CM .t/ D hai D Z.G/, it follows that CM .xt/ D hai and so

.xt/2 2 hai. Set .xt/2 D a0 and assume that ha0i ¤ hai. This implies that there is anelement a00 2 hai�ha0i such that .a00/2 D .a0/�1. We get .xt a00/2 D .xt/2.a00/2 D 1

and so x.ta00/ (with ta00 2 S ) is an involution inG �S , contrary to�1.G/ D �1.S/.It follows that ha0i D hai and so replacing a with a0 (and writing again a instead ofa0), we may assume from the start that .xt/2 D a. From the last relation and tx D tb

we get:a D .xt/2 D xt xt D x2.x�1tx/t D x2tbt D x2b�1

and so x2 D ab. The structure of G is uniquely determined and we have obtained thegroup stated in part (d) of Theorem 81.1.(ii) Finally, assume that �2.G/ � M . Note that M is abelian of rank 2 and soM

is metacyclic. HenceG is also metacyclic. If G has a cyclic subgroup of index 2, thenG is either of maximal class or G Š M2n , n � 4, and these are the groups stated inparts (a) and (b) of Theorem 81.1. In what follows we assume that G has no cyclicsubgroups of index 2. We have U D �1.M/ D �1.G/ Š E4, whereM D CG.U / isan abelian maximal subgroup of G.LetH be a normal cyclic subgroup with cyclic G=H so that jH j � 4 and jG=H j �

4. We have U \ H D hzi Š C2 and z 2 Z.G/ so that if u 2 U � hzi D U � H ,then M D CG.u/, where jG W M j D 2 and M is abelian. Suppose that u does notcentralizeH . Then jH W .H \M/j D 2 and thereforeM coversG=H . Letm 2 M be


such that hmi coversM=M\H and note that CG.H/ D H since u does not centralizeH . Let h 2 H �M so thatH D hhi and hm D hz. Then hm2

D .hz/m D hz z D h.This is a contradiction since jG=H j � 4 and so m2 62 H .We have proved that u centralizes H and so M > H . Let g 2 G � M so that

hgi covers G=H , g2 2 M and g2 centralizesH and therefore g induces on H D hhi

an involutory automorphism. Also, ug D uz since Z.G/ s cyclic. If hg D hz, thenG0 D hzi and G is minimal nonabelian. In that case G is splitting metacyclic, i.e.,there is g0 2 G � M such that hg0i covers G=H and hg0i \ H D f1g. It followsthat �1.hg

0i/ � Z.G/ and so Z.G/ � hz;�1.hg0i/i Š E4, a contradiction. We

have proved that hg D h�1z� , � D 0; 1 and jH j � 8. (Indeed, if jH j D 4, thenhg D h�1 D hz and we have again G0 D hzi, as above.) In particular, CH .g/ D hzi

and so hgi\H � hzi. However, if hgi\H D f1g, then CG.�1.hgi// � hM;gi D G

and so E4 Š hz;�1.hgi/i � Z.G/, a contradiction.We have proved that hgi \ H D hzi and so o.g/ � 8 and Z.G/ D hg2i. If

hg D h�1z, then we replace h with h0 D hu, where Œh; u� D 1 and so o.h0/ D o.h/

and.h0/g D .hu/g D h�1z uz D h�1u D .hu/�1 D .h0/�1

and hh0; gi D hhu; gi D hh; gi D G since u 2 U � ˆ.G/. (Indeed, ˆ.G/ � M isabelian and so if ˆ.G/ D Ã1.G/ were cyclic, then jG W ˆ.G/j D 4 implies that Gwould have a cyclic subgroup of index 2.) Writing h instead of h0, we see that we mayassume from the start that hg D h�1. We have obtained the group stated in part (e) ofTheorem 81.1.

Theorem 81.2. A \ B D Z.G/ for any two distinct maximal abelian subgroups A;Bif and only if CG.x/ is abelian for each x 2 G � Z.G/.

Proof. Let x 2 G � Z.G/ and suppose that CG.x/ is nonabelian. Let A be a maximalabelian subgroup of CG.x/ so that A ¤ CG.x/ and A � Z.G/hxi > Z.G/. Letb 2 CG.x/ � A and let B be a maximal abelian subgroup of CG.x/ containing hbi sothat A ¤ B and B also contains the abelian subgroup Z.G/hxi. Obviously, A and Bare also maximal abelian subgroups of G but A \ B � Z.G/hxi > Z.G/.Conversely, let C ¤ D be maximal abelian subgroups of G such that C \ D >

Z.G/. Let y 2 .C \ D/ � Z.G/ so that CG.y/ � hC;Di, where hC;Di is non-abelian.

Theorem 81.3. Let G be a nonabelian 2-group such that A\B D Z.G/ for every twodistinct maximal abelian subgroups A and B . Then one of the following holds:

(a) G has an abelian subgroup of index 2.

(b) G is of class 2, G0 is elementary abelian and ˆ.G/ � Z.G/.

Proof. Let A be a maximal normal abelian subgroup of G. Then G=A ¤ f1g actsfaithfully on A and f1g ¤ Z.G/ < A. Let K be a G-invariant subgroup such thatZ.G/ < K � A and jK W Z.G/j D 2. Let x be any element in G � A. Then


CA.x/ D Z.G/ and so hxi \A � Z.G/. Indeed, let B be a maximal abelian subgroupcontaining the abelian subgroup CA.x/hxi. ThenA ¤ B andA\B � CA.x/ D Z.G/.Let k 2 K � Z.G/ so that k2 2 Z.G/ and kx D kl with some 1 ¤ l 2 Z.G/. We get

k2 D .k2/x D .kx/2 D .kl/2 D k2l2;

and so l2 D 1 and therefore l is an involution in Z.G/. This gives

kx2

D .kx/x D .kl/x D kxl D .kl/l D kl2 D k;

and so (by the above) x2 2 A and (since hxi \ A � Z.G/) x2 2 Z.G/. In particular,G=A is elementary abelian. Let a 2 A � Z.G/ and set ax D a0 2 A � Z.G/ so that.a0/x D .ax/x D ax2

D a (since x2 2 Z.G/). Therefore, .aa0/x D a0a D aa0 whichimplies that aa0 D z 2 Z.G/ and a0 D ax D a�1z and so x inverts A=Z.G/.Suppose that jG=Aj � 4. Then there are elements x; y 2 G � A such that xy 2

G � A. In this case x and y both invert A=Z.G/ and so xy centralizes A=Z.G/. Butxy also must invert A=Z.G/ which implies that A=Z.G/ is elementary abelian. Henceˆ.G/ � Z.G/ (noting that for each x 2 G � A, x2 2 Z.G/) and so G is of class 2.For each g; h 2 G, Œg; h�2 D Œg2; h� D 1 and so G0 is elementary abelian.

Theorem 81.4. Let G be a nonabelian 2-group such that A=Z.G/ is cyclic for eachmaximal abelian subgroup A of G. Then one of the following holds:

(a) G has an abelian subgroupM of index 2 and we have either G D HZ.G/ withH minimal nonabelian or G=Z.G/ Š D2n , n � 3, is dihedral of order 2n withG0 cyclic of order � 4, G0 \ Z.G/ Š C2, and if x 2 G �M , then x2 2 Z.G/and x inverts G0.

(b) G is of class 2, G0 is elementary abelian of order � 8, ˆ.G/ � Z.G/ andwhenever A is a maximal abelian subgroup of G, then jA W Z.G/j D 2.

Proof. Suppose that there is an element a 2 G � Z.G/ such that H D CG.a/

is nonabelian. Let A be a maximal abelian subgroup of G containing hai. ThenZ.G/hai � A < H < G. By our assumption, A=Z.G/ ¤ f1g is cyclic. Assumethat H=Z.G/ contains a subgroup of order 2 distinct from �1.A=Z.G//. In thatcase there is x 2 H � A such that x2 2 Z.G/. Since Œa; x� D 1, ha; xi is abelianbut ha; xiZ.G/=Z.G/ is noncyclic, a contradiction. We have proved that H=Z.G/has only one subgroup of order 2 and so H=Z.G/ Š Q2n , n � 3, is generalizedquaternion of order 2n. Indeed, if H=Z.G/ were cyclic, then H is abelian, a con-tradiction. Since H=Z.G/ Š Q2n , it follows that a2 2 Z.G/, Z.H/ D Z.G/hai,jZ.H/ W Z.G/j D 2 and for each y 2 Z.H/ � Z.G/, CG.y/ D H D CG.a/. SetjZ.G/j D 2m, m � 1, so that jH j D 2mCn. Let A0=Z.G/ be a cyclic subgroup of in-dex 2 inH=Z.G/ so that A0 is abelian and Z.H/ < A0. Let A1=Z.G/ D .H=Z.G//0

so that Z.H/ � A1 < A0 and jH W A1j D 4 and therefore jA1j D 2mCn�2.Since A1 D H 0Z.G/, H 0 covers A1=Z.G/ and so jH 0j � 2n�2 D jA1=Z.G/j.


By Lemma 1.1, we get jH j D 2jZ.H/jjH 0j and so 2mCn D 2 2mC1jH 0j andtherefore jH 0j D 2n�2. This gives H 0 \ Z.G/ D f1g and so H 0 is cyclic withH 0 \ Z.H/ D hyi Š C2. It follows that hyi is characteristic in H and so if T isa subgroup of G such that H < T � G and jT W H j D 2, then hyi is central in T ,contrary to the above fact that CG.y/ D H , where y 2 Z.H/�Z.G/. We have provedthat for each a 2 G � Z.G/, CG.a/ is abelian.By Theorem 81.2, A \ B D Z.G/ for any two distinct maximal abelian subgroups

A and B of G. We may use Theorem 81.3 and so either G has an abelian subgroup ofindex 2 or G0 � Z.G/, �.G/ � Z.G/ and G0 is elementary abelian.

(i) First we consider the case, where G has an abelian subgroup M of index 2.Then Z.G/ < M and for each x 2 G � M , CM .x/ D Z.G/ and so x2 2 Z.G/.By our assumption, M=Z.G/ ¤ f1g is cyclic. If G has another abelian maximalsubgroup N , thenM \N D Z.G/ and this implies that jM W Z.G/j D 2. Conversely,suppose that M=Z.G/ D 2. In that case G=Z.G/ Š E4 (because G=Z.G/ Š C4

would imply thatG is abelian) and soG has more than one abelian maximal subgroup.We analyze this case further. Let H be a minimal nonabelian subgroup of G. ThenjH W .H \ Z.G//j D 4, H covers G=Z.G/ and so G D HZ.G/ and G0 Š C2. Wehave obtained the first possibility stated in part (a) of our theorem.It remains to consider the case, whereM=Z.G/ Š C2n , n � 2, whereM is a unique

abelian maximal subgroup of G. We know that for each x 2 G �M , x2 2 Z.G/. Itfollows that x inverts the cyclic groupM=Z.G/ of order� 4 and soG=Z.G/ Š D2nC1

is dihedral of order 2nC1. Set jZ.G/j D 2m, m � 1, and .G=Z.G//0 D L=Z.G/ sothatG=L Š E4 andL D G0Z.G/. SinceG has an abelian maximal subgroup, we mayuse Lemma 1.1 and we get jGj D 2mCnC1 D 2 2mjG0j and so jG0j D 2n. HenceG0 \ Z.G/ D hzi Š C2, G0=hzi is cyclic of order 2n�1 and G0 is abelian.Suppose thatG0 is not cyclic. ThenG0 splits over hzi D G0\Z.G/. SinceÃ1.G

0/ isnormal in G andÃ1.G

0/\Z.G/ D f1g, it follows thatÃ1.G0/ D f1g and soG0 Š E4

and G=Z.G/ Š D8. Let a 2 M � .Z.G/G0/ so that hai coversM=Z.G/ Š C4 and soa2 62 Z.G/. For an x 2 G �M , we have Œa; x� D t 2 G0 � hzi. Then we get

Œa2; x� D Œa; x�aŒa; x� D tat D t2 D 1:

But then CG.a2/ D hM;xi D G and so a2 2 Z.G/, a contradiction. We have proved

that G0 is cyclic of order � 4.For any x 2 G�M and anym 2 M �L (whereL D G0Z.G/), we have x2 2 Z.G/

and Œm; x� D g with hgi D G0 Š C2n . This gives mx D mg and so m D mx2

D

.mg/x D mggx and this implies gx D g�1 and therefore x inverts G0. We haveobtained the second possibility in part (a) of our theorem.(ii) Now we consider the case, where G has no abelian subgroups of index 2, G0 �

Z.G/, ˆ.G/ � Z.G/ and G0 is elementary abelian. It follows that jA W Z.G/j D 2 foreach maximal abelian subgroup A of G. If jG W Z.G/j D 4, then G would have anabelian subgroup of index 2, a contradiction. Hence, G=Z.G/ Š E2m , m � 3, and so


there exist elements g; h; i 2 G �Z.G/ such that hg; h; iiZ.G/=Z.G/ Š E8. We haveŒg; h� ¤ 1, Œg; i � ¤ 1, and Œh; i � ¤ 1.Suppose that jG0j D 2. Then Œg; h� D Œg; i � and so Œg; hi � D Œg; h�Œg; i � D 1 and

therefore hg; hiiZ.G/=Z.G/ Š E4, a contradiction.Suppose that G0 Š E4. In that case Œg; h� D t1, Œg; i � D t2 and Œh; i � D t3, where

t1, t2, t3 are pairwise distinct involutions in G0. In this case,

Œgh; gi � D Œg; i �Œh; g�Œh; i � D t2t1t3 D 1;

and so hgh; giiZ.G/=Z.G/ Š E4, a contradiction. We have proved that jG0j � 8.

Corollary 81.5. Let G be an arbitrary nonabelian 2-group. Let A and B be any twodistinct maximal abelian subgroups in G with and intersectionA\B of maximal pos-sible order. Then the nonabelian subgroup H D hA; Bi either possesses an abeliansubgroup of index 2 or H is of class 2 and H 0 is elementary abelian.

Proof. Obviously, A \ B D Z.H/. If C andD are any two distinct maximal abeliansubgroups inH , then C \D � Z.H/ and the maximality of jA\Bj forces C \D D

Z.H/. Then our result follows from Theorem 81.3.

�82

A classification of 2-groupswith exactly three involutions

This section is written by the second author. The first author inserted in proofs oftheorems a number of explanations and added a few exercises.

According to Sylow–Frobenius, the number of involutions in a 2-group (also in anyfinite group of even order) is odd. The 2-groups with exactly one involution are clas-sified (Proposition 1.3). Therefore, it is natural to classify the 2-groups with exactlythree involutions. It appears that this problem is enormously difficult.

We finish here a classification of 2-groups with exactly three involutions (Theorems82.1, 82.2, 82.8, 82.16, and 82.18). A.D. Ustjuzaninov [Ust2] has proved that if a2-group G has exactly three involutions and Z.G/ is noncyclic, then G has a normalmetacyclic subgroupM of index at most 4 and G=M is elementary abelian. M. Kon-visser [Kon2] goes one step further and proves that if a 2-group G has exactly threeinvolutions and Z.G/ is cyclic, then G has a metacyclic subgroupM of index at most4 and “M is normal in G in most of the cases”. But he is not very precise in whichcases M is not normal in G and what is the structure of G in these cases. Here weclear up this remaining difficult situation and determine completely the structure of Gin terms of two generators and relations.

Of course, if a 2-group G has exactly three involutions, it has no normal elementaryabelian subgroups of order 8. However, it is impossible to deduce the classificationof such groups as G from results of �50 where the groups without normal elementaryabelian subgroups of order 8 are treated. It follows from �50 that our group G has anormal metacyclic subgroupM such that G=M is isomorphic to a subgroup of D8.Now we describe the obtained results in some detail.

Let a 2-group G has exactly three involutions and jGj > 4. Then it possesses anormal abelian subgroup W which is either of type .4; 2/ or .4; 4/. Set C D CG.W /;then C is metacyclic since �2.C / D W . (i) Suppose that G has no normal abeliansubgroups of type .4; 4/. Then C is abelian of type .2n; 2/, n � 2 and jG=C j � 4.If n > 2 and jG=C j > 2, then G=C Š E4. If G=C Š C4, then n D 2, C D W

and G is uniquely determined. (ii) Now let W be abelian of type .4; 4/. (ii1) Supposethat �1.W / � Z.G/. In that case G=C is elementary abelian of order � 8 andC � M � G, where M is metacyclic with jG W M j � 4. (ii2) Next assume that�1.W / — Z.G/, i.e., Z.G/ is cyclic. (ii2.1) If jG=C j > 23, then C D W and G is

�82 A classification of 2-groups with exactly three involutions 369

one of eight two-generator groups of order 28. (ii2.2) Now let jG=C j D 8. Then Gpossesses a metacyclic subgroup which is normal in G except in two cases describedin detail; in both these cases G=C Š D8, C < ˆ.G/, jZ.G/j D 2 and G is given interms of generators and defining relations.

Theorem 82.1. Let G be a metacyclic 2-group. Then:

(i) G contains exactly one involution if and only ifG is either cyclic or generalizedquaternion.

(ii) G contains more than three involutions if and only if G is either dihedral orsemidihedral.

(iii) All other metacyclic 2-groups contain exactly three involutions.

Proof. If G is cyclic or of maximal class, then the number of involutions in G is 1

.mod 4/.Suppose that G is metacyclic but neither cyclic nor of maximal class. Let f1g ¤

Z D hbi ¤ G be a cyclic normal subgroup of G, where G=Z is cyclic. Let H=Z bethe subgroup of order 2 inG=Z. Then all involutions inG lie inH andH is not cyclic(otherwise,Z � ˆ.H/ � ˆ.G/ soG=ˆ.G/ is cyclic andG is cyclic). IfH is abelianor isomorphic to M2n , n � 4, then �1.H/ Š E4 and G has exactly three involutions.If H is of maximal class, then o.b/ � 4 and taking an element x 2 H � Z, we havebx D b�1 or bx D b�1C2n�1

, where o.b/ D 2n, n � 3. But such an automorphismof order 2 induced with x on hbi is not a square in Aut.hbi/ and so we have G D H isof maximal class, a contradiction.

For another proof of Theorem 82.1, one can use Proposition 10.19(a).

Theorem 82.2. Let G be a nonmetacyclic 2-group with exactly three involutions. IfW is a maximal normal abelian noncyclic subgroup of exponent � 4 in G, then W Š

C4 � C2 orW Š C4 � C4, W D �2.CG.W // and CG.W / is metacyclic.Suppose that G has no normal subgroups isomorphic to C4 � C4. Then G has a

normal subgroup W Š C4 � C2, C D CG.W / is abelian of type .2n; 2/, n � 2,and G=C is isomorphic to a proper (!) nontrivial subgroup of D8. If n > 2, thenG=C Š C2 or E4. If G=C Š C4, then n D 2 and G is the unique 2-group of order 25

with �2.G/ Š C2 � Q8 (see �52 ):

G D hg; v j g8 D v4 D 1; g4 D v2 D z; Œv; g� D u; u2 D Œu; v� D 1; Œu; g� D zi:

Here Z.G/ D hzi is of order 2, W D hv; ui Š C4 � C2 is a maximal normal abeliansubgroup of G, G=W Š C4, �1.G/ D �1.W / D hz; ui Š E4, ˆ.G/ D hg2; ui Š

C4 � C2, G0 D hz; ui, and �2.G/ D hW;g2i Š C2 � Q8.

Proof. Suppose that G is a nonmetacyclic 2-group with exactly three involutions. LetW be a noncyclic maximal abelian normal subgroup of exponent � 4 in G. SincejGj > 8, we get W 6Š E4 (otherwise, CG.W / D W so jGj D 8). Therefore we


have either W Š C4 � C2 or W Š C4 � C4. By a result of Alperin (Corollary 10.2),W D �2.C /, where C D CG.W /, and a result of N. Blackburn (Theorem 41.1,Remark 2) implies that C is metacyclic.(See Exercise 49.1.) Suppose that W Š C4 � C2. Then Theorem 42.1 (noting that

j�2.C /j D 8 and Z.C / is noncyclic) implies that C Š C2n �C2 with n � 2. If n > 2,thenW contains a cyclic subgroup Z of order 4 which is characteristic in C and so Zis normal in G. Each y 2 G acts on Z and onW0 D �1.W / Š E4 and so (noting thatAut.Z/ Š C2 and a Sylow 2-subgroup of Aut.W0/ Š C2) G=C Š C2 or G=C Š E4

(take into account that C D CG.Z/\CG.�1.W //. We know that Aut.W / Š D8 andtherefore if G=C Š C4 or G=C Š D8, then n D 2 so C D W .It remains to treat the casesG=W Š C4 or D8, whereW D C D CG.W /. We shall

show (as a surprise!) thatG=W Š D8 actually cannot occur. We setW D hv; u j v4 D

u2 D Œv; u� D 1i and z D v2 so that hzi D Ã1.W / � Z.G/. In any case, there is anelement g 2 G inducing on W the following automorphism of order 4:

vg D vu; ug D uz; .vg�1

D vug�1

D vuz/;(1)

which implies

vg2

D .vu/g D .vu/.uz/ D vz D v�1;

ug2

D .uz/g D .uz/z D u D u�1;(1a)

so that g2 inverts each element of W so g4 2 CG.W / D W , and CW .g/ D hzi.Hence

g4 D z;(2)

since �1.G/ D �1.W / D W0 D hz; ui; in particular, o.g/ D 8.We compute for each w 2 W :

.g2w/2 D g2wg2w D g4wg2

w D zw�1w D z;

and so all elements in g2W are of order 4 and therefore all elements in hW;gi�hW;g2i

are of order 8 (it is easy to check, using the previous displayed formula, that .gw/4 D

.g3w/2 D z). We get �2.hW;gi/ D hW;g2i D hui � hv; g2i Š C2 � Q8, and sohW;gi is the unique group of order 25 with �2.hW;gi/ Š C2 � Q8 according to theresults in �52. If hW;gi D G, we are done.We assume, by the way of contradiction, that hW;gi ¤ G. Then, by the above,

G=W Š D8 so all elements in G=W � .hg;W i=W / are involutions. Therefore, thereis h 2 G � hW;gi such that h induces on W the following involutory automorphism:

uh D uz; vh D v;(3)

and so CW .h/ D hvi which together with h2 2 W forces h2 2 hvi. Indeed, sinceW has exactly two cyclic subgroups of order 4 and hvi is not normal in G, by (1), it


follows that jG W NG.hvi/j D 2, so each h 2 CG.hvi/�W satisfies (3). We also haveG D hW;g; hi. If h2 2 hzi, then hW0; hi Š D8 has 5 > 3 involutions, a contradiction.Thus, replacing v with v�1 (if necessary, and noting that (1) remains valid), we mayassume from the start:

h2 D v:(4)

Indeed, from what has just been said and (3), o.h2/ > 2 and h2 2 W . From (3)follows hu D hz D hh4 D h1C4 and so hW;hi Š M24 is a metacyclic nonnormalsubgroup of index 4 in G with �1.hW;hi/ D hu; zi and �2.hW;hi/ D W .If�2.G/ D �2.hW;gi/ D hW;g2i Š C2 �Q8, then by the results in �52 (and the

fact that G > �2.G/) we get jGj D 25, contrary to our assumption that hW;gi ¤ G.It follows that G � hW;gi must contain elements of order 4 and they must lie in thecoset .hg/W of the group G=W Š D8 (note that .hg/W is a noncentral involution inG=W ). By (3) and (4), we have

uhg D .uz/g D .uz/z D u; vhg D vg D vu;(4a)

and so CW .hg/ D W0 which implies:

.hg/2 D w0 2 W0 and w0 ¤ 1:(5)

From (5) and (4) follows hghg D w0, h2ghg D w0, vgh D w0g�1, and so we obtain:

gh D .v�1w0/g�1:(6)

For each uivj 2 W we compute (take into account that all elements in the set G �W

are not involutions)

1 ¤ ..hg/uivj /2 D .hg/2.uivj /hguivj D w0ui.vu/juivj D w0.uz/

j

which implies, if we take j D 1 in the displayed formula, that w0 D .hg/2 ¤ uz.Thus, .hg/2 D u or z. However, if .hg/2 D u, then (by the above) .h.gv//2 D

..hg/v/2 D u.uz/ D z and so replacing g with gv (if necessary), we may assumefrom the start:

.hg/2 D w0 D z and gh D .v�1w0/g�1 D .v�1v2/g�1 D vg�1(7)

(in the second formula we used (6)). Here we note that replacing g with gv, theprevious relations (1) and (2) remain unaltered.We observe that

L D hW;g2; hgi D �2.G/; hL;hi D G; L0 D hu; zi D W0;

Z.L/ D L0; ˆ.L/ D L0


so that L is a special 2-group of order 25. From (a4) and (a1) follows Œv; hg� D u andŒv; g2� D z, respectively. We compute using (1), (2), (4), and (7):

.g2/h D .vg�1/2 D vvgg�2 D v.vu/g2z D v2ug2z D g2u;

(since g2 centralizes W0 D hu; zi and v2 D z) and so Œg2; h� D u. This implies thatŒg2; hg� D Œg2; g�Œg2; h�g D ug D uz.We claim that S D hg2.hg/; v.hg/i Š C4 � C4 is normal in G. This gives us our

final contradiction since we have assumed that G has no normal subgroup isomorphicto C4 � C4. Indeed,

Œg2.hg/; v.hg/� D Œg2; v�Œg2; hg�Œhg; v� D z.uz/u D 1;

.g2.hg//2 D g4.hg/2Œhg; g2� D zzuz D uz;

.v.hg//2 D v2.hg/2Œhg; v� D zzu D u;

and so S Š C4 � C4 and S is normal in L (since S � W0 D L0). Finally, using (1),we get (recall that g4 D z, by (2))

.hg/h D hvg�1 D .hg/.g�1vg�1/ D .hg/g�2.gvg�1/ D .hg/g2zvg�1

D .hg/g2z.vuz/ D .hg/g2vu;

so that, using the facts that .hg/g2 D g2.hg/uz and .hg/v D v.hg/u, we get

.g2.hg//h D .g2u/.hg/g2vu D g4.hg/uzv D .hg/uv D v.hg/;

.vg2/h D .vg2/u;

which shows that S is normal in G since v.hg/ g2.hg/ vg2 .mod W0/.

If, in Theorem 82.2, jG W C j D 2, then G possesses a cyclic subgroup of index 4 soits structure is described in �74. Next, one may assume, in this case, that G � C hasan element of order 4 (otherwise, G has exactly two cyclic subgroups of order 4, i.e.,belongs to the groups classified early in this book; see �43). This fact can help to writeout the defining relations for G.In what follows G will denote a nonmetacyclic 2-group containing exactly three

involutions and let W be a maximal normal abelian noncyclic subgroup of exponent� 4 inG and assume thatW Š C4 �C4. By Theorem 82.2,�2.C / D W , where C D

CG.W / is metacyclic. We set W0 D �1.W / D �1.G/ and denote with AG.W / Š

G=C the automorphism group ofW induced by G.

Lemma 82.3. Let v be an element of order 4 in G � W . Then v centralizes W0 andW=W0 and v inverts some element of order 4 in W .


Proof. We have v2 ¤ 1 is an involution in W0. If v does not centralize W0, thenhW0; vi Š D8 has exactly 5 > 3 involutions, a contradiction. Thus v centralizes W0.Let us prove that v also centralizesW=W0. The subgroup H D hv;W i is nonabelianand Z.H/ < W . It suffices to show thatH=W0 is abelian. Assuming that this is false,we get H=W0 Š D8. If L=W0 is a cyclic subgroup of order 4 in H=W0, then H hastwo distinct abelian maximal subgroups W and L so Z.H/ D W \ L has index 4in H . It follows that jH 0j D 2 (Lemma 1.1) so H 0 < W0. Then H=W0 is abelian,contrary to the assumption. For each w 2 W , vw 2 H �W so 1 ¤ .vw/2 D v2w2s

with s 2 W0 (since v commutes with w mod W0). Note that if x 2 vW , then eachof the four elements in xW0 have the same square x2 ¤ 1 2 W0 since hx;W0i isabelian of type .4; 2/ and o.x/ D 4. Since jvW j D 16 and W0 has exactly threeinvolutions, it follows that there is x 2 vW and w 2 W �W0 such that x2 D .xw/2.Hence x2 D .xw/2 D xwxw D x2wxw, which gives wx D w�1. But then alsowv D w�1, where w is some element of order 4 in W (since x 2 vW and v act thesame way on the abelian group W ).

Lemma 82.4. Suppose that G contains an element g which induces an automorphism� of order 2 on W and for each element w of order 4 inW , w� ¤ w. Then o.g/ D 4.

Proof. Clearly, g 2 G �W so o.g/ > 2. Since o.�/ D 2, g2 2 C . If o.g/ > 4, then�2.hgi/ � �2.C / D W hence the group hg0i D hgi \ W is cyclic of order 4, sog

�0 D g0, a contradiction. Thus o.g/ D 4.

Lemma 82.5. Suppose that g 2 G �C with g2 2 C and the automorphism � inducedby g on W inverts no element of order 4 in W . Then hg;C i is metacyclic (and so gcentralizes some element of order 4 in W ).

Proof. Each element in gC induces the involutory automorphism � onW . By Lemma82.3, no element in gC is of order 4. Thus �2.hg;C i/ D W and so hg;C i is meta-cyclic (�41, Remark 2). The last assertion holds since o.g/ > 4 and �2.hgi/ �

�2.C / D W .

Lemma 82.6. The automorphism group Aut.W / of W D hu; y ju4 D y4 D Œu; y� D

1i Š C4 � C4 is of order 25 3. The subgroup A of Aut.W / of all automorphismsnormalizing the subgroup Y D hu2; yi Š C2 � C4 is of order 25 and so is a Sylow2-subgroup of Aut.W /. We have A0 D Z.A/ D ˆ.A/ Š E4 and so A is a special2-group. Set W0 D �1.W / D hu2; y2i. Then the stabilizer A0 of the chain W >

W0 > f1g is elementary abelian of order 24. The subgroup A0 contains the “special”subset S D f�2; �; �; ; �g of five automorphisms defined by:

u� D uy; y� D yI u� D u; y� D yu2I

u� D uu2y2; y� D yu2y2I u D uu2y2; y D yu2I

u D uy2; y D yu2y2:


Each 2 S has the property that does not invert any element of order 4 inW . If X is any maximal subgroup of A0, then X \ S is nonempty. In addition,the “superspecial” automorphisms and � have also the property that they do notcentralize any element of order 4 in W .Finally, A is Q8-free but is not D8-free.

Proof. See Proposition 50.5. According to the Main Theorem 79.7, A is a Wilkensgroup of type (b) (= Wb-group) and so A is Q8-free but is not D8-free.

Remark. The group A of Lemma 82.6 has class number 14. Indeed, if is an ir-reducible character of A, then .1/2 � jA W Z.A/j D 8 so .1/ 2 f1; 2g. SincejA W A0j D 8, A has exactly 8 distinct linear characters and 1

4.jAj � jA W A0j/ D 6

distinct nonlinear characters. It follows that the class number of A equals 8C 6 D 14,as was to be shown. Since each element of A0 � Z.A/ has exactly two A-conjugates,A0 is the union of jZ.A/jC 1

2.jA�Z.A/j/ D 10 A-classes. It follows from jZ.A/j D 4

that four A-classes contained in A � A0, have size 4.

Lemma 82.7. No element ofG can induce either of the “superspecial” automorphisms or � on W D hu; yi (from Lemma 82.6).

Proof. Suppose that g 2 G induces or � on W . Then g2 2 C and g neither invertsnor centralizes any element of order 4 inW . By Lemma 82.3, o.g/ > 4 and by Lemma82.4, o.g/ D 4, a contradiction.

Theorem 82.8 ([Ust2]). Let G be a 2-group containing exactly three involutions anda normal subgroup W Š C4 � C4. If �1.W / � Z.G/, then G contains a normalmetacyclic subgroupM of index at most 4 and exp.G=M/ divides 2.

Proof. Since �1.W / D W0 � Z.G/, G=C stabilizes the chain W > W0 > f1g (seeLemma 82.6) and so G=C is elementary abelian of order � 24. By Lemma 82.7,jG=C j � 23. Suppose that jG=C j D 23. Then Lemma 82.6 implies that there isg 2 G � C inducing a “special” automorphism on W (which does not invert anyelement of order 4 in W ). By Lemma 82.5, the subgroup M D hg;C i is metacyclicwith jG W M j D 4 andM is normal in G and G=M is elementary abelian.

In what follows we assume, in addition, that W0 D �1.W / 6� Z.G/.

Lemma 82.9. If�1.W / 6� Z.G/, then with the appropriate choice of generators u andy of W , one of the following four automorphisms is contained in AG.W / D G=C :�.i; j / W u �! uy, y �! u2iyj , where i D 0; 1 and j D 1;�1. These automor-phisms lie in four distinct conjugate classes in A (recall that A D NAut.W /.hu

2; yi/ 2

Syl2.Aut.W /); see Lemma 82.6), where each one is of length 4. We have Z.A/ D

ˆ.A/ D A0 D h�2.0; 1/; �2.1; 1/i D h�2.0; 1/; �2.1;�1/i and no two of the el-ements of order 4 in f�.0; 1/; �.1; 1/; �.1;�1/g are permutable and �.0;�1/ is oforder 2.


Proof. Let � be an element of AG.W / which does not centralize �1.W /. Each ele-ment in�1.W / is a square inW . Let u2 2 �1.W / be such that .u2/� ¤ u2 (u 2 W ).Set y D u�1u�; then y2 D u2.u2/� 62 f1; u2g and o.y/ D 4 so that u� D uy,W D hui � hyi and y� D u2iyj , i D 0; 1, j D 1;�1.Recall that A0 is the stabilizer of the chain f1g < W0 < W ; then A0 is elementary

abelian of order 24 (Lemma 82.6). For each � 2 A � A0, CA.�/ D h�iA0 (sinceA0 D Z.A/ is of order 4) and so the conjugacy class of � has exactly four elementslying in f�A0g. Also, �2.0; 1/, �2.1; 1/, and �2.1;�1/ are three distinct involutions inA0. In particular, �.0; 1/ (order 4), �.1; 1/ (order 4), �.1;�1/ (order 4), and �.0;�1/(order 2) lie in four distinct conjugacy classes in A. Since C4 � C4 is not a sub-group of A (because jZ.A/j D 4), no two elements in f�.0; 1/; �.1; 1/; �.1;�1/g arepermutable (recall that, by the above, any two cyclic subgroups, generated by theseautomorphisms, have trivial intersection).

Lemma 82.10. In the notation of Lemma 82.9, we have �.1; 1/ 62 AG.W /.

Proof. If g 2 G induces the automorphism �.1; 1/ (of order 4) on W given by ug D

uy, yg D u2y, then g4 2 C D CG.W / and CW .hgi/ D hy2i. Hence g4 D y2 sinceg2 cannot be an involution. But then .g2u/2 D g4ug2

u D y2u�1y2u D 1, which isa contradiction since G has only three involutions.

Lemma 82.11. If �.0;�1/ 2 AG.W /, then CG.W / > W , and if g 2 G induces�.0;�1/ on W , then g2 2 CG.W / �W and hg2i > hu2yi.

Proof. Let g 2 G induce �.0;�1/ on W so that ug D uy, yg D y�1. Then ug2

D

uyy�1 D u, yg2

D y so g2 2 C D CG.W /. Next, it follows from uiyj D

.uiyj /g D .uy/iy�j D uiyi�j that 2j i .mod 4/, and so CW .g/ D hu2yi. Ifg2 D .u2y/i , then .gu�i /2 D gu�igu�i D g2.u�i /gu�i D .u2y/i .uy/�iu�i D 1,a contradiction. So g2 62 W and therefore hg2i > hu2yi and C > W .

Lemma 82.12. Setting C D CG.W /, we have either j�3.C /j � 25 in which caseC Š C4 � C2n , n � 2, or j�3.C /j D 26 in which case C=W is metacyclic withexactly three involutions.

Proof. We have W D �2.C / and C is metacyclic so that C=W is also metacyclic.Suppose that C=W is of maximal class. Let D=W be a cyclic maximal subgroupof C=W . Then D is an abelian maximal subgroup of C . Set R=W D .C=W /0 sothat R > W , jD=Rj D 2, and C=R Š E4. Now, C 0 covers R=W and C 0 is cyclicsince C is metacyclic, and so C 0 \ W Š C4. Indeed, C 0 \ W � 2, and, assumingthat there we have equality, we see that a metacyclic group C=.C 0 \ W / containsa nonmetacyclic subgroup C 0=.C 0 \ W / � W=.C 0 \ W /, which is a contradiction.This implies jR W C 0j D 4 and so jC W C 0j D jC W RjjR W C 0j D 24. Noting thatC is nonabelian but possesses an abelian maximal subgroup, we get, by Lemma 1.1,jZ.C /j D 1

2 jC W C 0j D 23. This is a contradiction sinceW � Z.C / and jW j D 24.


We have proved that C=W is not of maximal class. If C=W is cyclic, then, since�2.C / D W , C is abelian of type .22; 2n/, n � 2, and 24 � j�3.C /j � 25. If C=Wis noncyclic, then C=W is metacyclic with exactly three involutions (Theorem 82.1)and so�1.C=W / Š E4 which implies j�3.C /j D 26.

Lemma 82.13. If j�3.C /j D 26, then jAG.W /j � 8.

Proof. Here C D CG.W /, W0 D �1.W /, and D D �3.C /. If X is any subgroupof D such that W < X < D, then X Š C8 � C4 is abelian since �2.C / D W �

Z.C /. But D=W Š E4 (Lemma 82.12) and D is metacyclic and so ˆ.D/ D W . Itfollows that D is either abelian or minimal nonabelian. Since exp.D/ D 8, we haveD D ha; d j a8 D d8 D 1; ad D a1C4�; � D 0; 1i, where W D ha2; d2i.Suppose that jG=C j > 8 (recall that AG.W / D G=C ). We identify G=C with a

subgroup S of the group A of all 25 automorphisms of W D hu; yi keeping hy; u2i

fixed (see Lemma 82.6 and 82.9). In particular, S contains the involutory automor-phism ˇ D �2.1; 1/ (since S � ˆ.A/), where uˇ D u�1y2, yˇ D y�1 and we notethat ˇ centralizesW0 (andW=W0) and wˇ ¤ w for eachw 2 W �W0. Let g 2 G in-duce the automorphism ˇ onW . Then g2 2 C and since g fixes no element inW �W0,we have g2 2 W0 implying that g2 2 Z.C / and o.g/ D 4. It follows that g induces aninvolutory automorphism onD D �3.C / D ha; d i. We claim that g fixes each of thethree maximal subgroups S1 D ha;W i, S2 D hd;W i, S3 D had;W i of D each ofwhich is abelian of type .8; 4/. If, for example, Sg

1 D S2, then .Ã1.S1//g D Ã1.S2/

and so ha2;W0ig D hd2;W0i which contradicts the fact that ˇ centralizes W=W0.Hence g induces an involutory automorphism on each of three abelian subgroups S1,S2, S3 and CSi

.g/ D W0 for each i D 1; 2; 3. By Proposition 51.2, g inverts each ofthree subgroups Ã1.S1/, Ã1.S2/, and Ã1.S3/ (which are three maximal subgroups ofW ) and so g invertsW , contrary to the fact that uˇ D u�1y2 ¤ u�1.

Lemma 82.14. Suppose there is an element s 2 G which induces the automorphism� D �2.0; 1/ onW and j�3.C /j D 25, where C D CG.W /. Then C D hs2i � hui Š

C2n � C4, where n � 3 and the element u 2 W is as in Lemma 82.9. If S D hs; C i,then Ã2.S/ D hs4i and S is minimal nonabelian metacyclic.

Proof. Set againW0 D �1.W /, whereW D hu; yi, u� D uy2, y� D y and note that� is “special” in the sense of Lemma 82.6. Also set S D hs; C i so that jS W C j D 2

since s2 2 C . Each element in sC induces the “special” automorphism � on W andsince � does not invert any element inW �W0, Lemma 82.3 implies that there are noelements of order 4 in S � C . Hence�2.S/ D W , and Remark 2 following Theorem41.1 shows that S is metacyclic.We know from Lemma 82.12 that C is abelian of type .2n; 22/, n � 3, so that

�3.C / is abelian of type .8; 4/ and C=W ¤ f1g is cyclic.Suppose that there is x 2 S � C with x2 2 W (in which case x2 2 W � W0

since �2.S/ D W ). Set T D h�3.C /; xi, where j�3.C / W W j D 2, by hypothesis,T=W Š E4, and all elements in T � W are of order 8 (noting that W � �2.T / �


�2.S/ D W so�2.T / D W ). Since T is metacyclic as a subgroup of S and T=W Š

E4, we get ˆ.T / D W . Also x, the element of T , induces on W the involutoryautomorphism � (indeed, x�1s 2 C in view of jS W C j D 2) and so T is nonabelian.If jT 0j D 2, then this fact together with d.T / D 2 implies that T is minimal nonabelianwhich forces W D ˆ.T / D Z.T /, a contradiction since, by the choice, x does notcentralizes W . Hence jT 0j > 2. On the other hand, exp.T / D 8, jT j D 26, and T ismetacyclic. Thus T possesses a cyclic normal subgroupZ of order 8 such that T=Z Š

C8. Since T 0 < Z, we get T 0 Š C4. But Aut.Z/ Š E4 and so jT W CT .Z/j D 2 andif t 2 T � CT .Z/, then t induces an involutory automorphism on Z. Next, CT .Z/

is abelian of type .8; 4/, t2 2 Z.T /. Assume that T 0 < Z.T /. Then jT W Z.T /j D 4

so jT 0j D 2 (Lemma 1.1), a contradiction. Then t does not centralize T 0 so t invertsT 0 (in view of T 0 Š C4, the unique nonidentity automorphism inverts T 0). Since tinduces on W the “special” automorphism � which does not invert any element oforder 4 in W , we get a contradiction since T 0 < W .We have proved that for each x 2 S � C , x2 2 C �W . In particular, �3.C /=W

is the unique subgroup of order 2 in S=W , and so S=W is either cyclic or generalizedquaternion.Suppose that S=W is generalized quaternion. Set C0=W D .S=W /0 so that jS W

C0j D 4, S 0 < C0 (<, by Taussky’s theorem, since S is not of maximal class inview of W < S ), S 0 covers C0=W , S 0 is cyclic since S is metacyclic and �2.S

0/ <

�2.S/ D W . Hence S 0 \W Š C4 which implies jC0 W S 0j D 4 and jS W S 0j D jS W

C0jjC0 W S 0j D 16. Noting that S has the abelian maximal subgroup C Š C2n � C4,we get, by Lemma 1.1, jZ.S/j D 1

2jS W S 0j D 23. On the other hand, W0 � Z.S/ and

if x 2 S�C , then x2 2 �3.C /�W and so o.x2/ D 8. We get CS .x2/ � hC; xi D S

and so x2 2 Z.S/. Hence Z.S/ � hW0; x2i and so jZ.S/j � 24, a contradiction.

We have proved that S=W is cyclic and so, if s 2 S � C , then hsi covers S=Wand hsi \W Š C4. Since CW .s/ D hW0; yi, we have hsi \W D hyi or hsi \W D

hyu2i and so in any case C D hs2i � hui. Also, hsiW0 is another abelian maximalsubgroup of S (distinct from C ) and so jS 0j D 2 (Lemma 1.1). It follows that S (beingmetacyclic) is minimal nonabelian (see also Lemma 65.2(a)). For each siuj 2 S , weget .siuj /4 D .s4/i .u4/j Œuj ; si �6 D .s4/i , which implies Ã2.S/ D hs4i, and we aredone.

Recall that the unique minimal nonabelian nonmetacyclic group of order 24 is iso-morphic to hx; y j x4 D y2 D z2 D 1; z D Œx; y�; Œx; z� D Œy; z� D 1i (Lemma65.1). Recall also that AG.W / is the group of automorphisms induced by G on itsnormal abelian subgroup W D hui � hyi of type .4; 4/.

Lemma 82.15. If jAG.W /j > 23, then jAG.W /j D 24,W D CG.W / and AG.W / D

h�.0; 1/; �.1;�1/i which is the minimal nonabelian nonmetacyclic group of order 24.

Proof. The “superspecial” automorphisms and � and �.1; 1/ cannot be containedin AG.W / (Lemma 82.7 and 82.10) and so jAG.W /j D 24 and AG.W / is a max-imal subgroup in the automorphism group A (of order 25) of all automorphisms of


W fixing the subgroup hW0; yi, where W0 D �1.W /. This implies that � and��.1; 1/ are contained in AG.W /. We check that ��1.0; 1/ D ��.1; 1/ and soh �; �.0; 1/i � AG.W /, where we also see that � D �2.1;�1/ and Z.A/ D

ˆ.A/ D h�2.0; 1/; �2.1;�1/i (see Lemma 82.6 and Lemma 82.9).There are exactly three maximal subgroups of A containing h �, �.0; 1/i (which is

abelian of type .4; 2/ and contains ˆ.A/) and they are candidates for AG.W /:

h�.0; 1/; �.1; 1/i; h�.0; 1/; �.0;�1/; �2 .1; 1/i; h�.0; 1/; �.1;�1/i:

We claim that the last subgroup coincides with AG.W /. Since �.1; 1/ 62 AG.W /, itwill suffice to show that AG.W / ¤ h�.0; 1/; �.0;�1/; �2 .1; 1/i.Suppose, by the way of contradiction, that AG.W / D h�.0; 1/; �.0;�1/; �2.1; 1/i.

Then using Lemmas 82.11, 82.12, and 82.13, we see that j�3.C /j D 25 and C Š

C2n � C4, n � 3. Let t 2 G induce �.0; 1/ on W so that ut D uy, yt D y andCW .t/ D hyi. Using Lemma 82.14, we get hti \W Š C4 and hti \W D Œt;W � D

hŒt; u�i D ht2n

i for some n � 3. We have C D CG.W / D ht4i � hui and settingT D ht; C i we get T 0 D Œt;W � D hyi � Z.T / and so T is of class 2. Considerthe element � D � �.0; 1/�.0;�1/ 2 AG.W / and check that u D u�1 andy D y so that .t2

n

/ D t2n

. On the other hand, ��1�.0; 1/� D ��1.0; 1/ (whichis verified by direct application of both sides on u and y) and so, if r 2 G induces �on W , then tr t�1 .mod C/ which gives tr D t�1C4iuj for some i; j 2 Z. Thus.t2

n

/r D .t�1C4iuj /2n

D t�2n

, where we have used the fact that T is of class 2 andn � 3. But this contradicts the above result .t2

n

/r D t2n

.We have proved that AG.W / D h�.0; 1/; �.1;�1/i. Hence AG.W / is generated by

two elements �.0; 1/ and �.1;�1/ of order 4 and we have

Œ�2.0; 1/; �.1;�1/� D Œ�.0; 1/; �2.1;�1/� D 1; Œ�.0; 1/; �.1;�1/� ¤ 1;

and AG.W / is Q8-free (Lemma 82.6 and 82.9). This implies that AG.W / is theuniquely determined minimal nonabelian nonmetacyclic group of order 24 (Lemma79.2).Suppose C > W so that C is abelian of type .2m; 22/,m � 3 (Lemma 82.12, 82.13,

and 82.14). Set S D ht2; C i, where t 2 G induces �.0; 1/ on W . By the structure ofG=C Š AG.W /, S is normal inG and by Lemma 82.14, Ã2.S/ D hs4i (with s D t2)is a cyclic characteristic subgroup of S of order at least 4. Hence W contains a cyclicsubgroup of order 4 which is normal in G. However, �.1;�1/ does not normalize anyof the six cyclic subgroups of order 4 in W . Hence C D W .

Recall that automorphisms �.i; j / of W are defined in Lemma 82.9 as follows:

�.i; j / W u �! uy; y �! u2iyj ; where i D 0; 1 and j D 1;�1:

Theorem 82.16. Let G be a 2-group containing exactly three involutions and a nor-mal subgroup W Š C4 � C4. Suppose that jG=CG.W /j > 23. Then CG.W / D W ,


�1.W / — Z.G/ so that Z.G/ is cyclic, G=W is the unique minimal nonabelian non-metacyclic group of order 24, andG is isomorphic to one of the following eight groupsof order 28:

G D hg; h j g16 D 1; g4 D y; h4 D y2; .h2g/2 D uiy2�g2;

u4 D Œu; y� D 1; .gh/2 D u2�y2; ug D uy; yg D y;

uh D uy; yh D u2y�1i; i D ˙1; �; � D 0; 1:

(�)

Here we have hu; yi D W , Z.G/ D hy2i Š C2, ˆ.G/ D hW;g2; h2i, T D hW;gi

is a nonnormal metacyclic subgroup of class 2 with jG W T j D 4, S D hW;g2i isnormal in G and G=S Š D8.Conversely, each of the eight groups of order 28 given with (�) has exactly three

involutions and satisfies the assumptions of our theorem.

Proof. (i) Using Lemma 82.15, we have W D CG.W / and G=W Š AG.W / D

h�.0; 1/; �.1;�1/i is the minimal nonabelian nonmetacyclic group of order 24 so thatˆ.AG.W // D Z.AG.W // D h�2.0; 1/; �2.1;�1/i, �.0; 1/ �.1;�1/ is an involutionnot contained in Z.AG.W //, and .AG.W //

0 D h�2.0; 1/ �2.1;�1/i, where

u�.0;1/ D uy; y�.0;1/ D y; u�.1;�1/ D uy; y�.1;�1/ D u2y�1:

Let g 2 G be an element inducing �.0; 1/ on W D hu; yi so that

ug D uy; yg D y; ug2

D uy2; yg2

D y; ug�1

D uy�1; yg�1

D y:

Since CW .g/ D hyi, we have g4 2 hyi. Also, ŒW; g� D hyi and since g2 does notinvert any element inW �W0, g2W does not contain any element of order 4 (Lemma82.3). Hence �2.hW;gi/ D W and so T D hW;gi is metacyclic, hg4i D hyi, ando.g/ D 16. We have T 0 D hyi D hg4i � Z.T / and so T is of class 2, hgi is normalin T and jG W T j D 4. Set S D hW;g2i. By the structure of G=W , T is a non-normal metacyclic subgroup of index 4 in G and S is normal in G with G=S Š D8. Ifg4 D y�1, then we replace u; y with u0 D u�1; y0 D y�1 so that (noting that T is ofclass 2):

Œu0; g� D Œu�1; g� D Œu; g��1 D y�1 D y0;

g4 D y0; .u0/g D u0y0; .y0/g D y�1 D y0;

.u0/�.1;�1/ D u�1y�1 D u0y0; .y0/�.1;�1/ D u2y D .u0/2.y0/�1:

Thus, writing again u; y instead of u0; y0, we may assume from the start that Œu; g� D

y D g4.For each uiyj 2 W , we compute:

.g2uiyj /2 D g4.ug2

/i .yg2

/juiyj D yuiy2iyjuiyj D y1C2.iCj /u2i ;

which shows that o.g2uiyj / D 8 and so indeed�1.hW;gi/ D hu2; y2i D W0.


Let h 2 G be an element such that h induces �.1;�1/ on W so that G D hW;g; hi

and

uh D uy; yh D u2y�1; uh2

D u�1;

yh2

D y�1; uh�1

D u�1y�1; yh�1

D u2y;

which implies CW .h/ D hy2i and therefore h4 D y2. Since h2 inverts W , we have,for each w 2 W . .h2w/2 D h4.w/h

2

w D y2w�1w D y2, and so �1.hW;hi/ D

W0 D hy2; u2i.Since �.0; 1/ �.1;�1/ is an involution in AG.W /, we have .gh/2 2 W and

ugh D u�1; ygh D y.u2y2/; .uy/gh D .uy/y2;

so that CW .gh/ D W0 and .gh/2 D w0 2 W0, w0 ¤ 1. All elements in ghW mustbe of order 4. For each uiyj 2 W , we get

..gh/uiyj /2 D .gh/2.ugh/i .ygh/juiyj D w0u�iyju2jy2juiyj D w0u

2j ;

and so w0 ¤ u2 which implies:

.gh/2 D u2�y2; � D 0; 1:(8)

As result, we have obtained the group G satisfying (�).

(ii) It remains to show that any group given by (�), has exactly three involutions.We have �1.G=W / D hgh; g2; h2iW=W Š E8 and so we must show that there

are no involutions in .hgh; g2; h2iW / �W . We have already shown that there are noinvolutions in cosets g2W , h2W , and .gh/W . Since .G=W /0 D hg2h2iW=W and

.gh/g ghg gh.g2h2/ .mod W /;

..gh/h2/g ghgh2 gh.g2h2/h2 .gh/g2 .mod W /;

it follows that we still have to require (and show) that there are no involutions in cosets.gh/h2W and g2h2W .From (8) follows:

.hg/2 D hghg D g�1.ghgh/g D ..gh/2/g D .u2�y2/g

D u2�y2�y2 D u2�y2.�C1/;

and so we get:

.hg/2 D u2�y2.�C1/:(9)

From (8) and (9) (since g commutes with y) follows:

.ghgh/.hghg/ D u2�y2 u2�y2.�C1/ D y2�;

gh�1.h2gh2g/hg D y2�; h�1.h2g/2h D y2�g�2 D y2�g�4g2 D y2��1g2;


and so

.h2g/2 D .y2��1/h�1

.g2/h�1

D u2y2��1.g2/h�1

:(10)

Again from (8) and (9) follows: .hghg/.ghgh/ D y2� , hg.hg2h�1/.h2g/h D y2� ,and so multiplying the last relation with h from the left and with h�1 from the rightwe get:

.h2g/.g2/h�1

.h2g/ D hy2�h�1 D .yh�1

/2� D .u2y/2� D y2�;

and so

.g2/h�1

D .h2g/�1y2�.h2g/�1 D .y2�/h2g.h2g/�2 D y�2�.h2g/�2;

which together with (10) gives .h2g/2 D u2y2��1 y�2�.h2g/�2, and so we obtainfinally,

.h2g/4 D u2y�1:(11)

Since W hh2; gi=W is abelian of type .4; 2/, we get .h2g/2 D uiyjg2 for somei; j 2 Z. This gives together with (11):

.h2g/4 D u2y�1 D uiyjg2 uiyjg2 D uiyjg4.uiyj /g2

D uiyjyuiy2iyj D u2iy2.iCj /C1;

which implies that i is odd and j D 2� is even and so we get the fundamental relation:

.h2g/2 D uiy2�g2; i D ˙1; � D 0; 1:(12)

Since .h2g/2 2 Ã1.G/, y2�g2 2 Ã1.G/, (12) gives that u 2 Ã1.G/. This facttogether with g4 D y shows that Ã1.G/ � W and so d.G/ D 2 (since d.G=W / D 2).From (12) follows at once h2gh2g D uiy2�g2, h2.gh2g�1/ D uiy2�, which

together with h4 D y2 gives:

.h2/g�1

D h2uiy2.�C1/:(13)

From (13), we get directly the following two relations which we shall need:

.h2/g D h2u�iy�i�2.�C1/;(14)

.h2/g�2

D h2u2iy�i D h2u2y�i :(15)

Using (8) and (14), we get (since h4 D y2)

..gh/h2/2 D ghh2 ghh2 D ghg.g�1h2g/h3 D ghg h2u�iy�i�2��2h3

D .ghg/uiyiC2�C2h5 D .ghgh/.h�1uiyiC2�C2y2h/

D .gh/2uiyi .u2y�1/iC2� D u2��iy2.�C1/;


and so for each urys 2 W (r; s 2 Z), we get

..gh/h2urys/2 D ..gh/h2/2.ugh3

/r.ygh3

/surys D u2��iC2rC2sy2.�C1/C2s;

which is an element of order 4 in W (since i D ˙1) and so we have shown that thereare no involutions in .gh/h2W .Using (15), we get (since g4 D y)

.g2h2/2 D g2h2g2h2 D .g2h2g�2/g4h2 D .h2/g�2

yh2

D h2u2y�iyh2 D h4.h�2.u2y�iC1/h2/ D y2u2yi�1 D u2yiC1;

and so for each urys 2 W (r; s 2 Z), we get

.g2h2urys/2 D .g2h2/2.ug2h2

/r.yg2h2

/surys D u2y2rCiC1;

which is an involution in W0 and so we have also shown that there are no involutionsin g2h2W . The proof is complete.

Lemma 82.17. Let G be a 2-group with exactly three involutions and a normal sub-group W Š C4 � C4 so that W0 D �1.W / 6� Z.G/. Suppose that AG.W / Š

G=CG.W / is of order 8. Then G contains a metacyclic subgroupM of index � 4 andM is normal in G except in the case, where AG.W / D h�.1;�1/; �.0;�1/i Š D8.

Proof. We consider X D AG.W / as a subgroup of the group A (of order 25) of allautomorphisms of W D hu; yi normalizing hW0; yi (see Lemmas 82.6 and 82.9). LetA0 be the stabilizer of the chainW > W0 > f1g so that A0 Š E24 and X 6� A0. Thenwe can choose a basis fu; yg of W so that X contains one of the elements of order4: �.0; 1/, �.1; 1/, �.1;�1/, or the involution D �.0;�1/ (Lemma 82.9), where allthese elements lie in A � A0.Set C D CG.W /. If �.0; 1/ 2 X and g 2 G induces �.0; 1/ on W , then

jG W hC; gij D 2 and hC; gi is metacyclic (of index 4 in G) since �2.hC; gi/ D

�2.hC; g2i/ D W noting that g2 (inducing �2.0; 1/ on W ) does not invert any ele-

ment inW �W0 and so there are no elements of order 4 in g2C (Lemma 82.3). Hencewe may assume that �.0; 1/ 62 X .Since �.1; 1/ 62 X (Lemma 82.10), it follows that �.1;�1/ 2 X or 2 X . Note

that for each x 2 A � A0, jCA.x/j D 8 and CA.x/ D hxiA0, where A0 D Z.A/ D

ˆ.A/ D h�2.0; 1/; �2.1; 1/i Š E4. If X is abelian, we have in any case �2.0; 1/ 2 X .But in that case, if t 2 G induces �2.0; 1/ on W , then jG W hC; tij D 4, hC; ti isnormal in G, and hC; ti is metacyclic (by the previous paragraph).Suppose thatX is nonabelian so thatX Š D8 (sinceA is Q8-free). SinceX\A0 Š

E4, there are elements of order 4 inX �A0 and so we may assume that �.1;�1/ 2 X .But there are also involutions in X � A0 and so either 2 X or �2.1; 1/ 2 X . (Werecall that has exactly fourA-conjugates which lie in the set A0.) But, if �2.1; 1/ 2

X , then � D �.1;�1/�2.1; 1/ 2 X . We check that u� D uu2y2, y� D yu2, so


that � D is “superspecial” (Lemma 82.6). According to Lemma 82,7, 62 AG.W /

and so this case cannot occur. HenceX D h�.1;�1/; �.0;�1/ D i Š D8, where theinvolution �.0;�1/ inverts the element �.1;�1/ of order 4.Suppose that t 2 G induces �.0;�1/ on W . By Lemma 82.3 (noting that t2 2 C ),

there are no elements of order 4 in tC and since tC cannot contain involutions, wehave�2.hC; ti/ D W and so hC; ti is a nonnormal metacyclic subgroup of index 4 inG and we are done.

Theorem 82.18. Let G be a 2-group with exactly three involutions and a normal sub-group W Š C4 � C4 so that W0 D �1.W / 6� Z.G/. Set C D CG.W / and assumethat jG=C j D 8. Then G contains a metacyclic subgroup M of index at most 4 in Gand M is normal in G except in the following cases .˛/ and .ˇ/, where G=C Š D8,jC j D 22nC1, n � 2, C D ha; b j a2nC1

D b2n

D 1; Œa; b� D a2n�; � D 0; 1i with� D 0 if n D 2, Z.G/ is of order 2, ˆ.G/ > C , and the structure of G is determinedwith two generators and relations:

(˛) n D 2, jGj D 28, and G D hg; t jg8 D t16 D 1; g4 D t8 D z; t2 D a; ag D

au�1; u4 D Œa; u� D 1; ug D u�1a2; ut D u�1a2; .gt/2 D zui ; i D 0; 1; 2; 3i

with Z.G/ D hzi Š C2, C D hai � hui Š C8 � C4, W D ha2; ui Š C4 � C4,�.G/ D hg2iC , where g2 inverts each element in C , and �1.G/ D �1.W / D

hz; u2i Š E4. Finally, hC; ti is a non-normal metacyclic subgroup of index 4 in G.

(ˇ) n > 2, jGj D 22nC4, and G D hg; ti, where

g8 D t2nC2

D 1; g4 D t2nC1

D z; t2 D a;

ag D ab; b2n

D 1; Œa; b� D z�; � D 0; 1:

(ˇ1) If � D 0, then C D ha; bi is abelian of type .2nC1; 2n/ and

bg D b�1a�2; bt D b�1a�2; .tg/2 D z.a2b/s with s 2 Z (any integer) :

(ˇ2) If � D 1, then C D ha; bi is a minimal nonabelian metacyclic group, bg D

b�1a�2z�b2n�1

, � D 0; 1, and either:

(ˇ21) .tg/2 D zr .a2b/s with bt D b�1a�2z1C�b2n�1

, where s is any oddinteger and if s 1 .mod 4/, then r D 1 and if s �1 .mod 4/,then r D 0; or

(ˇ22) .tg/2 D zr .a4b2/s with bt D b�1a�2z�b2n�1

, where in case that sis odd, then r D 0 and in case that s is even, then r D 1.

In all these cases Z.G/ D hzi Š C2, ˆ.G/ D hC; g2i, W D ha2n�1

; b2n�2

i D

�2.C / Š C4 �C4, hC; ti is a nonnormal metacyclic subgroup of index 4 in G, and inall above groups �1.G/ D �1.W / D hz; b2n�1

i Š E4.


Proof. We continue with the situation in Lemma 82.17, where AG.W / D h�.1;�1/,�.0;�1/i Š D8 and

u�.1;�1/ D uy; y�.1;�1/ D u2y�1; u�.0;�1/ D uy; y�.0;�1/ D y�1:

We recall that �.1;�1/ (of order 4) does not fix any of the six cyclic subgroups oforder 4 in W and �.0;�1/ is an involution inverting �.1;�1/.Let g be an element in G which induces �.1;�1/ (of order 4) on W so that ug D

uy, yg D u2y�1, and CW .g/ D hy2i. This implies that CC .g/ D hy2i, whereC D CG.W / since �2.C / D W D hu; yi. There are no involutions in g2C and sog4 D y2 and we set y2 D z. We have Z.G/ � C which together with CC .g/ D hzi

implies Z.G/ D hzi.Let t be an element in G which induces the involution �.0;�1/ on W so that ut D

uy, yt D y�1. By Lemma 82.11, t2 2 C �W and therefore C > W .Replace y with y0 D yu2. Then .y0/2 D y2 D z, y D y0u2, and so

ug D uy D uy0u2 D u�1y0; .y0/g D .yu2/g D u2y�1u2y2 D y D u2y0;

ut D uy D uy0u2 D u�1y0; .y0/t D .yu2/t D y�1u2y2 D u2y D u2y0u2 D y0:

Since g4 D y2 D z D .y0/2 and W D hu; y0i, we may write again y instead of y0 sothat we get from the start the following fundamental relations:

ug D u�1y; yg D u2y; ut D u�1y; yt D y; g4 D y2 D z;(��)

and so we see that CW .t/ D hyi which implies ht2i > hyi since t2 2 C �W . Alsonote that ug2

D u�1, yg2

D y�1 and so g2 inverts each element inW .Since g does not normalize any cyclic subgroup of order 4 in C , Proposition 50.4

implies (together with the facts that W0 62 Z.G/, C > W , and W � Z.C /) that wehave the following cases for the structure of C :

jC j D 25; C D ha; b j a8 D b4 D Œa; b� D 1i;(16)

jC j D 22n; n � 3; C D ha; b j a2n

D b2n

D 1; Œa; b� D a2n�1�; � D 0; 1i;(17)

jC j D 22nC1; n � 3; C D ha; b j a2nC1

D b2n

D 1; Œa; b� D a2n�; � D 0; 1i:(18)

Case (a). In this case C is given in (16). Since t2 2 C � W and ht2i > hyi, wemay set C D hai � hui, where a2 D y and t2 2 hai � hyi. Replacing t with tai

for a suitable i 2 Z, we may assume from the start that t2 D a and we know thatut D u�1y D u�1a2. Also, we know from (��) that ug D u�1y, yg D u2y, andg4 D y2 D a4 D z.We have ugt D u, ygt D y.u2y2/ so that CW .gt/ D hW0; ui, where W0 D

hz; u2i. Since g2 induces an involutory automorphism on C and g2 inverts each ele-ment in W , we get CC .g

2/ D W0 and so g2 inverts C=W0 (Proposition 51.2) whichgives ag2

D a�1w0 with some w0 2 W0.


For each aiuj 2 C , we compute

.taiuj /2 D taiuj taiuj D t2.aiuj /taiuj D aaiu�ja2j aiuj D a1C2.iCj /;

which is an element of order 8. Thus �2.hC; ti/ D W and so hC; ti is a non-normalmetacyclic subgroup of index 4 in G.We set ag D aiuj for some i; j 2 Z, where i is odd since ag must be also an

element of order 8 in C . We have

yg D u2y D u2a2 D .a2/g D .ag /2 D a2iu2j D u2jyi ;

which implies i 1 .mod 4/, j 1 .mod 2/. Since o.a/ D 8 and o.u/ D 4, wemay set i D 1C 4�, j D 1C 2�, where �; � D 0; 1.By the above result ag2

D a�1w0 withw0 2 W0, we get (also using the expressionsfor i; j ):

ag2

D a�1w0 D .aiuj /g D .aiuj /iu�jyj D ai2C2juj.i�1/ D a1C2j ;

which gives a2.j C1/ D w0 and sow0 D a4.1C�/ D z1C�. If � D 0, then ag2

D a�1z

and so ha; g2i is semidihedral of order 16, contrary to our assumption that G hasexactly three involutions. Thus, � D 1, w0 D 1, ag2

D a�1, g2 inverts C and ag D

a1C4�u3 with � D 0; 1. For each c 2 C , we have .g2c/2 D g4cg2

c D zc�1c D z,and so all elements in g2C are of order 4.Suppose that � D 1 so that ag D aa4u�1 D azu�1 D a.zu/�1. Set u0 D zu

so that u D zu0 and ag D a.u0/�1. We see that all the previous relations remainunchanged if we replace u with u0. Indeed,

yg D .a2/g D a2.u0/2 D y.u0/2;

.u0/g D .zu/g D zu�1y D .u0/�1y;

.u0/t D .zu/t D zu�1y D .u0/�1y:

Writing again u instead of u0, we see that we may assume from the start that � D 0 andso we get ag D au�1. Since t2 D a and Œa; g� D u�1, we get C D ha; ui � ˆ.G/

and so ˆ.G/ D hg2; C i, d.G/ D 2 and G D hg; ti. We also get agt D .au�1/t D

auy�1 D a�1u, ugt D u.We compute for each arus 2 C ,

arus D .arus/gt D a�rurus if and only if a2r D ur ; yr D ur ; r 0 .mod 4/;

and so CC .gt/ D hz; ui D CW .gt/ which forces .gt/2 D w1 2 hz; ui.For each apuq 2 C , we compute

..gt/.apuq//2 D .gt/2.apuq/gtapuq D w1a�pupuqapuq D w1u

pC2q;

and so we must have w1upC2q ¤ 1 for all p; q 2 Z. This forces w1 D zui and


so .gt/2 D zui with i D 0; 1; 2; 3. The structure of G is determined and we haveobtained the groups .˛/ of our theorem.

Cases (b) and (c), whereC D CG.W / is abelian or minimal nonabelian with n � 3,given in (17) and (18). Since ht2i > hyi > hzi D Z.G/, t2 2 C , and C 0 � hzi, wehave Ã2.C / � C 0 and so C is a powerful 2-group. Obviously, C is D8-free and Q8-free since C is “ordinary metacyclic” (Proposition 26.27). In particular, C is modular.By Proposition 26.24, Ã1.C / D ha2; b2i � Z.C / and W D hu; yi D �2.C /. By

Proposition 26.23, for each x 2 Ã1.C /, there is r 2 C with x D r2. If r 2 Ã1.C /,then there is s 2 C so that s2 D r and so x D s4 and so on. Hence for each c 2 C ,there is a generator l 2 C�Ã1.C / such that hci � hli. Obviously, all the above resultscould be also obtained directly (without quoting the results about powerful 2-groups)from the structure of C .It is possible to prove the crucial result that t2 2 C � Ã1.C / is a generator of C .

Suppose that this is false. Then there is s 2 C with s2 D t2. If hti normalizes hsi, thenht; si has two distinct cyclic subgroups hti and hsi of index 2, where 2l D o.t/ D o.s/,l � 4. Thus ht; si is noncyclic and so ht; si is either abelian of type .2l ; 2/ or ht; si Š

M2lC1 . In any case, there is an involution in ht; si � hsi, where ht; si \ C D hsi,a contradiction. Hence t does not normalize hsi D S , where jS j D 2l , l � 4, andjS W ht2ij D 2. Set S0 D S t D hst i, where S0 � C , S0 � ht2i, jS0 W ht2ij D 2

since t normalizes (centralizes) ht2i. Since S is normal in C (noting that S � hzi andC 0 � hzi), T D SS0 is of order 2lC1 and .st /t D st2

D s implies that t normalizesT and therefore jhT; tij D 2lC2 with hT; ti \C D T . Also, S and S0 are two distinctcyclic subgroups of index 2 in T and so T is not cyclic. We have o.t2/ � 8 andht2i � Z.T / so that T cannot be generalized quaternion. Hence T has more than oneinvolution and so T � �1.C / D W0. But W0 2 Z.C / and W0 \ S D hzi so thatT D hsi � hu2i is abelian of type .2l ; 2/.Since o.st / D o.s/ D 2l , we may set st D st2iu2 for some integer i . We have

s D st2

D .st2iu2/t D st2iu2t2iu2z, which gives t4i D z and t8i D 1. Sinceo.t/ D 2l .l � 4/, we get i 0 .mod 2l�3/ and we may set i D 2l�3i 0. On the otherhand, z D t4i D t2

l�1i 0

D zi 0

and so i 0 is odd. We get Œs; t � D t2iu2 D t2l�2i 0

u2 D

yju2, where j D ˙1. In particular, Œs; t � is of order 4.Since t acts non-trivially on T=ht2i Š E4, we have ht; si=ht2i Š D8, where

ht; si D hT; ti. Let A=ht2i be the cyclic subgroup of index 2 in ht; si=ht2i so thatA is abelian since ht2i � Z.ht; si/. But then ht; si possesses two distinct abelianmaximal subgroups T and A. By a well-known result of A. Mann (Lemma 64.1 (u)),.ht; si/0 is of order 2, contrary to the above result that o.Œs; t �/ D 4. We have provedthat t2 2 C � Ã1.C /.Let x 2 C � Ã1.C /. Since x is a generator of C , there is v 2 C � Ã1.C / so that

hx; vi D C . Since C is modular, we have hxihvi D C and so jC j D .o.x/o.v// W

.jhxi \ hvij/. If jC j D 22n, then exp.C / D 2n and in that case o.x/ D o.v/ D 2n

and hxi \ hvi D f1g so that hxi has a complement in C . If jC j D 22nC1, thenexp.C / D 2nC1 and in that case o.x/ D 2n or o.x/ D 2nC1.


Case (b), where jC j D 22n, exp.C / D 2n, n � 3, is given in (17). As a surprise,we shall show that this case cannot occur at all! Set ht2i D hai so that o.a/ D 2n

since we have proved that t2 2 C � Ã1.C /. We choose the generator a of hai sothat a2n�2

D y, where y2 D z with hzi D Z.G/. We have t2 D aa2i (i 2 Z)and replacing t with t 0 D ta�i we get .t 0/2 D t2a�2i D aa2ia�2i D a. Note thatt 0 operates the same way on W as t does (since W � Z.C /) and so writing again tinstead of t 0, we may assume from the start that t2 D a.Let hbi be a cyclic subgroup of order 2n of C which contains hui so that hai \

hbi D f1g and we may choose the generator b of hbi so that b2n�2

D u. Hencejha; bij D o.a/o.b/ D 22n so that ha; bi D C . If C is nonabelian, then jC 0j D 2 andso C 0 � Z.G/ which implies C 0 D hzi. Hence, we have in any case, ab D az� , where� D 0; 1.We consider first the special case n D 3, � D 1 so that o.a/ D o.b/ D 8 and

ab D az. Set bt D amb (for some m; � 2 Z) and so

ut D u�1y D .b2/t D .amb/2 D a2mb2 Œb; a�m D ymuzm

D ymuy2m D uym.1C2/;

which gives � �1 .mod 4/, m.1C 2�/ 1 .mod 4/, m �1 .mod 4/. We mayset m D �1C 4m0, � D �1 C 4�0 so that bt D a�1zm0

b�1u20

D a�1b�1w1 withw1 2 W0. However,

bt2

D ba D bz D .a�1b�1w1/t D a�1w�1

1 bawt1 D a�1abzw�1

1 wt1 D bzw�1

1 wt1;

and so wt1 D w1, which gives w1 D z� , � D 0; 1. We get .tb/2 D t2btb D

aa�1b�1z�b D z� , so that htb;W0i Š D8 since tb acts non-trivially on W0. But thenG has more than three involutions, a contradiction.We have proved that the case n D 3, � D 1 cannot happen. In general, we set again

bt D amb (for some m; � 2 Z) and so:

ut D u�1y D .b2n�2

/t D .amb/2n�2

D .a2n�2

/m.b2n�2

/ D uym;

since either n > 3 or n D 3, � D 0. This gives � �1 .mod 4/, m 1 .mod 4/ andso setting m D 1C 4m0, � D �1C 4�0, we get bt D .ab�1/a4m0

b40

.We compute:

bt2

D ba D bz� D .a1C4m0

b�1C40

/t D a1C4m0

.a1C4m0

b�1C40

/�1C40

D a1C4m0

b1�40

a�1�4m0

a40C16m00

b�40C16.0/2

D a40.1C4m0/bz�b80.�1C20/;

and so a40.1C4m0/b80.�1C20/ D 1, which implies �0 0 .mod 2n�2/ and bt D

a1C4m0

b�1.


Now, we get atb D ab D az� D aa2n�1� D a1C2n�1�, and so tb normalizes hai

and .tb/2 D t2btb D aa1C4m0

b�1b D a2.1C2m0/ is of order 2n�1. Set R D ha; tbi

so that hai and htbi are two distinct cyclic maximal subgroups (of order 2n, n � 3) ofR. It follows that R is either abelian of type .2n; 2/ or R Š M2nC1 . In any case, thereare involutions in R� hai D R�C , a contradiction. We have proved that the Case (b)cannot occur.

Case (c), where jC j D 22nC1, exp.C / D 2nC1, n � 3 is given in (18). We knowthat t2 2 C � Ã1.C / is a generator of C and, since ht2i > hyi > hzi D Z.G/, wemay set hai D ht2i so that hai is normal in C (since C 0 � hzi). Let d 2 C be anothergenerator for C so that ha; d i D C . By Proposition 26.24, Ãn.C / D ha2n

; d2n

i

and by the structure of C (in Case (c)), jÃn.C /j D 2. If a is of order 2n, thenÃn.C / D hd2n

i Š C2 is normal in G with o.d/ D 2nC1 and hai \ hd i D f1g (sincejC j D 22nC1), contrary to the fact that Z.G/ D hzi � hai. Hence a is of order 2nC1

and we may choose a generator a of hai so that a2n�1

D y. We have t2 D aa2i (forsome i 2 Z) and replacing t with t 0 D ta�i , we get .t 0/2 D t2a�2i D aa2i�2i D a.Note that t 0 acts the same way on W D �2.C / D hu; yi as t does since W � Z.C /and so we may assume from the start that t2 D a, where o.t/ D 2nC2, n � 3.Let hbi be a cyclic subgroup of C containing hui such that b 2 C � Ã1.C /. We

know that o.b/ D 2n or o.b/ D 2nC1. But hai \ hbi D f1g since hyi D �2.hai/ andhui D �2.hbi/ and hyi \ hui D f1g. We get ha; bi D haihbi is of order o.a/o.b/ andso o.b/ D 2n, jha; bij D 22nC1 and ha; bi D C with ab D az� . We may choose agenerator b of hbi so that b2n�2

D u.We recall (from (��)) that ug D u�1y, yg D yu2, g2 inverts W , ut D u�1y,

at D a (since t2 D a), and g4 D z. We have S D �n.C / D ha2; bi is abelian(of type .2n; 2n/) g2-invariant maximal subgroup of C and g2 induces an involutoryautomorphism on S . Since CS .g

2/ D CW .g2/ D W0, g2 inverts each element in

Ã1.S/ D ha4; b2i (Proposition 51.2). This fact will be used often in the sequel.It is now easy to determine the action of t on C by lifting up the action of t on W .

We set bt D a2ibj (i; j 2 Z) since bt 2 S and compute (noting that a2i 2 Z.C /):

u�1y D ut D .b2n�2

/t D .a2ibj /2n�2

D a2n�1ib2n�2j D yiuj ;

which implies i 1 .mod 4/, j �1 .mod 4/ so that we may set i D 1 C 4i 0,j D �1C 4j 0, and bt D a2a8i 0

b�1b4j 0

. Since t2 D a, we get:

bt2

D ba D bz� D .a2a8i 0

b�1b4j 0

/t

D a2a8i 0

a�2a�8i 0

bb�4j 0

a8j 0

a32i 0j 0

b�4j 0

b16.j 0/2

D a8j 0.1C4i 0/b1�8j 0.1�2j 0/;

and so j 0 0 .mod 2n�3/. We may set j 0 D 2n�3j 00 and so z� D a2nj 00.1C4i 0/ D

zj 00

and j 00 � .mod 2/. We obtain bt D a2.1C4i 0/b�1b2n�1j 00

D a2.1C4i 0/b�1u2�.We consider new elements t 0 D a2i 0

t and a0 D a1C4i 0

and see that ha0; bi D C ,a0 is of order 2nC1, .a0/2

n�1

D a.1C4i 0/2n�1

D a2n�1

a2nC1i 0

D a2n�1

D y, .a0/b D


a0z�, .t 0/2 D a4i 0

t2 D a1C4i 0

D a0, and, finally, bt 0

D ba2i 0t D bt D .a0/2b�1u2�.

Hence, writing again a and t instead of a0 and t 0, respectively, we get bt D a2b�1u2�.We check that all elements taibj 2 tC are of order 2nC2:

.taibj /2 D t2.aibj /taibj D aaia2j b�ju2j�aibj D a1C2iC2ju2j�zij� ;

and so �2.ht; C i/ D �2.C / D W , which implies that ht; C i is a non-normal meta-cyclic subgroup of index 4 in G.It is very difficult to lift up the action of g from the action onW to the action on C .

We set ag D amb, bg D a2pbq for some integers m; �; p; q, where we have usedthe fact that b 2 �n.C / D ha2; bi. We get (since n � 3):

yg D yu2 D .a2n�1

/g D .amb/2n�1

D .a2n�1

/m.b2n�1

/ D ymu2 ;

so that m 1 .mod 4/, � 1 .mod 2/. Further (since a2 2 Z.C /):

ug D u�1y D .b2n�2

/g D .a2pbq/2n�2

D ypuq;

so that p 1 .mod 4/, q �1 .mod 4/. Therefore we may set:

m D 1C4˛; � D 1C2ˇ; p D 1C4�; q D �1C4ı; ˛; ˇ; �; ı 2 Z;

and obtain:

ag D a1C4˛b1C2ˇ ; bg D a2.1C4�/b�1C4ı :

We set b0 D a4˛b1C2ˇ so that we get a simple relation ag D ab0, where o.b0/ D

2n, ha; b0i D C , .b0/2n�2

D z˛uu2ˇ D z˛u˙1 D u0, .u0/2 D u2, .b0/2n�1

D u2,

.b0/g D .a4˛b1C2ˇ /g D .ab0/4˛.a2.1C4�/b�1C4ı/1C2ˇ

D a4˛.b0/4˛a2.1C2ˇ/.1C4�/b.1C2ˇ/.�1C4ı/

D a4˛.�1C4ı/a4˛.1�4ı/a4˛.b0/4˛a2.1C2ˇ/.1C4�/b.1C2ˇ/.�1C4ı/

D .b0/�1C4ı.b0/4˛a2.2˛C.1C2ˇ/.1C4�/C2˛.1�4ı// D .b0/�1C4�a2� ;

where �; � 2 Z and � is odd,

.b0/t D .a4˛b1C2ˇ /t D a4˛.a2b�1u2�/1C2ˇ D a4˛a2.1C2ˇ/b�.1C2ˇ/u2�

D a8˛a2.1C2ˇ/.a�4˛b�.1C2ˇ//u2� D .b0/�1a2.1C2ˇC4˛/u2�

D .b0/�1a2�u2�;

where � is an odd integer. We write again b instead of b0 and also write � D i ,


� D 1C 2j , � D 1C 2k (with i; j; k 2 Z) so that we obtain the important relations:

o.b/ D 2n; b2n�1

D u2; b2n�2

D z˛u˙1;

t2 D a; g4 D z; a2n�1

D y;

y2 D z; ag D ab; bg D b�1C4ia2C4j ;

bt D b�1a2C4ku2�;

(19)

where ˛; � D 0; 1 and i; j; k 2 Z.The fact that g4 D z gives us more information about the action of g on C . Indeed,

ag2

D .ab/g D abb�1C4ia2C4j D a1C2C4j b4i D a�1C4C4j b4i

D a�1C4.1Cj /b4i ;

and so (noting that g2 inverts each element in ha4; b2i)

a D ag4

D .a�1C4.1Cj /b4i/g2

D .ag2

/�1a�4.1Cj /b�4i

D a1�4.1Cj /b�4ia�4.1Cj /b�4i D a1�8.1Cj /b�8i ;

which gives 1 C j 0 .mod 2n�2/ and i 0 .mod 2n�3/, and so we may seti D 2n�3i 0, j D �1C 2n�2j 0, which gives us a simple expression for bg :

bg D b�1b2n�1i 0

a2�4C2nj 0

D b�1u2i 0

a�2zj 0

D b�1a�2w0;

where w0 D u2i 0

zj 0

2 W0. We have obtained:

bg D b�1a�2w0; with w0 2 W0 D ha2n

D z; b2n�1

D u2i:(20)

From (19) and (20) we get the action of g2 on C :

ag2

D .ab/g D abb�1a�2w0 D a�1w0;

bg2

D .b�1a�2w0/g D ba2w0a

�2b�2z�wg0 D b�1z�w0w

g0 ;

and the fact that there are no involutions in g2C gives us some information about w0:

.g2a/2 D g4ag2

a D za�1w0a D zw0 and so w0 ¤ z;

.g2b/2 D g4bg2

b D zb�1z�w0wg0 b D zz�w0w

g0 ¤ 1:

This gives us the following information about w0 2 W0.

If � D 0; then C is abelian, w0 D 1 and g2 inverts each element in C I(21)

If � D 1; then C is minimal nonabelian, w0 2 W0 � hzi(22)

and so we may set w0 D u2z� , � D 0; 1.


Using the above results, it is easy to check that each element g2c (c D aibj 2 C )is of order 4. Indeed, if � D 0, then .g2c/2 D g4cg2

c D zc�1c D z and, if � D 1,then (noting that aibj D bjaizij and w0 D u2z� , wg

0 D u2z�C1)

.g2aibj /2 D g4.ag2

/i .bg2

/j aibj D za�iwi0b

�j zjwj0 .w

g0 /

j aibj

D z1Ci.j C�/u2i ;

and so if i is even, then .g2aibj /2 D z and if i is odd, then

.g2aibj /2 D z1Ci.j C�/u2 ¤ 1:

It remains to determine .tg/2 2 C , where .tg/C is an involution inG=C Š D8 andwe require that there are no involutions in .tg/C and note that .tg/2 2 CC .tg/. Using(19) and (20), we obtain the action of tg on C :

atg D ag D ab;

btg D .b�1a2C4ku2�/g D a2bw0.ab/2C4ku2�z�

D a2bw0a2C4kb2C4kz�u2�z� D a4.1Ck/b�1C4.1Ck/w0u

2�:

All elements in C � S (where S D �n.C / D ha2; bi) are of order 2nC1 and supposethat an element as (s 2 S ) is centralized by tg. Then .as/2

n�1

D a2n�1

s2n�1

D yx0

with x0 2 W0 is also centralized by tg. But tg centralizes W0 and so ytg D y,contrary to the relations in (��) which give ytg D yu2. Thus, CC .tg/ � ha2; bi andso .tg/2 2 ha2; bi which is an abelian group and therefore b.tg/2

D b.This gives

b D b.tg/2

D .a4.1Ck/b�1C4.1Ck/w0u2�/tg

D .ab/4.1Ck/.a4.1Ck/b�1C4.1Ck/w0u2�/�1C4.1Ck/w0u

2�

D a4.1Ck/b4.1Ck/a�4.1Ck/b1�4.1Ck/w0u2�a16.1Ck/2

b�4.1Ck/C16.1Ck/2

w0u2�

D a16.1Ck/2

b1C4.1Ck/.�1C4.1Ck//;

and so 1 C k 0 .mod 2n�2/ and we may set 1 C k D 2n�2k0 for some k0 2 Z.Hence, from the above and using (19), we get

btg D a2nk0

b�1b2nk0

w0u2� D b�1w0z

k0

u2�;

bt D b�1a2C4ku2� D b�1a�2a4.1Ck/u2� D b�1a�2zk0

u2�;

and so

atg D ab; btg D b�1w0zk0

u2�;

bt D b�1a�2zk0

u2�; .tg/2 2 ha2; bi; k0 D 0; 1:(23)


Assume at the moment that � D 0. Then (21) implies that C D ha; bi is abelian,w0 D 1, and g2 inverts each element in C . Since .tg/2 2 C , we get from (23):a D a.tg/2

D .ab/tg D ab b�1zk0

D azk0

and so k0 D 0. We obtain from (23),btg D b�1, bt D b�1a�2 and compute, using (19):

.a2b/tg D .ab/2b�1 D a2b; .a2b/2n�1

D a2n

b2n�1

D zu2;

so that CC .tg/ D W0ha2bi D hzi � ha2bi and .tg/2 D zr .a2b/s with r D 0; 1 andsome s 2 Z. For each aibj 2 C , we compute

..tg/aibj /2 D .tg/2.atg /i .btg /j aibj D .tg/2a2ibi

D zr .a2b/sa2ibi D a2.iCsCr2n�1/biCs D x:

We see that x D 1 if and only if i �s .mod 2n/ and r D 0. Hence we must haver D 1 and so .tg/2 D z.a2b/s , s 2 Z, in which case no element in .tg/C is aninvolution. Also, from (20) we get bg D b�1a�2. We have obtained the groups statedin parts .ˇ/ and .ˇ1/ of our theorem.In what follows, we always assume � D 1. Then (22) implies that C D ha; bi is

minimal nonabelian, ab D az� , and w0 D u2z� with � D 0; 1. Note that (19), (20),and (23) imply

u2 D b2n�1

; bg D b�1a�2z�u2; atg D ab;

btg D b�1zk0C�; bt D b�1a�2zk0

u2; k0; � D 0; 1:

We compute CC .tg/ � ha2; bi (noting that .ab/2i D .a/2i .b/2izi )

a2ibj D .a2ibj /tg D .ab/2ib�j z.k0C�/j

D .a/2i .b/2izib�j z.k0C�/j D a2ib2i�j ziC.k0C�/j ;

and so a2ibj 2 CC .tg/ if and only if b2.i�j /ziC.k0C�/j D 1 and this is satisfied ifand only if i � j 0 .mod 2n�1/ and i C .k0 C �/j 0 .mod 2/. We may setj D i C 2n�1˛ (˛ 2 Z) and then

a2ibj D a2ibiC2n�1˛ D a2ibiu2˛; and i.1C k0 C �/ 0 .mod 2/:

Since u2˛ 2 W0 � CC .tg/, we get CC .tg/ D fW0; .a2b/i , where i.1C k0 C �/ 0

.mod 2/g. It follows that we have exactly two possibilities for the structure of CC .tg/

(noting that .a2b/2n�1

D zu2):

If 1C k0 C � 0 .mod 2/; then CC .tg/ D hzi � ha2biI(i)

If 1C k0 C � 1 .mod 2/; then CC .tg/ D hzi � ha4b2i:(ii)


Suppose that (i) holds. In that case we get k0 1C � .mod 2/ and so

btg D b�1z; bt D b�1a�2z1C�b2n�1

; .tg/2 D zr .a2b/s; r D 0; 1; s 2 Z:

If s is even, then a.tg/2

D azr .a2b/s

D a D .ab/tg D ab b�1z D az, a contradiction.Hence s must be odd and we show that if s 1 .mod 4/, then r D 1 and if s �1

.mod 4/, then r D 0. Indeed, we require that ..tg/aibj /2 ¤ 1 for all i; j 2 Z.Suppose that for some i; j , ..tg/aibj /2 D 1. Then

1 D ..tg/aibj /2 D .tg/2.atg /i .btg /jaibj D zra2sbs.ab/ib�j zj aibj ;

which gives at once

biCs D 1 and so i C s 0 .mod 2n/; n � 3:(iii)

If s 1 .mod 4/, then (iii) gives i �1 .mod 4/ and i D �1C 4i 0, so that

.ab/i D .ab/�1C4i 0

D b�1a�1a4i 0

b4i 0

D b�1C4i 0

a�1C4i 0

D biai

and (noting that zij D zj since i is odd), we get 1 D zra2sbsbiaib�j zj bjaizij D

zr , and so r D 0. Thus, ..tg/aibj /2 ¤ 1 for all i; j 2 Z if and only if r D 1. Ifs �1 .mod 4/, then (iii) gives i 1 .mod 4/ and i D 1C 4i 0, so that

.ab/i D .ab/1C4i 0

D aba4i 0

b4i 0

D a1C4i 0

b1C4i 0

D aibi

and (noting again that zij D zj )

1 D zra2sbsaibib�j zj bjaizij D zrC1

and so r D 1. Thus, ..tg/aibj /2 ¤ 1 for all i; j 2 Z if and only if r D 0. We haveobtained the groups stated in parts .ˇ/, .ˇ2/, and .ˇ21/ of our theorem.Suppose that (ii) holds. In that case we get k0 � .mod 2/ and so

btg D b�1; bt D b�1a�2z�b2n�1

; .tg/2 D zr.a4b2/s; r D 0; 1 and s 2 Z:

We show that if s is odd, then r D 0 and if s is even, then r D 1. Indeed, we requirethat ..tg/aibj /2 ¤ 1 for all i; j 2 Z. Suppose that for some i; j , ..tg/aibj /2 D 1.Then

1 D ..tg/aibj /2 D .tg/2.atg /i .btg /j aibj D zra4sb2s.ab/ib�jaibj ;

which gives at once biC2s D 1 and so i C 2s 0 .mod 2n/, n � 3, and this impliesthat i D 2i 0 must be even and

.ab/i D .ab/2i 0

D ..ab/2/i0

D ..a2b2z//i0

D .a/2i 0

.b/2i 0

zi 0

D aibizi 0

;


which together with ai 2 Z.C / gives

1 D zra4sb2saibizi 0

b�jbjai D zra4sC2izi 0

b2sCi D zrCi 0

and rC i 0 0 .mod 2/. Since iC 2s 0 .mod 2n/, we get 2i 0 C 2s 0 .mod 2n/

and so i 0 C s 0 .mod 2n�1/. Since ..tg/aibj /2 must be ¤ 1 for all i; j 2 Z, fromthe above we get that if s is odd, then i 0 is odd and so r D 0 and if s is even, then i 0

is even and so we must have r D 1. We have obtained the groups stated in parts .ˇ/,.ˇ2/, and .ˇ22/ of our theorem which is now completely proved.

Exercise 1. The following conditions for a nonmetacyclic 2-group G are equivalent:(a) G has exactly three involutions, (b) every minimal nonmetacyclic subgroup of Ghas exactly three involutions.

Solution. It suffices to prove that (b) ) (a). Assume that G has more than threeinvolutions. Clearly, G is not of maximal class. There is E4 Š R G G. Since R hasexactly three involutions, there exists an involution x 2 G � R and we have H D

hx;Ri Š D8. Since G is not of maximal class, CG.H/ — H (Proposition 10.17). Lety 2 CG.H/ �H has the minimal possible order. Since G has no elementary abeliansubgroups of order 8, we get o.y/ > 2. It follows that exp.CG.H// > 2 so we musthave o.y/ D 4. In that case, hH;yi D H � hyi of order 16 is minimal nonmetacyclicwith exactly 7 > 3 involutions (see Appendix 16), a contradiction.

It follows from Exercise 1 and Theorem 66.1 that if a nonmetacyclic 2-group G hasexactly three involutions and has no subgroups Š Q8, then all its minimal nonmeta-cyclic subgroups have order 25. Hence, in any case, G has a sectionŠ Q8.

Exercise 2. If G has exactly three involutions, then all its minimal nonabelian sub-groups are metacyclic.

It follows from Exercise 2 that if G of that exercise has no subgroups Š Q8, itinvolves M2n , n > 3.

Exercise 3. Suppose that G is a 2-group with exactly three involutions and 2

Aut.G/ has an odd prime order p. Find all possible values of p.

Hint. Since d.G/ � 4 (see �50), we get p 2 f3; 5; 7g since o. / divides the number.24�1/.24�2/.24�22/.24�23/ D 26 32 57. We claim that p ¤ 7. Assume that thisis false. LetW D h i G be the natural semidirect product. Without loss of generalityone may assume that W is a minimal nonnilpotent group. Since G has exactly threeinvolutions, it is nonabelian. Since the minimal natural b such, that 2b 1 .mod 7/,equals 3 and this number is odd, it follows from the structure of minimal nonnilpotentgroups thatG is abelian, which is a contradiction. We do not know if p D 5 is possible.

Exercise 4. Describe Aut.A/, where A is a group of Lemma 82.6.


According to Theorem 1.17(a), if a 2-groupG is neither cyclic nor of maximal class,then the number of involutions in G is 3 .mod 4/. Therefore, the next interestingcase presents 2-groups with exactly seven involutions. To finish the classification of2-groups G with exactly 7 involutions, it remains to consider the case j�1.G/j D 8

(in this case involutions generate elementary abelian subgroup of order 8; see Theorem64.17). There exist 2-groups G with exactly seven involutions and such that d.G/ D 6

(for example, the direct product of three generalized quaternion groups), however, wedo not know if it is possible d.G/ > 6 for such groups.

�83

p-groups G with�2.G/ or��2.G/ extraspecial

In this section we classify the p-groups with the properties given in the title. Thissolves problems 157 and 1429 (see Research problems and themes, I and II, respec-tively). All results of this section are due to the second author.

Theorem 83.1. If G is a p-group with extraspecial�2.G/, then �2.G/ D G.

Proof. Suppose that the theorem is false.

Case 1. Let p D 2. Set E D �2.G/ and hzi D Z.E/ so that jEj D 22nC1,n � 1, where n is the width of E, and o.z/ D 2. Let F be a subgroup of G containingE such that jF W Ej D 2. We use induction on n. Suppose that n D 1. ThenE Š D8 or Q8. If CF .E/ 6� E, then CF .E/ � hzi contains an element of order 2 are4 since jE \ CF .E/j D jZ.E/j D 2, a contradiction. Hence CF .E/ � E and then Fis of maximal class (Proposition 10.17). But then �2.F / D F , a contradiction.Now we assume that n > 1. Since jEj � 25 and exp.E/ D 4, F has no cyclic

subgroup of index 2 and so F is not of maximal class. It follows that F possessesa normal four-subgroup R (Lemma 1.4). We have R � �2.F / D E. In particular,z 2 R and we may setR D hz; ui for some involution u 2 .E�hzi/ so that CF .R/ D

CF .u/. Since jE W CE .R/j D 2, it follows that CF .R/ covers F=E. By the structureof E, CE .u/ D hui � E0, where E0 is extraspecial of order 22.n�1/C1. Set F0 D

CF .u/ so that jF0 W .hui �E0/j D 2 and consider the factor-group F0=hui D NF0 (barconvention), where j NF0 W NE0j D 2 and NE0 is extraspecial of width n � 1.By induction, there is an element x 2 F0 � .hui � E0/ such that o. Nx/ � 4. If

o. Nx/ D 2, then x2 2 hui and so o.x/ � 4, a contradiction. Hence o. Nx/ D 4 and sox4 2 hui but x2 62 hui. We have x2 2 CE .u/. If x2 is an involution, then o.x/ D 4, acontradiction since�2.F0/ D F0 \E D hui �E0. Hence x2 is an element of order 4in CE .u/ and so, in view of Ã1.CE .u// D hzi, we get x4 D z, contrary to x4 2 hui.Thus, F does not exist so G D E. The theorem is proved for p D 2.

Case 2. Now let p > 2 and let E D �2.G/ be extraspecial. Since G > E , we haveexp.E/ D p2. By the structure of E, we may set E D E1 � E2 � � Em, m � 1,whereEm Š Mp3 and, in casem > 1, the subgroups E1; : : : ; Em�1 are nonabelian oforder p3 and exponent p. We have S D �1.E/ D �1.G/ D E1 : : : Em�1�1.Em/ sothat jE W S j D p, R D �1.Em/ D Z.S/ Š Ep2 , R is normal in G and CE .R/ D S .Set C D CG.R/ so that jG W C j D p and therefore G D CE with C \ E D S and

�83 p-groups G with �2.G/ or ��2.G/ extraspecial 397

C > S . But then C � S contains an element of order p or p2, a contradiction. Thiscompletes Case 2. The theorem is proved.

Recall that if G is a p-group, then��n.G/ D hx 2 G j o.x/ D pni if exp.G/ � pn

and ��n.G/ D f1g if exp.G/ < pn.

Theorem 83.2. Let G be a 2-group such that ��2.G/ is extraspecial. Then either

��2.G/ D G or G Š SD16.

Proof. Set E D ��2.G/, jEj D 22nC1, n � 1, Z.E/ D hzi, and assume G > E.

Suppose that CG.E/ > hzi. There are no elements of order 4 in CG.E/ � hzi and soCG.E/ is elementary abelian. Let i 2 CG.E/ � hzi so that i is an involution. If v isan element of order 4 in E, then vi is an element of order 4 in G �E, a contradiction.We have proved that CG.E/ D hzi and so Z.G/ D hzi.By Theorem 83.1, there is an involution t 2 G � E. Set F D Ehti so that jF j D

22nC2. If t centralizes an element v of order 4 in E, then tv is an element of order 4and tv 2 F �E, a contradiction. Hence CE .t/ is elementary abelian and each elementin F �E is either an involution or an element of order 8 so that exp.F / D 8.If all elements in F �E are involutions, then t inverts each element in E and so E

would be abelian, which is not the case. Let r be an element of order 8 in F �E. Thenv D r2 generates a cyclic subgroup of order 4 in E so that v2 D z and hvi is normalin E. Thus, NG.hvi/ � hE; ri D F so hvi is normal in F . Since r 2 CF .v/ � E,it follows that CF .v/ covers F=E. As above, all elements in CF .v/ � E are notinvolutions so have order 8. Since t 62 CF .v/ and hvi is normal in F , we get

vt D v�1 D vz:(1)

Act with the involution t on the elementary abelian group E=hzi of order 22n. Itis easy to see that E=hzi is a direct product of t -invariant four-subgroups. SettingE1=hzi D CE=hzi.t/, we conclude that jE1=hzij � 2n, E2 D CE .t/ < E1 is el-ementary abelian, and v 2 E1 � E2, by (1). Consider any element x 2 E1 � E2.If x is an involution, then xt D xz implies that hx; ti Š D8. But then o.xt/ D 4

and xt 2 F � E, a contradiction. Hence o.x/ D 4 and therefore x2 D z andxt D xz D x�1. It follows that t inverts each element in E1 and so E1 is abelian,�1.E1/ D E2 and all elements in the coset E1t are involutions. If x; y 2 E1 � E2,then .xy/2 D x2y2 D zz D 1 and so xy 2 E2 which implies jE1 W E2j D 2. It isknown (see �4) that maximal abelian subgroups of the extraspecial group E are of or-der 2nC1. Since jE1j � 2nC1, we get jE1j D 2nC1 and jE2j D jE1=Ã1.E1/j D 2n.Note that te .e 2 E/ is an involution if and only if .te/2 D tete D 1 or equivalentlyet D e�1 and so e 2 E1. Hence E1t is the set of all involutions in F � E whichimplies that E1t is a normal subset in F . But F1 D hE1; ti D hE1ti and so F1 (oforder 2nC2) is normal in F . On the other hand, F2 D CF .t/ D hE2; ti is of order2nC1 and jF W F2j D 2nC1. Since jE1t j D 2nC1, it follows that the set E1t of allinvolutions in F �E forms a single conjugacy class in F .


The nonabelian subgroup F1 has exactly three abelian maximal subgroups: E1

(of type .4; 2; : : : ; 2/) and elementary abelian subgroups F2 D hE2; ti and F �2 D

hE2; tvi. Since E1t D E2t [E2tv and all involutions in E1t form a single conjugacyclass in F , we get that F2 is not normal in F . Therefore, acting with F=F1 on the setfF2; F

�2 g, we get F3 D NF .F2/ D NF .F

�2 / and jF W F3j D 2 with F3 � F1.

We have jF j D 22nC2 so that jF3j D 22nC1. On the other hand, jF1j D 2nC2

and so, if n > 1, we get F3 > F1. In that case take an element y 2 F � E so thaty 2 F3 � F1. Since F3=F2 and F2 are elementary abelian, we get 1 ¤ y2 2 F2 andso o.y/ D 4 (noting that y is not an involution because all involutions in F �E lie inF1 �E). This is a contradiction.We have proved that n D 1. In that case E Š Q8. But the Sylow 2-subgroup of

Aut.Q8/ is isomorphic to D8 and CG.E/ D hzi imply that G D F , jG W Ej D 2, andso G is a group of maximal class and order 24 (Proposition 10.17). Since there areinvolutions in G �E, we get G Š SD16 (Theorem 1.2), and we are done.

Recall that, given a p-group G, Hp.G/ D hx 2 G j o.x/ > pi if exp.G/ > p andHp.G/ D f1g if exp.G/ � p.

Theorem 83.3. Let G be a p-group, p > 2, such that ��2.G/ is extraspecial. Then

��2.G/ D Hp.G/ and so, in the case G > ��

2.G/, all elements in G ��2.G/ are of

order p.

Proof. Set E D ��2.G/, Z D Z.E/ Š Cp and note that exp.E/ D p2. Assuming

G > E, we have to prove that all elements of the set G �E are of order p.Let F=E be a subgroup of order p in G=E ¤ f1g. Let x be any element in F �E.

Since xp 2 E, we have either o.x/ D p or o.x/ D p3. Suppose that o.x/ D p3.Then hxpi D P is a cyclic subgroup of order p2 in E. Since P > Z, P is normal inE and so NF .P / � hx;Ei D F , and we conclude that P is normal in F . BecausejE W CE .P /j D p, we get that CF .P / covers F=E and each element in CF .P / � E

must be of order p3 (otherwise, if u 2 CF .P / � E is of order p, then uxp 62 E hasorder p2, contrary to the hypothesis). By Theorem 83.1, �2.F / D E is not possibleand so�2.F / D F and so there is an element y of order p in F-E. Since y 62 CF .P /,we get hP; yi Š Mp3 . But then o.yxp/ D p2 and yxp 62 E, a contradiction. Wehave proved that all elements in F �E are of order p.Suppose that G=E is not of exponent p. Let S=E be a cyclic subgroup of order p2

in G=E. Then S D Ehli for some l 2 G. It follows that Ehlpi=E is a subgroup oforder p inG=E and so by the above o.lp/ D p and o.l/ D p2, l 62 E, a contradiction.We have proved that exp.G=E/ D p and so all elements in G � E are of order p.

We get E D Hp.G/, completing the proof.

�84

2-groups whose nonmetacyclic subgroupsare generated by involutions

Involutions play an important role in 2-groups. Here we classify nonmetacyclic 2-groups all of whose nonmetacyclic subgroups are generated by involutions. Moreprecisely, we prove the following

Theorem 84.1 ([BozJ4]). Let G be a nonmetacyclic 2-group all of whose nonmeta-cyclic subgroups are generated by involutions. Suppose that G is not elementaryabelian. Then G is nonabelian and each abelian subgroup of G is either metacyclicor a self-centralizing elementary abelian group of order 8. Moreover, we have thefollowing possibilities.

(a) If G has no elementary abelian subgroups of order 8, then G D D � C , whereD Š D2n , n � 3, C Š C4 and D \ C D Z.D/.

(b) If G has a normal elementary abelian subgroup of order 8, thenG is isomorphicto one of the following groups:

(b1) G Š D8 � C2;

(b2) G Š Q8 � Q8 (the central product of two quaternion groups), which isextraspecial of order 25 and type “+”;

(b3) G D M hti, where M Š C4 � C4 and t is an involution which invertseach element inM .

(c) If G has no normal elementary abelian subgroups of order 8 but G has an el-ementary abelian subgroup E of order 8, then E 6� ˆ.G/ and G is one of thegroups appearing in Theorems 51.11 to 51.15.

Proof. Let G be a nonmetacyclic 2-group all of whose nonmetacyclic subgroups aregenerated by involutions. Also, we suppose that G is not elementary abelian. Then Gis nonabelian since G is nonmetacyclic and so G is generated by its involutions.By Lemmas 65.1 and 65.2, each minimal nonabelian subgroup of G is metacyclic.

Let A be a maximal abelian subgroup of G so that CG.A/ D A. Then A is eithermetacyclic (i.e., A is of rank � 2) or A is of rank � 3 in which case A must beelementary abelian of order� 8 and we consider the second case. Let T be a subgroupof G such that A < T � G and jA W T j D 2. Let v be an element of order 4 in T � A

so that v2 2 A and therefore v induces on A an involutory automorphism. In that case


it is well known that T 0 � CA.v/, CA.v/ D Z.T /, jT 0j D jA=CA.v/j which gives thatjCA.v/j � jA=CA.v/j. Since S D hviCA.v/ is abelian but not elementary abelian,S is not generated by its involutions and so S must be metacyclic. It follows thatU D CA.v/ Š E4 and v2 2 U and so A Š E8 or A Š E16. Suppose that jAj D 16

in which case T 0 D U D Z.T / Š E4. Hence, there is an involution a 2 A � U suchthat b D Œa; v� 2 U � hv2i. Set M D ha; vi so that M 0 D hbi and therefore M isnonmetacyclic minimal nonabelian (of order 24) because �1.M/ D ha; b; v2i Š E8,a contradiction. We have proved that each abelian subgroup of G is either metacyclicor a self-centralizing elementary abelian group of order 8.

Assume that G does not possess any elementary abelian subgroup of order 8. SinceG is neither cyclic nor of maximal class,G has a normal four-subgroup U . If�1.G/D

U , then G is not generated by involutions, a contradiction. Set T D CG.U / so thatthere is an involution t 2 G � T , jG W T j D 2 and �1.T / D U . We have CG.t/ D

hti � CT .t/ and since CT .t/ contains only one involution z, where hzi D CU .t/, itfollows that CT .t/ is either cyclic or generalized quaternion. If CT .t/ is generalizedquaternion, then CG.t/ is nonmetacyclic but CG.t/ is not generated by involutions, acontradiction. Hence, CT .t/ is cyclic. If jCT .t/j D 2, then G is of maximal classand so G is metacyclic, a contradiction. We have proved that CT .t/ is cyclic of order� 4, where t 62 ˆ.G/. By Theorem 48.1, �1.G/ D D � Z, whereD Š D2n , n � 3,Z Š C4 and D \ Z D Z.D/. Since in our case �1.G/ D G, we have obtained agroup stated in part (a) of the theorem.

Assume that G possesses a normal elementary abelian subgroup E of order 8. LetF=E be a cyclic subgroup of order 4 in G=E . Let T=E be a subgroup of order 2in F=E. Then �1.F / � T but F is nonmetacyclic. This is a contradiction and sof1g ¤ G=E is elementary abelian of order � 4 since CG.E/ D E and Aut.E/ Š D8.Let H=E be any subgroup of order 2 in G=E. Since H is nonmetacyclic, there isan involution t 2 H � E. Then CE .t/ D V Š E4 so that the coset Et consistsof the set V t of four involutions and the set V tu of four elements of order 4, whereu is an element in E � V . Since Œu; t � ¤ 1, D D hu; ti is dihedral of order 8.Because jD \ V j D 2, there is an involution z 2 V �D, where V D Z.H/. HenceH D hzi �D Š C2 � D8. If H D G, then we have obtained the group stated in part(b1) of the theorem.

Assume that G=E Š E4 so that jGj D 25. By the previous paragraph (since G=Ehas exactly three subgroups of order 2), G � E consists of 12 involutions and 12elements of order 4. Hence, G has exactly six cyclic subgroups of order 4. Supposethat ��

2.G/ D hx 2 G j o.x/ D 4i is of order > 24. Then ��2.G/ D G is of order 25

and, by Lemma 89.6, G Š Q8 � Q8 and so we have obtained the group in part (b2) ofTheorem 85.1. Suppose that ��

2.G/ is of order � 24 so that j��2.G/j D 24 since G

has exactly 12 elements of order 4. SetM D ��2.G/, where all elements in G �M

must be involutions. If t 2 G � M is one of them, then t inverts each element inMwhich implies thatM is abelian. SinceM has exactly 12 elements of order 4, the onlypossibility isM Š C4 � C4 and G D M hti is the group of part (b3) of the theorem.

�84 2-groups whose nonmetacyclic subgroups are generated by involutions 401

Finally, suppose that G has no normal elementary abelian subgroups of order 8but G has an elementary abelian subgroup E of order 8. We know that E is self-centralizing in G and suppose that E � ˆ.G/. By Theorem 51.6, G possesses anormal metacyclic subgroup N such that G=N Š C4 or G=N Š D8. Let T=N bea cyclic subgroup of order 4 in G=N so that jG W T j � 2. Then ˆ.G/ � T andso E � T and therefore T is nonmetacyclic. Let T0=N be the subgroup of order 2in T=N . Then �1.T / � T0 and therefore T is not generated by its involutions, acontradiction. We have proved that E 6� ˆ.G/. Therefore G satisfies the assumptionsof the last part of �51 and so G must be a group appearing in Theorems 51.11 to51.15.

�85

2-groups with a nonabelianFrattini subgroup of order 16

According to a classical result of Burnside, if G is a finite 2-group, then the Frattinisubgroup ˆ.G/ of G cannot be a nonabelian group of order 8. Here we study “thenext possible case”, where G is a 2-group and ˆ.G/ is nonabelian of order 16 (seeResearch problems and themes II, #994). We show that in that caseˆ.G/ Š M �C2,whereM Š D8 orM Š Q8 and we shall classify all such groups G (Theorem 85.1).(As it follows from Lemma 1.4, if G is p-group and N is a G-invariant subgroup ofˆ.G/, then N is cyclic if and only if Z.N / is cyclic.)To facilitate the proof, we make the following

Remark. Let G be a p-group and let N be a nonabelian G-invariant subgroup ofˆ.G/ of order p4. We claim that either N D M � C , where M is nonabelian oforder p3 or p > 2 and N is metacyclic of exponent p2. Indeed, if d.N / D 2, then Nis metacyclic (see Theorem 44.13). Since Z.N / is noncyclic, we get exp.N / D p2.Assume, in addition, that p D 2. Then N D ha; b j a4 D b4 D 1; ab D a3i. In thatcase, L D ha2b2i is characteristic in N so normal in G, and N=L Š Q8, contrary toBurnside’s result. Thus, if d.N / D 2, then p > 2 and N is metacyclic of exponentp2. Now let d.N / D 3. Then N contains a nonabelian subgroup M of order p3 andN D MZ.N /. By Lemma 1.4, Z.N / Š Ep2 so N D M � C , and we are done.

Theorem 85.1 (Z. Bozikov). LetG be a finite 2-group such that ˆ.G/ is a nonabeliangroup of order 16. Then ˆ.G/ D M � C2, where M Š D8 or Q8. The group Gpossesses normal subgroups H and C such that G D HC , A D H \ C Š C4 � C2

or E8, CG.A/ D C , d.H/ D 2, A < ˆ.H/ D ˆ.G/, and ˆ.C/ � Z.ˆ.G// Š E4 sothat C is a group of class � 2 and exponent � 4. For the structure of H we have thefollowing possibilities:

(a) IfM Š D8, thenA Š E8 andH is isomorphic to one of the four groups of order26 given in Theorem 51.4.

(b) IfM Š Q8, then A Š C4 �C2 andH is isomorphic to one of the following twogroups of order 26:

hx; y j x8 D y8 D 1; x4 D y4 D z; x2 D a; y2 D b; Œa; b� D z;

ay D at; t2 D Œt; a� D Œt; b� D 1; tx D ty D tz; bx D btz;

.xy/2 D tz�; � D 0; 1i:

�85 2-groups with a nonabelian Frattini subgroup of order 16 403

Here ha; bi Š Q8, ˆ.H/ D ha; bi � hti Š Q8 � C2, Z.H/ D hzi Š C2, H 0 D

hab; ti Š C4 � C2, Z.ˆ.H// D hz; ti Š E4, T1 D CH .hz; ti/ D ˆ.H/hxyi isa characteristic subgroup of index 2 in H with ˆ.T1/ D T 0

1 D Z.T1/ and Y D

hab; t; xyi is the unique abelian maximal subgroup of T1, where Y is of type .4; 4/.Finally, A D ha; ti and B D hb; ti are self-centralizing abelian normal subgroups ofH and both are of type .4; 2/.Both groups (for � D 0 and � D 1) exist as transitive subgroups of the alternating

group A16 and they are not isomorphic.

Proof. Let G be a finite 2-group such that ˆ.G/ is nonabelian of order 16. Then, bythe Remark, ˆ.G/ D M � hti, whereM 2 fD8;Q8g and t is an involution.We can now use a theorem of Nekrasov (see Proposition 4.9) stating that if X is

a 2-group with ˆ.X/ Š E4, then ˆ.X/ � Z.X/. Set U D Z.ˆ.G//. Then U D

Z.M/ � hti. We get ˆ.G=U / D ˆ.G/=U Š E4 and so ˆ.G/=U � Z.G=U /. Thisimplies that each subgroup S with U � S � ˆ.G/ is normal in G. This fact will beused often in our proof.We have CG.U / D CG.t/ and jG W CG.U /j � 2. Suppose that CG.U / D G.

Then ˆ.G=hti/ D ˆ.G/=hti Š D8 or Q8, contrary to the above result of Burnside. Itfollows that jG W CG.U /j D 2. Set T D CG.U / D CG.t/ so that jG W T j D 2.We want to investigate the structure of NT D T=hti and we use the bar convention.

We have NM Š D8 or Q8, NM 0 D NU and we know that each subgroup NX with NU < NX <NM is normal in NT which implies that no element in NT induces an outer automorphismon NM . Hence NN D CT .

NM/ covers NT = NM and therefore NT D NM� NN with NM\ NN D NU .If N is the inverse image of NN , then we get T D ˆ.G/N , N is normal in T andˆ.G/ \ N D U D ht; zi. Note that T=N Š ˆ.G/=U Š E4 and Œˆ.G/;N � � hti

and so ˆ.T / � U and exp.T / D 4.Suppose Œˆ.G/;N � D f1g. Since ˆ.G/ > U , there is y 2 G � T such that

y2 D l 2 ˆ.G/ � U . We have l2 2 hzi D .ˆ.G//0 and so hl; zi is a subgroup oforder 4 which is normal in ˆ.G/. But N centralizes l (and z) and so hl; zi is normalin T . Also, y 2 G � T centralizes l (and z) and so hl; zi is normal in G. In thatcase, jG W CG.hl; zi/j � 2 and so CG.hl; zi/ contains ˆ.G/, a contradiction. We haveproved that Œˆ.G/;N � D hti and so T 0 D ˆ.T / D U � Z.T /.Since ˆ.T / D U , there is an element x 2 G � T such that x2 D a 2 ˆ.G/ � U .

If ˆ.G/ Š D8 � C2, then ˆ.G/ has a maximal subgroup X Š C4 � C2 so thatall elements in ˆ.G/ � X are involutions. In that case we may assume that x2 2

ˆ.G/ � X so that x2 D a is an involution and so A D ha;U i Š E8. However, ifˆ.G/ Š Q8 � C2, then A D ha;U i Š C4 � C2.We have ˆ.G/ > A and so there is y 2 G � T such that y2 D b 2 ˆ.G/ � A. Set

B D hb;U i and note that both A and B are normal in G.The subgroup ha; bi covers ˆ.G/=U and ha; bi \ U ¤ f1g (because ha; bi \

U D f1g would imply that ˆ.G/ is elementary abelian). Hence jha; bij � 8. Sinced.ˆ.G// D 3, we have jha; bij D 8 and ha; bi > hzi because ˆ.ˆ.G// D hzi.It follows that ha; bi \ U D hzi and so ˆ.G/ D ha; bi � hti which implies that


F D ha; bi Š D8 or Q8. Set H D hx; yi so that H is a 2-generator subgroup ofG with F � ˆ.H/ � ˆ.G/. By the above result of Burnside, F D ˆ.H/ is notpossible. We have proved that F < ˆ.H/ D ˆ.G/ Š C2 � D8 or C2 � Q8 andtherefore jH j D 26 andH is normal in G.It is easy to see that A D ha;U i is self-centralizing in H . Set T1 D T \ H D

CH .U / and so x; y 2 H � T1, where jH W T1j D 2 and jT1 W ˆ.H/j D 2. Notethat A is self-centralizing in ˆ.H/ D ˆ.G/ and so CH .A/ > A and CH .A/ � T1

would imply that CH .A/ covers T1=ˆ.H/. In that case ha; zi is normal in H and sojH W CH .ha; zi/j � 2 and CH .ha; zi/ � ˆ.H/, a contradiction. Hence A is self-centralizing in H and so H=A Š D8. This implies that H 0 6� A. On the other hand,ha; zi is not normal in H and so there is s 2 H so that as D atz� , � D 0; 1 whichshows thatH 0 � hz; ti D U .In the same way we see that B D hb;U i is self-centralizing in H . Therefore

H=B Š D8 and soH 0 6� B . On the other hand,H 0 < ˆ.H/. Indeed, ifH 0 D ˆ.H/,then jH=H 0j D 4 and so (by a well-known result of O. Taussky) the group H wouldbe of maximal class, a contradiction. Since H 0 � U and H 0 6� A and H 0 6� B , wemust haveH 0 D hab;U i.Since A is normal in G and A is self-centralizing inH withH=A Š Aut.A/ Š D8,

we get that C D CG.A/ covers G=H and so G D HC , where H and C are bothnormal in G andH \C D A Š E8 or C4 �C2. On the other hand, C � T D CG.U /

and we know that ˆ.T / D T 0 D U � Z.T / and so ˆ.C/ � U and consequently Cis of class � 2 and exponent � 4.IfM Š F Š D8, then A Š E8 and so A is a self-centralizing elementary abelian

normal subgroup of order 8 in H with A < ˆ.H/. In this case H is isomorphic toone of four groups of order 26 given in Theorem 51.4.We shall determine the structure ofH D hx; yi in the case where F Š Q8 so that

x2 D a; y2 D b; a2 D b2 D z; z2 D 1; and Œa; b� D z; where ha; bi D F:

Since x; y 2 H � T1, where T1 D CH .ht; zi/, we get tx D ty D tz. We act withhyi on the abelian group A D ha; ti of type .4; 2/. Since y2 D b 62 A and A isself-centralizing in H , y induces an automorphism of order 4 on A and so we haveay D at or ay D atz. However, if ay D atz, we replace t with t 0 D tz so thatay D at 0. Writing again t instead of t 0, we may assume from the start that ay D at .Similarly, acting with hxi on the self-centralizing normal abelian subgroup B D hb; ti

(of type .4; 2/) in H and noting that x2 D a 62 B , we see that bx D btz or bx D bt .However, if bx D bt , we replace x with x0 D x�1 and a with a0 D a�1 so that

.x0/2 D a0; .a0/y D .a�1/y D .ay/�1 D .at/�1 D a�1t D a0t

and b D .bt/x�1

D bx�1

tz which gives bx�1

D btz and so bx0

D btz. Hence,writing again x and a instead of x0 and a0, respectively, we may assume from the startthat bx D btz. We know thatH 0 D hab; ti Š C4 � C2 and we compute:

.ab/xy D .abtz/y D atbtzz D ab:

�85 2-groups with a nonabelian Frattini subgroup of order 16 405

Hence Y D hab; t; xyi is abelian since txy D t and so xy 2 T1. Since T1 is a specialgroup with T 0

1 D ˆ.T1/ D Z.T1/ D ht; zi Š E4, we get .xy/2 2 ht; zi and Y is theunique abelian maximal subgroup of T1 since Z.T1/ D ht; zi is of order 4.It remains to show that .xy/2 D tz� , � D 0; 1. Suppose that this is false. Since

.xy/2 2 ht; zi, we have in that case .xy/2 D z� , � D 0; 1. Recall that H 0 D hab; ti

and so G=hz; ti is nonabelian. Therefore Œx; y� 2 H 0 � hz; ti and so we may setŒx; y� D abz˛tˇ , ˛; ˇ D 0; 1. Note that xy 2 T1 and so xy centralizes hz; ti. Wecompute:

1 D Œx; .xy/2� D Œx; xy�Œx; xy�xy D Œx; y�Œx; y�xy D abz˛tˇ .abz˛tˇ /xy

D abz˛tˇ .at/.btz/yz˛tˇ D abz˛tˇ .at/.btzz/z˛ tˇ D .ab/2;

and this is a contradiction since .ab/2 D z. We have proved that .xy/2 D tz� ,� D 0; 1.Finally, we see that both groups H D hx; yi for � D 0 and � D 1 exist as transitive

subgroups of A16. Indeed, for � D 0 we set:

x D .1; 2; 3; 4; 5; 6; 7; 8/.9; 13; 15; 12; 11; 14; 16; 10/;

y D .1; 9; 10; 6; 5; 11; 12; 2/.3; 16; 14; 8; 7; 15; 13; 4/;

and for � D 1 we set:

x D .1; 2; 3; 4; 5; 6; 7; 8/.9; 14; 15; 10; 11; 13; 16; 12/;

y D .1; 9; 10; 6; 5; 11; 12; 2/.3; 15; 14; 4; 7; 16; 13; 8/;

and we verify that all the defining relations for H are satisfied. The first group (for� D 0) has exactly 3 involutions and the second group (for � D 1) has exactly 11involutions and so they are not isomorphic. Our theorem is proved.

�86

p-groups G with metacyclic��2.G/

The main result of this section, Theorem 86.2, is due to the second author. The threeproofs of Theorem 86.1 are given by the first author.We classify here the p-groups G with ��

2.G/ metacyclic. We begin with the casep > 2.

Theorem 86.1 (Berkovich). Let G be a nonmetacyclic p-group of exponent > p, p >2, such that H D ��

2.G/ is metacyclic. Then G is a 3-group of maximal class. IfjH j D 33, then all elements of the set G � H have order 3. If jGj > 34, then allelements of the set G � CG.�1.H// have order 3.

Proof. Since G is noncyclic, it has � p cyclic subgroups of order p2 (Theorem1.10(b)). In that case, in view of regularity of H , we have p3 � jH j � p4 and�1.H/ Š Ep2 . If jH j D p3, then c2.G/ D c2.H/ D p, If jH j D p4, then

c2.G/ D c2.H/ D jH��1.H/j

'.p2/D p4�p2

p.p�1/D p2 C p. In both cases, c2.G/ p

.mod pp�1/. It follows from Theorem 13.2(b) that G is absolutely regular or irregu-lar of maximal class.Let G is absolutely regular; then H is regular, by Hall’s regularity criterion (Theo-

rem 9.8(a)). In that case, ��2.G/ D �2.G/ so �2.G/ D H is metacyclic. Then G is

metacyclic since it has no minimal nonmetacyclic subgroups (Theorem 41.1).Now let G be of maximal class. Then G possesses an absolutely regular subgroup

M of index p with j�1.M/j D pp�1, and exp.M/ > p in view of jM j D 1p

jGj �1p ppC1 D pp (Theorems 9.5 and 9.6). As in the previous paragraph, �2.M/ D

��2.M/ .� H/ so �1.M/ Š Ep2 , It follows that p � 1 D 2 so p D 3. If jH j D 33,

thenM D H (indeed, if jM j � 34, then j�2.M/j D 34, by Theorem 9.6) so jGj D

34; in this case, all elements of the set G �H have order 3 since exp.G/ D 32. Nowwe let jGj > 34; thenM D CG.�1.H// (Theorem 9.6). Since all elements of G oforder > 32 lie in M (Theorem 13.19), it follows that all elements of the set G � M

have order 3.

We turn now to more difficult case p D 2 and prove the following classificationresult.

Theorem 86.2. Let G be a nonmetacyclic 2-group of exponent > 2 such that H D

��2.G/ is metacyclic. Then one of the following holds:

�86 p-groups G with metacyclic��2.G/ 407

(a) H Š C4 � C2 is the unique abelian subgroup of G of type .4; 2/ and G isisomorphic to one of the groups given in (b) and (c) of Theorem 52.7.

(b) H Š C4 � C4 and G is isomorphic to one of the groups given in (c) of Theorem55.1.

(c) G D ht; c j t2 D c2nC1

D 1; n � 2; tc D b; b4 D Œb2; c� D 1i; wherejGj D 2nC3, n � 2, H D ��

2.G/ D hc2; bi with .c2/b D c�2 and H is asplitting metacyclic maximal subgroup , hb2i�hci is the unique abelian maximalsubgroup (of type .2; 2nC1/), Z.G/ D hb2; c2n

i Š E4,G0 D hc2b2i Š C2n ,andht; b2; c2n

i Š E8 (so that G is nonmetacyclic).

The two families of groups in (a) are U2-groups (except the smallest members ineach of these two families) and also all groups in (c) are U2-groups. However, nogroup given in (b) is a U2-group.

Proof. Let G be a nonmetacyclic 2-group of exponent > 2 such that the subgroupH D ��

2.G/ is metacyclic. If H D �2.G/, then a result of N. Blackburn (Theorem41.1, Remark 2) implies that G is metacyclic, a contradiction. Hence �2.G/ > H

and so there exist involutions in G �H .Suppose that H is cyclic. Then H Š C4 and so c2.G/ D 1. But then Theorem

1.17(b) implies that G is dihedral so metacyclic, a contradiction. Hence H is non-cyclic.Assume that H is abelian (of rank 2). Since H D ��

2.H/, we have either H Š

C4 � C2 orH Š C4 � C4.Suppose H Š C4 � C2. In that case H is the unique abelian subgroup of type

.4; 2/ in G since each such subgroup is generated by elements of order 4 so coincideswithH .Suppose H Š C4 � C4. In that case c2.G/ D 6 and �2.G/ > H and so G is

isomorphic to a group given in the part (c) of Theorem 55.1.From now on we assume that H is nonabelian. Suppose in addition that H has

a cyclic subgroup of index 2. Since ��2.H/ D H , we get H Š Q2n , n � 3. Let

H0 Š Q8 be a quaternion subgroup ofH so that CH .H0/ D Z.H0/ D Z.H/ Š C2. IfCG.H0/ � H0, thenG is of maximal class (Proposition 10.17) and soG is metacyclic,a contradiction. Hence D D CG.H0/ 6� H0 so that D \H D Z.H0/, D > Z.H0/,and D must be elementary abelian. Let d 2 D � Z.H0/ and s 2 H0 with o.s/ D 4.Then o.ds/ D 4 and ds 62 H , a contradiction.Our subgroupH D ��

2.G/ is metacyclic nonabelian andH has no cyclic subgroupsof index 2 and so, by Theorem 82.1,H has exactly three involutions and�1.H/ Š E4.Let Z D hai be a cyclic normal subgroup of H such that H=Z is cyclic and we havejH=Zj � 4. Let K=Z be the subgroup of index 2 inH=Z. Since ��

2.H/ D H , thereis an element b of order 4 in H � K. This implies jH=Zj D 4, H D haihbi withhai\hbi D f1g and soH is splitting overZ. We set o.a/ D 2n with n � 2 sinceH isnonabelian. Since K D haihb2i contains exactly three involutions, K is either abelian


of type .2; 2n/, n � 2 or K Š M2nC1 , n � 3. In the last case, hbi Š C4 acts faithfullyon hai and so in that case n � 4.First assume K Š M2nC1 , n � 4, where hbi acts faithfully on Z D hai. We have

ab D av or ab D a�1v, where v is an element of order 4 in hai. Set v2 D z, wherez 2 Z.H/.Suppose ab D av so that H 0 D hvi and .a4/b D .av/4 D a4. Since hvi � ha4i,

we have H 0 � Z.H/. If x; y 2 H with o.x/ � 8 and o.y/ � 8, then .xy/8 D

x8y8Œy; x�28 D 1 and so �3.H/ < H because o.a/ � 24. This is a contradictionsince we must have ��

2.H/ D H but ��2.H/ � �3.H/.

Assume ab D a�1v so that .a2/b D .a�1v/2 D a�2z and .a4/b D .a�1v/4 D

a�4. Therefore b inverts ha4i and so vb D v�1. Also,

ab2

D .a�1v/b D .a�1v/�1v�1 D av�2 D az; and a D .ab�1

/�1vb�1

;

which gives ab�1

D a�1v�1 and ab�

D a�1v� , where � D ˙1. We compute:

.ba2/2 D ba2ba2 D b2.a2/ba2 D b2a�2za2 D b2z;

and so o.ba2/ D 4. This implies ��2.H/ � hb; a2i, where L D hb; a2i is a maximal

subgroup of H . We claim that the setH � L contains no elements of order 4 and thisgives us a contradiction. Indeed, each element in H � L has the form .bja2i /a D

bja2iC1 (i; j are integers). If j D 2, then

.b2a2iC1/2 D b2a2iC1b2a2iC1 D b4.ab2

/2iC1a2iC1

D .az/2iC1a2iC1 D .a4iz/a2;

which is an element of order � 8. If j D � D ˙1, then

.b�a2iC1/2 D b�a2iC1b�a2iC1 D b2�.ab�

/2iC1a2iC1

D b2.a�1v�/2iC1a2iC1 D b2.v�/2iv� D b2ziv�;

which is an element of order 4 since Œb2; v� D 1.We have proved that K D hb2; ai must be abelian of type .2; 2n/, n � 2, E4 Š

�1.H/ D hb2; zi � Z.H/, where we have set z D a2n�1

. The element b induces onhai an involutory automorphism and so we have either ab D az, n � 3 or ab D a�1z� ,� D 0; 1, n � 2 (and if � D 1, then n � 3).First assume ab D az, n � 3, where H 0 D hzi and so H is of class 2. In that

case, if x; y 2 H with o.x/ � 4 and o.y/ � 4, then .xy/4 D x4y4Œy; x�6 D 1

and so exp.�2.H// D 4. But o.a/ D 2n � 8 and so ��2.H/ � �2.H/ < H , a

contradiction.We have proved that ab D a�1z�, � D 0; 1, n � 2, and if � D 1, then n � 3.

Assume n D 2 so that H D ha; b j a4 D b4 D 1; ab D a�1i. By Theorem 55.1(b),G is isomorphic to the following (uniquely determined) group of order 25:

G D hb; t j b4 D t2 D 1; bt D ab; a4 D 1; ab D a�1; at D a�1i;(1)

where��2.G/ D ha; bi, ˆ.G/ D ha; b2i Š C4 � C2, and �2.G/ D G.


It remains to study the case n � 3, where

H D ha; b j a2n

D b4 D 1; n � 3; ab D a�1z�; � D 0; 1; z D a2n�1

i;

H 0 D ha2i Š C2n�1 , �1.H/ D Z.H/ D hb2; zi Š E4, and K D hb2; ai is theunique abelian maximal subgroup (of type .2; 2n/ ) of H .Let t be an involution inG�H and setL D H hti. Since hzi D �1.H

0/, z 2 Z.G/and let hvi be the cyclic subgroup of order 4 inH 0 so that hvi is normal inG. Note thatvb D v�1 and CH .v/ D K so that C D CG.v/ covers G=H . Set C0 D CL.v/ andwe see that jG W C j D jL W C0j D 2, L D C0hbi, G D C hbi, C \H D K. If t doesnot centralize Z.H/ D hb2; zi, then ht;Z.H/i Š D8 and tb2 is an element of order4 in L �H , a contradiction. Thus t centralizes Z.H/ and so Z.H/ � Z.L/. Also, tdoes not centralize any element of order 4 inH and so CH .t/ D hb2; zi D �1.H/.Since hvi is central in C , there are no involutions in C � K. But there are no

elements of order 4 in C � K and so �2.C / D �2.K/ D hb2i � hvi Š C2 � C4.The fact that CK.t/ D Z.H/ also implies CC .t/ D Z.H/ D �1.C /. Note thatZ.H/ � Z.L/ implies that Z.C0/ � hb2; zi and so Z.C0/ is noncyclic. By Lemma42.1, C0 is abelian of type .2; 2nC1/.We act with the involution t on the abelian group C0 and apply Proposition 51.2.

It follows that t inverts on C0=hb2; zi. We get at D a�1s, where s 2 hb2; zi. Then

.ta/2 D tata D ata D a�1sa D s and so s D 1 since ta 62 H and ta cannot be anelement of order 4. We get at D a�1 and so t inverts K. On the other hand, b D tc0

with c0 2 K and so ab D atc0 D .a�1/c0 D a�1 because C0 is abelian. We haveproved that � D 0 and so b also inverts K.We show that the involution b2z is not a square inH . Indeed, for any x 2 K, we get

.bx/2 D bxbx D b2xbx D b2x�1x D b2. On the other hand, b2 and z are squaresin H and so hb2zi is a characteristic subgroup of H and therefore b2z 2 Z.G/. Itfollows that Z.H/ D hb2; zi � Z.G/.We use again Lemma 42.1 and get that C is also abelian (of type .2; 2k/, k � nC1).

If C ¤ C0, then there is an element d 2 C0 �K such that d 2 Ã1.C /. By Proposition51.2, t inverts Ã1.C / and so d t D t�1. But then t inverts each element in C0 whichimplies that all elements in tC0 D L�C0 are involutions. This is a contradiction sinceb 2 L � C0 and o.b/ D 4.We have proved that C D C0 and so G D L. Since t inverts K, all elements in

tK are involutions. But b is not an involution and so b D tc with a suitable elementc 2 C0 �K so that o.c/ D 2nC1. Since C0 is abelian, we have Œb2; c� D 1. We haveobtained the following group of order 2nC3:

G D hc; t j c2nC1

D t2 D 1; n � 3; tc D b; b4 D Œb2; c� D 1i;(2)

where��2.G/ D hc2; bi with .c2/b D c�2.

If we set n D 2 in (2), we get a groupG of order 25 with��2.G/ D hc2; b j .c2/4 D

b4 D 1; .c2/b D c�2i and �2.G/ D G and so this group G (because Theorem


55.1(b) implies the uniqueness of such a group) must be isomorphic to the group givenin (1). We have obtained the groups given in part (c) of our theorem for all n � 2.

The second proof of Theorem 86.1. Assume thatG is neither metacyclic nor a 3-groupof maximal class. Then G, by Theorem 13.7, possesses a normal subgroup S of orderp3 and exponent p. Let R be a minimal G-invariant subgroup of S not containedin H . Then E D �1.H/R is of order p3 and exponent p. Set T D HE. Thesubgroup T is irregular (otherwise, ��

2.T / D �2.T / D T > H , which is not thecase). In that case, by Theorem 12.1(b), �1.T / D E. Since exp.T / D p2, we get��

2.T / D hT � Ei D T > H , a contradiction. Thus, S does not exist so G is a3-group of maximal class (Theorem 13.7), and the proof is complete.

The third proof of Theorem 86.1. We retain the notation of Theorem 86.1.If G is regular, then H D �2.G/ so G is metacyclic since it has no minimal

nonmetacyclic subgroups (Theorem 41.1).Next letG be irregular. SetR D �1.H/; then R Š Ep2 . Set T D CG.R/. Assume

that H — T ; then H is nonabelian of order p3. If CG.H/ — H and K=H is asubgroup of order p in HCG.H/=H , then d.K/ D 3 so K is regular and ��

2.K/ D

�2.K/ D K > H , a contradiction. Thus, H � T .Assume that there is no a G-invariant subgroup L Š Ep3 such that R < L � T .

Then T is metacyclic, by Theorems 13.7 and 10.4. By Theorem 12.1, G D T�1.G/,where �1.G/ is of order p3 and exponent p, and so p D 3. Assume that jGj > 34.Let Z � Z.�1.G//. Set C D CG.�1.G/=Z/ and let U=�1.G/ be a subgroup oforder 3 in C=�1.G/. Then U is of class � 2 so regular, and exp.U / D 9, and weconclude that ��

2.U / D U , a contradiction since U is nonmetacyclic. Thus, G is oforder 34 so it is of maximal class.Now let T contains a G-invariant subgroup L Š Ep3 . Considering LR, one may

assume that R < L. Then K D HL is of class at most 2 so regular (indeed, K=Ris elementary abelian). In that case, ��

2.K/ D �2.K/ D K > H , a final contradic-tion.

Exercise 1. Let p > 2, n > 1 and ��n.G/ is absolutely regular.

(a) If n D 2, then G is either absolutely regular or irregular of maximal class.

(b) Is it true that, if n > 2 and ��n.G/ is absolutely regular, then G is either abso-

lutely regular or of maximal class?

Exercise 2. Let G be a metacyclic 2-group of order > 24 such that ��2.G/ D G and

G 6Š Q2n for all n 2 N. Then the following holds: (a) R D �1.G/ Š E4. (b) G=Ris dihedral. (c) R � Z.G/. (d) If T=R < G=R is cyclic of index 2, then �1.T / D R,i.e., G is a U2-group (see �67).

Solution. By hypothesis, G is not of maximal class so (a) is true, by Proposition 1.19.Let x 2 G be of order 4. Then o.xR/ D 2 so G=R is generated by involutions. It


follows that G=R is dihedral, proving (b). Let K=R < G=R be of order 2. Since Gis not of maximal class, K is abelian so that CG.R/ � K. It follows that R � Z.G/sinceG is generated by such subgroupsK. Now let T=R be a cyclic subgroup of index2 in G. The subgroup T is metacyclic so T has a cyclic subgroup of index 2, and weconclude that �1.T / D R. (Obviously, T is abelian.)

Problem. Let n > 2. Classify the p-groups G such that ��n.G/ is metacyclic.

�87

2-groups with exactly onenonmetacyclic maximal subgroup

LetG be a nonmetacyclic 2-group. If all maximal subgroups ofG are metacyclic, thenG is minimal nonmetacyclic and then d.G/ D 3, jGj � 25, and there are exactly foursuch groups (see Theorem 64.1(l) or ��66, 69).

It is natural to ask what happens if all maximal subgroups except one are metacyclic.In that case the situation is essentially more complicated since there exist many infinitefamilies of such 2-groups.

We determine here the structure of all 2-groups G which have exactly one nonmeta-cyclic maximal subgroup. All such groups G will be given in terms of generators andrelations but we shall also describe many important subgroups of these groups. It iseasy to see that we must have d.G/ � 3.

If d.G/ D 3, then the problem is simpler because in this case the group G has sixmetacyclic maximal subgroups. Such groups are given in Theorem 87.8 and we seethat there are exactly five infinite families of these groups. It is interesting to note thatin all such groups the commutator subgroup G0 is elementary abelian of order � 4.

Now assume that d.G/ D 2. This is essentially more difficult. In this case we showthatG=G0 is abelian of type .2; 2m/,m � 2, andG0 ¤ f1g is abelian of rank � 2. IfGhas a normal elementary abelian subgroup of order 8, then these groups are determinedin Theorems 87.9 and 87.10. IfG has no normal elementary abelian subgroups of order8, then many properties of such groups are described in details in Theorem 87.11. Infact, this theorem is a key result for further case-to-case investigations depending onthe structure of G0 and G=ˆ.G0/. It is interesting to note that if G0 is noncyclic butG=ˆ.G0/ has no normal elementary abelian subgroups of order 8, thenG0 has a cyclicsubgroup of index 2 and m D 2 (i.e., G=G0 is abelian of type .2; 4/) and such groupsare determined in Theorems 87.14 and 87.15, where we get an exceptional group oforder 25 and two infinite classes. However, if G0 has no cyclic subgroups of index 2,then m D 2, Z.G/ is elementary abelian of order � 4 (Theorem 87.18) and all suchgroups are completely determined in Theorems 87.17, 87.19, 87.20, and 87.21 (wherewe get infinite classes of groups in each case). If G0 is cyclic or if G0 is noncyclic butG0 has a cyclic subgroup of index 2 and G=ˆ.G0/ has a normal elementary abeliansubgroup of order 8, then such groups are determined in Theorems 87.12 and 87.16.This exhausts all possibilities.

�87 2-groups with exactly one nonmetacyclic maximal subgroup 413

The most impressive result is Corollary 87.13, where it is shown that in each casewith d.G/ D 2 such a group G D AB is a product of two suitable cyclic subgroupsA and B . The converse of the last result is Theorem 87.22 which was also provedindependently by the first author. Here was proved that if G D AB is a nonmetacyclic2-group, where A and B are cyclic, then G has exactly one nonmetacyclic maximalsubgroup and so all such groups have been completely determined in our previoustheorems for d.G/ D 2.In each infinite class of 2-groups (given in terms of generators and relations) we

have checked several smallest groups with a computer (coset enumeration program)and so we have proved that they exist. Actually, we have obtained faithful permutationrepresentations for these groups.From a description of the structure of the obtained groups (and their distinct orders

in each series), we also see that such groups are pairwise nonisomorphic.Finally, it is easily checked that all 2-groups given in our theorems have exactly one

nonmetacyclic maximal subgroup.

1o. We assume in this section that G is a 2-group with exactly one nonmetacyclicmaximal subgroupM and d.G/ D 3.

Lemma 87.1. We have d.M/ D 3 and all other six maximal subgroups of G aremetacyclic.

Proof. Suppose at the moment that d.M/ D 2 so that all maximal subgroups of Gare two-generator. Obviously, M is nonabelian and so G is nonabelian. If G is ofclass 2, we may apply Theorem 70.1. It follows that either each maximal subgroup ofG is metacyclic or G has more than one nonmetacyclic maximal subgroup. This is acontradiction and soG is of class> 2. In that case we may apply Theorem 70.2 whichimplies that all maximal subgroups of G are nonmetacyclic, a contradiction. Henced.M/ > 2 and, consideringM \F , where F is a metacyclic maximal subgroup of G,we get d.M/ D 3.

Lemma 87.2. The abelian group G=G0 is of type .2; 2; 2m/, m > 1.

Proof. Since G0 � ˆ.G/, we have d.G=G0/ D 3 and we want to show that G0 <

ˆ.G/. Assume that this is false. Then NG D G=ˆ.M/ is nonabelian of order 16. LetND be a minimal nonabelian subgroup of NG; then ND Š D8 since ND \ NM Š E4. ByProposition 10.17, NG D NDZ. ND/ D ND � NC , where j NC j D 2. In that case, NG has twodistinct elementary abelian subgroups of order 8, contrary to the hypothesis.

On the other hand, abelian groups of type .2; 2; 2m/,m > 1, satisfy the assumptionsof this section. Therefore, we assume in the sequel that G0 > f1g.

Remark 1. By Lemma 87.2, G=G0 is abelian of type .2m; 2; 2/, m > 1. Therefore,G=G0 D .E=G0/ � .F=G0/, where E=G0 Š E4 and F=F 0 Š C2m (Lemma 4(b),Introduction, Volume 1). Let R � G0 be G-invariant. We claim that G=R satisfies


the hypothesis and, ifM is the maximal subgroup of G, containing E, thenM is notmetacyclic. Indeed, writeH=G0 D �1.G=G

0/ .Š E8/; then H=G0 is not metacyclic.Then E < H so, if E < H < M 2 �1, then M=G0 is not metacyclic. Clearly,M=G0 is the unique nonmetacyclic maximal subgroup of G=G0 sinceM is the uniquenonmetacyclic maximal subgroup of G. It follows that each of three maximal sub-groups, say X , of G containing F , is metacyclic but not of maximal class (otherwise,jG=G0j D 8). It follows that �1.X/ Š E4. Assume that F is noncyclic. We claimthat then�1.G/ D �1.F / .Š E4). Indeed, if i 2 G�F is an involution, then, takingX D F hii, we get �1.X/ Š E4 Š �1.F /, which is absurd.

Lemma 87.3. If the nonabelian group G has a normal subgroup E Š E8, then G Š

C2 �M2mC2 , m � 1.

Proof. By Lemma 87.2 and our assumption that G is nonabelian, we have jGj � 25.Since G has exactly one nonmetacyclic maximal subgroup, G=E must be cyclic oforder � 4. Let a 2 G �E be such that hai coversG=E. Then f1g < G0 D ŒE; hai� <

E and ˆ.G/ D G0ha2i. But d.G/ D 3 implies jG0j D 2 so jGj D 2mC3. LetU be minimal nonabelian subgroup of G not containing E (U exists since G has atmost three abelian maximal subgroups). Since U is metacyclic and is not of maximalclass, it has exactly three involutions so we get �1.U / D U \ E Š E4. It followsfrom U=�1.U / Š G=E that U Š M2mC2 . Since G0 D U 0, one may assume thatG0 < A D hai < U so A G G and ˆ.G/ D Ã1.A/ � Z.G/. Let U D U1, U2, U3 beall maximal subgroups of G containing A, and assume that all of them are nonabelian;then they are isomorphic with M2mC2 (Theorem 1.2). Let Ui D ha; bi j a2mC1

D

b2i D 1; abi D a1C2m

i, i D 1; 2; 3. We have bi 2 E, all i , and b3b�12 is an involution

centralizing a, so ha; b3b�12 i is an abelian maximal subgroup of G containing A, a

contradiction. Let, say U2, be abelian. Then CG.b2/ � hA;Ei D G. In that case,G D U1 � hb2i.

In the sequel we assume that G has no normal elementary abelian subgroups oforder 8.

Lemma 87.4. Suppose thatG is nonabelian andG does not have a normal elementaryabelian subgroup of order 8. Then the following two assumptions are equivalent:

(a) ˆ.G/ is cyclic.

(b) G has two distinct normal four-subgroups.

If G satisfies (a) or (b), then G D D � Z with D Š D8, Z Š C2n , n � 3, andD \Z D Z.D/.

Proof. Suppose that ˆ.G/ is cyclic. Since ˆ.G/ D Ã1.G/, it follows that G has acyclic subgroup A D hai of index 4, and ˆ.G/ D ha2i. Let A D U \ V , where Uand V are distinct maximal subgroups of G. It follows from Lemma 87.2 that G hasno subgroups of maximal class and index 2. Therefore, U , V have G-invariant four-subgroups R1, R2, respectively (Lemma 1.4). Clearly, R1 ¤ R2 so (b) is proved and


D D R1R2 Š D8. IfH=D < G=D is maximal, thenH is not metacyclic (otherwise,H is of maximal class) so G=D must be cyclic. Next,ˆ.G/ � CG.R1/\ CG.R2/ D

CG.D/. Since CG.D/ \ D D Z.D/, we get, by the product formula, G D D � C ,where C D CG.D/. Note that C G G. Assuming that C has a G-invariant four-subgroup R, we obtain a G-invariant subgroup R1R Š E8, a contradiction. Thus, Cis either cyclic or of maximal class (Lemma 1.4). The second alternative is impossiblesince d.G/ D 3.Suppose that G has two distinct normal four-subgroups. By Theorem 50.2, G D

D�Z withD Š D8,D\Z D Z.D/, andZ is cyclic (see the previous paragraph).

In the sequel we assume that ˆ.G/ is noncyclic which is equivalent with the as-sumption that G has a unique normal four-subgroup W (Lemma 87.4).

Lemma 87.5. Let G be nonabelian without a normal E8 and having a unique normalfour-subgroup W . If X is any metacyclic maximal subgroup of G, then �1.X/ D W

and X is not of maximal class.

Proof. Since ˆ.G/ is metacyclic but noncyclic, it follows that the subgroup W D

�1.Z.ˆ.G/// Š E4 (Lemma 64.1(v)). Let X be a metacyclic maximal subgroup ofG. Let i 2 X � W be an involution. Since i cannot centralize W (because X ismetacyclic), it follows hW; ii Š D8. By Proposition 10.19, X is of maximal class,contrary to Lemma 87.2.

Lemma 87.6. Let the group G be nonabelian without normal elementary abelian sub-groups of order 8 and having a unique normal four-subgroup. If G0 is cyclic, thenG0 Š C2 and G D Q �Z, whereQ Š Q8 and Z Š C2m , m � 2.

Proof. Let G0 Š C2r , r � 1. By Lemma 87.2, G D EF with normal subgroups EandF , whereE\F D G0,E=G0 Š E4, andF=G0 Š C2m ,m � 2. Let a 2 F be suchthat hai covers F=G0. Then G D Ehai, ˆ.G/ D G0ha2i and W D �1.Z.ˆ.G/// isa unique normal four-subgroup of G. Since W 6� E, E does not have a G-invariantfour-subgroup. Because E is noncyclic, E is of maximal class with jEj D 2rC2 andE 0 D G0 D ˆ.E/ (Lemma 1.4). We note thatM D Eˆ.G/ D Eha2i is the uniquenonmetacyclic maximal subgroup of G since Eha2m�1

i=G0 Š E8 (see also Remark1). Hence each maximal subgroup of G containing F is metacyclic. Suppose thatthere is an involution i 2 E �G0. Then X D F hii is a metacyclic maximal subgroupof G with �1.X/ > W , contrary to Lemma 87.5. Since there are no involutions inE � G0, we conclude that E Š Q2rC2 , r � 1.Suppose r > 1. Let y be an element of order 4 in E � G0 so that y2 2 �1.G

0/

and Y D F hyi is a metacyclic maximal subgroup of G. Since jG0j � 4, there is anelement v of order 4 in G0 so that hy; vi Š Q8 is a nonabelian subgroup of order 8contained in Y . By Proposition 10.19, Y is of maximal class, contrary to Lemma 87.5.We have proved that r D 1 and soG0 Š C2 and E Š Q8. Since ˆ.G/ is noncyclic,

hai splits over G0, and so we get F D G0 � hai with o.a/ D 2m, m � 2. We


have CG.E/ \ E D G0 so G=CG.E/ is an abelian subgroup of D8. It follows thatG=CG.E/ Š E4 Š Inn.E/ so, by the product formula, G D E � CG.E/. But then.G0 �/ ŒE; hai� Š C4, a contradiction. Hence a induces an inner automorphism on Ewhich implies with E \ C D G0 and C=G0 Š C2m . Since ˆ.G/ D ˆ.C/ and ˆ.G/is noncyclic, C splits over G0.

From now on we assume that G0 is noncyclic. Then, by Lemma 87.3, G has nonormal elementary abelian subgroups of order 8 and G has a unique normal four-subgroup W .

Lemma 87.7. Suppose thatG0 is a four-group. ThenG D Qhai, whereQ D hx; yi Š

Q8, o.a/ D 2n, n � 3, Q \ hai D f1g, a2 centralizes Q, Œa; x� D 1, and Œa; y� D

a2n�1

D z. Here G0 D hu; zi Š E4, where hui D Z.Q/, ˆ.G/ D Z.G/ D ha2i �

hui Š C2n�1 � C2, andM D Q � ha2i is a unique nonmetacyclic maximal subgroupof G.

Proof. Using Lemma 87.2, we haveG D EF with normal subgroups E and F , whereE \ F D G0, E=G0 Š E4, and F=G0 Š C2m , m � 2. Let M be the maximalsubgroup of G containing E so that d.M/ D 3 (Remark 1). Since F 6� M , eachmaximal subgroup of G containing F is metacyclic but not of maximal class (Lemma87.5). In particular,�1.F / D G0. Since F=G0 is cyclic, F has a cyclic subgroup hai

of order 2mC1 (and index 2). Set z D a2m

.Since F 0 � hai \ G0 D hzi, F is either abelian of type .2mC1; 2/ or F Š M2mC2 .

In any case, ˆ.G/ D G0ha2i is abelian of type .2m; 2/, m � 2.Since F is noncyclic, we get �1.G/ D W (Remark 1) so G has only three involu-

tions.Let x 2 E �G0 such that x does not centralizeG0. Then x2 2 G0 and so hG0; xi Š

D8. But in that case there are involutions in hG0; xi � G0, a contradiction. We haveproved that G0 � Z.E/.Set v D a2m�1

so that o.v/ D 4 and v2 D z. Since v 2 ˆ.G/, v centralizes G0.If X is any maximal subgroup of G containing F , then X is metacyclic and thereforeX 0 is cyclic of order at most 2 (since X 0 � G0 Š E4). In particular, X is of class � 2.For any x 2 E � G0, F hxi < G so Œx; a2� D Œx; a�2 D 1 which gives ŒE; a2� D f1g.If for some y 2 E �G0, y2 D z, then .yv/2 D y2v2Œv; y� D zz D 1 and so yv is aninvolution in G � E, a contradiction. We have proved that z is not a square in E. Inparticular, E is nonabelian (since E has exactly three involutions and exp.E/ D 4).Since z 2 ˆ.ˆ.G// D ha4i, we have z 2 Z.G/. Take an x 2 E � G0 so that

x2 2 G0 � hzi; then cl.F hxi/ � 2. This gives Œx2; a� D Œx; a�2 D 1 and CG.x2/ �

hE; ai D G. We have proved that G0 � Z.G/ so cl.G/ D 2 and F is abelian of type.2mC1; 2/.We have ˆ.G/ D G0ha2i � Z.G/. If Z.G/ > ˆ.G/, then G=Z.G/ Š E4, and so

jG0j � 2 (Lemma 64.1(u)), a contradiction. Thus, Z.G/ D ˆ.G/. If E is minimalnonabelian, then it follows from�1.E/ D G0 Š E4 thatE is metacyclic of exponent 4


(Lemma 65.1) and so there are elements x and y of order 4 in E � G0 so that E D

hx; y j x4 D y4 D 1; xy D x�1i, where E 0 D hx2i, x2 ¤ y2, x2 2 G0 � hzi,y2 2 G0 � hzi, and x2y2 D z because z is not a square in E.If E is not minimal nonabelian, then the fact that E has only three involutions and

z is not a square in E implies E D Q � hzi withQ Š Q8.

Suppose that E is minimal nonabelian given above. We note that v D a2m�1

cen-tralizes E and v2 D z. Replace y with y0 D vy so that .y0/2 D .vy/2 D v2y2 D

zy2 D x2 and xy0

D xvy D xy D x�1, and therefore hx; y0i D Q Š Q8

and E� D Q � hzi is another complement of F modulo G0. Indeed, note thatE� > G0 D �1.G/, E�=G0 Š E4, E� G G and E� \ F D G0. Hence, replac-ing E with E� (if necessary), we may assume from the start that E is not minimalnonabelian and so E D Q � hzi, Q Š Q8, and setting hui D Q0 D Z.Q/, we haveG0 D hui � hzi Š E4,M D Q � ha2i Š Q8 � C2m , and ˆ.M/ D hui � ha4i, whereha4i � hzi.Write Q D hx; yi j x2 D y2 D Œx; y� D ui. We have G D Qhai with Q Š Q8,

Q \ hai D f1g, o.a/ D 2mC1, m � 2, and a2 centralizesQ.Let l 2 G � M so that l D aiq, where i is an odd integer and q 2 Q. Then

(noting that G is of class 2), we get l2 D a2iq2Œq; ai �, where q2Œq; ai � 2 G0. Henceo.l2/ D 2m and l4 D .a4/i and so hl4i � hzi. Hence each element l 2 G �M is oforder 2mC1 and hli � hzi.Since G0 is noncyclic, Q is not normal in G. Since jE W Qj D 2 and E G G, we

have Q \ Qa D hxi Š C4 and .Q \ Qa/a D Qa \ Qa2

D Qa \ Q (since a2

centralizes Q) so that hxia D hxi. If xa D x�1, then we replace a with a0 D ay,where y 2 Q � hxi. We get xa0

D xay D .x�1/y D x and so a0 centralizes x,o.a0/ D 2mC1, and ha0i � hzi. Hence we may assume from the start that xa D x andthe maximal subgroup A D hxi�hai is abelian of type .4; 2mC1/. By Lemma 64.1(u),A is a unique abelian maximal subgroup of G. If y 2 Q � hxi, then Œy; a� 2 G0 � hui

(otherwiseQ would be normal in G). Suppose that Œy; a� D uz. Then replace a witha� D ax (noting that a� centralizes x, o.a�/ D 2mC1, and ha�i � hzi), we getŒy; a�� D Œy; ax� D Œy; a�Œy; x� D uz u D z. Thus, we may assume from the startthat Œy; a� D z.

In the rest of this section we assume that G0 is noncyclic and G0 6Š E4. SinceG0 < ˆ.G/, G0 is metacyclic and so G0=ˆ.G0/ Š E4 and ˆ.G0/ > f1g. Let Rbe a G-invariant subgroup of index 2 in ˆ.G0/. We want to study the structure ofG=R. To this end, in view of Remark 1, we may assume that R D f1g. In that caseˆ.G0/ Š C2 and G0 is abelian of type .4; 2/. Here W D �1.G/ (see Remark 1) isa unique normal four-subgroup of G and hzi D Ã1.G

0/ � Z.G/. By Lemma 87.2,G D EF with normal subgroups E and F , where E \ F D G0, E=G0 Š E4, andF=G0 Š C2m , m � 2. Let a 2 F be such that hai covers F=G0 and ˆ.G/ D G0ha2i.Also,M D Eha2i is the unique nonmetacyclic maximal subgroup of G (Remark 1),and so any proper subgroup of G which is not contained inM is metacyclic.


Suppose thatW � Z.G/. In that case, take an involution s 2 W � hzi and considerthe group G=hsi. We have .G=hsi/0 Š C4, which contradicts our previous results(which show that a cyclic commutator group is of order at most 2). HenceW 6� Z.G/so that CG.W / is a maximal subgroup of G and �1.Z.G// D hzi which implies thatZ.G/ is cyclic.Let v be an element of order 4 in G0 and let u 2 W � hzi. Then v2 D z and the

set fhvi; hvuig is the set of cyclic subgroups of order 4 in G0. Suppose that hvi is notnormal in G (and then also hvui is not normal in G). Let fX1; X2; X3g be the setof maximal subgroups of G containing F . Since Xi is metacyclic, X 0

i is cyclic foreach i D 1; 2; 3. By our assumption (and noting that W 6� Z.G/), we get X 0

i � hzi.However, by Lemma 64.1(u), this gives a contradiction.We have proved that hvi and hvui are normal subgroups in G. This implies that

ˆ.G/ � CG.v/ \ CG.vu/ and so G0 � Z.ˆ.G// because G0 D hv; vui. Butˆ.G/=G0 is cyclic and so ˆ.G/ is abelian. In particular, a2 centralizes G0.We want to determine the subgroup ha2m

i � G0. If a2m

D 1, then a2m�1

is aninvolution in F �G0, contrary to our result that �1.G/ D W � G0. Hence a2m

¤ 1.If a2m

D z, then a2m�1

2 ˆ.G/ �G0, o.a2m�1

/ D 4, and a2m�1

v is an involution inˆ.G/ � G0, a contradiction. Suppose that a2m

D u 2 W � hzi and so F D haihvi,hai \ hvi D f1g and a normalizes hvi (since hvi is normal in G). We get va D vz� ,� D 0; 1 and so F 0 2 hzi which implies that Ã2.F / D ha4i � hui. It followsthat hui is a characteristic subgroup in F and so u 2 Z.G/, a contradiction. Hence,replacing hvi with hvui and v with v�1 (if necessary), we may assume that a2m

D v.Thus, o.a/ D 2mC2 and so hai is a cyclic subgroup of index 2 in F . Since F is notof maximal class (W Š E4 is normal in F and jF j � 25), F is either abelian orF Š M2mC3 . In any case, F 0 � hzi and a2 2 Z.F /.It is easy to see that v 2 Z.G/. If q is an element in E, then Œq; a� 2 G0 and so

Œq; a2� D Œq; a�Œq; a�a D Œq; a�Œq; a�z� D Œq; a�2z� D z� ; �; � D 0; 1;

since Œq; a�2 2 Ã1.G0/ D hzi. This gives Œq; a4� D Œq; a2�Œq; a2�a

2

D z�.z�/a2

D

.z�/2 D 1. But a2m

D v, m � 2, and so ha4i � hvi which implies that v centralizesE. It follows CG.v/ � hE; ai D G and we are done.Now we use Lemma 87.7 for the group G=hzi since .G=hzi/0 D G0=hzi Š E4. It

follows that G=hzi possesses a quaternion subgroup QQ=hzi Š Q8. Suppose that v 2QQ. Then ˆ. QQ/ D hvi and QQ possesses a cyclic subgroup of index 2. But such groupscannot have a proper homomorphic image QQ=hzi isomorphic to Q8. Hence v 62 QQ andso QQ \ hvi D hzi. If j QQ0j D 4, then a result of O. Taussky (Lemma 64.1(s)) impliesthat QQ is of maximal class. This is again a contradiction since QQ=hzi Š Q8. HenceQQ0 D hui is of order 2 and u ¤ z since QQ=hzi is nonabelian. We get hu; vi D G0 andtherefore E� D QQ � hvi is normal in G. But .E�/0 D QQ0 D hui is a characteristicsubgroup of E� and so u 2 Z.G/. This givesW D hu; zi � Z.G/ and this is our finalcontradiction. We have proved that such a group G does not exist. We conclude withthe following result which sums up all results of this section.


Theorem 87.8. Let G be a 2-group which possesses exactly one nonmetacyclic max-imal subgroup M . Then d.G/ � 3 and we assume here d.G/ D 3. In that cased.M/ D 3, G=G0 is abelian of type .2; 2; 2m/, m � 2, G0 is elementary abelian oforder � 4, and we have exactly the following five possibilities:

(a) G is abelian of type .2; 2; 2m/, m � 2.

(b) G Š C2 � M2nC1 , n � 3, where M2nC1 D ha; v j a2n

D v2 D 1; Œv; a� D

a2n�1

i.

(c) G D Q �Z, whereQ Š Q8, Z Š C2n , n � 3, andQ \Z D Z.Q/.

(d) G D Q �Z, whereQ Š Q8, Z Š C2n , n � 2.

(e) G D QZ, where Q D hx; yi Š Q8, Z D hai Š C2n , n � 3, Q \ Z D f1g,a2 centralizesQ, Œa; x� D 1, and Œa; y� D a2n�1

D z. Setting Z.Q/ D hui, wehave here G0 D hu; zi Š E4, ˆ.G/ D Z.G/ D ha2i � hui Š C2n�1 � C2, andM D Q � ha2i.

Conversely, it is easily checked that all groups G given in (a) to (e) have exactly onenonmetacyclic maximal subgroup and d.G/ D 3.

2o. We assume in this section that G is a 2-group with exactly one nonmetacyclicmaximal subgroupM and d.G/ D 2. By Lemma 64.1(n) follows at once that d.M/ D

3, G is nonmetacyclic and so G0 ¤ f1g.First we treat the easy case jG0j D 2. By Lemma 65.2, G is minimal nonabelian.

By Lemma 65.1, we have

G D ha; b j a2m

D b2n

D c2 D 1; Œa; b� D c; Œa; c� D Œb; c� D 1;

m � n � 1; m � 2i;

where jGj D 2mCnC1,�1.G/ D ha2m�1

; b2n�1

; ci Š E8, andG=�1.G/ Š C2m�1 �

C2n�1 . Since there is only one maximal subgroup of G containing �1.G/, G=�1.G/

must be cyclic and this implies n D 1 so that G=G0 is abelian of type .2m; 2/, m � 2.We have jG W haij D 4, .ab/2 D a2c, .ab/4 D a4, and so jhabi W .habi \ hai/j D 4

which gives (by the product formula) G D haihabi. Also, hci D G0 is a maximalcyclic subgroup of G. We have proved:

Theorem 87.9. Let G be a 2-group with exactly one nonmetacyclic maximal subgroupand d.G/ D 2. If jG0j D 2, then G D ha; b j a2m

D b2 D c2 D 1; m � 2; Œa; b� D

c; Œa; c� D Œb; c� D 1i, which is a nonmetacyclic minimal nonabelian group withG=G0 being abelian of type .2m; 2/, m � 2, G0 D hci is a maximal cyclic subgroupof G, G D haihabi, and �1.G/ Š E8 so that G has a normal elementary abeliansubgroup of order 8.

Now assume that G has a normal elementary abelian subgroup E of order 8 butjG0j > 2. Then G=E ¤ f1g must be cyclic and since G0 < E, we have G0 Š E4.


Let a 2 G � E be such that hai covers G=E. Since G0 D ŒE; hai�, a induces on Ean automorphism of order 4 which implies jG=Ej � 4. We have ˆ.G/ D G0ha2i andso E \ hai � G0 (noting that jG W ˆ.G/j D 4). The maximal subgroup M D Eha2i

is nonmetacyclic and so the maximal subgroup X D G0hai is metacyclic (of order� 24) with a normal four-subgroup G0. This implies that X is not of maximal class.If i is an involution in X � G0, then i cannot centralize G0 (since X is metacyclic).But in that case G0hii Š D8 and so, by Proposition 10.19, X is of maximal class, acontradiction. Hence �1.X/ D G0 and so G0 \ hai D hzi Š C2. Let v 2 E �G0 sothat Œv; a� D u 2 G0 �hzi and Œu; a� D z. This gives va D vu, ua D uz, hzi � Z.G/,o.a/ D 2n, n � 3, and a2n�1

D z. The structure of G is uniquely determined. Wecompute: .av/2 D avav D a2vav D a2.vu/v D a2u and .av/4 D .a2u/2 D a4.Thus, havi \ hai D ha4i and since jG W haij D 4 and jhavi W .havi \ hai/j D 4, weget havihai D G. If n � 4, then CG.E/ > E and therefore �1.G/ D E. If n D 3,then CG.E/ D E and�1.G/ D Eha2i D hui � hv; a2i Š C2 �D8. We have proved:

Theorem 87.10. Let G be a 2-group with exactly one nonmetacyclic maximal sub-group and d.G/ D 2. Suppose that G has a normal elementary abelian subgroup Eof order 8 and jG0j > 2. Then G0 Š E4 and we have G D EZ, Z D hai is oforder 2n, n � 3, E \ Z D hzi Š C2, z D a2n�1

, and setting E D hu; v; zi, wehave ua D uz, va D vu. We have G0 D hu; zi Š E4, Z.G/ D ha4i Š C2n�2 ,ˆ.G/ D hui � ha2i Š C2 � C2n�1 , and G D havihai. If n > 3, then �1.G/ D E

and if n D 3, then �1.G/ D Eha2i Š C2 � D8.

In the rest of this section we assume that G has no normal elementary abelian sub-groups of order 8. We prove the following key result which will be used (with theintroduced notation and with all details) in the rest of this section.

Theorem 87.11. Let G be a 2-group with exactly one nonmetacyclic maximal sub-group and d.G/ D 2. Assume in addition that G has no normal elementary abeliansubgroups of order 8. Then the following holds:

(a) jG0j > 2 and jGj � 25.

(b) G has exactly one normal four-subgroup W D �1.Z.ˆ.G///.

(c) For each metacyclic maximal subgroup X of G, �1.X/ D W .

(d) Let R be a G-invariant subgroup of index 2 in G0. Then R is unique and G=R isisomorphic to a group of Theorem 87.9. In particular, G=G0 is abelian of type .2m; 2/,m > 1,�1.G=R/ Š E8, and if y is any element in G such that y2 2 G0, then y2 2 R.Also, each proper characteristic subgroup of G0 is contained in R.

(e) G0 is abelian of rank � 2.

(f) There are normal subgroups E and F of G such that G D EF , E \ F D G0,F=G0 Š C2m , m � 2, E=G0 Š C2, and there is an element x 2 E � G0 of order� 4 and we fix such an element x. Let a 2 F � G0 be such that hai covers F=G0.


Then ˆ.G/ D G0ha2i, �1.G=R/ D .Eha2m�1

i/=R Š E8, M D Eha2i is theunique nonmetacyclic maximal subgroup of G, F D G0hai and F1 D G0haxi are twodistinct metacyclic maximal subgroups of G, and F 0F 0

1 D R. We have G D ha; xi

and v D Œa; x� 2 G0 �R.

(g) Assuming in addition that G0 is noncyclic, we have the following properties:

(g1) All elements in G0 � R are of order 2e D exp.G0/. In particular, o.v/ D 2e.(g2) If R is cyclic, then jRj D 2 and G0 Š E4.(g3) We have G0=hvi Š R=hv2i is cyclic of order � 2e and if y is any element in

R��.G0/, then hyi coversR=hv2i and hv2i has a cyclic complement of order� 2e in R.

(g4) If exp.R/ D exp.G0/ D 2e , then G0 Š C2e �C2e is homocyclic of rank 2 andif exp.R/ < exp.G0/ D 2e, then exp.R/ D 2e�1.

(g5) We have a2m

2 R �ˆ.G0/ and .ax/2m

2 R � ˆ.G0/.(g6) If G=ˆ.G0/ has a normal elementary abelian subgroup of order 8, then

ˆ.G0/ ¤ f1g and we may assume that E=ˆ.G0/ Š E8 so that our fixedelement x 2 E � G0 with o.x/ � 4 has in this case the additional propertyx2 2 ˆ.G0/.

(g7) We have x2 2 W � G0 and if G0 6Š E4, then W � R.(g8) We have vx D v�1z� , where � D 0; 1 and � D 1 if and only if x2 2 W �Z.G/

in which caseW 6� Z.G/ and hzi D E \ Z.G/.(g9) We have F D haihvi, F1 D haxihvi, and ˆ.G/ D ha2ihvi.(g10) Setting b D Œv; a� (which is equivalent with va D vb) and b1 D Œv; ax�,

we have F 0 D hbi, F 01 D hb1i with b; b1 2 R � ˆ.G0/, hbihb1i D R,

o.b/ D exp.R/, and b1 D v�2z�b�1.(g11) We have bx D b�1, bx

1 D b�11 so that x inverts each element in R. Also,

b�11 ba

1 D .bba/�1 2 hbi \ hb1i.(g12) We have .ˆ.G//0 D hbbai and ˆ.G/ is powerful.(g13) We have G D haxihai and so G is a product of two cyclic subgroups.

Proof. SinceG has no normal subgroups isomorphic to E8, we get jG0j > 2 (Theorem87.10) and jGj � 25 because jG=G0j � 23 (O. Taussky). If G has two distinct normalfour-subgroups, then Theorem 50.2 implies that d.G/ > 2, a contradiction. Hence, byLemma 1.4, G has exactly one normal four-subgroup W .Let X be a metacyclic maximal subgroup of G so that jX W ˆ.G/j D 2. If ˆ.G/

is cyclic, then G has a cyclic subgroup of index 2, a contradiction. Hence ˆ.G/ isnoncyclic and a result of Burnside (Lemma 64.1(v)) implies that Z.ˆ.G// is noncyclic.This gives W D �1.Z.ˆ.G/// noting that ˆ.G/ is metacyclic. Since jX j � 24 andX has a normal four-subgroup, X is not of maximal class. Let i be an involution inX � W . Since X is metacyclic, i cannot centralize W . It follows hW; ii Š D8 andthen Proposition 10.19 implies that X is of maximal class, a contradiction. Hence wehave �1.X/ D W for each metacyclic maximal subgroup X of G.


Let R be a G-invariant subgroup of index 2 in G0. By Lemma 44.1, we haveR D ˆ.G0/K3.G/ and so such a subgroup R is unique. Since G is nonmetacyclic,NG D G=R is also nonmetacyclic (Proposition 44.2). If X is a metacyclic maximalsubgroup of G, then NX (bar convention) is metacyclic. If M is the unique nonmeta-cyclic maximal subgroup of G, then NM is also nonmetacyclic (otherwise, Lemma64.1(n) would imply that NG is metacyclic). Since j NG0j D 2, NG must be isomorphicto a group of Theorem 87.9. In particular, G=G0 is abelian of type .2m; 2/, m > 1,�1.G=R/ Š E8, and if y is any element in G such that y2 2 G0, then y2 2 R.The uniqueness of R also implies that each proper characteristic subgroup of G0 iscontained in R.Let X1 ¤ X2 be two metacyclic maximal subgroups of G so that X 0

1 and X02 are

cyclic normal subgroups ofG. SinceG=R is minimal nonabelian, we haveX 01X

02 � R.

By a result of A. Mann (Lemma 64.1(u)), we get R D X 01X

02. On the other hand,

G=CG.X01/ and G=CG.X

02/ are abelian groups and so G

0 centralizesX 01X

02 D R. But

jG0 W Rj D 2 and therefore G0 is abelian (of rank � 2).Now we use the structure of G=R. There are normal subgroups E and F of G such

that G D EF , E \F D G0, F=G0 Š C2m , m � 2, and E=G0 Š C2. Let a 2 F �G0

be such that hai covers F=G0. Then ˆ.G/ D G0ha2i and �1.G=R/ D S=R Š E8,where S D Eha2m�1

i (because E=R Š E4 and a2m

2 R). It follows that M D

Eha2i is the unique nonmetacyclic maximal subgroup of G (noting that already S isnonmetacyclic). Let x be any element in E � G0 so that G D ha; xi, F D G0hai

and F1 D G0haxi are two metacyclic maximal subgroups of G, where we use thefacts that haxi also covers G=E Š C2m and .ax/2 2 ˆ.G/. Set S0 D G0ha2m�1

i

and E1 D G0ha2m�1

xi so that S0 is a metacyclic maximal subgroup of S , E1=G0

is another complement of F=G0 in G=G0, and S � S0 D .E � G0/ [ .E1 � G0/.By Lemma 64.1(l), �2.S/ 6� S0 and so there are elements of order � 4 in S � S0.Interchanging E and E1 (if necessary), we may assume from the start that there is anelement x 2 E � G0 with o.x/ � 4 and we choose and fix such an element x. We setv D Œa; x� so that v 2 G0 � R. Indeed, we have G D ha; xi and so if v 2 R, thenG=R would be abelian.In what follows we assume thatG0 is noncyclic so thatˆ.G0/ < R andG0=ˆ.G0/ Š

E4. Let 2e D exp.G0/, e � 1, be the exponent ofG0. If there is an element inG0�R oforder < 2e (in which case e > 1), then �e�1.G

0/ is a proper characteristic subgroupof G0 which is not contained in R, a contradiction. Hence all elements in G0 � R areof order 2e. In particular, the element v 2 G0 � R of the previous paragraph is oforder 2e . By Lemma 4 (Introduction, Volume 1), hvi has a cyclic complement hsi oforder � 2e (noting that G0 is of rank 2). Hence G0=hvi Š R=hv2i is cyclic of order� 2e . Since v2 2 ˆ.G0/, we have ˆ.G0/ D hs2i � hv2i and if y is any element inR �ˆ.G0/, then hyi covers R=hv2i.Suppose that R is cyclic of order > 2. Since G0 is noncyclic, there is an involution

in G0 � R, contrary to the fact that all elements in G0 �R are of order 2e D exp.G0/.Hence, if R is cyclic, then jRj D 2 and G0 Š E4.


Suppose that exp.R/ D exp.G0/ D 2e , e > 1. Let y be an element of order 2e inR. Suppose also that hyi \ hv2i ¤ f1g. Then we have jhyi W .hyi \ hvi/j D jhvi W

.hyi \ hvi/j D 2e0

, e0 < e, and so there is an element y0 of order 2e in hyi such that

.y0/2e0

D v�2e0

. But then .y0v/2e0

D 1 and y0v 2 G0 �R, a contradiction. Thus, hyi

splits over hv2i, R D hyi � hv2i, and so G0 Š C2e � C2e is homocyclic of rank 2.But if G0 is not homocyclic, then exp.R/ D 2e�1 and so (by Lemma 4 (Introduction,Volume 1)) hv2i has a cyclic complement in R. It follows that in any case hv2i has acyclic complement in R.Since a2m

2 G0, we know that a2m

2 R. Suppose that a2m

2 ˆ.G0/. We lookat F=ˆ.G0/ D NF so that NG0 D G0=ˆ.G0/ Š E4 is a normal four-subgroup of the

metacyclic group NF (of order � 24) and so NF is not of maximal class. But a2m�1 is an

involution in NF � NG0 and a2m�1 cannot centralize NG0. It follows that h NG0; a2m�1i Š D8

and this is a contradiction (by Proposition 10.19). We have proved that a2m

2 R �

ˆ.G0/. With the same argument (working in NF1 D F1=ˆ.G0/), we get .ax/2

m

2

R �ˆ.G0/.Assume for a moment that G=ˆ.G0/ has a normal elementary abelian subgroup

S�=ˆ.G0/ of order 8 so that S� < S and jS W S�j D 2 (where S=R D �1.G=R/ Š

E8). Since S� is nonmetacyclic, there is only one maximal subgroup of G containingS� and so G=S� must be cyclic. In particular, G0 � S�. Hence S� is equal toone of the three maximal subgroups of S containing G0. They are E, E1, and S0 D

G0ha2m�1

i. But S0 is metacyclic (as a subgroup of F ) and so S0=ˆ.G0/ cannot be

elementary abelian of order 8. It follows that S� is equal to E or E1. Interchanging Eand E1 (if necessary), we may assume that S� D E and so E=ˆ.G0/ Š E8. Since Eis not metacyclic and G0 is a metacyclic maximal subgroup of E, there is (by Lemma64.1(l) ) an element x of order � 4 in E�G0 (as before) and we have here (in our casewhere G=ˆ.G0/ has a normal E8) in addition that x2 2 ˆ.G0/.We have W � G0 and if G0 is not a four-group, then also W � R. If W 6� Z.G/,

then we always set hzi D W \ Z.G/. Since x2 2 G0 and o.x/ � 4, we have x2 2 W

and therefore Œa; x2� D z�, where � D 0; 1 and � D 1 if and only if a does notcentralize x2 (in which case W 6� Z.G/). We have z� D Œa; x2� D Œa; x�Œa; x�x D

vvx , and so vx D v�1z� .We know that a2m

2 R � ˆ.G0/, .ax/2m

2 R � ˆ.G0/ and so ha2m

i and .ax/2m

cover G0=hvi and therefore G0 D ha2m

ihvi D h.ax/2m

ihvi. But hai covers F=G0

and haxi covers F1=G0 and so F D haihvi and F1 D haxihvi. Set b D Œv; a� and

b1 D Œv; ax� so that F 0 D hbi and F 01 D hb1i, where we have used the facts that F

and F1 are metacyclic and F D ha; vi, F1 D hax; vi. Since hbihb1i D R, we mayassume (interchanging F D G0hai with F1 D G0haxi D G0hxai, if necessary) thatb 2 R �ˆ.G0/. Indeed, we have Œxa; x� D Œa; x� D v. Then we compute

b1 D Œv; ax� D Œv; x�Œv; a�x D v�1.x�1vx/bx

D v�1.v�1z�/bx D v�2z�bx 2 R �ˆ.G0/;


since v�2 2 ˆ.G0/, bx 2 R � ˆ.G0/, and z� 2 ˆ.G0/. Indeed, if ˆ.G0/ ¤ f1g

and W 6� Z.G/, then hzi D W \ Z.G/ � ˆ.G0/. If ˆ.G0/ D f1g, then jRj D 2

and R � Z.G/ and so the fact that x2 2 R gives � D 0. Hence, in any case we getb 2 R � ˆ.G0/ and b1 2 R � ˆ.G0/ and (interchanging F and F1, if necessary) wemay assume that o.b/ D exp.R/.Conjugating the relation Œv; a� D b (which gives va D vb and .v�2/a D v�2b�2)

with x we get:

bx D Œv�1z�; ax � D Œv�1; a.a�1x�1ax/� D Œv�1; av� D Œv�1; v�Œv�1; a�v

D Œv�1; a� D v.a�1v�1a/ D v.va/�1 D v.vb/�1 D b�1;

and so we get bx D b�1. From the above we also get: b1 D v�2z�b�1, and sobx

1 D v2z�b D b�11 . Thus, x inverts R. We compute

ba1 D .v�2/az�.ba/�1 D v�2b�2z�.ba/�1 D b�1.v�2z�b�1/.ba/�1

D b�1b1.ba/�1

and so b�11 ba

1 D .bba/�1 2 hb1i\hbi, since hbi and hb1i are normal subgroups ofG.We show that ˆ.G/ is a powerful 2-group and .ˆ.G//0 D hbbai. Indeed, we have

F D haihvi, a2; v 2 ˆ.G/, jF W ˆ.G/j D 2 and so ˆ.G/ D ha2ihvi. This gives.ˆ.G//0 D hŒv; a2�i and since Œv; a2� D Œv; a�Œv; a�a D bba, we get .ˆ.G//0 D

hbbai. But hbi D F 0 is normal in G and so hbi D hbai and therefore hbbai �

Ã1.hbi/. On the other hand, F is metacyclic and therefore b is a square in F andso b D y2 for some y 2 F . But F=G0 is cyclic of order � 4 and b 2 G0 and soy 2 ˆ.G/. It follows that bba 2 Ã2.hyi/ and so ˆ.G/0 � Ã2.ˆ.G// and this meansthat ˆ.G/ is powerful.We have Œa; x2� D Œa; x�Œa; x�x D vvx D v.v�1z�/ D z� , and so

.ax/2 D axax D ax.xa/Œa; x� D ax2av D a2x2z�v D a2v.x2z�/:

It is easy to see that x2z� 2 ˆ.ˆ.G//. Indeed, x2z� 2 W and the facts that a2m

2

R � ˆ.G0/ and jR W ˆ.G0/j D 2 give R D ha2m

; ˆ.G0/i � ˆ.ˆ.G// becausea2m�1

2 ˆ.G/ .m � 2/. Hence, if W � R, we are done. If W 6� R, then R is cyclicand we know that in that case G0 Š E4 and so jRj D 2. But then x2 2 R � Z.G/and so � D 0 and again a2m

2 R � ˆ.G0/ D R � f1g and therefore ha2m

i D R �

ˆ.ˆ.G//. We get again x2z� D x2 2 R � ˆ.ˆ.G//. We have proved that in anycase x2z� 2 ˆ.ˆ.G//.Since .ax/2 D a2v.x2z�/ and x2z� 2 ˆ.ˆ.G//, we get

ˆ.G/ D hv; a2i D ha2v; a2i D ha2v.x2z�/; a2i D h.ax/2; a2i:

But ˆ.G/ is powerful and so Theorem 26.25 implies ˆ.G/ D h.ax/2iha2i. Weconclude:

G D haxiF D haxi.ˆ.G/hai/ D haxi.h.ax/2iha2i/hai

D .haxih.ax/2i/.ha2ihai/ D haxihai:


In the rest of this section we make case-to-case investigations depending on thestructure of G0 andG=ˆ.G0/. We shall use freely the notation and the results stated inTheorem 87.11.

Theorem 87.12. Let G be a 2-group with exactly one nonmetacyclic maximal sub-group M and d.G/ D 2. Suppose that G0 is cyclic of order 2n, n > 1. ThenG D EZ, where E is normal in G and E D hv; x j v2n

D 1; n > 1; x2 2 hzi; z D

v2n�1

; vx D v�1i is dihedral or generalized quaternion, Z D hai, Z \ E � hzi D

Z.E/, jZ=.Z \ E/j D 2m, m > 1, Œa; x� D v, va D v�1C4i (i integer), andŒv; a2m�1

� D 1. Here G0 D hvi Š C2n , n > 1, ˆ.G/ D G0ha2i, M D Eha2i,G D haxihai is a product of two cyclic subgroups and CG.E/ 6� E.

Proof. By Theorem 87.10, G has no normal subgroups isomorphic to E8 and so wemay use Theorem 87.11 (a) to (f). Since v D Œa; x� 2 G0�R, we haveG0 D hvi. Also,R D F 0F 0

1 implies that interchanging F D G0hai and F1 D G0haxi D G0hxai (andnoting that Œxa; x� D Œa; x� D v), if necessary, we may assume that F 0 D R D hbi,where b D Œv; a� which gives va D vb. Set W \ G0 D hzi D �1.R/ D �1.G

0/ �

Z.G/. Since o.x/ � 4, x2 2 hzi and so x2 D z�, � D 0; 1. Therefore 1 D Œa; x2� D

Œa; x�Œa; x�x D vvx and so vx D v�1. It follows that E is dihedral or generalizedquaternion. Since

.v2n�2

/a D .vb/2n�2

D v2n�2

b2n�2

D v2n�2

z D v�2n�2

;

where o.v2n�2

/ D 4, it follows that hai \E � hzi D Z.E/.Since hv�2i D hbi, we may set b D v�2C4i for some integer i . We have va D

v�1C4i and vax D v1�4i D vv�4i so that F1 D hvihaxi is ordinary metacyclic(since F1 centralizes hvi=Ã2.hvi/ D hvi=hv4i). By Proposition 26.27, F1 is powerful.Since .ax/2 D axax D axxaŒa; x� D ax2av D a2vz� , where z� 2 ˆ.F1/, we getF1 D hax; vi D hax; a2i D haxiha2i, where we have used Proposition 26.25. Butthen G D F1hai D .haxiha2i/hai D haxihai, and so G is a product of two cyclicsubgroups.

Consider E1 D G0hxa2m�1

i, where E1 \ F D G0 and so W 6� E1. Hence E1

is a normal subgroup of G which does not possess a G-invariant four-subgroup. ByLemma 1.4,E1 is of maximal class (sinceE1 cannot be cyclic). But then .xa2m�1

/2 2

hzi and therefore xa2m�1

also inverts G0 D hvi. Indeed, setting xa2m�1

D x0, we getG D ha; x0i and so Œa; x0� D v0 2 G0 �R and G0 D hv0i. This gives 1 D Œa; .x0/2� D

Œa; x0�Œa; x0�x0

D v0.v0/x0

and .v0/x0

D .v0/�1. It follows that a2m�1

centralizes hvi.Since a2m�1

does not fuse x and vx (noting that a fuses x and vx), there is g 2 hvi

with ga2m�1

centralizing x and ga2m�1

centralizesE and so CG.E/ 6� E.

From Theorems 87.9, 87.10, 87.11(g13), and 87.12, we get the following importantresult.


Corollary 87.13. Let G be a 2-group with exactly one nonmetacyclic maximal sub-group and d.G/ D 2. Then G D AB with cyclic A and B .

The next two results are devoted to the case, where G=ˆ.G0/ has no normal ele-mentary abelian subgroups of order 8.

Theorem 87.14. Let G be a 2-group with exactly one nonmetacyclic maximal sub-group M and d.G/ D 2. Assume that G0 Š E4 and G has no normal elementaryabelian subgroups of order 8. Then G is a unique group of order 25:

G D ha; x j a8 D x4 D 1; a4 D x2 D z; Œa; x� D v; v2 D 1; Œv; a� D zi;

where Z.G/ D hzi Š C2, G0 D hz; vi Š E4,M D hvi � ha2; xi Š C2 � Q8 (and infact this group is isomorphic to the group of Theorem 74.2(f)).

Proof. We may use Theorem 87.11 (a) to (g) (except (g6)). Here jRj D 2 so thatR D hzi � Z.G/ and G0 D hzi � hvi with v D Œa; x�. Since x2 2 R, we havevx D v�1 D v and so E D G0hxi is abelian. But (by our assumption) E is notelementary abelian and so x2 D z. We know that b D Œv; a� 2 R � f1g and soŒv; a� D z, W D G0 6� Z.G/ and CG.G

0/ D M D Eha2i. Also, a2m

2 R �

f1g and so a2m

D z. Since .Eha2m�1

i/=R Š E8, we have Œa2m�1

; x� � R. IfŒa2m�1

; x� D 1, then i D xa2m�1

is an involution inM �F and so hii�G0 is a normalelementary abelian subgroup of order 8, a contradiction. Hence Œa2m�1

; x� D z and soEha2m�1

i D hvi � hx; a2m�1

i Š C2 � Q8. We compute:

Œa2; x� D Œa; x�aŒa; x� D vav D .vz/v D z; Œa4; x� D Œa2; x�a2

Œa2; x� D zz D 1;

and so ifm > 2, then ha4i � ha2m�1

i and in that case Œa2m�1

; x� D 1, a contradiction.Hencem D 2 and the structure of G is uniquely determined.

Theorem 87.15. Let G be a 2-group with exactly one nonmetacyclic maximal sub-group and d.G/ D 2. Assume that G0 is noncyclic, ˆ.G0/ ¤ f1g, and G=ˆ.G0/ hasno normal elementary abelian subgroups of order 8. Then G0 has a cyclic subgroupof index 2, G=G0 Š C4 � C2, ˆ.G/ is abelian, and we have one of the followingpossibilities (depending on whether Z.G/ is noncyclic or cyclic):

G D ha; x j a8 D x4 D 1; x2 D u; a4 D uz� ; � D 0; 1; Œa; x� D v;(a)

v2n

D 1; n � 2; Œu; a� D 1; v2n�1

D z; vx D v�1;

Œv; a� D uv�2z� ; � D 0; 1i;

where jGj D 2nC4, Z.G/ D hu; zi Š E4, and G0 D hu; vi Š C2 � C2n .

G D ha; x j a16 D x4 D 1; x2 D u; a4 D uv2n�2

z� ; � D 0; 1; Œa; x� D v;(b)

v2n

D 1; n � 4; ua D uz; v2n�1

D z; vx D v�1z; Œu; v� D 1;

Œv; a� D uv�2C2n�2

z� ; � D 0; 1i;

where jGj D 2nC4, Z.G/ D hzi Š C2, and G0 D hu; vi Š C2 � C2n .


Proof. We may use Theorem 87.11(a) to (g) (except (g6)). Indeed, since jG0j > 4,Theorem 87.10 implies that G has no normal subgroups isomorphic to E8. ApplyingTheorem 87.14 on the factor-group G=ˆ.G0/, we get at oncem D 2, i.e., F=G0 Š C4

and x2 2 R � ˆ.G0/. But x2 D u is an involution and if C is a maximal subgroupof G0 not containing u, then C \ R D ˆ.G0/, G0 D hui � C and therefore C iscyclic of order 2n, n � 2 (since G0 is of rank 2). Hence, G0 has a cyclic subgroupof index 2 and since Œa; x� D v 2 G0 � R, we have G0 D hui � hvi, o.v/ D 2n,n � 2, and ˆ.G0/ D hv2i with R D hui � hv2i. Let hzi D Ãn�1.G

0/ so thatz D v2n�1

, W D hu; zi Š E4 and z 2 Z.G/. It follows that E centralizes W and soCG.W / � M D Eha2i.We know that a4 2 R�ˆ.G0/ and the element b D Œv; a� is of order exp.R/ D 2n�1

so that b D uv2i for an odd integer i . Suppose that Chv2i.a/ > hzi which forces

n � 3. From va D vb, we get in this case .v2n�2

/a D .vb/2n�2

D v2n�2

b2n�2

D

v2n�2

z D v�2n�2

since v2n�2

is an element of order 4 in hv2i and v2n�1

D z. This isa contradiction and so Chv2i.a/ D hzi. In particular, a8 2 hzi and so we have either

a4 D uz� (first case) or n � 3 and a4 D uv2n�2

z� (second case), where � D 0; 1.Since F 0 D hbi and F is metacyclic, there is y 2 F such that y2 D b. But b 2

G0�ˆ.G0/ and so y 2 F �G0. The fact that F=G0 Š C4 implies that y 2 G0a2 � M .SinceG0 is abelian, CG0.y/ D CG0.a2/ and so a2 centralizes b. But b D uv2i (i odd)and y 2 M and so a2 centralizes u which gives that a2 centralizes hv2i i D hv2i. Onthe other hand, hv2i D ˆ.G0/ is normal in G and so in case n > 2, a induces aninvolutory automorphism on hv2i. From va D vb, we get .v2/a D .vb/2 D v2b2 andso Œv2; a� D b2 D v4i (i odd). Hence, in case n > 2, a induces on hv2i an involutoryautomorphism such that .hv2; ai/0 D hv4i and so .v2/a D v�2z� , � D 0; 1, where� D 1 is possible only if n > 3. Thus b2 D v�4z� which gives v4i D v�4v�2n�1

andso 4i �4C �2n�1 .mod 2n/ and therefore 2i �2C �2n�2 .mod 2n�1/. We getb D uv2i D uv�2C�2n�2C�2n�1

( � an integer) and so b D uv�2v�2n�2

z� , � D 0; 1,� D 0; 1, and � D 1 is possible only if n > 3.In the first case, where a4 D uz�, we have CG.W / � hM;ai D G and so W �

Z.G/ which implies � D 0 and vx D v�1. It is easy to see that in this case Z.G/ D

W Š E4 (since x inverts G0, Z.G/ � ˆ.G/ D G0ha2i and Œa2; x� D vav D .vb/v D

v2b ¤ 1). Suppose that in this case � D 1, i.e., b D uv�2v2n�2

z� , n � 4. ThenŒa2; x� D Œa; x�aŒa; x� D vav D v2b D uv2n�2

z� , and noting that a2 centralizes v2,we get

1 D Œuz�; x� D Œa4; x� D Œa2; x�a2

Œa2; x� D .uv2n�2

z�/a2

uv2n�2

z� D v2n�1

D z;

a contradiction. Hence in this case � D 0.Suppose that we are in the second case, where n � 3 and a4 D uv2n�2

z� . In thiscase we show first that ua D uz and so W 6� Z.G/, x2 D u 2 W � Z.G/ and � D 1,vx D v�1z. Indeed, u D a4v�2n�2

z� and so

ua D a4.v�2n�2

/az� D a4v2n�2

z� D .a4v�2n�2

z�/z D uz:


Also, it is easy to show that in this case n � 4. If n D 3, then

b1 D v�2zb�1 D v�2z.uv�2i / D u.zv�2.1Ci// 2 W � hzi

since 1 C i is even and so v�2.1Ci/ 2 hzi. But then hb1i D F 01 � Z.G/ and W �

Z.G/, a contradiction. Assume that � D 0 so that b D uv�2z� . But then b1 D

v�2zb�1 D uz�C1 is an involution in W � hzi and hb1i D F 01 � Z.G/ which gives

W � Z.G/, a contradiction. Hence, in this case we must have � D 1. It is easy to seethat in this case Z.G/ D hzi.In both cases, using the obtained relations, we compute bba D 1 and so .ˆ.G//0 D

hbbai D f1g and ˆ.G/ is abelian.

In what follows we may assume that G0 is noncyclic of order > 4 andG=ˆ.G0/ hasa normal elementary abelian subgroup of order 8.

Theorem 87.16. Let G be a 2-group with exactly one nonmetacyclic maximal sub-group and d.G/ D 2. Assume that G0 Š C2 � C2n , n � 2, and G=ˆ.G0/ has anormal elementary abelian subgroup of order 8. Then we have:

G D ha; x j Œa; x� D v; v2n

D 1; n � 2; v2n�1

D z; x2 2 hzi;

Œv; a� D uv2C4s .s integer/; u2 D Œv; u� D 1; ux D u; vx D v�1;

a2m

D uz� or a2m

D uv2n�2

z�; .� D 0; 1/i;

wherem � 2 and, in the second case, n � 4, 1C s 6 0 .mod 2n�3/, and n � mC 2.Here jGj D 2nCmC2, n � 2, m � 2, G0 D hui � hvi Š C2 � C2n , where incase o.a/ D 2mC1 we have hu; zi � Z.G/ and so Z.G/ is noncyclic and in caseo.a/ D 2mC2 we have hu; zi 6� Z.G/ and so Z.G/ is cyclic in which case n � 4.

Proof. Since jG0j � 8, Theorem 87.10 implies thatG has no normal normal subgroupsisomorphic to E8 and so we may use Theorem 87.11 (a) to (g). Since Œa; x� D v isof order 2n, we get G0 D hui � hvi for some involution u, ˆ.G0/ D hv2i, andR D hui � hv2i. By Theorem 87.11(g6), E=ˆ.G0/ Š E8 and x2 2 hzi, where weset v2n�1

D z and so hzi D �1.ˆ.G0// � Z.G/. This gives � D 0 and vx D v�1

and therefore x inverts G0. We have W D hui � hzi and since jG W CG.W /j � 2 andW D Z.E/, we have CG.W / � M D Eha2i. Also, F=G0 Š C2m , m � 2.Since b D Œv; a� 2 R � ˆ.G0/ is of order exp.R/ D 2n�1, we may set b D uv2i

with an odd integer i and we may also write b D uv2C4s (s integer). Suppose thatChv2i.a/ > hzi which implies n � 3. From b D Œv; a� we get va D vb and so

.v2n�2

/a D .vb/2n�2

D v2n�2

.uv2i /2n�2

D v2n�2

z D v�2n�2

;

where o.v2n�2

/ D 4. This is a contradiction and so Chv2i.a/ D hzi. But we know

that a2m

2 R � ˆ.G0/ and so a2mC1

2 hzi which gives either a2m

D uz� or n � 3


and a2m

D uv2n�2

z� , where � D 0; 1. If a2m

D uz� , then o.a/ D 2mC1, CG.W / �

hM;ai D G and soW � Z.G/ and Z.G/ is noncyclic.Suppose that we are in the second case, where n � 3, a2m

D uv2n�2

z�, � D 0; 1,and o.a/ D 2mC2. In this case u D a2m

v�2n�2

z� and so

ua D a2m

v2n�2

z� D .a2m

v�2n�2

z�/z D uz;

which implies thatW 6� Z.G/ and so Z.G/ is cyclic. In this case we must have n � 4.Indeed, if n D 3, then b1 D v�2b�1 D v�2.uv�2i / D uv�2.1Ci/, and so the factthat 1C i is even and o.v/ D 8 implies v�2.1Ci/ 2 hzi. Hence b1 is an involution inW � hzi and since hb1i D F 0

1 � Z.G/, we get W � Z.G/, a contradiction. Since

b1 D v�2b�1 D v�2uv�2�4s D uv�4.1Cs/

and b1 2 R � ˆ.G0/ cannot be an involution (Z.G/ is cyclic), it follows 1 C s 6 0

.mod 2n�3/. We have G D haihaxi, where a2m

D uv2n�2

z� and a2mC1

D z. Sinceo.a2m

/ D 4 and a2m

is inverted by x, it follows that ha2m

i 6� Z.G/ and so ha2m

i 6�

hai \ haxi � Z.G/. Since .ax/2m

2 R � ˆ.G0/ cannot be an involution (becauseW 6� Z.G/), it follows that h.ax/2

m

i being distinct from ha2m

i, o..ax/2m

/ � 8 and soh.ax/2

mC1

i � hv2i and h.ax/2mC1

i > hzi. This implies that hai \ haxi D hzi and soo.a/ D 2mC2 and jGj D 2nCmC2 gives o.ax/ D 2mC1Cr , where o..ax/2

mC1

/ D 2r ,r � 2. This implies (by the product formula) mC r D n and so n � mC 2.

In the rest of this section we consider the remaining case, where G0 has no cyclicsubgroups of index 2. By Theorems 87.10 and 87.15, G has no normal elementaryabelian subgroups of order 8 but G=ˆ.G0/ has a normal elementary abelian subgroupof order 8. We shall use freely the notation and all results from Theorem 87.11.

Theorem 87.17. Let G be a 2-group with exactly one nonmetacyclic maximal sub-group and d.G/ D 2. Assume that G0 Š C2r � C2r , r � 2, is homocyclic. ThenG=G0 Š C2 � C4 and

G D ha; x j a2rC2

D 1; r � 2; Œa; x� D v; Œv; a� D b; v2r

D b2r

D Œv; b� D 1;

v2r�1

D u; b2r�1

D z; x2 2 hu; zi; bx D b�1; vx D v�1z�;

� D 0; 1; and � D 1 if and only if x2 62 hzi; ba D b�1z� ; � D 0; 1;

a4 D v�2b�1u�z� ; � D 0; 1i:

Here jGj D 22rC3, r � 2, G0 D hvi � hbi Š C2r � C2r , Z.G/ D hzi Š C2, and.ˆ.G//0 D hz�i, where ˆ.G/ D ha2ihvi.

Proof. The element v D Œa; x� 2 G0 � R is of order 2r and if hb0i Š C2r is acomplement of hv2i in R, then R D hb0i � hv2i, ˆ.G0/ D h.b0/2i � hv2i and all


elements in R �ˆ.G0/ are of order 2r . Hence b D Œv; a� 2 R � ˆ.G0/ is of order 2r

and so G0 D hbi � hvi, R D hbi � hv2i, ˆ.G0/ D hb2i � hv2i. We set v2r�1

D u,b2r�1

D z so that W D hu; zi, hzi D Ãr�1.R/ � Z.G/ and we know that x invertsR. From b D Œv; a� follows va D vb, .v2r�1

/a D v2r�1

b2r�1

and ua D uz so thatW 6� Z.G/ and W \ Z.G/ D hzi. SinceW � Z.E/, we get CG.W / D M D Eha2i.We have x2 2 W and vx D v�1z� , � D 0; 1, where � D 1 if and only if x2 62 hzi. Itfollows Z.E/ D W .We have b1 D v�2z�b�1 2 R � ˆ.G0/, where o.b1/ D 2r , hb1i D F 0

1, F1 D

G0haxi, and b2r�1

1 D .b�1/2r�1

D z. But R D hbihb1i, jRj D 22r�1, and so (by theproduct formula) hbi \ hb1i D hzi. Since bba 2 hbi \ hb1i (Theorem 87.11(g11)),we get ba D b�1z� , � D 0; 1, and .ˆ.G//0 D hbbai D hz�i. Hence, ˆ.G/ D

G0ha2i D ha2ihvi is either abelian or minimal nonabelian (Lemma 65.2). Also,b�1

1 .b1/a D .bba/�1 D z� and .b1/

a D b1z� which gives .b1u

�/a D b1z�.uz/� D

b1u� D v�2b�1u�z� , so that CG0.a/ D hv�2b�1u�z�i is of order 2r (noting that

G0 D hb1u�i � hvi and ua D uz forces Chvi.a/ D f1g). But a2m

2 R � ˆ.G0/ is of

order 2r and so ha2m

i D hv�2b�1u�z�iwhich gives ha2mC1

i D hv�4b�2i D hv4b2i.We claim that for all s � 1, Œa2s

; x� D v2s

b2s�1

. Indeed, Œa2; x� D Œa; x�aŒa; x� D

vav D .vb/v D v2b and, using the facts that v2 2 ˆ.ˆ.G// � Z.ˆ.G// andba2

D .b�1z�/a D b, we get Œa4; x� D Œa2; x�a2

Œa2; x� D .v2b/a2

v2b D v4b2.Assuming s > 2 and using the induction on s (since a2s�1

2 Z.ˆ.G//), we get

Œa2s

; x� D Œa2s�1

a2s�1

; x� D Œa2s�1

; x�a2s�1

Œa2s�1

; x�

D .v2s�1

b2s�2

/a2s�1

.v2s�1

b2s�2

/ D .v2s�1

b2s�2

/2 D v2s

b2s�1

:

Since a2m

2 R � ˆ.G0/, m � 2, and x inverts R, we get .a2m

/x D a�2m

, and soŒa2m

; x� D a�2mC1

. By the above relation, Œa2m

; x� D v2m

b2m�1

D a�2mC1

and sousing a result from the previous paragraph we get hv2m

b2m�1

i D hv4b2i which forcesm D 2. We have proved that G=G0 is abelian of type .4; 2/ and so a4 2 R �ˆ.G0/.Because ha4i D hv�2b�1u�z�i, we get a4 D v�2b�1u�z�.v�2b�1u�z�/2i for

some integer i , and so a4 D v�2�4ib�1�2iu�z�. Then (noting that a8 D v�4b�2)a8 D v�4�8ib�2�4i D v�4b�2 implies i 0 .mod 2r�2/ and so we have a4 D

v�2b�1z�iu�z� and a4 D v�2b�1u�z� , � D 0; 1. We see also Z.G/ D hzi.

Theorem 87.18. Let G be a 2-group with exactly one nonmetacyclic maximal sub-group and d.G/ D 2. Assume that G0 has no cyclic subgroups of index 2. Then Z.G/is elementary abelian of order at most 4, Z.G/ � �1.G

0/, and G=G0 Š C4 � C2.

Proof. We consider G=Ã2.G0/, where G0=Ã2.G

0/ D .G=Ã2.G0//0 Š C4 � C4 and

so by Theorem 87.17, G=G0 Š C4 � C2. We may use Theorem 87.11 with m D 2.We have Z.G/ � ˆ.G/ D G0ha2i, where j.G0ha2i/ W G0j D 2 and W � ˆ.G0/.Note that x inverts R. If x commutes with an element y 2 G0 �R, then y2 2 R must


be an involution and so exp.G0/ D 4 and G0 Š C4 � C4. But in that case (Theorem87.17), Z.G/ Š C2 and Z.G/ � W and we are done. Hence, we may assume thatCG0.x/ D CR.x/ D W and so Z.G/ \ G0 � W . Suppose that there is an elementl 2 ˆ.G/ � G0 such that l 2 Z.G/. We have l2 2 G0 and so l2 2 R and therefore l2

must be an involution inW (since�1.ˆ.G// D W ). ButW � ˆ.G0/ and so there isan element k 2 G0 such that k2 D l2. In that case, kl is an involution in ˆ.G/ � G0,a contradiction. Hence, Z.G/ � W and we are done.

Theorem 87.19. Let G be a 2-group with exactly one nonmetacyclic maximal sub-group and d.G/ D 2. Assume that G0 Š C2r � C2rC1 , r � 2. Then we have:

G D ha; x j a2rC2

D 1; r � 2; Œa; x� D v; Œv; a� D b; v2rC1

D b2r

D Œv; b� D 1;

v2r

D z; b2r�1

D u; x2 2 hu; zi Š E4; bx D b�1; vx D v�1;

ba D b�1; a4 D v�2b�1w; w 2 hu; zii:

Here jGj D 22rC4, r � 2, G0 D hbi � hvi Š C2r �C2rC1 , Z.G/ D hu; zi Š E4, andˆ.G/ D G0ha2i is abelian.

Proof. By Theorem 87.18, we have m D 2 in Theorem 87.11. The element v D

Œa; x� 2 G0 � R is of order 2rC1 so that R is homocyclic of rank 2 and exponent 2r .It follows R D hbi � hv2i D hb1i � hv2i D hbi � hb1i, and so � D 0, W D hv2r

i �

hb2r�1

i D Z.G/, x inverts G0, x2 2 W , ba D b�1 and ba1 D b1, where va D vb

and b1 D v�2b�1. We set v2r

D z and b2r�1

D u. Since Chvi.a/ D hzi, we haveCG0.a/ D hb1i � hzi. On the other hand, a4 2 R � ˆ.G0/ is of order 2r and so ha4i

is a cyclic subgroup of index 2 in CG0.a/. This gives ha4i D hb1z� i D hv�2b�1z� i,

� D 0; 1. Also, .ˆ.G//0 D hbbai D f1g and therefore ˆ.G/ D G0ha2i is abelian.We get a4 D v�2b�1z� .v�2b�1z� /2i D v�2�4ib�1�2iz� , where i is an in-

teger. On the other hand, Œa2; x� D Œa; x�aŒa; x� D vav D v2b and Œa4; x� D

Œa2; x�a2


.v2b/ D .v2b/2 D v4b2. Since x inverts G0, Œa4; x� D

a�8 and so a8 D v�4b�2. This gives a8 D v�4b�2 D v�4�8ib�2�4i and �2i 0

.mod 2r�1/, which implies a4 D v�2b�1w with w 2 hu; zi since v�4i 2 hzi andb�2i 2 hui.

Somewhat more difficult is the next special case, where G0 Š C2r � C2rC2 , r � 2.After that we shall be able to investigate the general case.

Theorem 87.20. Let G be a 2-group with exactly one nonmetacyclic maximal sub-group and d.G/ D 2. Assume that G0 Š C2r � C2rC2 , r � 2. Then we have:

G D ha; x j a2rC2

D 1; r � 2; Œa; x� D v; Œv; a� D b; v2rC2

D b2rC1

D Œv; b� D 1;

v2rC1

D b2r

D z; v2r

b2r�1

D u; x2 2 hu; zi Š E4;

bx D b�1; vx D v�1; ba D b�1; a4 D v�2b�1w; w 2 hu; zii:


Here jGj D 22rC5, r � 2, G0 D hb; vi Š C2r � C2rC2 , Z.G/ D hu; zi Š E4, andˆ.G/ D G0ha2i is abelian.

Proof. We use freely Theorem 87.18 and 87.11 withm D 2. The element v D Œa; x� 2

G0�R is of order 2rC2 and we set z D v2rC1

so that z 2 Z.G/ since hzi D ÃrC1.G0/.

The element b D Œv; a� 2 R �ˆ.G0/ is of order exp.R/ D 2rC1 and since hbi coversR=hv2i Š C2r , we get hbi \ hv2i D hbi \ hvi D hzi. We have b1 D Œv; ax� 2

R � ˆ.G0/ and hb1i also covers R=hv2i. We know that b1 D v�2z�b�1 and sob2r

1 D v�2rC1

b�2r

D zz D 1 and therefore b1 is of order 2r . Thus hb1i \ hv2i D f1g

and so R D hb1i � hv2i D hb1i � hbi. Set u D b2r�1

1 D v�2r

b�2r�1

D v2r

b2r�1

sothat W D hu; zi Š Z.G/ (noting that hb1i D F 0

1 is normal in G and so u 2 Z.G/),which gives � D 0 and vx D v�1. We have x2 2 W and we know that x invertson R and so bx D b�1. We have also b�1

1 ba1 D .bba/�1 2 hbi \ hb1i D f1g and

so ba D b�1, ba1 D b1, and .ˆ.G//0 D hbbai D f1g. Since va D vb, we get

.v2r

/a D v2r

b2r

D v2r

z D v�2r

(since .v2r

/2 D z) and so CG0.a/ D hb1i � hzi.On the other hand, a4 2 R � ˆ.G0/, ha4i covers R=hv2i and a4 2 CG0.a/ whichgives ha4i D hb1z

�i D hv�2b�1z�i, � D 0; 1.We compute (noting that x inverts a4):

Œa2; x� D Œa; x�aŒa; x� D vav D v2b; Œa4; x� D a�8

D Œa2; x�a2


.v2b/ D v4b2

and so a8 D v�4b�2. Using the last result from the previous paragraph, we get

a4 D v�2b�1z�.v�2b�1z�/2i D v�2�4ib�1�2iz� .i integer/;

a8 D v�4b�2 D v�4�8ib�2�4i and so v�8ib�4i D 1. This gives i 0 .mod 2r�2/

since hvi \ hbi D hzi and v2rC1

b2r

D z2 D 1. We may set i D �2r�2 (� integer) andthen

a4 D v�2b�1z�.v�2r

b�2r�1

/� D v�2b�1z�u� ; � D 0; 1:

Finally, we consider the general case, where G0 Š C2r � C2rCsC1 with r � 2 ands � 2.

Theorem 87.21. Let G be a 2-group with exactly one nonmetacyclic maximal sub-group and d.G/ D 2. Assume that G0 Š C2r � C2rCsC1 , r � 2, s � 2. Then we haveone of the following two possibilities (depending on Z.G/ being noncyclic or cyclic):

G D ha; x j a2rC2

D 1; r � 2; Œa; x� D v; Œv; a� D b;(a)

v2rCsC1

D b2rCs

D Œv; b� D 1; s � 2; b2r

D v�2rC1

;

v2rCs

D z; a2rC1

D u; x2 2 W D hu; zi Š E4;

ba D b�1; bx D b�1; vx D v�1; a4 D v�2b�1w; w 2 W i:


Here jGj D 22rCsC4, r � 2, s � 2, G0 D hb; vi Š C2r � C2rCsC1 , hbi \ hvi Š C2s ,Z.G/ D W D hu; zi Š E4, and ˆ.G/ D G0ha2i is abelian.

G D ha; x j a2rC3

D 1; r � 2; Œa; x� D v; Œv; a� D b;(b)

v2rCsC1

D b2rCs

D Œv; b� D 1; s � 2;

v2rCs

D a2rC2

D z; b2r

D v�2rC1

z;

u D v�2r .1C2s�1/b�2r�1

; ua D uz; x2 2 W D hu; zi Š E4;

ba D b�1zı ; ı D 0; 1; bx D b�1; vx D v�1z�; � D 0; 1;

� D 0 if and only if x2 2 hzi; a4 D v�2b�1uız� ; D 0; 1i:

Here jGj D 22rCsC4, r � 2, s � 2, G0 D hb; vi Š C2r � C2rCsC1 , hbi \ hvi Š C2s ,Z.G/ D hzi Š C2, ˆ.G/ D G0ha2i , .ˆ.G//0 D hzı i and so ˆ.G/ is either abelian(ı D 0) or minimal nonabelian (ı D 1).

Proof. We use freely Theorem 87.11 with m D 2 and Z.G/ � W (see Theorem87.18). The element v D Œa; x� 2 G0 � R is of order 2rCsC1 and the element b D

Œv; a� 2 R � �.G0/ is of order exp.R/ D 2rCs . Since hbi covers R=hv2i Š C2r ,we have c D b2r

2 hv2i, o.c/ D 2s and hv2i=hci Š C2r . Set z D v2rCs

so thathzi D ÃrCs.G

0/ � Z.G/ and hzi < hci. We know that the element x 2 E �G0 (withx2 2 W D �1.G

0/) inverts each element in R and vx D v�1z�, where � D 0; 1 and� D 1 if and only if x2 2 W � Z.G/. We note that W � Z.E/ and so CG.W / �

M D Eha2i. Also, b1 D Œv; ax� D v�2b�1z� , bx D b�1 and bx1 D b�1

1 .Set S D hci and S� D Ãr.G

0/ D hv2r

i so that S� is normal inG and jS� W S j D 2.We have .v2rCs�1

/a D v2rCs�1

b2rCs�1

D v2rCs�1

z D v�2rCs�1

since v2rCs�1

is anelement of order 4. This gives Chvi.a/ D hzi. Also, .v2r

/a D .bv/2r

D cv2r

. Sincea4 2 R�ˆ.G0/ and ha4i coversR=hv2i, we have either o.a/ D 2rC2 with hai\hvi D

f1g or o.a/ D 2rC3 with hai \ hvi D hzi. On the other hand, .ax/4 2 R � ˆ.G0/

and h.ax/4i coversR=hv2i so that o.ax/ D 2rC2Ct , where 2t D jhaxi\ hv2ij. SincejGj D 22rCsC4, G D haihaxi and o.a/ � 2rC3, it follows that o.ax/ D 2rC2Ct �

2rCsC1 and so t � s � 1. By our assumption, s � 2 and so t � 1, which implies thathaxi � hzi. If hai\hvi D f1g, then hai\haxi D f1g, o.ax/ D 2rCsC2 and thereforet D s. If hai \ hvi D hzi, then hai \ haxi D hzi, o.a/ D 2rC3 and so again t D s.In any case, haxi \ hv2i D S D hci and so ax centralizes c. But x inverts c and soca D c�1. From .v2r

/a D cv2r

follows .v2r

/a2

D ca.v2r

/a D c�1.cv2r

/ D v2r

.Hence a induces an involutory automorphism on S� D hv2r

i with ca D c�1, whereo.c/ � 4 and jS� W hcij D 2. This gives .v2r

/a D cv2r

D v�2r

z� , � D 0; 1, and soc D b2r

D v�2rC1

z� , where o.v2r

/ D 2sC1 � 8.

We compute b2r

1 D .v�2b�1z�/2r

D v�2rC1

v2rC1

z� D z� and so hbi \ hb1i D

hz� i. From Theorem 87.11(g11) follows b�11 ba

1 D .bba/�1 2 hz� i, and so ba D


b�1zı , ba1 D b1z

ı , ı D 0; 1, where � D 0 implies ı D 0. Also, .ˆ.G//0 D hbbai D

hzıi and therefore ˆ.G/ is either abelian or minimal nonabelian.We get ba2

D .b�1zı /a D b and .v2/a2

D v2 since v2 2 ˆ.ˆ.G// � Z.ˆ.G//,where ˆ.G/ D G0ha2i. Also we know that x inverts R and a4 2 R � ˆ.G0/ and allthis gives:

Œa2; x� D Œa; x�aŒa; x� D vav D .vb/v D v2b;

Œa4; x� D a�8 D Œa2; x�a2


.v2b/ D .v2b/2 D v4b2;

and so a8 D v�4b�2. From this result also follows

.a8/2r�1

D a2rC2

D v�2rC1

b�2r

D v�2rC1

v2rC1

z� D z�

and, since haxi \ hvi D hci Š C2s , we get haxi \ hai D hz� i.Suppose � D 0. In that case b2r

D v�2rC1

, hbi \ hb1i D F 0 \ F 01 D f1g, ı D 0,

o.b1/ D 2r , a2rC1

D u 2 Z.G/ and soW D hu; zi D Z.G/, � D 0, vx D v�1, ba D

b�1, ba1 D b1, and .ˆ.G//0 D f1g. Since CG0.a/ D hb1i � hzi and b1 D v�2b�1,

we get ha4i D hv2bz�i, � D 0; 1. Hence a4 D v2bz�.v2bz�/2i D v2C4ib1C2iz� (iinteger), and so, using a result from the previous paragraph, we get

a8 D v�4b�2 D v4C8ib2C4i ; v8C8ib4C4i D 1; i C 1 0 .mod 2r�2/;

and we set i D �1C t2r�2 (t an integer). This gives

a4 D v�2Ct2r

b�1Ct2r�1

z� D v�2b�1.v2r

b2r�1

/tz�

and since .v2r

b2r�1

/2 D v2rC1

b2r

D 1, we get .v2r

b2r�1

/tz� D w 2 W D Z.G/.Suppose � D 1. In that case we have hbi \ hb1i D F 0 \F 0

1 D hzi, b2r

D v�2rC1

z,b2r

1 D z, .ˆ.G//0 D hzı i, ı D 0; 1, ba D b�1zı , ba1 D b1z

ı . We set u0 D

b2r�1

1 c2s�2

so that u20 D b2r

1 c2s�1

D zz D 1 and

ua0 D .b2r�1

1 c2s�2

/a D b2r�1

1 c�2s�2

D b2r�1

1 c2s�2

z D u0z;

where we have used the facts that a inverts c and a centralizes an element of order 4in hb1i. Hence Z.G/ D hzi. This implies that a2rC2

D z and so o.a/ D 2rC3. Sincec D v�2rC1

z and b1 D v�2b�1z�, where vx D v�1z� , � D 0; 1 (and � D 0 if andonly if x2 2 hzi), we get:

u0 D .v�2b�1z�/2r�1

.v�2rC1

z/2s�2

D .v�2r .1C2s�1/b�2r�1

/z2s�2

D uz2s�2

;

where we have set u D v�2r .1C2s�1/b�2r�1

. We see that u2 D 1, ua D uz (sinceua

0 D u0z), and so W D hu; zi Š E4.Since .b1u

ı/a D b1zı.uz/ı D b1u

ı , .b1uı/2

r

D b2r

1 D z, and Chvi.a/ D hzi,we have CG0.a/ D hb1u

ıi, and so ha4i D hb1uıi D hv�2b�1z�uıi D hv2buıi since


z� 2 ˆ.hb1uıi/. This gives a4 D v2buı.v2buı/2i D v2C4ib1C2iuı (i integer), and

therefore

a8 D v�4b�2 D v4C8ib2C4i ; v8C8ib4C4i D 1; and so i C 1 0 .mod 2r�2/:

We set i D �1C t2r�2 (t an integer) and compute:

1 D vt2rC1

bt2r

D .v2rC1

b2r

/t D .v2rC1

v�2rC1

z/t D zt

and this forces t 0 .mod 2/. Hence we may set t D 2 ( an integer) and theni D �1C 2r�1 so that

a4 D v2�4C�2rC1

b1�2C�2r

uı D v�2b�1uı.v2rC1

b2r

/�

D v�2b�1uı.v2rC1

v�2rC1

z/� D v�2b�1uız� ; D 0; 1;

and we are done.

3o. The result of this section was also proved independently by the first author (seeSupplement to Corollary 36.11).

Theorem 87.22. Let G D AB be a nonmetacyclic 2-group, where the subgroups Aand B are cyclic. If fU;V;M g is the set of maximal subgroups of G, where A < U

and B < V , then U and V are metacyclic and d.M/ D 3. Hence,M is a unique non-metacyclic maximal subgroup ofG and these groups have been completely determinedin section 2o.

Proof. Assume, for example, that U is nonmetacyclic. Then U=Ã2.U / is nonmeta-cyclic (Lemma 64.1(o)) and so, in particular, jU=Ã2.U /j � 24. We set A D hai

and B D hbi so that U D haihb2i, jA W .A \ B/j � 4, and jhb2i W .A \ B/j � 4

(otherwise, U would be metacyclic). Since a4 2 Ã2.U /, b8 2 Ã2.U /, and jU W

ha4; b8ij D 24 (noting that [Hup2, Satz 2] implies that ha4; b8i D ha4ihb8i), we getÃ2.U / D ha4ihb8i and so jU W Ã2.U /j D 24. We want to investigate the structureof G=Ã2.U / and so we may assume that Ã2.U / D f1g. In that case G D haihbi is agroup of order 25 with o.a/ D 4, o.b/ D 8, hai \ hbi D f1g, and G has a nonmeta-cyclic subgroup U D haihb2i of order 24 and exponent 4 which is a product of twocyclic subgroups hai and hb2i of order 4.The subgroup U is nonabelian and U is not of maximal class (otherwise, U would

be metacyclic). By a result of O. Taussky (Lemma 64.1(s)), jU 0j D 2 and so U isminimal nonabelian (Lemma 65.2). By Lemma 65.1, Z.U / D ˆ.U / D ha2i�hb4i Š

E4 and U 0 D ha2b4i � Z.G/ since a2b4 is not a square in U . But b4 2 Z.U / and soŒb4; a� D 1 which gives b4 2 Z.G/. Hence a2 2 Z.G/ and we get E4 Š ha2; b4i �

Z.G/. We have G0 � ˆ.G/ D ha2i � hb2i Š C2 � C4 and G0 � U 0 D ha2b4i. Wehave G0 > U 0 because in case G0 D U 0, G would be minimal nonabelian and then Uwould be abelian, which is not the case. By the result of O. Taussky (and noting thatG


is not of maximal class), we get jG=G0j � 8 and so jG0j D 4. Hence G0 is a maximalsubgroup of ˆ.G/ and so G0 � Ã1.ˆ.G// D hb4i. Hence E4 Š G0 D ha2; b4i �

Z.G/ and so G is of class 2. By Lemma 44.1, G0 must be cyclic and this is our finalcontradiction.We have proved that U and V are metacyclic. If d.M/ � 2, then Lemma 64.1(n)

implies that G is metacyclic, a contradiction. Hence d.M/ D 3 and M is a uniquenonmetacyclic maximal subgroup of G.

�88

Hall chains in normal subgroups of p-groups

This section supplements Theorem 24.1. We assume throughout this section that

H > f1g is a normal subgroup of a p-group G.(�)

We begin with the following definitions.

Definition 1. Given k 2 N, let

C W f1g D L0 < L1 < < Ln D H

be a chain (of length jC j D n) of G-invariant subgroups in (a normal subgroup) Hsuch that exp.Li=Li�1/ D p and jLi=Li�1j � pk , i D 1; : : : ; n. Then C is calleda k-admissible chain in H . Given a k-admissible chain C , set i0.C/ D max fi � 0 j

jLi j D pki g.

Thus, if i0.C/ D 0, thenH has no G-invariant subgroups of order pk and exponentp. We have jLi0.C/j D pki0.C/ and, if n > i0.C/, then jLi0.C/C1j < pk.i0.C/C1/. Inwhat follows, C is such as in Definition 1. We have exp.Li / � pi for all i .

Definition 2. A k-admissible chain C in H dominates over a k-admissible chain C1 W

f1g D M0 < M1 < < Ms D H if, with respect to lexicographic ordering, thesequence fjL1j; jL2 W L1j; : : : ; jLn W Ln�1jg is greater or equal than the sequencefjM1j; jM1 W M0j; : : : ; jMs W Ms�1jg. In that case, we write C � C1.

In the sequel, C1 is such as in Definition 2.

Definition 3. A k-admissible chain C in H is said to be dominating if, for each k-admissible chain C1 inH , we have C � C1.

Thus, two k-admissible dominating chains inH have the same sequence of indices,and such chains exist. We also consider k-admissible chains in other G-invariant sub-groups A of H and inH=A.

Definition 4. A k-admissible chain C inH s said to be a Hall chain (orHk-chain, forbrevity), if Li0Cj D �i0Cj .H/, where i0 D i0.C/ and j > 0.

In contrast to k-admissible dominating chains which exist always, this is not the casefor Hk-chains (the group D16 has no H2-chains). It follows that if, in Definition 4,Li0C1 < H , then exp.Li0C1/ D pi0C1, and so exp.Li / D pi for all i � n. A1-admissible chain inH is anH1-chain.


Clearly, if i0.C/ � n � 1, then a k-admissible dominating chain C must be anHk-chain. As Lemma 88.3(a) shows,Hk-chains are k-admissible dominating chains,however, the converse is true under additional assumptions only.It is asserted in Theorem 24.1 and Supplement 1 to Theorem 24.1 that there exists

in H an Hk-chain for k � p � 1 (the proof of the Supplement is the same as theproof of Theorem 24.1), however, this is not true for k D p as any p-group H ofmaximal class and order � p2p shows (Theorem 9.6). Moreover, ifH is a p-group ofmaximal class and order > ppC1 with j�1.H/j > pp�1, then H has no Hp-chains.So, there is an Hp-chain in H provided it satisfies additional conditions, and somesuch conditions are stated in Theorem 88.10. Theorem 88.11 shows, in particular, thatthere is in regularH anHk-chain for any k. Corollary 88.12 asserts that an abelian p-group G has only oneHk-chain if and only if j�1.G/j � pk . Note that Theorem 24.1is not a consequence of Theorem 88.10. In Proposition 88.13 we study the p-groupswithout Hp-chains.

Example. Each 2-group of order � 23 has anH2-chain. If G is a 2-group of maximalclass and order � 24 then, as we have noticed, G has no H2-chains. (i) We claimthat if a group G of order 24 is not of maximal class, it has an H2-chain. Indeed,take E4 Š R G G. If G=R is noncyclic, then f1g < R < G is an H2-chain. Nowsuppose that G=R is cyclic. If G has a cyclic subgroup of index 2, then R D �1.G/

and f1g < R < �2.G/ < G is the desired chain. If G has no cyclic subgroups ofindex 2, then G D C R, where C is cyclic of order 4. Let U � R \ Z.G/ beof order 2 and R1 D U � �1.C /; then G=R1 Š E4, and so f1g < R1 < G is anH2-chain. (ii) A 2-group of order > 24 with cyclic subgroup of index 2, which isnot of maximal class, has the unique H2-chain f1g < �1.G/ < �2.G/ < < G

(Theorem 1.2). (iii) We claim that a 2-group G of order 25, which is not of maximalclass, has an H2-chain. One may assume that G has no cyclic subgroups of index 2.Let E4 Š R GG. IfH=R < G=R is abelian of type .2; 2/, then f1g < R < H < G isan H2-chain. Now assume that G=R has no four-subgroups. Then G=R 2 fC8;Q8g.(iii1) Let G=R be cyclic. Then G D Z R is a semidirect product, where Z is cyclicof order 8. In that case,R1 D .R\Z.G//��1.Z/GG is a four-subgroup and G=R1

has a four-subgroup; then G has an H2-chain, as above. (iii2) Now let G=R Š Q8.If �1.G/ D R, then f1g < �1.G/ < �2.G/ < G is the unique H2-chain in G.It remains to consider the case where �1.G/ D U Š E8; then exp.G/ D 4. LetF=R � T=R be of order 4, where T D CG.R/; then F D L R, where L is cyclic oforder 4 and R \ L D f1g. Let K � R \ Z.G/ be of order 2. Set R1 D K � ˆ.F /;then f1g < R1 < F < G is the desired H2-chain in G. We suggest to the reader tocheck whether the following assertion is true. If H is a normal subgroup of order 25

in a 2-group G, then there is no H2-chain inH if and only ifH is of maximal class.

We are interested in the following statements concerning a p-group G and all n:

1. Each element of Ãn.G/ is a pn-th power.

2. exp.�n.G// � pn.

�88 Hall chains in normal subgroups of p-groups 439

3. j�n.G/j D jG W Ãn.G/j.

Definition 5 (see �11). For i 2 f1; 2; 3g, a p-group G is called a Pi -group, if allsections of G satisfy condition (i) for all n, and G is called a P -group, if it is a Pi -group for i D 1; 2; 3 simultaneously. (According to Mann, the following unexpectedresult holds: P3 � P2 � P1.)

Remark 1. For a P -group G, the following assertions hold:

j�1.G/j � j�2.G/=�1.G/j � j�3.G/=�2.G/j � : : : ;(i)

jG=Ã1.G/j � jÃ1.G/=Ã2.G/j � jÃ2.G/=Ã3.G/j � : : : ;(ii)

so that G is pyramidal (see �8). We have exp.�1.G// D exp.�2.G/=�1.G// D p

so Ã1.�2.G// � �1.G/ and

j�1.G/j D j�1.�2.G//j D j�2.G/=Ã1.�2.G//j � j�2.G/=�1.G/j:

We have, for k > 1, �k.G/=�k�1.G/ D �1.G=�k�1.G//. Therefore, the inequal-ity j�k.G/=�k�1.G/j � j�kC1.G/=�k.G/j follows by induction on k, completingthe proof of (i). As to (ii), we have

jG=Ã1.G/j D j�1.G/j � j�2.G/=�1.G/j DjG W Ã2.G/j

jG W Ã1.G/jD jÃ1.G/=Ã2.G/j:

We suggest to the reader to finish the proof of (ii). The groups satisfying (i) and (ii), arecalled upper and lower pyramidal, respectively. Next we prove that (iii) if A < G, thenjA=Ã1.A/j � jG=Ã1.G/j. Indeed, jA=Ã1.A/j D j�1.A/j � j�1.G/j D jG=Ã1.G/j

since G is a P3-group.

In what follows we use freely the following fact: if exp.G/ D pe and k < e, thenexp.G=�k.G// � pe�k .

Lemma 88.1. LetF E G, whereG is a p-group, and letK be aG-invariant subgroupof order p in F . Write NG D G=K.

(a) Suppose that fN1g D NF0 < NF1 < < NFn D NF is an Hp-chain in NF such thatj NFi j D ppi for i D 1; : : : ; n. If all sections of F of order ppC1 are P -groups, thenthere exists in F an Hp-chain f1g D L0 < L1 < < Ln < LnC1 D F such thatjFi W Li j D p for i D 0; 1; : : : ; n so that Ln is of order ppn and exponent � pn.

(b) Let k be fixed and suppose that all sections of F of order pkC1 are P3-groups.Suppose that fN1g D NF0 < NF1 < < NFn D NF is an Hk-chain in NF such thatj NFi j D pki for i D 1; : : : ; n. Then there exists in F an Hk-chain f1g D L0 < L1 <

< Ln < LnC1 D F such that jFi W Li j D p for i D 0; 1; : : : ; n so that Ln is oforder pkn and exponent � pn.


Proof. We proceed by induction on jF j.

(a) By hypothesis, F1 is of order ppC1 and F1=K is of order pp and exponent p.Suppose that F1 is irregular. Then exp.F / D p2 so K D Ã1.F1/ and, since F1 isa P3-group, we get j�1.F1/j D jF1=Ã1.F1/j D pp , and hence �1.F1/ is of orderpp and exponent p. In that case, we set L1 D �1.F1/. Now suppose that F1 isregular. Then Ã1.F1/ � K so j�1.F1/j D jF1=Ã1.F1/j � jF1=Kj D pp , and weconclude that �1.F1/ is of order � pp and exponent p. As L1 we take, in this case,an arbitrary G-invariant subgroup of order pp in �1.F1/. If F1 D F , we are done,so we assume that F1 < F . The group F=L1 is an extension of F1=L1 of order pby F=F1 of order pp.n�1/. By induction, there is an Hp-chain L1=L1 < L2=L1 <

< Ln=L1 < F=L1 such that j.Fi=L1/ W .Li=L1/j D p for all i D 2; : : : ; n. Thenf1g D L0 < L1 < < Ln < F is the desired chain. (LetK D F 0 < F < G, whereG is dihedral of order 16 and F is dihedral of order 8. Then there is an H2-chain inF=K but there is noH2-chain in F . Notice that F is not a P -group.)

(b) is proved in the same way as (a).

Lemma 88.2. Let F D �n.H/ and let C1 W f1g D L0 < L1 < < Ln D F be anHk-chain of length n in F .

(a) Suppose that

C2 W f1g < LnC1=Ln < LnC2=Ln < < LnCm=Ln D H=Ln;

where fLnCi=Ln D �i.H=Ln/g is anHk-chain in the quotient groupH=Ln D

H=F . Then

C W f1g D L0 < L1 < Ln D F < LnC1 < < LnCm D H

is an Hk-chain in H .

(b) If, in addition, k D p and H=F is absolutely regular, then the chain C from (a)is an Hp-chain.

Proof. (b) follows from (a) immediately sinceH=F has anHp-chain withLnCi=Ln D

�i.H=Ln/. It remains to prove (a).One may assume that F < H ; then exp.H/ > pn so exp.F / D pn. For j � n, we

have �j .F / D �j .�n.H// D �j .H/. To prove that C is anHk-chain, it suffices toprove that LnCi D �nCi.H/ for 1 � i � m. Take x 2 H with o.x/ � pnCi . Wehave to prove that x 2 LnCi . It follows from F D Ln D �n.H/ and exp.Ln/ D pn

that hxi \ Ln D �n.hxi/ so (in H=Ln) we have o.xLn/ � pi , and hence xLn 2

�i.H=Ln/ D LnCi=Ln.

Lemma 88.3. Let C be an Hk-chain in H . Then:

(a) C is a k-admissible dominating chain,

(b) all k-admissible dominating chains in H are Hk-chains.


Proof. Suppose that all considered chains are k-admissible.

(a) Let C1 be a dominating chain inH . We have to prove that jLi j D jMi j for all i .Let t be such that jMt j < pkt . We have jLt j � jMt j < pkt and Lt D �t.H/ sinceC is anHk-chain. Since exp.Mt/ � pt , we getMt � �t .H/ D Lt so Lt D Mt .

(b) Let C1 be an arbitrary dominating chain in H ; we have to show that C1 is alsoan Hk-chain. By (a), the chain C is dominating so jLi j D jMi j for all i , and we geti0 D i0.C/ D i0.C1/. Since C is an Hk-chain and exp.Mi0Cu/ � pi0Cu, we getLi0Cu D �i0Cu.H/ � Mi0Cu for all u > 0 so Mi0Cu D �i0Cu.H/, and we aredone.

The point of Lemma 88.3(b) is that if we want to prove that all k-admissible domi-nating chains inH areHk-chains, it suffices to show that at least one of these chains isanHk-chain. It follows from this lemma and Theorem 24.1 that all .p�1/-admissibledominating chains in H are Hp�1-chains. If C and C1 are Hk-chains in H , thenLi0Cu D �i0Cu.H/ D Mi0Cu, where i0 D i0.C/ and u � 1.

Lemma 88.4. Let C be a k-admissible dominating chain in H . Set NG D G=L1. ThenC W fN1g < NL2 < < NLn D NH is a k-admissible dominating chain in NH .

Proof. Indeed, suppose that C1 is a k-admissible dominating chain in NH and assumethat C1 > C ; then the chain C1, which is the ‘inverse image’ of the chain C1, isk-admissible and satisfies C1 > C , a contradiction.

If C is a k-admissible dominating chain in NG D G=N (here N is of order pk andexponent p), then its inverse image C is k-admissible but can be not dominating inG. Indeed, let G D U � V � W , where U;V and W are cyclic of orders p;p; p2 ,respectively. Let k D 2 and L1 D U � V , NG D G=L1. Then the chain C with Li D

L1 ��i�1.W /, i D 1; 2; 3, is not an H2-chain in G although fN1g < NL2 < NL3 D NG

is anH2-chain in the cyclic p-group NG. Indeed, ifM1 D U ��1.W / andM2 D G,then the chain C1 W f1g < M1 < M2 D G dominates strongly over C .Lemma 88.5 follows easily from Theorems 9.5 and 9.6.

Lemma 88.5. Let G be an irregular p-group of maximal class. Then:

(a) All sections of G of order ppC1 are P3-groups if and only if jGj D ppC1 andj�1.G/j D pp.

(b) All sections of G of order ppC1 are P2-groups if and only if jGj D ppC1 and�1.G/ < G.

Assuming, for example, that in (b), jGj > ppC1, we get exp.G=Z.G// > p and�1.G=Z.G// D G=Z.G/, which is a contradiction

Lemma 88.6. If H has a G-invariant subgroup B of order pp and exponent p, theneach maximal G-invariant subgroup ofH of exponent p is of order � pp .


Proof. Let A be a maximal G-invariant subgroup of exponent p in H . We have toprove that jAj � pp . Assume that this is false; then A 6� B . Let R < B be a leastG-invariant subgroup such that R 6� A. Then jAj < jARj D pjAj � pp so AR isregular. It follows from �1.AR/ D AR that exp.AR/ D p, contrary to the choiceof A.

Lemma 88.7 (= Theorem 13.5). Let a p-group G be neither absolutely regular nor ofmaximal class. Then the number of subgroups of order pp and exponent p inG is 1

.mod p/.

Let G be a p-group. Set Ã1.G/ D Ã1.G/. If Ãi.G/ has been defined, we setÃiC1.G/ D Ã1.Ã

i .G//. Since exp.G=Ãk.G// � pk , we get Ãk.G/ � Ãk.G/ forall k > 0. If G is a P -group of exponent pe , then Ãi.G/ D Ãi.G/ for all i .

Lemma 88.8. Let jGj D pm, exp.G/ � pe and m � pe. Then Ãe�1.G/ is eitherabsolutely regular or of order pp and exponent � p. In either case, Ãe�1.G/.�

Ãe�1.G// is of exponent p. If, in addition, m < pe, then we have j�1.Ãe�1.G//j <

pp so Ãe�1.G/ is of order � pp�1 and exponent p.

For a proof, see �24.The following two assertions hold:

(a) Let G be an absolutely regular p-group. If jGj > p.p�1/k , then exp.G/ > pk .

(b) Let G be a p-group of maximal class and order pm, m > p C 1. If m � 1 D

.p � 1/k, then exp.G/ D pk . If m � 1 > .p � 1/k, then exp.G/ > pk .Assertion (i) follows since absolutely regular p-groups are pyramidal. As to (ii),a p-group G of maximal class has an absolutely regular subgroup G1 of orderpm�1 (the fundamental subgroup of G), and exp.S/ D exp.G/ so the resultfollows from (i).

Lemma 88.9. (a) LetH be of order ppe and exponent � pe . Then all indices of anyp-admissible dominating chain inH equal pp so it is an Hp-chain of length e.

(b) Let H � G be a pyramidal P2-subgroup of order pke and exponent � pe. Thenall indices of any k-admissible dominating chain in H equal pk so it is an Hk-chainof length e.

Proof. By agreement, H E G. One may assume that e > 1. We use induction onjH j.

(a) By the paragraph preceding the lemma, H is neither absolutely regular nor ofmaximal class. Therefore, by Lemma 88.7, H has a G-invariant subgroup, say R1, oforder pp and exponent p. Suppose that exp.H/ < pe . Then exp.H=R1/ � pe�1 andjH=R1j D pp.e�1/. Therefore, by induction, there is in H=R1 an Hp-chain f1g D

R1=R1 D T1=R1 < T2=R1 < < Te=R1 D H=R1, and all indices of this chain areequal pp . In that case, f1g < T1 < < Te D H is the desiredHp-chain. Therefore,


one may assume, in what follows, that exp.H/ D pe . In that case, Ãe�1.H/ is oforder � pp and exponent � p (Lemma 88.8 soÃe�1.H/ � R, where R < H is a G-invariant subgroup of order pp and exponent p (Lemma 88.6). ThenH=R is a normalsubgroup of order pp.e�1/ and exponent � pe�1 in G=R. By induction, there existsan Hp-chain f1g D R=R D L1=R < L2=R < < Le=R D H=R in H=R suchthat all indices of this chain equal pp , and so f1g < R D L1 < L2 < < Le D H

is the desiredHp-chain.

(b) Since H is lower pyramidal, jÃe�1.H/j � pk . Since Ãe�1.H/ is generatedby elements of order � p, we get exp.Ãe�1.H// � exp.�1.H// D p. Since H isupper pyramidal, we get j�1.H/j � pk . Let Ãe�1.H/ � R � �1.H/, where R isG-invariant of order pk . Then H=R is of order pk.e�1/ and exponent � pe�1. Nowthe result follows by induction inH=R, as in (a).

Remark 2. LetH be a normal subgroup of order� ppe and exponent pe in a p-groupG. Suppose that there exists an Hp-chain C W f1g D L0 < L1 < < Ln D H

in H . We claim that then n D e. Indeed, this is trivial for e D 1 so we assume thate > 1. Clearly, n � e. Assume that n > e. Then jLn�1j < jH j � ppe � pp.n�1/

so �n�1.H/ D Ln�1 < H since C is an Hp-chain. It follows that exp.Ln�1/ D

pn�1 � pe, a contradiction since �n�1.H/ D H .

Suppose that G is a group of order pn possessing an Hk-chain, say C W f1g D

L0 < L1 < < Li0 < < G. Let n D Œn=k�k C s, where s < k. Assume thatj�t.G/j � pkt for all t � Œn=k�. We claim that then i0 D i0.C/ D Œn=k�, where Œx�is the integer part of the real number x. Clearly, i0 � Œn=k�. Assume that i0 < Œn=k�.Then j�i0C1.G/j < p

k.i0C1/ so, since i0C1 � Œn=k�, we get j�Œn=k�.G/j < pŒn=k�k ,

contrary to the assumption.Let C be anHk-chain of length n in H with i0 D i0.C/. Suppose that exp.Li0/ <

pi0 . We claim that then n � i0 C 1. Assume that n > i0 C 1. We have exp.Li0C1/ <

pi0C1. Since �i0C1.H/ D Li0C1 < H , it follows that exp.Li0C1/ D pi0C1, con-trary to what has just been said.Let C be a k-admissible chain in H . We use freely the following assertions. (i)

If jC j D i0.C/ C 1, then C is an Hk-chain. (ii) If i < j and jLi j < pki , thenjLj j < pkj .

Remark 3. Suppose that G is irregular p-group of maximal class (i) If G has an Hp-chain, then either jGj D ppC1 or ppC1 < jGj < p2p and j�1.G/j D pp�1.(ii) Conversely, if j�1.G/j D pp�1 and jGj < p2p , then G has the unique Hp-chain f1g < �1.G/ < G of length 2. Let us prove these assertions using results�9. Let C be an Hp-chain in G and assume that jGj > ppC1. Since G has nonormal subgroups of order pp and exponent p, we get j�1.G/j D pp�1. SincejL2j < p2p , we get L2 D �2.G/. Since �2.G/ D G, we have jGj < p2p . In thatcase,C W f1g < �1.G/ < G. Next, any groupG of order ppC1 has anHp-chain. Thisis obvious providedG is regular. IfG is irregular,ˆ.G/ is of order pp�1 and exponent


p. If �1.G/ D ˆ.G/, then f1g < ˆ.G/ < G is an Hp-chain. If �1.G/ > ˆ.G/,thenG has a maximal subgroupM of exponent p; then f1g < M < G is anHp-chain.Now assertion (ii) is obvious.

If a subgroupH has aG-invariant subgroup R of order pk and exponent p such thatexp.H=R/ D p, then H has an Hk-chain. Indeed, H=R has an Hk-chain R=R D

L1=R < L2=R < < Ln=R D H=R. Then f1g < L1 < L2 < < Ln D H isanHk-chain inH .

Theorem 88.10. Let H > f1g be a normal P2-subgroup of a p-group G. Suppose,in addition, that every irregular section of H of order ppC1 has a (characteristic)subgroup of order pp and exponent p. Then there exists in H a chain

C W f1g D L0 < L1 < < Ln D H

of G-invariant subgroups with the following properties (i D 1; : : : ; n):

(a) Li=Li�1 is of order � pp and exponent p, and

(b) either jLi j D ppi or else Li D �i .H/.

In other words, there is inH anHp-chain. Then it follows from Lemma 88.3(b) thateach p-admissible dominating chain in H is an Hp-chain. Next, if, in our theorem,p D 2, then the subgroup H is powerful, i.e., H=Ã2.H/ is abelian (see subsection2o of �26). To prove this, we suppose that H is a minimal counterexample. By hy-pothesis, all sections of H of order 8 are abelian so H is also modular (Iwasawa).One may assume that exp.H/ D 4. Then H is minimal nonabelian so jH j � 25.It follows from Redei’s classification of minimal nonabelian 2-groups that H has anonabelian section of order 8, contrary to the hypothesis. In what follows, we do notuse this result. Also note that if U is an irregular section of G of order ppC1, thenj�1.U /j D jU=Ã1.U /j D pp . Indeed, jU=Ã1.U /j D pp (Theorem 9.5). Next,j�1.U /j � pp , by hypothesis. Since U is a P2-group and exp.U / > p, we getj�1.U /j < U so j�1.U /j D pp , completing the proof.

Proof of Theorem 88.10. If f1g < N � H is G-invariant, then the pairs H=N �

G=N andN � G satisfy the hypothesis. We use induction on jH j. Let exp.H/ D pe .

(i) Assume thatH has no G-invariant subgroups of order pp and exponent p. ThenH is either absolutely regular so f1g D �0.H/ < �1.H/ < < �e.H/ D H isanHp-chain inH , or of maximal class and order ppC1 (Theorem 12.1(a) and Lemma88.5), contrary to the assumption (by hypothesis, every irregular section ofH of orderppC1 has a characteristic subgroup of order pp and exponent p).

In the sequel we assume that H has a G-invariant subgroup of order pp and expo-nent p. Since the theorem is trivial for e D 1, we also assume that e > 1.LetF0 < H be aG-invariant subgroup of orderp and set NG D G=F0. By induction,

there exists an Hp-chain C0

W fN1g D NF0 < NF1 < < NFn D NH in NH . Write


i0 D i0.C0/. By Lemma 88.1, there exists an Hp-chain C

00 W f1g D L0 < L1 < <

Li0 < Fi0 in Fi0 with jFi0 W Li0 j D p, so that

(˛) Fi0 contains a G-invariant subgroup Li0 of order ppi0 and exponent � pi0 .

Therefore, one may assume that i0 < n (otherwise, C 00 is the desired chain; see theparagraph preceding Remark 3) so Fi0 < Fi0C1. Next, H=Fi0.Š

NH= NFi0/ has no G-invariant subgroups of order pp and exponent p (otherwise, if U=Fi0 � H=Fi0 is sucha subgroup, then the p-admissible chain fN1g D NF0 < NF1 < < NFi0 <

NU < < NH

dominates strongly over C0, contrary to Lemmas 88.4 and 88.3(a)). Therefore, by

Lemma 88.7,H=Fi0 is either absolutely regular or irregular of maximal class. Assumethat H=Fi0 is irregular of maximal class. Then jH=Fi0 j D ppC1 (Lemma 88.5(b)).In that case,H=Fi0 has a characteristic subgroup U=Fi0 of order p

p and exponent p,contrary to what has just been said. Thus, H=Fi0 is absolutely regular so H=Li0 isregular, by Remark 7.2. We have

(ˇ) H=Fi0 is absolutely regular,H=Li0 is regular (Remark 7.2) and

j�1.H=Li0 /j � p j�1.H=Fi0 /j � p pp�1 D pp ;

and so:

(� ) If j � i0 C 1, then j NFj j < ppj so that jFj j � ppj .

Since C0is anHp-chain, one has�i0C1. NH/ D NFi0C1, so, by Remark 24.3,

(ı) �i0C1.H/ � Fi0C1 hence �i0C1.H/ D �i0C1.Fi0C1/ so, if Fi0C1 < H , thenexp. NFi0C1/ � pi0C1.

Since exp. NFi / � pi , we get exp.Fi/ � p exp. NFi/ � piC1 for all i so

(�) exp.Fi0C1/ � pi0C2. Therefore, the following three possibilities must be con-sidered: exp.Fi0C1/ � pi0 , exp.Fi0C1/ D pi0C1, exp.Fi0C1/ D pi0C2.

Suppose that i0 D 0. Then NF1 D �1. NH/ is of order < pp, by (ı), so F1 D �1.H/

is of order pp and exponent p since, by assumption,H has a G-invariant subgroup oforder pp and exponent p, and H=�1.H/ D H=F1 is absolutely regular, by (ˇ). Inthis case, by Lemma 88.2(b), there exists anHp-chain inH .Next we assume that i0 > 0; then jF1j D ppC1 and exp.F1/ � p2.

(ii) Suppose that exp.Fi0C1/ < pi0C1. Then, by (ı), Fi0C1 D H so i0 C 1 D n.We also have jH j D jFi0C1j � pp.i0C1/. Let f1g D L0 < L1 < < Li0 < Fi0 beanHp-chain in Fi0 with jFi0 W Li0 j D p (Lemma 88.1).If exp.H=Li0 / D p, then f1g D L0 < L1 < < Li0 < H is an Hp-chain in

H since all indices of this chain but last one are equal to pp and jH W Li0 j � pp .Since exp.H=Li0 / � p2, one may assume that exp.H=Li0 / D p2. Therefore, sinceH=Fi0.D Fi0C1=Fi0/ is of order � pp�1 and exponent p, the G-invariant subgroupU=Li0 D �1.H=Li0/ is of exponent p and index p in H=Li0 . In that case, f1g D

L0 < L1 < < Li0 < U is anHp-chain in U since only the last index of this chainis < pp . Since exp.H=Fi0/ D p and jH=Fi0 j � pp�1, we get Ã1.H/ < Fi0 since


H is not absolutely regular, by Hall’s regularity criterion (Theorem 9.8(a)). It followsthat jÃ1.H/j � p�p jH j � ppi0 . Therefore, there exists a G-invariant subgroupTi0 satisfying Ã1.H/ � Ti0 < Fi0 and jTi0 j D ppi0 . Since, in addition, we also haveexp.Ti0/ � exp.H/ � pi0 , there exists an Hp-chain f1g D T0 < T1 < < Ti0

in Ti0 of length i0 and all indices of that chain equal pp (Lemma 88.9(a)). Then

f1g D T0 < T1 < < Ti0 < H is an Hp-chain in H since jH=Ti0 j � pp andexp.H=Ti0/ D p.

(iii) Let exp.Fi0C1/ D pi0C1. By (ı), �i0C1.H/ D Fi0C1, and H=Fi0C1 is abso-lutely regular, by (ˇ).

Suppose that Fi0C1 < H . To prove that there is an Hp-chain in H , it sufficesto show, in view of Lemma 88.2(b), that Fi0C1 has an Hp-chain of length i0 C 1.Let U D �i0.Fi0C1/ .D �i0.H//; then exp.U / D pi0 since H is a P2-group andexp.Fi0C1/ D pi0C1. Since Li0 � U , we get jU j � jLi0 j D ppi0 . We haveexp.Fi0C1=U / D p since the P2-group Fi0C1=U is generated by elements of orderp. It follows that Ã1.Fi0C1/ � U . Since i0 > 0, we get jFi0C1=Ã1.Fi0C1/j � pp

so jÃ1.Fi0C1/j � p�pjFi0C1j � ppi0 . It follows that there is a G-invariant subgroupTi0 of order p

pi0 such that Ã1.Fi0C1/ � Ti0 � U ; then exp.Ti0/ � exp.U / D pi0 .By Lemma 9(a), there is an Hp-chain f1g D T0 < T1 < < Ti0 in Ti0 of length i0,and all indices of that chain equal pp . Then f1g D T0 < T1 < < Ti0 < Fi0C1 is anHp-chain in Fi0C1 of length i0 C1 since exp.Fi0C1=Ti0/ D p and jFi0C1=Ti0 j � pp .

Now we assume that Fi0C1 D H ; then exp.H/ D pi0C1 and ppi0 D jLi0 j <

jH j � pp.i0C1/. Write U D �i0.H/; then exp.U / D pi0 since H is a P2-group,exp.H=U / D p (see the paragraph following Remark 1) and jU j � jLi0 j D ppi0

so jH W U j � pp . In that case, there exists a G-invariant subgroup Ti0 of orderppi0 such that Ã1.H/ � Ti0 � U since jH W Ã1.H/j � pp : H is not absolutelyregular. We have exp.Ti0/ � exp.U / D pi0 and exp.H=Ti0 / D p. Therefore, ifC 0 W f1g D T0 < T1 < < Ti0 is an Hp-chain in Ti0 all of whose indices equal p

p

(Lemma 88.9(a)), then f1g D T0 < T1 < < Ti0 < H is an Hp-chain in H sincejH=Ti0 j � pp .

(iv) It remains to consider the possibility exp.Fi0C1/ D pi0C2; then exp.Fi0/ D

pi0C1 since exp.Fi0/ � pi0C1 and pi0C2 D exp.Fi0C1/ � p exp.Fi0/. By (ı),�i0C1.H/ D �i0C1.Fi0C1/. By (� ), jFi0C1j � pp.i0C1/.

(iv1) First suppose that Fi0C1 < H and let C0 W f1g D L0 < L1 < < Li0 <

< Fi0C1 be anHp-chain in Fi0C1 existing by induction. By (ˇ), H=Li0 is regularand j�1.H=Li0 /j � pp. We have Li0C1 D �i0C1.Fi0C1/.D �i0C1.H// since C 0

is an Hp-chain. Also, Li0C1 < Fi0C1 in view of exp.Li0C1/ � pi0C1 and (ı). Next,Li0C1=Li0 D �1.H=Li0 / (indeed, if D=Li0 D �1.H=Li0 /, then exp.D/ � pi0C1

so D � �i0C1.H/ D �i0C1.Fi0C1/ D Li0C1), so Fi0 � Li0C1. It follows thatH=Li0C1 is absolutely regular as an epimorphic image of H=Fi0 (see (ˇ)). In thatcase, there is an Hp-chain in H since f1g D L0 < L1 < < Li0 < Li0C1 is anHp-chain in Li0C1 D �i0C1.H/ of length i0 C 1 (Lemma 88.2(b)).


(iv2) Now let Fi0C1 D H ; then jH j � pp.i0C1/ and exp.H/ D pi0C2. SetU D �i0C1.H/; then exp.U / D pi0C1 since H is a P2-group, exp.H=U / D p

since �1.H=U / D H=U and U contains a G-invariant subgroup Li0 of order ppi0

and exponent pi0 . By induction, there is in U an Hp-chain, say C0 W f1g D L0 <

L1 < < Li0 < < U . We have Li0C1 D �i0C1.U / D U so jC 0j D i0 C 1. Itfollows that f1g D L0 < L1 < < Li0 < Li0C1 D U < H is an Hp-chain in HsinceH=U is of order < jH=Li0 j � pp and exponent p.

Theorem 88.11. Let H > f1g be a normal P -subgroup of a p-group G and let k befixed. Then there exists in H a chain

C W f1g D L0 < L1 < < Ln D H

of G-invariant subgroups with the following properties (i D 1; : : : ; n):

(a) Li=Li�1 is of order � pk and exponent p, and

(b) either jLi j D pik or Li D �i .H/.

It is possible to prove Theorem 88.11 in the same way as Theorem 88.10, however,the offered proof is shorter and more elementary.

Proof of Theorem 88.11. We proceed by induction on jH j. Set exp.H/ D pe andassume that e > 1 and k > 1 (otherwise, there is nothing to prove).

(i) Suppose thatH has noG-invariant subgroups of order pk and exponent p. Thenj�1.H/j < p

k . SinceH is a pyramidal (Remark 1) P -group, f1g < �1.H/ < <

�e.H/ D H is the unique Hk-chain inH .

In what follows we assume that H has a G-invariant subgroup of order pk andexponent p so jH W Ã1.H/j D j�1.H/j � pk sinceH is a P3-group.

(ii) Suppose that H is of order ptk with e � t . In that case, the theorem is true, byLemma 88.9(b).

(iii) Suppose thatH is of order ptkCs with t � e and 1 � s < k; then jÃ1.H/j �

p�k jH j < ptk sinceH is aP3-group. Therefore, there exists aG-invariant subgroupU < H of order ptk such that Ã1.H/ < U . We have exp.U / � pe � pt so, byLemma 88.9(b), there is an Hk-chain f1g D U0 < U1 < < Ut D U of lengtht ; since all indices of that chain are equal to pk and H=U is of order ps < pk andexponent p, it follows that f1g D U0 < U1 < < Ut < H is anHk-chain inH .

(iv) Suppose that j�t.H/j D ptk for some t � e. If t D e, then there is an Hk-chain in H (Lemma 88.9(b) and Remark 1). Now let t < e; then exp.�t .H// D pt

since H is a P2-group. By Lemma 88.9(b), there is an Hk-chain f1g D L0 < L1 <

< Lt D �t .H/ of length t in�t.H/, and all indices of this chain are equal to pk .Set NG D G=Lt . By induction, there is an Hk-chain fN1g < NLtC1 < < NLtCm D NH

in NH . Then, by Lemma 88.2(a), f1g D L0 < L1 < < Lt < LtC1 < <

LtCm D H is anHk-chain inH .


(v) Suppose that j�t.H/j > ptk for all t � e. In particular, jH j > pek . ThenjH j D pt0kCs for some positive integers t0 and s < k. It follows that pek < pt0kCs

so t0 � e. As we have noticed, jH W Ã1.H/j � pk . Let Ut0=Ã1.H/ be a G-invariantsubgroup of index ps inH=Ã1.H/; then jUt0 j D pt0k and exp.Ut0/ � pe � pt0 . ByLemma 88.9(b), there is an Hk-chain f1g D U0 < U1 < < Ut0 in Ut0 with allindices equal pk ; then f1g D U0 < U1 < < Ut0 < H is anHk-chain inH .

(vi) Suppose that j�t.H/j < ptk for some positive t � e. Let t be minimalsubjecting to that inequality. Then t > 1 sinceH has a G-invariant subgroup of orderpk and exponent p. By the choice of t , we get j�t�1.H/j � p.t�1/k . In view of (iv),one may assume that j�t�1.H/j > p.t�1/k . It follows that j�t.H/=�t�1.H/j <

pk�1 so that j�1.H=�t�1.H//j < pk�1 (indeed, if A=�t�1.H/ � H=�t�1.H/

is of order pk�1 and exponent p, then A � �t.H/ and A is of order � ptk >

j�t.H/j, which is not the case). Thus, the quotient group H=�t�1.H/ is pyramidal(Remark 1) and has no normal subgroups of order pk and exponent p. So, settingNG D G=�t�1.G/, we conclude that fN1g < �1. NH/ < < NH is an Hk-chain in NH .Therefore, in view of Lemma 88.2, it suffices to prove that there is in �t.H/ an Hk-chain of length t � 1. Assume that this is false; then the length of our chain is > t � 1

so let f1g D T0 < T1 < < Tt < < �t�1.H/ be an Hk-chain in �t�1.H/.Since Tt�1 < �t�1.H/, we get jTt�1j D p.t�1/k . By assumption, H=Tt�1 has nonormal subgroup of order pk and exponent p. It follows from Lemma 88.2(b) thatthere is anHk-chain inH .

Let G be an arbitrary p-group of order pn. Then W D G � E, where E is theelementary abelian p-group of order pn.k�1/, has a chain of normal subgroups oflength n all of whose factors are of order pk and exponent p.LetG be an abelian p-group of exponent pe > p. We claim thatG is homocyclic if

and only if Ãe�1.G/ D �1.G/. Suppose that the last equality holds. Set j�1.G/j D

pd . Then d D d.G/ so G D Z1 � � Zd , where Zi are all cyclic. In that case,Ãe�1.G/ D Ãe�1.Z1/� � Ãe�1.Zd / is of order pd so jÃe�1.Zi /j D p for all i ,and so G is homocyclic. The converse assertion is obvious.

Remark 4. Given a normal subgroup H of a p-group G, let Chk.H/ be the numberof Hk-chains in H . We claim that if C W f1g D L0 < L1 < < Ln D G is anHk-chain in G, then Chk.G/ � Chk.Lj / for all j � n. This is true for j � i0 D i0.C/.Now let j > i0; then Lj D �j .G/. Assume that f1g D M0 < M1 < < Mj D Lj

is anHk-chain in Lj . Then

C0 W f1g D M0 < M1 < < Mj D Lj < Lj C1 < < Ln D G

is an Hk-chain in G. Let j > i0 and i0 < i1 � j . Then Mi1 D �i1.Mj / D

�i1.Lj / D �i1.�j .G// D �i1.G/. If i2 > j , the Li2 D �i2.G/ so C 0 is anHk-chain in G, and we are done.


Corollary 88.12. An abelian p-group G has exactly one Hk-chain if and only ifj�1.G/j � pk .

Proof. Let exp.G/ D pe . If j�1.G/j � pk , then f1g < �1.G/ < < �e.G/ D G

is the unique Hk-chain in G. It remains to prove that if G has exactly one Hk-chain,then j�1.G/j � pk . Assume that G is a counterexample of minimal order; thenk > 1, e > 1 and j�1.G/j > pk . It follows that jG=Ã1.G/j D j�1.G/j > pk . ByTheorem 88.11, there is in G an Hk-chain C W f1g < L1 < < Ln D G, and thischain is unique; by assumption, jL1j D pk . Since, by Remark 4, Chk.Lj / D 1, weget, by induction, j�1.Lj /j D pk for 1 � j < n. Set i0 D i0.C/.

(i) Suppose that n > i0 C 1. Then Li0C1 < G and �i0C1.G/ D Li0C1. By theprevious paragraph, j�1.Li0C1/j D pk , contrary to the assumption since �1.G/ D

�1.Li0C1/. Thus, n � i0 C 1.

(ii) Suppose that n D 2. Then jGj � p2k so jÃ1.G/j D jG=�1.G/j < pk whence

Ã1.G/ < �1.G/ and j�1.G/ W Ã1.G/j � p2. Let Ã1.G/ < L1 < �1.G/, wherejL1j D pk ; then f1g < L1 < G is anHk-chain. However, L1 can be chosen in morethat one way, a contradiction. Thus, n > 2 so i0 > 1, by (i).

(iii) Write NG D G=L1. Then C W f1g D NL1 < NL2 < < NLn D NG is theunique Hk-chain in NG (Lemma 88.4). By induction, L2=L1 D �1.G=L1/. Then�1.G/ D �1.L2/ and so j�1.G/j D pk since j�1.L2/j D pk , by Remark 4 andinduction. Thus, G is not a counterexample.

Let p > 3 and let P be a Sylow p-subgroup of the symmetric group of degree p2.Set G D P=Ã1.P /; then jGj D pp. Let H be the unique abelian subgroup of indexp in G and k > 1 a proper divisor of p � 1. Then there is only one Hk-chain in Hand �1.H/ D H is of order pp�1 > pk . Let p D 2 and let H be dihedral of order8. The group H has exactly two H2-chains. Now let H < G, where G is dihedral oforder 16. Then there are no H2-chains inH as a normal subgroup of G.LetG be a p-group of maximal class and order> p3 andRGG with jG W Rj D p4.

Then G=R has the unique abelian subgroup G1=R of index p. This G1 is called thefundamental subgroup of G. Clearly, G1 is characteristic in G.In conclusion we consider an arbitrary p-group G without Hp-chains. By Theorem

88.11, G must be irregular. Let M be the set of all normal subgroups H of G suchthat there is an Hp-chain in H . To every H 2 M we associate an Hp-chain CH inH as a normal subgroup of G. Given H;H1 G G and CH W f1g D L0 < L1 < <

Lm D H and CH1W f1g D M0 < M1 < < Mn, we write CH � CH1

pro-vided the sequence fjL1j; jL2 W L1j; : : : ; jLn W Lm�1jg dominates over the sequencefjM1j; jM2 W M1j; : : : ; jMn W Mn�1jg in lexicographic ordering. Let M0 be the setof all H 2 M such that, whenever H1 2 M, then CH � CH1

(so that we compareonly Hp-chains, possibly, in distinct G-invariant subgroups). If H;H1 2 M0, theni0.CH / D i0.CH1

/, jCH j D jCH1j, jH j D jH1j and corresponding indices of these

chains are equal. In what follows we use the notation introduced in this paragraph.


Proposition 88.13. Let a p-group G, p > 2, have no Hp-chains and H 2 M0.Suppose thatH has no normal subgroups of order pp and exponent p, or, what is thesame, i0.CH / D 0. Then G is of maximal class and order � p2p and H is eitherabsolutely regular or irregular of maximal class.

(a) If H is absolutely regular, then H D G1, the fundamental subgroup of G, andM0 D fH g.

(b) Suppose thatH is irregular of maximal class. Then jG W H j D p, jH j D p2p�1

and j�1.H/j D pp�1. In that case, G1 62 M0.

Proof. Since regular p-groups areP -groups, G is irregular (Theorem 88.11). We alsohave jGj > ppC1 (otherwise, G has an Hp-chain as Remark 3 shows). By Lemma88.7, H is either absolutely regular or irregular of maximal class. Assume that thereis R G G of order pp and exponent p. Then the Hp-chain CR W f1g < R satisfiesi0.CR/ D 1 > 0 D i0.CH / so CR > CH , contrary to the choice of H . Thus, Ghas no normal subgroups of order pp and exponent p so, by Lemma 88.7, G is ofmaximal class. Since G has a normal subgroup of order pp�1 and exponent p, we getj�1.H/j D pp�1. It is worth while to note that any normal subgroup K of index > pinG is contained inˆ.G/ so absolutely regular (Theorem 9.6); thenK � ˆ.G/ < G1.Since G has no Hp-chains, we get jGj > p2p�1 (Remark 3).Let H be absolutely regular. Then jG W H j D p (otherwise, H < G1 and there is

anHp-chain inG1/. SinceG1 is the unique regular maximal subgroup ofG (Theorem9.6), we get H D G1. Suppose, in addition, that H0 2 M0 � fH g. In that case, H0

is irregular of maximal class and index p in G. Since H0 has an Hp-chain, we getjH j � p2p�1 so we have jH0j D p2p�1 since jG W H0j D p and jGj > p2p�1. Inthat case, the last index of the Hp-chain of H0 equals pp so it is not equal to everyindex of the Hp-chain of the absolutely regular group H , and this is a contradiction.Thus, we have M0 D fHg, completing the proof of (a).Now suppose thatH is irregular of maximal class. Then, as we have noticed already,

jG W H j D p. By Remark 3, since jGj � p2p , we get jH j D p2p�1. Since G has nonormal subgroups of order pp and exponent pp , we get j�1.H/j D pp�1. Since theHp-chain of G1 has no indices D pp, we get G1 62 M0, completing the proof of (b)and thereby the proposition.

Proposition 88.14. Suppose that a p-group G has no Hp-chains. Let H 2 M0 andlet

C D CH W f1g D L0 < L1 < < Ln D H

be an Hp-chain in H with i0 D i0.C/ > 0 (the case i0 D 0 is considered in theprevious proposition). Write U D Li0 and NG D G=U . Then jC j > i0 and one of thefollowing holds:

(a) NG is absolutely regular and exp. NG/ > p. In that case, if jCH j > i0 C 1, thenexp.Li0C1/ D pi0C1, Li0C1 < �i0C1.G/ and Li0C2=Li0C1 < �1.G=Li0C1/.


(b) NG is irregular of maximal class. Then NH is either absolutely regular or irregularof maximal class.

(b1) If j NGj D ppC1, then all maximal subgroups of NG are absolutely regularso j�1. NG/j D pp�1. In that case, Li0 < �i0.G/.

(b2) Let NH be irregular of maximal class. Then jG W H j D p, j�1. NH/ D

pp�1, j NGj � p2p and jCH j D i0 C 2.

Proof. By hypothesis, H < G. Write i0 D i0.C/ and U D Li0 . As in the proof ofTheorem 88.10, NG D G=U has no normal subgroups of order pp and exponent p so itis either absolutely regular or irregular of maximal class (Lemma 88.7). Assume thatjC j D i0; then U D H . Take in NG a normal subgroup NF of order p. Then there is inF anHk-chain f1g D L0 < L1 < < Ln D H < F so H 62 M0, a contradiction.Thus, jC j > i0.

(a) Suppose that NG D G=U is absolutely regular; then exp. NG/ > p (otherwise,G has an Hp-chain f1g D L1 < L1 < < Li0 D U < G). Since Li0C1

has an Hp-chain of length i0 C 1 and G=Li0C1 is absolutely regular, it follows thatLi0C1 < �i0C1.G/ (otherwise, by Lemma 88.2(b), G has an Hp-chain). Next, sup-pose that jCH j > i0 C 1. Consider the subgroup W D Li0C2. We have �i0C1.H/ D

Li0C1 since C D CH is an Hp-chain, and so exp.Li0C1/ D pi0C1. Assume thatW=Li0C1 D �1.G=Li0C1/. Let x 2 G � Li0C1 be of minimal order; then, bywhat has been said already, o.x/ � pi0C1 and xp 2 Li0C1. Then, by assumption,x 2 W D Li0C2. However, x 2 �i0C1.W / D Li0C1, contrary to the choice of x.Thus,W=Li0C1 < �1.G=Li0C1/, completing this case.

(b) Suppose that NG D G=U is irregular of maximal class.

(b1) Let j NGj D ppC1. If NH1 < NG is of order pp and exponent p, then f1g D L0 <

L1 < < Li0 < H1 < G is anHp-chain in G (all indices of that chain, apart of thelast one, equal pp , the last index equals p), so, comparing indices of that chain withindices of the chain C , we getH 62 M0, a contradiction. Thus, all maximal subgroupsof NG are absolutely regular (Lemma 88.7) so �1. NG/ D ˆ. NG/ is of order pp�1 andexponent p. Since G has no Hp-chains, it follows that Li0 < �i0.G/ (otherwise,G has an Hp-chain, whose .i0 C 1/-th member coincides with the inverse image of�1. NG/ in G, and the following member is G), completing the proof of (b1).

(b2) Now let NH be irregular of maximal class; then jG W H j D j NG W NH j D p

(Theorem 9.6). Since NH has no G-invariant subgroups of order pp and exponent p,we get j�1. NH/j D pp�1 since NH has an Hp-chain. All remaining assertions followfrom Proposition 88.13.

Metacyclic p-groups, p > 2, are regular so they have exactly one H2-chain.

Proposition 88.15. The following conditions for a metacyclic 2-groupG of order� 24

are equivalent:

(a) G has no H2-chains.


(b) Either G is of maximal class or there is k > 0 such that j�i.G/j D 22i for alli � k and G=�k.G/ is of maximal class and order � 24.

Proof. Let the set M0 be such as in Propositions 13 and 14. Take H 2 M0 and setC D CH . Set i0 D i0.C/. Then U D Li0 D �i0.G/ so G=U is of maximal class,by the above and Lemma 88.2(b). If G=U Š Q8, then f1g < �1.G/ < < U <

�t0C1.G/ < G is an H2-chain in G, a contradiction. Now assume that G=U Š D8.Let F=U < G=U be abelian of type .2; 2/. Then f1g < �1.G/ < < U < F < H

is an H2-chain in G, a contradiction. Thus, jG=U j � 24. Clearly, j�i.G/j D 22i forall i � i0.It remains to show that G has no H2-chains. Assume that C is an H2-chain in G

as in Definition 1. Since G=U is of maximal class, we get Li0C2 D G. We havejLi0C1=Li0 j D 2 so �i0C1.G/ D Li0C1. It follows that G=Li0 is a generalizedquaternion group. Then jLi0C2j � 22kC3 < 22kC4 � G, a contradiction sinceG D �i0C2.G/ D Li0C2.

Supplement to Proposition 88.15. Let G be a metacyclic 2-group with Ch2.G/ > 1.Then there is k such that j�i.G/j D 22i for all i � k and G=�k.G/ is dihedral oforder 8 (in the last case, Ch2.G/ D 2).

This follows easily from the proof of Proposition 88.15.

Problems

BelowH > f1g is a normal subgroup of a p-group G.

Problem 1. Let p > 2. Study the structure of H if there exists only one Hp�1-chaininH .

Problem 2. Is it true that the number ofHk-chains in any abelian p-group is congruentwith 1 .mod p/?

Problem 3. Find an algorithm producing allHk-chains in abelian p-groups.

Problem 4. Classify the 2-groups which have noH2-chains.

Problem 5. Given a natural number k, a chain C0 W H D H0 > H1 > > Hn D

f1g of G-invariant subgroups is said to be a lower k-admissible chain in H providedHi�1=Hi is of order � pk and exponent p for i D 1; : : : ; n. The above chain is saidto be a lower Hk-chain in G if, whenever jH=Hi j < pki , then Hi D Ãi.H/. (i)Is it true that, whenever H is a lower pyramidal (see Remark 1), it possesses a lowerHp-chain? (ii) Study the p-groups without lowerHp-chains.

Problem 6. Suppose that H E G is a P2-subgroup such that all sections of H arepyramidal. Is it true that there exists in H anHk-chain for any k?


Problem 7. Does there exist in H anHp-chain if all sections ofH of order ppC1 areP2-groups?

Problem 8. Suppose that p-groups G and G0 are lattice isomorphic and G has anHp-chain. Is it true that also G0 has anHp-chain?

Problem 9. Suppose that a p-group H has an Hk-chain. Now let H G G, where Gis a p-group. Find sufficient conditions for existing an Hk-chain in H (as a normalsubgroup in G).

�89

2-groups with exactlysix cyclic subgroups of order 4

If c2.G/ D 6, then such 2-groups G have been determined only in the special casewhere j��

2.G/j D 24, where ��2.G/ D hx 2 G j o.x/ D 4i. Such 2-groups G with

jGj > 24 are determined in Theorem 52.1 when j�2.G/j D 24 (since in that case�2.G/ Š Q8 � C2 or �2.G/ Š C4 � C4) and in Theorem 55.1 when j�2.G/j >

24. Here we shall classify 2-groups G with c2.G/ D 6 and j��2.G/j > 24. First

we show that we must have j��2.G/j D 25 and we get three possibilities for the

structure of ��2.G/ (Lemma 89.6). The corresponding 2-groups G are determined up

to isomorphism in Theorem 89.7. This solves #425 for p D 2. The general case, wherec2.G/ 2 .mod 4/ and c2.G/ � 10 is very difficult and is still open. Note that, forn > 2, the 2-groups G with cn.G/ Š 2 .mod 4/ are classified in Corollary 18.7.At the end we consider 2-groups G which possess only one conjugate class of cyclic

subgroups of order 4 and we show that in that case G has only one cyclic subgroup oforder 4 and therefore G is either cyclic or dihedral (Theorem 89.8). This solves a partof #1379.In what follows G will denote a 2-group with c2.G/ D 6 and H D ��

2.G/ is oforder > 24. Since H has exactly six cyclic subgroups of order 4, H is neither cyclicnor a group of maximal class. It follows that H has a G-invariant four-subgroup W(Lemma 1.4).

Lemma 89.1. If a cyclic subgroup V of order 4 in G normalizes another cyclic sub-group U of order 4, then U normalizes V and either UV Š C4 � C2 or UV Š Q8.

Proof. First suppose U \ V D f1g. Then jUV j D 24 and .UV /0 < U and we haveeither UV D U � V Š C4 � C4 or .UV /0 Š C2 in which case UV is a metacyclicminimal nonabelian group of order 24 and exponent 4. In any case, c2.UV / D 6 andsoUV D H D ��

2.G/, contrary to our assumption that jH j > 24. Thus, U \V Š C2

and so jUV j D 23. In this case, U also normalizes V and the only possibilities areUV Š C4 � C2 or UV Š Q8.

Lemma 89.2. Suppose that H D ��2.G/ contains a quaternion subgroup Q Š Q8.

Then jH j D 25 and we have the following two possibilities: (a) H Š Q8 � Q8, (b)H Š Q16 � C4.

�89 2-groups with exactly six cyclic subgroups of order 4 455

Proof. First we determine the structure of S D WQ, where W is a normal four-subgroup in H and jW \ Qj � 2. Let z be an involution in W \ Z.S/. If z 62 Q,then c2.Q � hzi/ D 6 and thereforeQ � hzi D ��

2.G/ D H , a contradiction. HenceQ \W D hzi Š C2, jS j D 24, and ŒW;Q� D hzi. This gives S D Q � hvi, wherehvi Š C4 and hvi \ Q D hzi. We have c2.S/ D 4, hvi D Z.S/ and Q is a uniquequaternion subgroup of S soQ is characteristic in S (see Appendix 16).Assume that S 6E H and setK D NH .S/ so that jK W S j � 2 and jH W Kj � 2. Let

K < M < H be such that jM W Kj D 2 and take m 2 M �K. Since Qm ¤ Q, wegetQm 6� S , andQm � K. We have jQm \ S j � 4 and soQm � S contains at leasttwo cyclic subgroup of order 4. It follows that c2.K/ D 6. But ��

2.G/ D H > K, acontradiction. We have proved that S GH and soQ and Z.S/ D hvi are also normal inH . Since c2.S/ D 4, we have exactly two cyclic subgroups of order 4, not containedin S . Set C D CH .Q/ so that C GH and jH W .QC/j � 2 (since Aut.Q/ Š S4).If there is an involution u in C � hvi, then c2.Q � hui/ D 6 and Q � hui D H ,

a contradiction. Hence C is either generalized quaternion or cyclic. In the first case(sinceC�hvi can contain at most four elements of order 4), C Š Q8,QC Š Q8�Q8,c2.QC/ D 6 and thereforeQC D H is an extraspecial group of order 25 (case (a)).We may assume that C is cyclic so that jH W .QC/j D 2 because H � S must

contain exactly four elements of order 4. Since H=C Š D8, there is x 2 H � .QC/

such that x2 2 C and x induces an involutory outer automorphism on Q. There areelements a; b 2 Q such that ha; bi D Q, ax D a�1 and bx D ab.Suppose that hx;C i is cyclic so that hx;C i D hxi and o.x/ � 2jC j � 8. If

o.x/ � 16, then there are no elements of order 4 inH � .QC/, a contradiction. Henceo.x/ D 8 so that we may assume that x2 D v. In that case, T D hx;Qi is of maximalclass (Theorem 1.2) so x does no normalizes hai since hai is not normal in T D ha; xi.Hence hx;C i is noncyclic.Assume that hx;C i is abelian or hx;C i Š M2m , m � 4, so that in both cases we

may assume that x is an involution centralizing hvi. We have o.xv/ D o.xva/ D 4

and we see that c2.Shxi/ D 6 and so H D Shxi D hQ;xvi � hvi, where hQ;xvi Š

Q16 (see Proposition 10.17 and Theorem 1.2) andH Š Q16 � C4 (case (b)).Assume that hx;C i Š Q2n or hx;C i Š SD2n . In both cases we may assume that

x2 D z and hv; xi Š Q8 since Q8 is a subgroup of Q2n and SD2n and Q8 contains thesubgroup hvi of C . But then x inverts hvi and hai (see above) and so all eight elementsin ha; vix from H � .QC/ are of order 4, a contradiction. Indeed, we compute forany integers i , j :

.aivjx/2 D aivjxaivjx D aivjx2.aivj /x D aivj za�iv�j D z:

Finally, suppose that hx;C i Š D2n , n � 3, where x is an involution. But then allelements in ha;C ix from H � .QC/ are involutions since x inverts hai and C andall other elements in .QC � ha;C i/x from H � .QC/ are elements of order 8, acontradiction (sinceH � S does not contain any elements of order 4). Indeed, we setC D hci and we know that bx D ab so that for any integers i , j we compute (noting


that inQ we have bai D aibzi and bab D a):

.baicjx/2 D baicj .baicj /x D baicj aba�ic�j D aibzicjaba�ic�j

D aizi .bab/cja�ic�j D aizi a cja�ic�j D zia;

which is an element of order 4. Hence, all elements baicjx are of order 8, as claimed.Our lemma is proved.

In the next three lemmas we assume, in addition, that Q8 is not a subgroup of H .We turn out to prove that jH j D 25.

Lemma 89.3. Assuming that Q8 is not a subgroup of H , we have jH W NH .X/j � 2

for each cyclic subgroup X of order 4 inH .

Proof. Suppose that the lemma is false. Then there is a cyclic subgroup U1 of order 4in H such that K D NH .U1/ is of index 4 in H (take into account that c2.H/ D 6).LetM be a maximal subgroup of H containing K so that jH W M j D jM W Kj D 2,and let m 2 M � K. Then U2 D Um

1 ¤ U1, NH .U2/ D K and so A D U1U2 Š

C4 �C2 (Lemma 89.1) and A 6 GH since c2.A/ D 2. Let x 2 H �M so that Ax ¤ A

and Ax < M . We have c2.M/ 2 f3; 4; 5g becauseM < H . If c2.M/ is odd, thenMis of maximal class (Theorem 1.17), a contradiction sinceM has an abelian subgroupof type .4; 2/). Hence c2.M/ D 4 since c2.M/ > C2.A/ D 2.Suppose that jM j > 24. If j�2.M/j > 24, then (see [�53, Introduction]) U1 GM >

K D NH .U1/, a contradiction. Hence we must have j�2.M/j D 24. In that casewe may use Theorems 52.2, 52.4, and 52.5 since Q8 is not a subgroup of M andc2.�2.M// D 4. This implies that �2.M/ is abelian of type .4; 2; 2/ and there isa cyclic subgroup of order 4 which is normal in M . This is a contradiction since�2.M/ D AAx and so all four cyclic subgroups of order 4 inM are conjugate in Hand so no one of them could be normal inM .We have proved that jM j D 24 so that K D A D NH .U1/, AAx D M is of order

24 and jH j D 25. In this case A and Ax are two distinct abelian maximal subgroupsof M which implies jZ.M/j D 4, jM 0j D 2, cl.M/ D 2 and M is of exponent 4.Suppose that M is not minimal nonabelian. ThenM possesses a subgroup D Š D8

(H has no subgroups isomorphic to Q8) and, sinceM is not of maximal class, we haveCM .D/ 6� D (see Proposition 10.17). Since c2.M/ D 4, we get M D D � C withC Š C4 and D \ C D Z.D/. But D8 � C4 Š Q8 � C4, contrary to our assumption.Hence M is minimal nonabelian. If M is metacyclic, then M has a cyclic normalsubgroup of order 4, contrary to the fact that all four cyclic subgroups of order 4 inM are conjugate in H . Hence M is a uniquely determined nonmetacyclic minimalnonabelian group of order 24 and exponent 4 (see Lemma 65.1).Since NH .X/ < M for each cyclic subgroup X of order 4 in M , there are no

elements of order 8 in H � M . It follows that H � M consists of four elements oforder 4 and 12 involutions. Set E D �1.M/ so that E is elementary abelian of order


8, Z.M/ D ˆ.M/ Š E4 and Z.H/ � Z.M/. Let v 2 H �M be of order 4 such thatv2 2 E and CE .v/ Š E4 sinceH �M contains exactly four elements of order 4 andthey are all contained in .Ehvi/�E (here we used Proposition 1.8). All eight elementsinH � .M [Ehvi/ are involutions and so if u 2 H � .M [Ehvi/, then u centralizesE and so F D E � hui Š E16 (take into account that two noncommuting involutiongenerate a dihedral group). In particular, Z.M/ < E < F and so Z.M/ D Z.H/. Lety 2 M �E and we know that all cyclic subgroups of order 4 inM are conjugate inHto hyi. But y2 2 ˆ.M/ D Z.H/ and soÃ1.M/ D hy2i, contrary toˆ.M/ Š E4.

Lemma 89.4. Assuming that Q8 is not a subgroup of H , we have jH W NH .X/j D 2

for each cyclic subgroup X of order 4 inH .

Proof. Suppose that the lemma is false. Then there are at least two distinct cyclicsubgroups U1 and U2 which are normal inH (Lemma 89.3). Let fU1; U2; : : : ; U6g bethe set of six cyclic subgroups of H . Since each Ui , i D 1; 2; : : : ; 6, normalizes U1

and U2, it follows (Lemma 89.1) that A D hU1; U2i Š C4 � C2, UiUj is abelian oftype .4; 2/, i D 1; 2, j > 2, and so A � Z.H/. Therefore, for each Uj , j D 3; : : : ; 6,we have U1 \Uj D U2 \Uj D U1 \U2 D hzi D Ã1.A/. It follows that B D AU3 D

hU1; U2; U3i is abelian of order 24 and exponent 4 with Ã1.B/ D hzi and so B isabelian of type .4; 2; 2/. Since c2.B/ D 4, we nay assume that fU1; U2; U3; U4g is theset of cyclic subgroups of order 4 in B . Similarly, C D AU5 is abelian of type .4; 2; 2/withÃ1.C / D hzi so that fU1; U2; U5; U6g is the set of cyclic subgroups of order 4 inC . We have B \ C D A andH D hB;C i. Thus, H=A is generated with two distinctcyclic subgroups B=A and C=A of order 2, and soH=A Š E4 orH=A Š D2n , n � 3.In particular, B and C are not conjugate inH . Let t be an involution inH � .B [C/

and let v be an element of order 4 in A � Z.H/. Then tv is an element of order 4 inH � .B [ C/, a contradiction. Hence, all elements inH � .B [C/ are of order � 8.This implies that B and C are normal in H and soH D hB;C i D BC is of order 25

with two distinct abelian maximal subgroups B and C . It follows that jH 0j � 2 andso H is of class � 2. But H is generated by its elements of order 4 and so H is ofexponent 4, a contradiction.

Lemma 89.5. Assuming that Q8 is not a subgroup ofH , we have c2.NH .X// D 2 foreach cyclic subgroup X of order 4 inH .

Proof. Let U1 be a cyclic subgroup of order 4 in H so that jH W NH .U1/j D 2

(Lemma 89.4). Set M D NH .U1/ and, taking an element h 2 H � M , we getU2 D U h

1 ¤ U1, NH .U2/ D M , A D hU1; U2i Š C4 � C2 (Lemma 89.1), andA G H . Assume that M has a further cyclic subgroup U3 6� A of order 4 so thathU1; U3i Š hU2; U3i Š C4 � C2, and therefore B D hU1; U2; U3i is abelian of type.4; 2; 2/. Since c2.B/ D 4, we may assume that fU1; U2; U3; U4g is the set of allcyclic subgroups of order 4 in B . There is an element g of order 4 inH �M and sincejH W NH .hgi/j D 2, U5 D hgi and U6 D hgxi (with an x 2 H � NH .hgi/) give two


last cyclic subgroups of order 4 in H which give exactly four elements of order 4 inH �M . This implies that B D ��

2.M/ GH and U3 and U4 are conjugate inH .Set H0 D BU5. Since H0 is not of maximal class, we get c2.H0/ D 6 (Theorem

1.17(b)), and so H0 D BU5 D H . Set B0 D �1.B/ Š E8. Suppose that B \

U5 D f1g so that jH W Bj D 4. Since jH W NH .U5/j D 2, U5 centralizes a four-subgroup S in B0. But then all eight elements of order 4 in S � U5 lie in H � B , acontradiction. Hence B \ U5 Š C2 and so jH W Bj D 2, jH j D 25, B D M , andNH .U3/ D NH .U4/ D B . This implies that there are no elements of order 8 inH �B

and soH � B consists of four elements of order 4 and twelve involutions.We have jB W NB.hgi/j D 2, where hgi D U5 and NB.hgi/ cannot contain an

element x of order 4 (otherwise, that element x would centralize U5, contrary to thefact that CH .x/ D B). Hence NB.hgi/ D B0. If g centralizes B0, then there are eightelements of order 4 inH�B , a contradiction. Hence CB.g/ D CB0

.g/ D Z Š E4 andso Z.H/ D Z. The set B0g consists of four elements of order 4 and four involutions.Hence all eight elements in H � .B [ B0hgi/ are involutions and if t is one of them,then H � B D B0g [ B0t and B0g \ B0t D ¿ so that t must centralize B0 andtherefore B0 � Z.H/, contrary to the fact that Z.H/ D Z Š E4.

Lemma 89.6. Let G be a 2-group with exactly six cyclic subgroups of order 4 and letH D ��

2.G/ D hx 2 G j o.x/ D 4i be of order > 24. Then H is of order 25 and wehave the following three possibilities:

(a) H Š Q8 � Q8 is extraspecial;

(b) H Š Q16 �C4 with Q16 \C4 D Z.Q16/ (by Exercise A in Appendix 16, indeedc2.H/ D 6);

(c) H is a special group possessing a unique elementary abelian subgroup E oforder 24 and there is an involution t 2 H � E such that H D hE; ti andCE .t/ D Z.H/ Š E4.

Proof. In view of Lemma 89.2, we may assume that Q8 is not a subgroup ofH and sowe may use Lemmas 89.1, 89.4, and 89.5. Let U1 be a cyclic subgroup of order 4 inH . Set K D NH .U1/ so that jH W Kj D 2 and if h 2 H �K, then U2 D U h

1 ¤ U1,A D hU1; U2i D ��

2.K/ Š C4 � C2 is normal in H , NH .U2/ D K and so no one ofU1, U2 is characteristic in K. Note that jH j > 24 and so jKj > 23.We are in a position to use Proposition 53.2 which gives that K is a uniquely de-

termined group of order 25 or 24. We may assume that we have the following con-jugacy classes of our six cyclic subgroups of order 4 in H : U1 U2, U3 U4,and U5 U6. Assume that jKj D 25 in which case jH j D 26. It follows thatˆ.NH .U1// D hU1; U2i and similarly (since jH W NH .U3/j D jH W NH .U5/j D 2),NH .U3/ Š NH .U5/ Š K. But then ˆ.NH .U3// D hU3; U4i, ˆ.NH .U5// D

hU5; U6i and therefore ˆ.H/ � hU1; U2; U3; U4i D H , a contradiction.We have proved that K D NH .U1/ D NH .U2/ is of order 24 and then K Š

D8 � C2 (Proposition 53.2(b)) and jH j D 25. The subgroup K has exactly three


abelian maximal subgroups: F1 Š E8, F2 Š E8, and A D hU1; U2i Š C4 � C2,where F1 \ F2 D F0 D Z.K/ Š E4. There are no elements of order 8 in H � K

since NH .U1/ D NH .U2/ D K and so H � K consists of eight elements of order 4and eight involutions. Let r be an involution in H �K. Then the coset rK D H �K

has exactly eight involutions, and let rk be one of them. Then .rk/2 D 1 or, what isthe same, kr D k�1 so o.k/ � 2 since r does not normalize any cyclic subgroup oforder 4 in K, and we get k 2 CK.r/. It follows that jCK.r/j D 8 so CH .r/ Š E16.Since Z.K/ D F1 \ F2, it follows that Z.K/ D Z.H/. By Lemma 1.4, jH 0j D 4.Next, H=A Š E4 (if not and x 2 H � A is of order 4, then CH .x/ is abelian of type.4; 2; 2/, which is impossible since CH .r/ is the unique abelian maximal subgroup ofH in view jH 0j D 4). Thus, Z.H/ D �1.A/ and, sinceH is not minimal nonabelian,we getH=Z.H/ Š E8. Thus, H is special so it is the group given in (c).

Theorem 89.7 (Janko). Let G be a 2-group with exactly six cyclic subgroups of order4 and let H D ��

2.G/ be of order > 24. Then H is of order 25 and we have threepossibilities for the structure of H (Lemma 89.6). However, if G > H , then H Š

Q16 � C4, jG W H j D 2, jGj D 26, and we have the following two possibilities:

(i) G has a dihedral subgroup D D hf; � jf 16 D �2 D 1; f � D f �1i Š D32 ofindex 2 and an involution u 2 G �D so that Œu; �� D 1 and f u D f z, z D f 8.

(ii) G D ha; t j a16 D t2 D 1; a8 D z; a4 D v; at D a�1vu; u2 D 1; Œu; a� D

1; ut D uzi, where G is a U2-group with respect to U D hu; zi Š E4, G=U Š

SD16 and Z.G/ D huvi Š C4.

Proof. For the structure of H D ��2.G/ we use Lemma 89.6. We assume in addition

that G > H . If H Š Q8 � Q8 is extraspecial (Lemma 89.6(a)), then we have acontradiction, by Theorem 83.2.Suppose that H is a special group given in Lemma 89.6(c). Let H < L � G be

such that jL W H j D 2. Let E be a unique elementary abelian subgroup of order 16 inH so thatE GL. Let j 2 H �E be an involution so that CE .j / D E0 D Z.H/ Š E4

(see the proof of Lemma 89.6); then E0 G L and F D E0 � hj i D CH .j / Š E8 isnormal in L since all 12 elements in H � .E [ F / are of order 4 and E G L. Fourinvolutions in F � E0 form a single conjugate class in H and an L-invariant set soI D CL.j / covers L=H and I \ H D F hence I Š E16 since I is abelian, andE0 � Z.L/. Let i 2 I � F and consider the subgroup J D Ehii of order 25; thenJ\H D E. All 16 elements in J�E D J�H must be involutions since exp.J / � 4,and so J Š E32. We get CH .i/ � hE;F i D H . If v is an element of order 4 in H ,then o.vi/ D 4 and vi 62 H , a contradiction.Thus, by Lemma 89.6, we must have H Š Q16 � C4. Let H D Q � C , where

Q D hb; t j b8 D 1; t2 D b4 D z; bt D b�1i Š Q16, C D hvi Š C4, v2 D z, andQ \ C D hzi. SinceQ is generated by all (five) noncentral cyclic subgroups of order4 in H , we get Q G G. Set D D CG.Q/ so that D G G, D � C and D \ H D C .If there is an involution i 2 D � C , then o.b2i / D 4 and b2i 62 H , a contradiction.


Hence z is a unique involution inD. Since c2.D/ D 1,D is cyclic (Theorem 1.17(b)).Let d 2 D � C be an element of order 8. Then b4 D d4 D z and so o.bd/ D 4 withbd 62 H , a contradiction. We have proved that D D C D CG.Q/.The group Aut.Q/ is generated by Inn.Q/ Š D8 and two involutory outer automor-

phisms ˛ and ˇ induced by t˛ D tb, b˛ D b�1, tˇ D t , bˇ D bz, where Œ˛; ˇ� D ib2

(the inner automorphism of Q induced by conjugation with the element b2) and soAut.Q/=Inn.Q/ Š E4 (and in fact h˛; ˇi Š D8); see Theorem 34.8. The subgroupQ contains exactly two quaternion subgroups Q1 and Q2 and we have Q

ˇ1 D Q1,

Qˇ2 D Q2, andQ˛

1 D Q2. It follows that G=H ¤ f1g is elementary abelian of order� 4 (here we use the N/C-Theorem).Assume that L D NG.Q1/ > H so that jL W H j D 2. Since Q=hzi Š D8 is

isomorphic to a Sylow 2-subgroup of Aut.Q1/ Š S4, it follows that C0 D CL.Q1/ >

C and jC0 W C j D 2. If y 2 C0 � C is an involution, then o.b2y/ D 4 (becauseb2 2 Q1) and b2y 62 H , a contradiction. Since c2.C0/ D 1, we get that C0 D hci Š

C8 is cyclic with c4 D z. Now, c normalizes hbi (since Q G G and hbi is a uniquecyclic subgroup of index 2 in Q) and centralizes hb2i D hbi \ Q1, but c does notcentralize hbi (otherwise, c would centralize hb;Q1i D Q, a contradiction) and sowe get bc D bz, hb; ci0 D hzi, cl.hb; ci/ D 2, .bc/4 D b4c4Œc; b�6 D zzz6 D 1,o.bc/ D 4 (Theorem 1.2), and bc 62 H , a contradiction.We have proved that jG=H j D 2, jGj D 26, and if g 2 G �H , thenQg

1 D Q2. Inparticular, CG.t/ D ht; vi Š C4�C2 and so eight elements of order 4 inQ�hbi form asingle conjugate class inG. Set T D hbihvi Š C8�C2 which is normal inG and eightelements inH � .Q[T / are involutions which form a single conjugate class inG andso if tv is one of them, then CG.tv/ D ht; vi. In particular, if x 2 G�H , then x2 2 T .We have U D �1.T / D hz; b2vi Š E4 is normal in G, �2.T / D hb2; vi Š C4 �C2,and hb2i and hvi are normal in G.Suppose that there is an involution � 2 G �H . Then � inverts hvi and hb2i (oth-

erwise, � centralizes v or b2 and then �v or �b2 would be an element of order 4 inG �H , a contradiction). If b� D b�1z, then .�b/2 D b�b D z and o.�b/ D 4 with�b 62 H , a contradiction. Hence � inverts each element in T and so, in particular, �centralizes U . Since Q�

1 D Q2, we have t� D tbi , where i is odd. Set b2v D u and�t D f so that � centralizes the involution u,

f 2 D �t�t D t� t D .bi /t D b�i ; o.f / D 16; f 8 D .b�i /4 D z;

f � D .�t/� D �t� D �tbi D f bi D f �1.f 2bi / D f �1b�ibi D f �1;

hf; �i Š D32

and

f u D .�t/b2v D v�1b�2�tb2v D �.v�1b�2/� tb2v D �vb2tb2v

D .�t/.vb2/tb2v D .�t/vb�2b2v D �tv2 D f z:

We have obtained the group given in part (i) of our theorem.


It remains to investigate the case, where there are no involutions in G � H . Then32 elements in G �H are of order 8 or 16. If all 32 elements in G �H are of order8, then c3.G/ D 10 and therefore G is a U2-group (see Corollary 18.7). But then Gmust also have elements of order 16 which is not the case. If all 32 elements in G�H

are of order 16, then c4.G/ D 4 and c3.G/ D 2. Again, G is a U2-group. But a U2-group of order 26 has exactly two cyclic subgroups of order 16, a contradiction. HenceG � H contains elements of order 8 and 16. Since the number of cyclic subgroupsof order 16 must be even (otherwise, G would be of maximal class), it follows thatG � H has exactly 16 elements of order 16 (and so c4.G/ D 2) and exactly 16elements of order 8. Hence c3.G/ D 6 and so G is a U2-group with respect to Usince in a U2-group a normal four-subgroup is unique. If R=U is a cyclic subgroupof index 2 in G=U , then G � R contains exactly eight involutions, eight elements oforder 4, and 16 elements of order 8. Hence G=U Š SD16 and ˆ.R/ Š C8. SinceH is nonmetacyclic, G is also nonmetacyclic. We have ˆ.G/ � T and so there areexactly three maximal subgroups of G containing T . They are H , R and a certainsubgroup V with the property that all 16 elements in V � T are of order 8. Since�2.V / D �2.T / D hb2; vi Š C4 � C2, jV j D 25, and V has no elements of order16, V must be isomorphic to a group (c) given in Lemma 42.1 and so ˆ.V / D �2.V /

and Z.V / Š C4. We get ˆ.G/ � hˆ.R/;ˆ.V /i D T and so G is 2-generated, i.e.,d.G/ D 2. Also, Z.V / Š C4 implies that U 6� Z.V / and so CG.U / D R (becauseCG.U / must be a maximal subgroup of G containing T and also U 6� Z.H/). Sinceˆ.T / D hb2i and ˆ.V / D hb2; vi (and no involution in ˆ.V / � hzi could be asquare of an element in V � T because U 62 Z.V /), there is an element s 2 V � T

such that s2 D v. Hence, CG.v/ � hH; si D G and so Z.G/ Š C4. We haveobtained a nonmetacyclic U2-group G of order 26 with respect to U Š E4 such thatG=U Š SD16, d.G/ D 2, and Z.G/ Š C4. It follows that G must be isomorphic to aU2-group given in Theorem 67.3(c). We have obtained the group given in part (ii) ofour theorem.

Note that the group H of Lemma 89.6(c) has exactly 16 elementary abelian sub-groups of order 8 so there is in H an odd number of metacyclic subgroups of order 8(Sylow).

Theorem 89.8 (Janko). Let G be a 2-group of exponent > 2 all of whose cyclic sub-groups of order 4 are conjugate. Then G has exactly one cyclic subgroup of order 4and G is either cyclic or dihedral.

Proof. First suppose that G has more than one cyclic subgroup of order 4. Let Ube one of them and set K D NG.U / so that jG W Kj � 2 and let M be a maximalsubgroup of G containing K. Then each cyclic subgroup of order 4 is contained inMand if X is one of them, then NG.X/ � M (since X is conjugate in G to U ). Let x beany element in G �M . We know that x is not of order 4 and suppose that o.x/ � 8.But then x2 2 M and o.x2/ � 4 and so x centralizes a cyclic subgroup of order 4 in


M , a contradiction. Hence each element x in G �M is an involution and soM mustbe abelian and x invertsM . But then U is normal in G, a contradiction.We have proved that G has a unique cyclic subgroup V D hvi. Then, by Theorem

1.17(b), G is either cyclic or dihedral.

�90

Nonabelian 2-groups all of whoseminimal nonabelian subgroups are of order 8

In this section we determine the structure of the title groups and show that this class ofgroups coincides with the class of nonabelian 2-groups in which any two noncommut-ing elements generate a group of maximal class (Corollary 90.2). This solves Problem920 for p D 2.

Theorem 90.1 (Janko). Let G be a nonabelian 2-group all of whose minimal non-abelian subgroups are isomorphic toD8 orQ8. ThenG is one of the following groups:

(a) G is generalized dihedral (i.e., jG W H2.G/j D 2).

(b) G D HZ.G/, whereH is of maximal class and Ã1.Z.G// � Z.H/.

(c) G D HZ.G/, whereH is extraspecial and Ã1.Z.G// � Z.H/.

Proof. Let A be a maximal normal abelian subgroup of G. We see at once that G=Ais elementary abelian. Indeed, let x 2 G � A with o.x/ D 4 and A \ hxi D f1g.According to Lemma 57.1, let a 2 A be such that Œa; x� ¤ 1 and F D ha; xi beminimal nonabelian. Since jF j D 8, we have F \ A D hai is of order 2 and soŒa; x� D 1, a contradiction. We shall also use induction on jGj.

(i) First we deal with the case exp.G/ D 4. In this case we show that each cyclicsubgroup of order 4 is normal inG. Suppose that this is false. LetX D hxi be a cyclicsubgroup of order 4 which is not normal in G. Set M D NG.X/ so that M ¤ G.Let M0 > M be such that jM0 W M j D 2 and assume M0 ¤ G. By induction, thenonabelian groupM0 is a group of our theorem with exponent 4 and soX is normal inM0, a contradiction. Thus jG W M j D 2 and with the same argument we see thatM is aunique maximal subgroup ofG containingX . In particular, d.G/ D 2,X 6� ˆ.G/ andhx; yi D G for each y 2 G �M . Let y be a fixed element in G �M so that y2 2 M ,o.y2/ � 2, y2 normalizes X and set x2 D z. We have xy 62 hxi, NG.hx

yi/ D M

and A D hx; xyi D hxihxyi is normal in G since .xy/y D xy2

2 hxi. Since Ahyi

contains X , we have, by the above. G D Ahyi and M D Ahy2i. If Œx; xy � D 1,then A is abelian of type .4; 2/ or .4; 4/. If Œx; xy � ¤ 1, then Œx; xy � 2 hxi \ hxyi

and so Œx; xy � is a central involution in A which implies that A is minimal nonabelian.But A contains two distinct cyclic subgroups of order 4 and so in this case A Š Q8

is quaternion.


First suppose that y2 62 A so that o.y/ D 4, G=A Š C4 and M D Ahy2i ¤ A.Since hxi is normal inM , y2 either centralizes or inverts hxi. Because yx 2 G �M

and exp.G/ D 4, .yx/2 D yxyx D y2.xyx/ is an involution in M � A. Hencey2.xyx/y2.xyx/ D 1 and so .xyx/y

2

D .xyx/�1. First consider the possibilityhx; xyi D A Š Q8 so that hxi \ hxyi D hzi D Z.A/ and hxi, hxyi and hxyxi arethe three cyclic subgroups of order 4 in A. If y2 inverts hxi, then y2 also inverts hxyi.By the above, y2 inverts hxyxi and so y2 inverts A and therefore A must be abelian, acontradiction. Hence, y2 centralizes hxi and hxyi and so y2 centralizesA, contrary tothe above fact that y2 inverts hxyxi. We have proved that A is abelian. Assume thaty2 inverts hxi and then y2 also inverts hxyi which implies that y2 inverts A and soM D Ahy2i is generalized dihedral. Hence A is a maximal normal abelian subgroupin G (since CG.A/ D A), a contradiction because G=A is not elementary abelian. Wehave proved that y2 centralizes hxi and hxyi and so M D A � hy2i is abelian. Bythe above, y2 also inverts xyx and so u D xyx must be an involution, A is abelian oftype .4; 2/, jGj D 25, hz; ui D �1.A/ Š E4, and hxi \ hxyi D hzi D Ã1.A/. Since

uy D .xyx/y D xy2

xy D xxy D xyx D u;

we get CM .y/ D Z.G/ D hy2; z; ui Š E8 is of index 4 inG and therefore, by Lemma64.1(q), jG0j D 2. But G D hy; xi is two-generated and so, by Lemma 65.2(a), G isminimal nonabelian of order 25, a contradiction.We have proved that y2 2 A and so A D M and jG W Aj D 2. Suppose that

A Š Q8. If CG.A/ � A, then Proposition 10.17 implies that G is of maximal class(and order 24). This is not possible since exp.G/ D 4. Hence, CG.A/ 6� A andso d.G/ D 3, a contradiction. Thus, A is abelian of type .4; 2/ or .4; 4/. We have.yx/2 D y2.xyx/, where o.y2/ � 2. It follows that o.xyx/ � 2 since yx cannotbe of order 8. This forces hxi \ hxyi D hzi, A is of type .4; 2/ and jGj D 24. Setu D xyx so that u is an involution and hz; ui D �1.A/ Š E4. We have uy D

.xyx/y D xy2

xy D xxy D xyx D u and so CA.y/ D Z.G/ D hz; ui Š E4

is of index 4 in G. By Lemma 64.1(q), jG0j D 2. But G D hy; xi and so, byLemma 65.2(a), G is minimal nonabelian (of order 24), a contradiction. We haveproved that each cyclic subgroup of order 4 is normal in G.If ��

2.G/ ¤ G, then G is generalized dihedral. Therefore we may assume that��

2.G/ D G in which case there is a cyclic subgroup X D hxi of order 4 which isnot central in G (otherwise, G would be abelian). SetH D CG.x/ and x2 D z. ThenjG W H j D 2 and for each g 2 G �H , g2 2 hzi (otherwise, hgi \ hxi D f1g impliesŒg; x� D 1, a contradiction). We have proved that G=hzi is generalized dihedral. Letg 2 G � H with g2 D z and assume that there is h 2 H with h2 62 hzi. Thenhg D h�1 since G=hzi is generalized dihedral and hhi is normal in G. But thenhhi \ hgi D f1g implies Œh; g� D 1, a contradiction. Hence, for each h 2 H , h2 2 hzi

and so hzi D ˆ.G/ D G0. By Lemma 4.2, G D H1 � � HsZ.G/, s � 1,where each Hi is minimal nonabelian and so H D H1 � �Hs is extraspecial andÃ1.Z.G// � Z.H/ since Z.H/ D ˆ.G/.

�90 Minimal nonabelian subgroups are of order 8 465

(ii) We assume that exp.G/ > 4. Then Lemma 57.2 implies that jG W Aj D 2,where A is a maximal normal abelian subgroup of G with exp.G/ D exp.A/ > 4.

(ii1) Suppose �1.A/ 6� Z.G/ so that CG.�1.A// D A. Let x 2 G � A and v 2

�1.A/� CA.x/ so that 1 D Œv; x2� D Œv; x�Œv; x�x , Œv; x�x D Œv; x� and hv; xi Š D8.For any a 2 Awe have CA.ax/ D CA.x/, Œv; ax� D Œv; x� and therefore hv; axi Š D8

which implies that G=hŒv; x�i is generalized dihedral. The subgroup B D fb j b 2

A; bx D b�1g is of index at most 2 in A and in fact jB W Aj D 2 since vx ¤ v�1 D v.We have A D hv;Bi, where B covers A=�1.A/. Either x or vx is of order 4 and soif w 2 B with o.w/ D 4, then we consider the subgroups hx;wi and hvx;wi to getw2 D Œv; x�. Hence B D hbi�1.B/, where �1.hbi/ D hŒv; x�i and H D hb; xi is ofmaximal class. Let b0 be an element of order 4 in hbi and set z D Œv; x� so that b2

0 D z.We get .vb0/

x D .vz/.b0z/ D vb0 and so G D HZ.G/ with Z.G/ D �1.B/hvb0i

and Ã1.Z.G// D Z.H/ D hzi.

(ii2) Suppose�1.A/ � Z.G/. Let x 2 G�A so that x2 2 �1.A/ and x inverts eachelement inA=�1.A/ (Lemma 57.2). Thus, Œ�2.A/;G� � �1.A/ and Œ�2.A/; x� ¤ 1.Indeed, by Lemma 57.1, there is a 2 A with ha; xi minimal nonabelian and so a 2

�2.A/ and Œa; x� ¤ 1. Let v 2 �2.A/ � CA.x/ so that 1 ¤ Œv; x� 2 �1.A/ � Z.G/and therefore hv; xi is minimal nonabelian (of order 8) and so vx D v�1. Assumethat there is w 2 �2.A/ � �1.A/ with Œw; x� D 1. Then vw 2 �2.A/ � CA.x/

and therefore vw is inverted by x. We get v�1w�1 D .vw/�1 D .vw/x D v�1w

which implies w2 D 1, a contradiction. We have proved that x inverts each element in�2.A/ and so CA.x/ D �1.A/ D Z.G/. If each element in G � A is an involution,then G is generalized dihedral. Therefore, we may take a fixed element x 2 G � A

with o.x/ D 4. Then for each v 2 A with o.v/ D 4 we have vx D v�1 which forcesv2 D x2. Indeed, if v2 ¤ x2, then hv; xi would be metacyclic minimal nonabelianof order 24, a contradiction. We have proved that A D hbi�1.A/ with o.b/ � 8 andhx2i D hbi \ �1.A/. By Lemma 57.2, bx D b�1� with � 2 �1.A/. We compute.xb/2 D x2bxb D x2b�1�b D x2� and let b0 be an element of order 4 in hbi so thatbxb

0 D b�10 . If � 62 hbi, then hb0; xbi is metacyclic minimal nonabelian of order 24, a

contradiction. Hence � 2 hbi and so � 2 hx2i andH D hb; xi is of maximal class andG D HZ.G/ with Ã1.Z.G// D f1g.

Corollary 90.2. Let G be a nonabelian 2-group in which any two noncommuting el-ements generate a subgroup of maximal class. Then G is one of the groups (a), (b)or (c) from Theorem 90.1. Conversely, each group in (a), (b) and (c) of Theorem 90.1satisfies the assumption of our corollary.

Proof. Since each minimal nonabelian subgroup of G is isomorphic to D8 or Q8, theresult follows from Theorem 90.1. It is necessary to prove only for groups (b) and (c)of Theorem 90.1 that any two noncommuting elements generate a group of maximalclass. Indeed, let h1z1 and h2z2 be any noncommuting elements inG, where h1; h2 2

H and z1; z2 2 Z.G/. Then Œh1z1; h2z2� D Œh1; h2� ¤ 1 and soH0 D hh1; h2i � H

is a group of maximal class with H 00 � Z.H/. On the other hand, a 2-group hh1; h2i


is of maximal class if and only if Œh1; h2� ¤ 1, hŒh1; h2�i is normal inH0 and h21; h

22 2

hŒh1; h2�i. Hence H1 D hh1z1; h2z2i is of maximal class since Œh1z1; h2z2� D

Œh1; h2� ¤ 1, h1z1 and h2z2 normalize hŒh1z1; h2z2�i D hŒh1; h2�i and .h1z1/2,

.h2z2/2 are contained in hŒh1; h2�i (noting that z2

1 ; z22 2 Z.H/ � H 0

0 D H 01).

�91

Maximal abelian subgroups of p-groups

Abelian subgroups in 2-groups G play an important role. Therefore, it is not verysurprising that our assumption that every two distinct maximal abelian subgroups havecyclic intersection determines completely the structure of G. We obtain five classes of2-groups with this property. More precisely, we prove here the following result.

Theorem 91.1 (Janko). Let G be a nonabelian 2-group in which any two distinct max-imal abelian subgroups have cyclic intersection. Then Z.G/ is cyclic, each abeliansubgroup ofG is of rank at most 2, the intersection of any two distinct maximal abeliansubgroups is equal Z.G/, and G has (at least) one abelian subgroup of index 2. More-over, G is isomorphic to one of the following groups:

(a) Group of maximal class.

(b) M2n .

(c) G D D � C (central product), where D Š D2n , C Š C2m , m � 2, is cyclic oforder 2m andD \ C D Z.D/.

(d) G D hx; t j .xt/2 D a; a2m

D t2 D 1; m � 2; x2 D ab; b2n�1

D 1; n �

3; bt D b�1; Œa; x� D Œa; t � D 1; tx D tb; a2m�1

D b2n�2

i, where jGj D

2mCn,m � 2, n � 3, Z.G/ D hai Š C2m ,G0 D hbi Š C2n�1 , andM D hx; ai

is a unique abelian maximal subgroup of G. We have CG.t/ D hti � hai Š

C2 � C2m and hb; ti Š D2n .

(e) G D hg; h jg2n

D h2m

D 1; m � 3; n � 3; g2n�1

D h2m�1

; hg D h�1i,where G is metacyclic, jGj D 2mCn�1 since hgi \ hhi D hg2n�1

i Š C2. Also,Z.G/ D hg2i Š C2n�1 , G0 D hh2i Š C2m�1 and M D hh; g2i is a uniqueabelian maximal subgroup of G.

The more general problem to determine the structure of a nonabelian p-group Gsuch that A \ B D Z.G/ for any two distinct maximal abelian subgroups A and B isvery difficult. First we show that a p-group G has this property if and only if CG.x/

is abelian for each x 2 G � Z.G/ (Theorem 91.2). Then we show that such a 2-groupG has either an abelian subgroup of index 2 or G is of class 2 and G0 is elementaryabelian (Theorem 91.3). In Corollary 91.5 we get a new result for an arbitrary 2-group.We also classify 2-groups G such that A=Z.G/ is cyclic for each maximal abelian

subgroup A of G (a problem of Heineken–Mann). It is surprising that such groups


have the property that CG.x/ is abelian for each element x 2 G � Z.G/. Then wemay use our Theorems 91.2 and 91.3 to classify such groups (Theorem 91.4). In thisclassification we distinguish the cases, where G has an abelian subgroup of index 2and the case where jG W Aj > 2 for each maximal abelian subgroup A of G.

Proof of Theorem 91.1. LetG be a nonabelian 2-group in which any two distinct max-imal abelian subgroups have cyclic intersection. Since Z.G/ is contained in each max-imal abelian subgroup of G, it follows that Z.G/ is cyclic.Suppose that G possesses a subgroup E Š E8. Let A be a maximal abelian sub-

group containing E and set F D �1.A/ so that E � F . Let B � G be such thatA < B and jB W Aj D 2 and let x 2 B � A. Then x2 2 A and therefore x induces onF an automorphism of order � 2. It follows that jCF .x/j � 4 (Proposition 1.8) andthe abelian subgroup CF .x/hxi is contained in a maximal abelian subgroup C whichis distinct from A since x 62 A. But A \ C � CF .x/ and so A \ C is noncyclic, acontradiction. We have proved that each abelian subgroup of G is of rank � 2.We may assume that G is not of maximal class (case (a) of Theorem 91.1) and so

there is E4 Š U GG. SetM D CG.U / so that jG W M j D 2 since Z.G/ is cyclic. LetA be a maximal abelian subgroup of G which contains U so that A � M . Supposethat A ¤ M and let y 2 M � A be such that y2 2 A. Let B be a maximal abeliansubgroup of G containing the abelian subgroup U hyi. Then B ¤ A (since y 62 A)and A \ B � U is noncyclic, a contradiction. Thus, M D A. We have proved thatwheneverU is a normal four-subgroup ofG, thenM D CG.U / is an abelian maximalsubgroup of G.If x is any element in G � M , then CM .x/ D Z.G/ and Z.G/hxi is a maximal

abelian subgroup of G. Thus, the intersection of any two distinct maximal abeliansubgroups of G is equal to Z.G/ and this is also true for 2-groups of maximal class.Suppose that G has two distinct normal four-subgroups. Then, by Theorem 50.2,

G D D � C with D Š D8, D \ C D Z.D/ and C is either cyclic of order � 4

or of maximal class 6Š D8. Let U be a four-subgroup in D; then U G G. By theabove, CG.U / is abelian and so C must be cyclic. We have obtained a group stated inpart (c). In the sequel we assume that G has a unique normal four-subgroup U and setM D CG.U / so thatM is an abelian maximal subgroup of rank 2 with�1.M/ D U .

(i) First assume �2.G/ 6� M . Then there is an element y 2 G � M of order � 4

so that y2 2 U (recall that U D �1.M/). We have U hyi Š D8 since y does notcentralize U , and so there is an involution t 2 G �M . Since t does not centralize Uand M is abelian of rank two, we get CG.t/ D hti � CM .t/, where CM .t/ is cyclicof order 2m, m � 2. Indeed, if m D 1, then G is of maximal class. Also, we havet 62 ˆ.G/, G has no elementary abelian subgroups of order 8 and G 6Š M2s , s � 4

(since M2s has only three involutions). We are now in a position to use Theorem 48.1.It follows that G has a subgroup S of index � 2, where S D AL, L is normal in G,L D hb; t j b2n�1

D t2 D 1; bt D b�1i Š D2n , n � 3, A D hai Š C2m , m � 2,A\L D Z.L/ D hzi, Œa; t � D 1, CG.t/ D hti � hai, �1.G/ D �1.S/ D �2.A/�L,

�91 Maximal abelian subgroups of p-groups 469

�2.A/ \ L D Z.L/ an if jG W S j D 2, then there is an element x 2 G � S such thattx D tb.Since hbi is a unique cyclic subgroup of index 2 in L, hbi is normal in G. Set

B D �2.A/ �L D �1.G/, �2.A/ D hli, l2 D z, and hvi D �2.hbi/ so that hvi andhli D Z.B/ are normal in G. Hence hl; vi Š C4 � C2 is normal in G. Set u D lv

so that U D hz; ui D �1.hl; vi/ Š E4 is a unique normal four-subgroup in G. Weknow that M D CG.U / is abelian and jG W M j D 2. Note that b centralizes U andut D .lv/t D lv�1 D lvz D uz. If va D v�1 D vz, then A D hai > �2.A/ D hli

and we replace a with a0 D at . In that case o.a0/ D o.a/, �2.ha0i/ D hli and

ua0

D .lv/at D .avz/t D av�1z D lv D u;

so that ha0i centralizes U and S D ha0iL. Writing again a instead of a0, we mayassume from the start that A D hai centralizes U . Hence CS .U / D ha; bi is of index2 in S and therefore M D CG.U / covers G=S and G D M hti. But M is abelianand t centralizes hai and so hai � Z.G/. On the other hand, CG.t/ D hti � hai andCM .t/ D hai so that A D hai D Z.G/.If G D S , then G D L � A, where L Š D2n , n � 3, A Š C2m , m � 2, and

L \ A D Z.L/. We have obtained groups stated in part (c) of Theorem 91.1. In whatfollows we assume that jG W S j D 2 and we know that in that case there is an elementx 2 G � S such that tx D tb. We may assume that x 2 M � S . Indeed, if x D tx0

with x0 2 M � S , then tb D tx D t tx0

D tx0

.SinceM is abelian and CM .t/ D hai D Z.G/, it follows that CM .xt/ D hai and so

.xt/2 2 hai. Set .xt/2 D a0 and assume that ha0i ¤ hai. This implies that there is anelement a00 2 hai�ha0i such that .a00/2 D .a0/�1. We get .xt a00/2 D .xt/2.a00/2 D 1

and so x.ta00/ (with ta00 2 S ) is an involution inG �S , contrary to�1.G/ D �1.S/.It follows that ha0i D hai and so replacing a with a0 (and writing again a instead ofa0), we may assume from the start that .xt/2 D a. From the last relation and tx D tb

we get a D .xt/2 D xt xt D x2.x�1tx/t D x2tbt D x2b�1, and so x2 D ab. Thestructure of G is uniquely determined and we have obtained the group stated in part(d) of Theorem 91.1.

(ii) Finally, assume that �2.G/ � M . Note that M is abelian of rank 2 and soMis metacyclic. HenceG is also metacyclic. If G has a cyclic subgroup of index 2, thenG is either of maximal class or G Š M2n , n � 4, and these are the groups stated inparts (a) and (b) of Theorem 91.1. In what follows we assume that G has no cyclicsubgroups of index 2. We have U D �1.M/ D �1.G/ Š E4, whereM D CG.U / isan abelian maximal subgroup of G.LetH be a normal cyclic subgroup with cyclic G=H so that jH j � 4 and jG=H j �

4. We have U \H D hzi Š C2 and z 2 Z.G/ so that if u 2 U � hzi D U �H , thenM D CG.u/, where jG W M j D 2 andM is abelian. Suppose that u does not centralizeH . Then jH W .H \M/j D 2 and thereforeM covers G=H . Letm 2 M be such thathmi coversM=M \H and note that CG.H/ D H since u does not centralizeH . Let


h 2 H �M so that H D hhi and hm D hz. Then hm2

D .hz/m D hz z D h. Thisis a contradiction since jG=H j � 4 and so m2 62 H .We have proved that u centralizes H and so M > H . Let g 2 G � M so that

hgi covers G=H , g2 2 M and g2 centralizesH and therefore g induces on H D hhi

an involutory automorphism. Also, ug D uz since Z.G/ is cyclic. If hg D hz, thenG0 D hzi and G is minimal nonabelian. In that case G is splitting metacyclic, i.e.,there is g0 2 G � M such that hg0i covers G=H and hg0i \ H D f1g. It followsthat �1.hg

0i/ � Z.G/ and so Z.G/ � hz;�1.hg0i/i Š E4, a contradiction. We

have proved that hg D h�1z� , � D 0; 1 and jH j � 8. (Indeed, if jH j D 4, thenhg D h�1 D hz and we have again G0 D hzi, as above.) In particular, CH .g/ D hzi

and so hgi\H � hzi. However, if hgi\H D f1g, then CG.�1.hgi// � hM;gi D G

and so E4 Š hz;�1.hgi/i � Z.G/, a contradiction.We have proved that hgi \ H D hzi and so o.g/ � 8 and Z.G/ D hg2i. If

hg D h�1z, then we replace h with h0 D hu, where Œh; u� D 1 and so o.h0/ D o.h/

and.h0/g D .hu/g D h�1z uz D h�1u D .hu/�1 D .h0/�1

and hh0; gi D hhu; gi D hh; gi D G since u 2 U � ˆ.G/. (Indeed, ˆ.G/ � M isabelian and so if ˆ.G/ D Ã1.G/ were cyclic, then jG W ˆ.G/j D 4 implies that Gwould have a cyclic subgroup of index 2.) Writing h instead of h0, we see that we mayassume from the start that hg D h�1. We have obtained the group stated in part (e).

Theorem 91.2. Let G be a nonabelian p-group. Then A \ B D Z.G/ for any twodistinct maximal abelian subgroups A;B if and only if CG.x/ is abelian for eachx 2 G � Z.G/.

Proof. Let x 2 G � Z.G/ and suppose that CG.x/ is nonabelian. Let A be a maximalabelian subgroup of CG.x/ so that A ¤ CG.x/ and A � Z.G/hxi > Z.G/. Letb 2 CG.x/ � A and let B be a maximal abelian subgroup of CG.x/ containing hbi sothat A ¤ B and B also contains the abelian subgroup Z.G/hxi. Obviously, A and Bare also maximal abelian subgroups of G but A \ B � Z.G/hxi > Z.G/.Conversely, let C ¤ D be maximal abelian subgroups of G such that C \ D >

Z.G/. Let y 2 .C \ D/ � Z.G/ so that CG.y/ � hC;Di, where hC;Di is non-abelian.

Theorem 91.3. Let G be a nonabelian 2-group such that A\B D Z.G/ for every twodistinct maximal abelian subgroups A and B . Then one of the following holds: (a) Ghas an abelian subgroup of index 2. (b) G is of class 2, G0 is elementary abelian andˆ.G/ � Z.G/.

Proof. Let A be a maximal normal abelian subgroup of G. Then G=A ¤ f1g actsfaithfully on A and f1g ¤ Z.G/ < A. Let K be a G-invariant subgroup such thatZ.G/ < K � A and jK W Z.G/j D 2. Let x be any element in G � A. ThenCA.x/ D Z.G/ and so hxi \A � Z.G/. Indeed, let B be a maximal abelian subgroup


containing the abelian subgroup CA.x/hxi. ThenA ¤ B andA\B � CA.x/ D Z.G/.Let k 2 K � Z.G/ so that k2 2 Z.G/ and kx D kl with some 1 ¤ l 2 Z.G/. We getk2 D .k2/x D .kx/2 D .kl/2 D k2l2, and so l2 D 1 and therefore l is an involutionin Z.G/. This gives

kx2

D .kx/x D .kl/x D kxl D .kl/l D kl2 D k;

and so (by the above) x2 2 A and (since hxi \ A � Z.G/) x2 2 Z.G/. In particular,G=A is elementary abelian. Let a 2 A � Z.G/ and set ax D a0 2 A � Z.G/ so that.a0/x D .ax/x D ax2

D a (since x2 2 Z.G/). Therefore, .aa0/x D a0a D aa0 whichimplies that aa0 D z 2 Z.G/ and a0 D ax D a�1z and so x inverts A=Z.G/.Suppose that jG=Aj � 4. Then there are elements x; y 2 G � A such that xy 2

G � A. In this case x and y both invert A=Z.G/ and so xy centralizes A=Z.G/. Butxy also must invert A=Z.G/ which implies that A=Z.G/ is elementary abelian. Henceˆ.G/ � Z.G/ (noting that for each x 2 G � A, x2 2 Z.G/) and so G is of class 2.For each g; h 2 G, Œg; h�2 D Œg2; h� D 1 and so G0 is elementary abelian.

Theorem 91.4. Let G be a nonabelian 2-group such that A=Z.G/ is cyclic for eachmaximal abelian subgroup A of G. Then one of the following holds:

(a) G has an abelian subgroupM of index 2 and we have either G D HZ.G/ withH minimal nonabelian or G=Z.G/ Š D2n , n � 3, is dihedral of order 2n withG0 cyclic of order � 4, G0 \ Z.G/ Š C2, and if x 2 G �M , then x2 2 Z.G/and x inverts G0.

(b) G is of class 2, G0 is elementary abelian of order � 8, ˆ.G/ � Z.G/ andwhenever A is a maximal abelian subgroup of G, then jA W Z.G/j D 2.

Proof. Suppose that there is an element a 2 G � Z.G/ such that H D CG.a/

is nonabelian. Let A be a maximal abelian subgroup of G containing hai. ThenZ.G/hai � A < H < G. By our assumption, A=Z.G/ ¤ f1g is cyclic. Assumethat H=Z.G/ contains a subgroup of order 2 distinct from �1.A=Z.G//. In thatcase there is x 2 H � A such that x2 2 Z.G/. Since Œa; x� D 1, ha; xi is abelianbut ha; xiZ.G/=Z.G/ is noncyclic, a contradiction. We have proved that H=Z.G/has only one subgroup of order 2 and so H=Z.G/ Š Q2n , n � 3, is generalizedquaternion of order 2n. Indeed, if H=Z.G/ were cyclic, then H is abelian, a con-tradiction. Since H=Z.G/ Š Q2n , it follows that a2 2 Z.G/, Z.H/ D Z.G/hai,jZ.H/ W Z.G/j D 2 and for each y 2 Z.H/ � Z.G/, CG.y/ D H D CG.a/. SetjZ.G/j D 2m, m � 1, so that jH j D 2mCn. Let A0=Z.G/ be a cyclic subgroup of in-dex 2 inH=Z.G/ so that A0 is abelian and Z.H/ < A0. Let A1=Z.G/ D .H=Z.G//0

so that Z.H/ � A1 < A0 and jH W A1j D 4 and therefore jA1j D 2mCn�2.Since A1 D H 0Z.G/, H 0 covers A1=Z.G/ and so jH 0j � 2n�2 D jA1=Z.G/j.By Lemma 1.1, we get jH j D 2jZ.H/jjH 0j and so 2mCn D 2 2mC1jH 0j andtherefore jH 0j D 2n�2. This gives H 0 \ Z.G/ D f1g and so H 0 is cyclic withH 0 \ Z.H/ D hyi Š C2. It follows that hyi is characteristic in H and so if T is


a subgroup of G such that H < T � G and jT W H j D 2, then hyi is central in T ,contrary to the above fact that CG.y/ D H , where y 2 Z.H/�Z.G/. We have provedthat for each a 2 G � Z.G/, CG.a/ is abelian.By Theorem 91.2, A \ B D Z.G/ for any two distinct maximal abelian subgroups

A and B of G. We may use Theorem 91.3 and so either G has an abelian subgroup ofindex 2 or G0 � Z.G/, ˆ.G/ � Z.G/ and G0 is elementary abelian.

(i) First we consider the case, where G has an abelian subgroup M of index 2.Then Z.G/ < M and for each x 2 G � M , CM .x/ D Z.G/ and so x2 2 Z.G/.By our assumption, M=Z.G/ ¤ f1g is cyclic. If G has another abelian maximalsubgroup N , thenM \N D Z.G/ and this implies that jM W Z.G/j D 2. Conversely,suppose that jM=Z.G/j D 2. In that case G=Z.G/ Š E4 (because G=Z.G/ Š C4

would imply thatG is abelian) and soG has more than one abelian maximal subgroup.We analyze this case further. Let H be a minimal nonabelian subgroup of G. ThenjH W .H \ Z.G//j D 4, H covers G=Z.G/ and so G D HZ.G/ and G0 Š C2. Wehave obtained the first possibility stated in part (a) of our theorem.It remains to consider the case, whereM=Z.G/ Š C2n , n � 2, whereM is a unique

abelian maximal subgroup of G. We know that for each x 2 G �M , x2 2 Z.G/. Itfollows that x inverts the cyclic groupM=Z.G/ of order� 4 and soG=Z.G/ Š D2nC1

is dihedral of order 2nC1. Set jZ.G/j D 2m, m � 1, and .G=Z.G//0 D L=Z.G/ sothatG=L Š E4 andL D G0Z.G/. SinceG has an abelian maximal subgroup, we mayuse Lemma 1.1 and we get jGj D 2mCnC1 D 2 2mjG0j and so jG0j D 2n. HenceG0 \ Z.G/ D hzi Š C2, G0=hzi is cyclic of order 2n�1 and G0 is abelian.Suppose thatG0 is not cyclic. ThenG0 splits over hzi D G0\Z.G/. SinceÃ1.G

0/ isnormal in G andÃ1.G

0/\Z.G/ D f1g, it follows thatÃ1.G0/ D f1g and soG0 Š E4

and G=Z.G/ Š D8. Let a 2 M � .Z.G/G0/ so that hai coversM=Z.G/ Š C4 and soa2 62 Z.G/. For an x 2 G �M , we have Œa; x� D t 2 G0 � hzi. Then we get

Œa2; x� D Œa; x�aŒa; x� D tat D t2 D 1:

But then G.a2/ D hM;xi D G and so a2 2 Z.G/, a contradiction. We have proved

that G0 is cyclic of order � 4.For any x 2 G�M and anym 2 M �L (whereL D G0Z.G/), we have x2 2 Z.G/

and Œm; x� D g with hgi D G0 Š C2n . This gives mx D mg and so m D mx2

D

.mg/x D mggx and this implies gx D g�1 and therefore x inverts G0. We haveobtained the second possibility in part (a) of our theorem.

(ii) Now we consider the case, where G has no abelian subgroups of index 2, G0 �

Z.G/, ˆ.G/ � Z.G/ and G0 is elementary abelian. It follows that jA W Z.G/j D 2 foreach maximal abelian subgroup A of G. If jG W Z.G/j D 4, then G would have anabelian subgroup of index 2, a contradiction. Hence, G=Z.G/ Š E2m , m � 3, and sothere exist elements g; h; i 2 G �Z.G/ such that hg; h; iiZ.G/=Z.G/ Š E8. We haveŒg; h� ¤ 1, Œg; i � ¤ 1, and Œh; i � ¤ 1.Suppose that jG0j D 2. Then Œg; h� D Œg; i � and so Œg; hi � D Œg; h�Œg; i � D 1 and

therefore hg; hiiZ.G/=Z.G/ Š E4, a contradiction.


Suppose that G0 Š E4. In that case Œg; h� D t1, Œg; i � D t2 and Œh; i � D t3, wheret1; t2; t3 are pairwise distinct involutions in G0. In this case,

Œgh; gi � D Œg; i �Œh; g�Œh; i � D t2t1t3 D 1;

and so hgh; giiZ.G/=Z.G/ Š E4, a contradiction. We have proved that jG0j � 8.

Corollary 91.5. Let G be an arbitrary nonabelian 2-group. Let A and B be any twodistinct maximal abelian subgroups in G with intersection A \ B of maximal possi-ble order. Then the nonabelian subgroup H D hA; Bi either possesses an abeliansubgroup of index 2 or H is of class 2 and H 0 is elementary abelian.

Proof. Obviously, A \ B D Z.H/. If C andD are any two distinct maximal abeliansubgroups inH , then C \D � Z.H/ and the maximality of jA\Bj forces C \D D

Z.H/. Then our result follows from Theorem 91.3.

�92

On minimal nonabelian subgroups of p-groups

1o. In studying the structure of nonabelian p-groups G, the minimal nonabelian sub-groups of G play an important role since they generate the group G (Theorem 10.28).More precisely, if A is a maximal normal abelian subgroup of G, then minimal non-abelian subgroups of G cover the set G � A (Lemma 57.1).It is an open problem to classify nonabelian p-groups which are covered by its

minimal nonabelian subgroups. Here we consider two special cases of this problem.First we determine in Theorem 92.1 nonabelian p-groups G with the property that

CG.x/ � H for any minimal nonabelian subgroupH ofG and each x 2 H�Z.G/. Itis easily seen that the assumption of Theorem 92.1 implies that each abelian subgroupof G is contained in a minimal nonabelian subgroup and so in this case G is coveredby its minimal nonabelian subgroups.In Theorem 92.2 we shall classify nonabelian p-groups G such that whenever A is

a maximal subgroup of any minimal nonabelian subgroup H in G, then A is also amaximal abelian subgroup of G. Again, it easily seen that the assumption of Theorem92.2 implies that each abelian subgroup of G is contained in a minimal nonabeliansubgroup in G and so the minimal nonabelian subgroups of G cover G.

Theorem 92.1 (Janko). Let G be a nonabelian p-group such that for each minimalnonabelian subgroup H of G and each x 2 H � Z.G/, we have CG.x/ � H . ThenG is one of the following groups:

(a) G is minimal nonabelian.

(b) p D 2, d.G/ D 3 and

G D ha; b; c j a4 D b4 D c4 D 1; Œa; b� D c2; Œa; c� D b2c2; Œb; c� D a2b2;

Œa2; b� D Œa2; c� D Œb2; a� D Œb2; c� D Œc2; a� D Œc2; b� D 1i;

where G is a special 2-group of order 26 with ha2; b2; c2i D G0 D Z.G/ D ˆ.G/ D

�1.G/ Š E8 and G is isomorphic to a Sylow 2-subgroup of the simple group Sz.8/.

(c) p > 2, d.G/ D 2, G is of order 25 and

G D ha; x j ap2

D xp2

D 1; Œa; x� D b; Œa; b� D y1; Œx; b� D y2;

bp D yp1 D y

p2 D Œa; y1� D Œx; y1� D Œa; y2� D Œx; y2� D 1;

ap D y˛1 y

ˇ2 ; x

p D y�1 y

ı2i;

�92 On minimal nonabelian subgroups of p-groups 475

where in case p > 3, 4ˇ�C .ı�˛/2 is a quadratic non-residue modp. Hereˆ.G/ D

G0 D hb; y1; y2i D �1.G/ Š Ep3 , Z.G/ D K3.G/ D Ã1.G/ D hy1; y2i Š Ep2 .

Conversely, all the above groups satisfy the assumptions of the theorem.

Proof. Let G be a nonabelian p-group such that for each minimal nonabelian sub-groups H of G and each x 2 H � Z.G/, we have CG.x/ � H . Let A be a maximalnormal abelian subgroup of G. Then for each x 2 G � A, there is an element x 2 A

such that H D ha; xi is minimal nonabelian (Lemma 57.1). Since a 62 Z.H/, it fol-lows that CG.a/ � H and so A � H . But A is a maximal abelian subgroup inH andso jH W Aj D p and xp 2 A. It follows that exp.G=A/ D p. If H D G, then G isminimal nonabelian so it is as in (a).In what follows we assume that jG=Aj � p2. For each x 2 G�A, xp 2 A, Ahxi is

minimal nonabelian and x 62 Z.Ahxi/ so that CG.x/ � Ahxi and CG.x/ D hxiCA.x/

is abelian and jA W CA.x/j D p. Therefore,A and CG.x/ (x 2 G�A) are all maximalabelian subgroups in G. Let K=A be a normal subgroup of order p in G=A and setA0 D Z.K/ so that A0 < A and jA W A0j D p. Suppose that there is g 2 G �K suchthat A1 D CA.g/ ¤ A0. Let y 2 A1 �A0 so that y 2 K �Z.G/ and so we must haveCG.y/ � K. This is a contradiction since g 2 G �K and g centralizes y.We have proved that for each g 2 G � K, CA.g/ D A0 and so A0 D Z.G/, and

A D hA0; ai with an element a 2 A � A0. For any g1; g2 2 G we have ag1 D az1,ag2 D az2 with z1; z2 2 A0 D Z.G/ so that ag�1

1 D az�11 and ag�1

2 D az�12 and

therefore aŒg1;g2� D a which gives Œg1; g2� 2 CG.A/ D A. Thus, G=A is elementaryabelian and also Ã1.G/ � A0 D Z.G/. In addition, A0 D Z.G/ D Z.K/ D ˆ.K/ �

ˆ.G/ which implies that either ˆ.G/ D Z.G/ or ˆ.G/ D A.Let S=A Š Ep2 be a subgroup of order p2 in G=A. Since Ã1.S/ � Z.G/ D Z.S/,

S=Ã1.S/ is of exponent p and order� p3 so that S is non-metacyclic. If jZ.G/j D p,then jAj D p2 and CG.A/ D A implies that jG W Aj D p, a contradiction. HencejZ.G/j > p so that jS j > p4.

(i) First assume thatˆ.S/ D A. In that case eachmaximal subgroup of S is minimalnonabelian and we are in a position to use Proposition 71.5. Since d.S/ D 2, we getp > 2, jS j D p5, A D S 0 D �1.S/ Š Ep3 and Z.S/ D K3.S/ D Ã1.S/ Š

Ep2 . Since CG.A/ D A and G=A is elementary abelian, we get (by the structure ofGL.3; p/) S D G and we have obtained the groups from (c).

(ii) Now assume thatˆ.S/ D Z.G/ so that d.S/ D 3. Each maximal subgroup of Scontaining A is minimal nonabelian. LetM be a maximal subgroup of S which doesnot contain A so thatM=M \A Š Ep2 andM \A D Z.G/ D ˆ.S/. Ifm 2 M �A,then CG.m/ D hmiZ.G/ with mp 2 Z.G/ so that M is nonabelian. Let M0 be aminimal nonabelian subgroup ofM . ThenM0 > Z.G/ and since jM0=Z.G/j � p2,we have M0 D M . Hence, each maximal subgroup of S is minimal nonabelian,jS j > p4, S is non-metacyclic and d.S/ D 3. By Proposition 71.5, p D 2, jS j D 26,Z.G/ D Z.S/ D �1.S/ D ˆ.S/ Š E8 and so A is abelian of type .4; 2; 2/. Leta 2 A � Z.G/ so that o.a/ D 4 and for each g 2 G, ag D az with z 2 Z.G/. Since


CG.A/ D A, we get jG=Aj � 8. Suppose that jG=Aj D 8. Then there is h 2 G � A

such that h2 2 Z.G/ and ah D aa2 D a�1. If h2 D a2, then hh; ai Š Q8 which isminimal nonabelian. If h2 ¤ a2, then hh; ai is metacyclic minimal nonabelian of order24. But in any case, Z.G/ 6� hh; ai, contrary to our assumption. Hence jG=Aj D 4

and so S D G and we have obtained the group stated in part (b) of our theorem.

Theorem 92.2. Let G be a nonabelian p-group such that whenever A is a maximalsubgroup of any minimal nonabelian subgroup H in G, then A is also a maximalabelian subgroup of G. Then each abelian subgroup of G is contained in a minimalnonabelian subgroup and one of the following holds:


(b) G is metacyclic.

(c) G is isomorphic to the group of order 26 defined in Theorem 92.1(b).

(d) G is isomorphic to a group of order p5 defined in Theorem 92.1(c).

Proof. LetG be a nonabelian p-group such that wheneverA is a maximal subgroup ofany minimal nonabelian subgroupH in G, then A is also a maximal abelian subgroupof G. First we note that our assumption is hereditary for all nonabelian subgroupsof G. Let A be any maximal abelian subgroup of G. Let B > A be a subgroupof G such that jB W Aj D p and let b 2 B � A. By Proposition 57.1, there isa 2 A such that H D ha; bi is minimal nonabelian. Then jH W H \ Aj D p andso H \ A must be a maximal abelian subgroup in G. This gives H \ A D A andso H D Ahbi D B is minimal nonabelian. We have proved that whenever A is amaximal abelian subgroup of G and a subgroup X contains A as a subgroup of indexp, then X is minimal nonabelian. We may assume that G is not minimal nonabelian(case (a) of our theorem) and so G does not possess any abelian maximal subgroup.Moreover, if U is a nonabelian subgroup of G and if U possesses an abelian maximalsubgroup, then U is minimal nonabelian.

(i) Assume p > 2. First we consider the case that G has no normal elementaryabelian subgroups of order p3 and use Theorem 13.7. In case (a) of that theorem G ismetacyclic so it is as in (b). The case (b) of Theorem 13.7 cannot occur. Indeed, in thatcase G is a 3-group of maximal class which possesses a self-centralizing subgroup Xof order 9. If Y=X < G=X is of order 3 and Y 6� G1, where G1 is the fundamentalsubgroup of G, then CG.X/ D G1 6� Y , a contradiction since Y is minimal non-abelian. It remains to consider part (c) of Theorem 13.7, whereG D EH ,E D �1.G/

is nonabelian of order p3 and exponent p, H is cyclic with E \H D Z.E/. Let E0

be a normal abelian subgroup of type .p; p/ contained in E so that CG.E0/ coversG=E. But CG.E0/=E0 is cyclic so that CG.E0/ is an abelian maximal subgroup ofG, a contradiction.

It remains to consider the case, where G has an abelian subgroup of type .p; p; p/.Let A be a maximal normal abelian subgroup of G with j�1.A/j � p3. We havejG=Aj � p2 since G does not have an abelian maximal subgroup.


(i1) Suppose that G=A is cyclic. Let H=A be the subgroup of order p in G=A sothatH is a non-metacyclic minimal nonabelian group of order � p4 and so�1.H/ D

�1.A/ Š Ep3 which implies that there are no elements of order p in G � A. LetK > H be a subgroup with jK W H j D p so that K=A Š Cp2 . Let k 2 K � H

so that hki covers K=A and since hki \ A ¤ f1g, we have exp.K/ � p3. Let M beany maximal subgroup of K which does not contain A. ThenM covers K=A. SinceK is not minimal nonabelian, it follows that M is nonabelian. On the other hand,M \H < H is an abelian maximal subgroup ofM which implies thatM is minimalnonabelian. We have proved that each maximal subgroup ofK is minimal nonabelian.By Proposition 71.5, jKj D p5 and exp.K/ D p2, a contradiction.

(i2) Assume that G=A is noncyclic. Let S=A be a normal elementary abeliansubgroup of order p2 in G=A. Then S is non-metacyclic and for each subgroupXi=A of order p in S=A, Xi is minimal nonabelian (i D 1; 2; : : : ; p C 1). SetZi D Z.Xi / D ˆ.Xi / � ˆ.S/ and we may assume that for a fixed i 0, Xi 0 isnormal in G. Let Y be a maximal subgroup of S which does not contain A sothat Y covers S=A. If Y is abelian, then S is minimal nonabelian, a contradic-tion. Hence Y is nonabelian with an abelian maximal subgroup Y \ Xi 0 < Xi 0

so that Y is minimal nonabelian. We use Proposition 71.5 and see that jS j D p5,S 0 D �1.S/ D A Š Ep3 and Z.S/ D Ã1.S/ D Zi 0 Š Ep2 . Note that CG.A/ D A

and so either S D G and we have obtained groups stated in (d) or G=A is non-abelian of order p3 and exponent p. In the second case, we consider another subgroupS�=A Š Ep2 in G=A (distinct from S=A). We have S \ S� D Xi 0 since Xi 0=A

is central in G=A and so Xi 0=A D ˆ.G=A/. We show (as above) that each maxi-mal subgroup of of S� is minimal nonabelian and so S� is isomorphic to a group ofProposition 71.5(b) which gives Zi 0 D Z.Xi 0/ D Z.S�/ and soZi 0 D Z.G/. But thenG=A stabilizes the chain A > Zi 0 > f1g which gives that G=A is elementary abelian,a contradiction.

(ii) We suppose p D 2 and we may assume that G is non-metacyclic. Let E be aminimal non-metacyclic subgroup of G. We use Theorem 66.1 and see that in cases(b), (c) and (d) of that theorem, E is nonabelian and possesses an abelian maximalsubgroup. It follows that in these cases E must be minimal nonabelian which is notthe case. Hence E Š E8 which is case (a) of Theorem 66.1. We claim that wheneverA is a maximal abelian subgroup of rank � 3 in G and K > A is a subgroup of Gsuch that jK W Aj D 4, then jKj D 26 andK is isomorphic to the group of Proposition71.5(a) so that K 0 D Z.K/ D ˆ.K/ D �1.K/ D �1.A/ Š E8 and A is abelianof type .4; 2; 2/. Indeed, if X > A is a maximal subgroup of K containing A, thenjX W Aj D 2 and therefore X is minimal nonabelian with E8 Š �1.A/ D �1.X/. LetY be a maximal subgroup of K distinct from X . Since K is not minimal nonabelian,Y is nonabelian. But Y \ X .< X/ is an abelian maximal subgroup of Y and so Yis minimal nonabelian. We have proved that K is a non-metacyclic group of order> 24 all of whose maximal subgroups are minimal nonabelian. It follows that K isisomorphic to the group of Proposition 71.5(a) and our claim is proved.


Suppose that K ¤ G and let L > K be a subgroup of G with jL W Kj D 2.Acting with L=K on seven subgroups Ai=�1.A/ (i D 1; 2; : : : ; 7) of order 2 inK=�1.A/ Š E8 (where all Ai are abelian of type .4; 2; 2/ and they are also maxi-mal abelian subgroups inG), we see that one of them Ai 0 is certainly normal in L. Wenote that L=Ai 0 is a noncyclic group of order 8 since K=Ai 0 Š E4. Let K�=Ai 0 bea maximal subgroup of L=Ai 0 distinct from K=Ai 0 . By the preceding paragraph, K�

is isomorphic to the group of Proposition 71.5(a). It follows �1.A/ D Z.K�/ so that�1.A/ D Z.L/. Set Ai 0 D ha;�1.A/i, where o.a/ D 4 and L=Ai 0 acts faithfully onAi 0 stabilizing the chain Ai 0 > �1.A/ > f1g. In particular, there is l 2 L such thatal D aa2 D a�1. Hence ha; li is a metacyclic minimal nonabelian subgroup so that�1.A/ 6� ha; li. This is a contradiction since�1.A/ centralizes ha; li. HenceK D G

and we have obtained the group of part (c) of our theorem.

2o. In Theorem 92.6 we classify nonabelian 2-groups all of whose minimal non-abelian subgroups are isomorphic to Q8 or H2 D ha; b j a4 D b4 D 1; ab D a�1i.This theorem generalizes a result of N. Blackburn [Bla7] concerning p-groups Gwhich possess nonnormal subgroups and such that the intersection of all nonnormalsubgroups is nontrivial (Corollary 92.7). Namely, it is easy to see that in such p-groupsG we must have p D 2 and each minimal nonabelian subgroup of G is isomorphic toQ8 or H2.Let G be a nonabelian 2-group all of whose minimal nonabelian subgroups are

isomorphic to Q8 or H2. Then we prove the following three key lemmas which playan important role in the proof of Theorem 92.6.

Lemma 92.3. We have �1.G/ � Z.G/.

Proof. Since D8 is not a subgroup of G, �1.G/ is elementary abelian. Let x 2 G

be any element of order 4 so that x2 2 �1.G/ and assume that G0 D �1.G/hxi isnonabelian. By Lemma 57.1 applied to the group G0, there is a 2 �1.G/ such thatha; xi is minimal nonabelian. But then a is a noncentral involution in ha; xi, contraryto the structure of Q8 or H2. Hence x centralizes �1.G/. Since G is generated byits minimal nonabelian subgroups, we have �2.G/ D G which implies �1.G/ �

Z.G/.

Lemma 92.4. Suppose that G possesses an element v of order 4 such that CG.v/

is nonabelian. Then CG.v/ has minimal nonabelian subgroups isomorphic to H2

and for each such subgroup H D ha; b j a4 D b4 D 1; ab D a�1i, we havev2 D a2b2 (which is a unique involution in H which is not a square in H ) andH hvi D ha; vbi � hvi Š Q8 � C4. In particular, G does not possess a subgroupisomorphic to H2 � C4.

Proof. Since CG.v/ possesses central elements of order 4, Corollary A.17.3 impliesthat CG.v/ has subgroups isomorphic to H2 and letH D ha; b j a4 D b4 D 1; ab D

a�1i be one of them. First suppose that hvi \ H D f1g. Then .av/2 D a2v2 62 H


and .av/b D abv D a�1v D .av/a2 so that hav; bi is a nonmetacyclic minimalnonabelian subgroup of order 25 and exponent 4, a contradiction. Hence v 62 H andv2 2 fa2; b2; a2b2g. If v2 D a2, then i D va 62 H is an involution and ib D va�1 D

ia2 so that i 62 Z.G/, contrary to Lemma 92.3. If v2 D b2, then j D vb 62 H isan involution and j a D vba D ja2 so that j 62 Z.G/, contrary to Lemma 92.3. Itfollows that we must have v2 D a2b2 which is a unique involution in H which is nota square inH . In that case .vb/2 D v2b2 D .a2b2/b2 D a2 and avb D ab D a�1 sothat ha; vbi Š Q8 andH hvi D ha; vbi � hvi Š Q8 � C4.

Lemma 92.5. The group G has no subgroups isomorphic to Q8 � C4 � C4.

Proof. Suppose thatK D Q� hci � hd i is a subgroup of G, whereQ D ha; bi Š Q8

and hci Š hd i Š C4. We have �1.K/ D ha2; c2; d2i Š E8, where Œac; bd � D

Œa; b� D a2 so that hac; bd i0 D ha2i and therefore hac; bd i is minimal nonabelian(Lemma 65.1). But .ac/2 D a2c2 and .bd/2 D a2d2 so that �1.hac; bd i/ D

ha2c2; a2d2; a2i D ha2; c2; d2i Š E8 and so hac; bd i is nonmetacyclic, a contra-diction.

Theorem 92.6. Let G be a nonabelian 2-group all of whose minimal nonabelian sub-groups are isomorphic to Q8 or H2 D ha; b j a4 D b4 D 1; ab D a�1i. Then thefollowing holds:

(a) IfG is of exponent > 4, thenG has a unique abelian maximal subgroup A, jG0j >

2, and all elements in G �A are of order 4. We have �1.A/ D �1.G/ � Z.G/ and ifx 2 G � A, then x inverts each element of A=�1.A/.

(b) If G is of exponent 4, then G D K � V , where exp.V / � 2 and for the group Kwe have one of the following possibilities:

(b1) K Š Q8 or K Š H2;

(b2) K D ha; b; c j a4 D b4 D c4 D Œa; b� D 1; c2 D a2; Œa; c� D b2; Œb; c� D

a2i is the minimal nonmetacyclic group of order 25;

(b3) K is a unique special group of order 26 with Z.K/ Š E4 given in Theorem57.3(b3) in which every maximal subgroup is isomorphic to the minimal non-metacyclic group of order 25 (from (b2));

(b4) K Š Q8 � C4;

(b5) K Š Q8 � Q8;

(b6) G D K�V has an abelian maximal subgroup B of exponent 4 and an elementv 2 G � B of order 4 which inverts each element of B;

(b7) K D Q � C is a central product of Q D ha; bi Š Q8 and C D hc; d j c4 D

d4 D 1; cd D c�1i Š H2 with Q \ C D hc2d2i D Z.Q/, where K isspecial of order 26 and Z.K/ D �1.K/ Š E4.

Conversely, in each of the above groups in part (b) every minimal nonabelian subgroupis isomorphic to Q8 or H2.


Proof. Let G be a nonabelian 2-group all of whose minimal nonabelian subgroups areisomorphic to Q8 or H2. Let A be a maximal normal abelian subgroup of G so that�1.G/ D �1.A/ � Z.G/ (Lemma 92.3) and �1.A/ < A. By Proposition 57.2, allelements in G � A are of order 4 and so G=A is elementary abelian.Suppose that exp.A/ > 4. Then by Proposition 57.2, jG W Aj D 2 and if x 2 G�A,

then o.x/ D 4 and x inverts each element in A=�1.A/. Let a 2 A with o.a/ D 8.Then ax D a�1z with z 2 �1.A/ so that Œa; x� D a�2z is of order 4 and so jG0j > 2.If G has more than one abelian maximal subgroup, then jG W Z.G/j D 4 and Lemma1.1 gives jGj D 2jZ.G/jjG0j and jG0j D 2, a contradiction. Hence A is a uniqueabelian maximal subgroup of G.In what follows we assume that exp.A/ D exp.G/ D 4. In view of Corollary

A.17.3 and Theorem 57.3 we may also assume that both Q8 andH2 are subgroups ofG. We have ˆ.G/ � �1.A/ � Z.G/ and jˆ.G/j � 4. If x; y are elements of order4 in G with Œx; y� ¤ 1, then Œx; y� is an involution in Z.G/ and so (Lemma 65.1)hx; yi Š H2 or Q8 which implies that in case x

2 ¤ y2, we have hx; yi Š H2 andyx 2 fy�1; yx2g.ConsideringG=ˆ.G/ we getG D K�V , where�1.A/ D ˆ.G/�V , exp.V / � 2,

�1.K/ D ˆ.K/ D ˆ.G/ � Z.K/ and A \K is a maximal normal abelian subgroupof K. We have to determine the structure of K and so in the sequel we may assumeK D G and so in that case�1.A/ D �1.G/ D ˆ.G/ � Z.G/.First assume jˆ.G/j D 4 so that our group G has exactly three involutions. By the

results stated in the introduction to �82, G has a metacyclic normal subgroupM suchthat G=M is elementary abelian of order � 4. But exp.M/ � 4 and so jM j � 24 andjGj � 26. Since both H2 and Q8 are subgroups of G, we have jGj � 25.Suppose that jGj D 25. In this case G is nonmetacyclic since exp.G/ D 4. If G

were minimal nonmetacyclic, then each minimal nonabelian subgroup of G is isomor-phic to H2 (see Theorem 66.1), contrary to our assumptions. But G has only threeinvolutions and so Theorem 66.1 implies that G has a minimal nonmetacyclic sub-group H D Q � hui with Q D ha; bi Š Q8, Z.Q/ D hzi and u is an involution.Since ˆ.G/ D hz; ui Š E4, we may assume (interchanging u and uz if necessary)that there is an element v 2 G � H such that v2 D u. If G0 D Q0 D hzi, then Qis normal in G, G=Q Š C4 and v induces an inner automorphism on Q. In that caseG D QCG.Q/ with Q \ CG.Q/ D hzi and CG.Q/=hzi Š C4. But exp.G/ D 4

and so CG.Q/ splits over hzi and we get G Š Q8 � C4 which is a group of part (b4)of our theorem. We assume that G0 D hz; ui D �1.G/ D ˆ.G/. If Œa; v� 2 hzi andŒb; v� 2 hzi, then G D ha; b; vi implies that G=hzi would be abelian and so G0 D hzi,contrary to our assumption. It follows that we may assume (interchanging a and b ifnecessary) that Œa; v� 2 hz; ui � hzi. Since a2 D z ¤ Œa; v�, we have ha; vi Š H2 andso Œa; v� D u D v2 and va D v�1. We have Œa; bv� D Œa; b�Œa; v� D zu ¤ a2 D z

and therefore ha; bvi Š H2. It follows that .bv/2 D zu or z. If zu D .bv/2 D

b2v2Œv; b� D zuŒv; b�, then Œv; b� D 1. If z D .bv/2 D b2v2Œv; b� D zuŒv; b�, thenŒv; b� D u which implies Œv; ab� D Œv; a�Œv; b� D uu D 1 and so in that case replacing


b with b0 D ab we get Œv; b0� D 1. Hence writing again b instead of b0 we may assumeagain Œv; b� D 1. In both cases we may assume Œv; b� D 1 so that hv; bi Š C4 � C4.We get G D hv; bihai, where a inverts hv; bi and so we have obtained a group ofpart (b6).

Suppose that jGj D 26. First assume that G has a normal subgroup Q D ha; bi Š

Q8 and set Z.Q/ D hzi. Since G0 � ˆ.G/ D �1.G/ � Z.G/, no element in G �Q

induces an outer automorphism onQ (if x 2 G�Q induces an outer automorphism onQ, then .Qhxi/0 Š C4). We get G D Q � C , where C D CG.Q/ andQ \ C D hzi.Also, z is not a square in C . Indeed, if x 2 C with x2 D z, then ha; bxi Š D8, acontradiction. We have exp.C / D 4, jC j D 24 and �1.C / D �1.G/ Š E4. If C isabelian, then C Š C4 � C4 in which case z is a square in C , a contradiction. HenceC is nonabelian. If C is minimal nonabelian, then C D hc; d j c4 D d4 D 1; cd D

c�1i Š H2. Since z is not a square in C , we must have z D c2d2 and this is the groupof part (b7) of our theorem. If C is not minimal nonabelian, it possesses a subgroupQ� Š Q8 withQ

� \ hzi D f1g. We get G D Q�Q� and this is a group of part (b5).

Now we treat the case, where no quaternion subgroup is normal in G. Let Q D

ha; bi Š Q8 be a quaternion subgroup of G with Z.Q/ D hzi. Set W D �1.G/ sothat ˆ.G/ D W Š E4, W � Z.G/ and W \ Q D hzi. Since Q is not normal inG, we have Q0 D hzi ¤ G0 which gives G0 D W . Since jGj D 26, exp.G/ D 4,and G has only three involutions, G does not have an abelian maximal subgroup.Set H D QW so that H is normal in G. Since Q is not normal in G, there isc 2 G � H such that ŒQ; c� 6� hzi. Interchanging a and b if necessary, we mayassume Œa; c� 62 hzi and so Œa; c� D u 2 W � hzi and H D Q � hui. Since ha; ci isminimal nonabelian and a2 D z ¤ Œa; c� D u, we have ha; ci Š H2 so that z and u arethe only involutions which are squares in ha; ci. If c2 D z, then .ac/2 D a2c2Œc; a� D

zzu D u and Œa; ac� D u so that replacing c with c0 D ac (if necessary), we mayassume (writing again c instead of c0) from the start that c2 D u. We consider thesubgroup ha; bci noting that Œa; bc� D zu and a2 D z. It follows that ha; bci Š H2

so that .bc/2 2 fzu; zg. If zu D .bc/2 D b2c2Œc; b� D zuŒc; b�, then Œc; b� D 1.If z D .bc/2 D b2c2Œc; b� D zuŒc; b�, then Œc; b� D u so that Œc; ab� D uu D 1.Therefore, replacing b with b0 D ab if necessary and writing again b instead of b0,we may assume from the start that Œc; b� D 1. Hence a inverts the abelian subgrouphb; ci Š C4�C4. SetM D ha; b; ci and consider the abelian subgroupX D ha;W i oftype .4; 2/. We have CM .X/ D X , X is normal in G, Aut.X/ Š D8 and G=X Š E8.It follows that there is d 2 G � M centralizing X so that ha; d iW is abelian oforder 24 and therefore ha; d iW Š C4 � C4. Replacing d with a suitable elementin ha; d i � X , we may assume from the start that d2 D u. If Œc; d � D 1, then cdwould be an involution in G �W , a contradiction. Hence Œc; d � ¤ 1, Œa; d � D 1 andG D ha; b; c; d i and so it remains to determine the action of d on hb; ci Š C4�C4. Wehave hc; d i Š Q8 or H2 and so Œc; d � 2 fu; z; uzg. Also, Œb; d � D 1 or hb; d i Š H2

and so we have Œb; d � 2 f1; z; ug. If Œb; d � D z and Œc; d � D u, then d inverts hb; ci.But in that case ad centralizes hb; ci and so hb; cihad i would be an abelian maximal


subgroup of G, a contradiction. It follows that we have to consider the following eightpossibilities for Œb; d � and Œc; d �:

(1) If Œb; d � D 1 and Œc; d � D u, then hc; d i Š Q8 and hc; d i is normal in G, acontradiction.

(2) If Œb; d � D 1 and Œc; d � D z, then Œb; cd � D 1 and .cd/2 D c2d2Œd; c� D z D

b2 so that bcd is an involution in G �W , a contradiction.

(3) If Œb; d � D 1 and Œc; d � D uz, then Œad; c� D z and .c ad/2 D 1 so that cad isan involution in G �W , a contradiction.

(4) If Œb; d � D z and Œc; d � D z, then hbc; ad i Š Q8 and hbc; ad i is normal in Gsince Œa; bc� D zu, Œd; bc� D 1, Œb; ad � D 1, Œc; ad � D zu. This contradicts ourassumption.

(5) If Œb; d � D z and Œc; d � D uz, then Œad; c� D z and .c ad/2 D 1 so that cad isan involution in G �W , a contradiction.

(6) If Œb; d � D u and Œc; d � D u, then hc; d i Š Q8 and hc; d i is normal in G sinceŒa; c� D u, Œb; c� D 1, Œa; d � D 1, Œb; d � D u. This contradicts our assumption.

(7) If Œb; d � D u and Œc; d � D z, then Œbc; ad � D 1 , .bc/2 D zu, .ad/2 D zu, sothat bcad is an involution in G �W , a contradiction.

(8) If Œb; d � D u and Œc; d � D uz, then Œad; c� D z and .c ad/2 D 1 so that cad isan involution in G �W , a contradiction.

In the rest of the proof we assume that jˆ.G/j > 4. First we consider the case thatG possesses an element v of order 4 such that CG.v/ is nonabelian. By Lemma 92.4,there is a quaternion subgroupQ D ha; bi contained in CG.v/ such thatQ\hvi D f1g

and we set Z.Q/ D hzi. Let A be a maximal abelian subgroup of G containinghvi � hai Š C4 � C4. It follows that A is a maximal normal abelian subgroup ofG since A > Z.G/ and G is of class 2. Moreover, we have �1.A/ D �1.G/ �

Z.G/ and ˆ.G/ � �1.A/. Suppose there is x 2 A such that x2 62 hv2; zi and sohx2; v2; zi Š E8. By Lemma 92.5, b does not commute with x and so hb; xi Š H2

since x2 ¤ b2 D z. But then hb; x; vi D hb; xi � hvi Š H2 � C4 which contradictsLemma 92.4. We have proved jÃ1.A/j D 4 so that A is of type .4; 4; 2; : : : ; 2/,AQ D Ahbi, CG.hv; ai/ D A, ˆ.AQ/ D hv2; zi Š E4 and .AQ/0 D hzi. Sincejˆ.G/j > 4, there is an element u 2 G � .AQ/ such that u2 62 hv2; zi and thereforehu2; v2; zi Š E8. Suppose that u commutes with an element x of order 4 in hv; ai.We may set hv; ai D hx; yi for a suitable element y of order 4 in hv; ai. The fact thatCG.hx; yi/ D A implies that Œu; y� ¤ 1 so that hu; yi Š H2 because u2 ¤ y2. But�1.hu; yi/ D hu2; y2i and so x2 62 hu2; y2i (because hu2; x2; y2i Š E8). We havehu; y; xi D hu; yi � hxi Š H2 � C4, contrary to Lemma 92.4. We have proved thatu does not commute with any element of order 4 in hv; ai. Suppose that u does notinvert an element x of order 4 in hv; ai. Then hu; xi Š H2 implies that xu D xu2. Ify is any element of order 4 in hv; ai with y2 ¤ x2 and yu ¤ y�1, then yu D yu2.


But then .xy/u D .xu2/.yu2/ D xy, a contradiction. Hence yu D y�1 and notingthat .xy/2 D x2y2 ¤ x2, we also get .xy/u D .xy/�1. But we get then

xu D ..xy/y�1/u D .xy/u.yu/�1 D .xy/�1y D x�1y�1y D x�1;

a contradiction. We have proved that u inverts hv; ai. If u commutes with any elements 2 Q�hai D Q�A, then s2 D z and hu; v; si D hu; vi�hsi Š H2 �C4, contrary toLemma 92.4. Hence u does not commute with any element ofQ�hai. Suppose that uinverts an element w 2 Q� hai so that Œu;w� D z. But then Œu; aw� D Œu; a�Œu;w� D

zz D 1, a contradiction. Hence u does not invert any element in Q � hai. Sincehu; bi Š hu; abi Š H2, the above results imply Œu; b� D Œu; ab� D u2. But thenu2 D Œu; ab� D Œu; a�Œu; b� D zu2 which gives z D 1, a final contradiction.We have proved that the centralizer of each element of order 4 is abelian. Let again

A be any maximal normal abelian subgroup of G so that we have �1.G/ D �1.A/ D

Z.G/ since for each x 2 G � A, CA.x/ D �1.A/. Also, �1.A/ < A and jA W

�1.A/j D jÃ1.A/j. Let x 2 G � A and consider any element y 2 A with y2 62 hx2i

and suppose that yx ¤ y�1 so that (noting that hx; yi Š H2) yx D yx2. Assumefurther that there is v 2 A with v2 62 hx2; y2i. If vx ¤ v�1, then vx D vx2 and.vy/x D .vx2/.yx2/ D vy, a contradiction since o.vy/ D 4. Thus vx D v�1.Since .vy/2 62 hx2; y2i, we also get .vy/x D .vy/�1 and this implies yx D y�1, acontradiction. Hence, yx D y�1 for all y 2 A with y2 62 hx2i and so x inverts A. Wehave proved that in case jA W �1.A/j � 8, each element x 2 G � A inverts A and sojG W Aj D 2 and G D Ahxi which gives groups in part (b6).Assume that jA W �1.A/j D jÃ1.A/j D 4 so that A D hy; zi�1.A/ with hy; zi Š

C4 � C4 and Ã1.A/ D hy2; z2i. Since jˆ.G/j > 4, there is an element u 2 G � A

such that u2 62 hy2; z2i. By the arguments of the previous paragraph, u inverts A andAhui is a group in part (b6). Assume that Ahui ¤ G and let x 2 G � .Ahui/. Ifx2 62 hy2; z2i, then x would invert A and then xu 62 A and xu would centralize A,a contradiction. Hence x2 2 hy2; z2i and we may set (say) x2 D z2. If x invertssome element s of order 4 in hy; zi, then xu centralizes s, a contradiction. Sincehy; xi Š hyz; xi Š H2, we have yx D yz2, .yz/x D .yz/z2 and then .yz/z2 D

.yz/x D yxzx D yz2zx , which gives zx D z, a contradiction.It remains to consider the case where jA W �1.A/j D 2 for each maximal normal

abelian subgroup A ofG. We haveA D hyi�1.A/ and note that eachmaximal abeliansubgroup of G is normal in G and so G does not possess a subgroup isomorphic toC4 � C4. For each x 2 G � A, we have Œx; y� ¤ 1. Suppose that each x 2 G � A

inverts y. Then jG W Aj D 2, G D Ahxi, hx; yi Š Q8 or H2 and ˆ.G/ D hx2; y2i

which is of order � 4, contrary to our assumption. Hence there is x 2 G �A such thatyx ¤ y�1 and so hx; yi Š H2. If x2 D y2, then replacing x with x0 D xy, we get.x0/2 ¤ y2 and x0 does not invert y. Writing again x instead of x0, we may assumefrom the start that x2 ¤ y2 and yx D yx2 and we have ˆ.Ahxi/ D hx2; y2i Š E4.Since jˆ.G/j > 4, there is u 2 G � .Ahxi/ with u2 62 hx2; y2i. Then we haveyu 2 fy�1; yu2g since u2 ¤ y2 and hu; yi Š H2. We have Œx; u� ¤ 1 and so


hx; ui Š H2 which gives .xu/2 2 fu2; x2g and so in any case .xu/2 ¤ y2. Firstsuppose yu D y�1. Then yxu D .yx2/u D y.y2x2/. Since hy; xui Š H2, we getyxu D yy2 or yxu D y.xu/2. But from the above, yxu D y.y2x2/ and so we musthave .xu/2 D y2x2, contrary to the above result that .xu/2 2 fu2; x2g. Hence wehave the second possibility yu D yu2 and this gives yxu D .yx2/u D y.u2x2/ andso u2x2 D y2 (which is not possible since u2 62 hx2; y2i ) or u2x2 D .xu/2. But thenwe have u2x2 D .xu/2 D x2u2Œu; x�which gives Œu; x� D 1, a final contradiction.

Corollary 92.7 ([Bla7]). Let G be a finite p-group which possesses nonnormal sub-groups and let R.G/ be the intersection of all nonnormal subgroups. If R.G/ > f1g,then p D 2, jR.G/j D 2 and G is one of the following groups:

(a) G Š Q8 � C4 � E2s , s � 0,

(b) G Š Q8 � Q8 � E2s , s � 0,

(c) G has an abelian maximal subgroup A of exponent > 2 and an element x 2

G � A of order 4 which inverts each element in A.

Proof. LetH be a minimal nonabelian subgroup of G which is not isomorphic to Q8.By Proposition 1.26, we get p D 2, H Š H2 D ha; b j a4 D b4 D 1; ab D a�1i

and so R.G/ D R.H/ D hb2i is or order 2. If all minimal nonabelian subgroupsof G are isomorphic to Q8, then Corollary A.17.3 implies that G D Q � V , whereQ Š Q2n is generalized quaternion of order 2n, n � 4, and exp.V / � 2 and so againR.G/ D Z.Q/ is of order 2.We are in a position to use Theorem 92.6. Suppose that G is a group of part (a) of

that theorem. For each a 2 A, o.ax/ D 4 .x 2 G � A/ and so haxi is not normal inG (since jG0j > 2). This gives x2 D .ax/2 D ax2ax and ax D a�1 and so we haveobtained a group of part (a) of our corollary. If G is a group of Theorem 92.6(b2),then we have hai \ hbi D f1g, and hai and hbi are nonnormal in K, a contradiction.If G is a group of Theorem 92.6(b7), then hd i \ had i D f1g and hd i and had i arenonnormal inK, a contradiction. Hence there remain groups given in parts (a), (b) and(c) of our corollary.

Appendix 16

Some central products

In this section we consider central products of some small 2-groups. Recall that agroup G is a central product of its subgroups A and B , if G D AB and ŒA;B� D f1g;we writeG D A�B . In that case,A\B � Z.G/. IfA\B D f1g, thenA�B D A�B .It is possible to define a central product of arbitrary number of groups. Central productsof a finite number of groups have appeared in �4.The following two exercises are cited many times in this book.

Exercise A. Suppose that a p-group G D A � C , where C is cyclic of order > p andA \ C D �1.C /. Let C be the set of cyclic subgroups of A of order p2 containing�1.C / and K the nonempty set of all subgroups of order p in G not contained in A.Write jC j D t . Then jKj D tp, and so

c1.G/ D jKj C c1.A/ D tp C c1.A/:(A)

In particular,

c1.D2n � C/ D 2 1C .2n�1 C 1/ D 2n�1 C 3;(B)

c1.Q2n � C/ D 2.2n�2 C 1/C 1 D 2n�1 C 3;(C)

c1.SD2n � C/ D 2.2n�3 C 1/C 2n�2 C 1 D 2n�1 C 3:(D)

We also have c1.M2n � C/ D 2 2 C 3 D 7. If A D H2 D ha; b j a4 D b4 D

1; ab D a3i and H2 \ C D ha2b2i, then (A) is not applicable (in that case, t D 0).If H2 \ C D hb2i, then c1.G/ D 2 4C 3 D 11. If H2 \ C D ha2i, then c1.G/ D

2 2C 3 D 7.

Solution. SinceG=A is cyclic, every element of G of order p is contained in A�2.C /,so one may assume from the start that jC j D p2. Take x 2 G � A of order p. Thenx D ac, where a 2 A � C and hci D C . It follows from apcp D .ac/p D xp D 1

that ap D c�p so o.a/ D p2. Therefore, every element x 2 G � A of order p iscontained in ZC , where �1.C / < Z < A, Z is cyclic of order p2 (in our concretecase, Z D hai). If V < A is another cyclic subgroup of order p2 with �1.C / < V ,then ZC \ VC D C and VC has exactly p subgroups of order p not containedin C (note that ZC Š VC Š Cp2 � Cp). If L1 be the set of subgroups of orderp in ZC that are not contained in C , and let L2 be defined similarly for VC ; then


jL1j D jL2j D p, L1 \ L2 D ¿ and members of sets L1 and L2 are not containedin A (indeed, A \ ZC D Z and A \ VC D V ). Let C and K be defined as in thestatement of exercise. Then jKj D

P

Z2CŒc1.ZC/ � 1� D tp, where t D jC j. Thus,

c1.G/ D jKj C c1.A/ D tp C c1.A/, where t D jC j, and this proves formula (A).

In particular, if A of our exercise is irregular of order ppC1, then �1.C / D Ã1.A/

so

c1.A � C/ D c1.A/C pc2.A/ D c1.A/C pppC1 � .p � 1/c1.A/ � 1

p.p � 1/

DppC1 � 1

p � 1D 1C p C C pp ;

and this number is independent of the structure of A.

Exercise B. Let a 2-group G D A � Q, where Q Š Q2n and A \ Q D Z.Q/.Find c1.G/ if we know c1.A/ and the number t of cyclic subgroups of A of order 4containing �1.Q/.

Solution. Let T be a subgroup of order 2 of G not contained in A. Then AT \Q D

C Š C4 since all subgroups ofQ of order 4 are cyclic. Clearly, T < AC . By (A),ACcontains exactly 2t subgroups of order 2 not contained in A. If C1 is another cyclicsubgroup of order 4 inQ, then AC \ AC1 D A. Since c2.Q/ D 2n�2 C 1, we get

c1.G/ D 2t.2n�2 C 1/C c1.A/:(E)

Here t is the number of cyclic subgroups ofA containing Z.Q/ (if t D 0, then c1.G/ D

c1.A/ which is obvious). In particular, we have

c1.D2m � Q2n/ D 2.2n�2 C 1/C 2m�1 C 1 D 2n�1 C 2m�1 C 3;(F)

c1.Q2m � Q2n/ D 2.2m�2 C 1/.2n�2 C 1/C 1;(G)

c1.SD2m � Q2n/ D 2.2m�3 C 1/.2n�2 C 1/C 2m�2 C 1:(H)

From this we have c1.D8 � D8/ D c1.Q8 � Q8/ D 2.2C 1/.2 C 1/C 1 D 19 (sinceD8 �D8 Š Q8 �Q8) and c1.D8 �Q8/ D 2.2C 1/C 22 C 1 D 11; both these resultsare known.

1o. Let G D Q � C , where Q D ha; b j a4 D 1; a2 D b2; ab D a�1i Š Q8

and C D hc j c4 D 1i Š C4, Q \ C D Z.Q/ D ha2i D hc2i, Then jGj D 24,ˆ.G/ D G0 D Q0 is of order 2, d.G/ D 3 so exp.G/ D 4 and j�1j D 7. Next,Z.G/ D C , G=C Š E4 so C is contained in exactly 3 members of the set �1, andall these members are abelian of type .4; 2/. All other members of the set �1 arenonabelian since G is nonabelian (see Exercise 1.6(a)). Set ab D d and A D hai,B D hbi, D D hd i. The following seven distinct elements are involutions: a2, ac,a3c, bc, b3c, dc, d3c. The cyclic subgroups A;B;D;C contain exactly 8 distinct

A.16 Some central products 487

elements of order 4: a, a3, b, b3, d , d3, c, c3. It follows that all involutions of G havelisted so that c1.G/ D 7. Since exp.G/ D 4, we get c2.G/ D 16�8

2D 4. Next we list

the set �1. As we know, there are exactly three distinct abelian maximal subgroupsof G: A1 D ha; ci, A2 D hb; ci, A3 D hd; ci; all of them contain C D Z.G/. Itremains to find 7 � 4 D 4 nonabelian maximal subgroups of G. One of these is Q.Since abc D ab D a�1 and o.bc/ D 2, the subgroup D1 D ha; bci Š D8. Similarly,D2 D hd; bci Š D8 andD3 D hb; dci Š D8 are two remaining nonabelian membersof the set �1. Since D1, D2 and D3 are dihedral, it follows that Q is characteristicin G; obviously, C D Z.G/ is also characteristic in G and it is the unique cyclicsubgroup of order 4 not contained in Q. We have also G D Di � C for i D 1; 2; 3.In particular, Q8 � C4 Š D8 � C4 if the both products are of order 16. It followsfrom c1.G/ D 7 that �1.G/ > D1 so �1.G/ D G. Since jG W Qj D 2, G hasno elementary abelian subgroups of order 8. It follows that all four-subgroups are G-invariant. Since all subgroups of order 4 contain G0 D Ã1.G/ and c2.G/ D 4, weconclude that G contains exactly 7� 4 D 3 normal four-subgroups.

2o. In this subsection,G D Q�C , where Q2n Š Q D ha; b j a2n�1

D 1; a2n�2

D

b2; ab D a�1i Š Q2n , n > 3, C D hc j c4 D 1i Š C4 and Q \ C D Z.Q/. ThenjGj D 2nC1, ˆ.G/ D G0 D Q0 so d.G/ D 3 and j�1j D 7. Set A D hai, ˇ D a2n�3

,˛ D ˇ2; then o.˛/ D 2, o.ˇ/ D 4. It follows from .cˇ/2 D c2ˇ2 D c4 D 1 that� D cˇ is an involution. Set A1 D ha; ci; then A1 D hai � h�i is abelian of type.2n�1; 2/. Since cl.G/ D cl.Q/ D n � 1 > 2, A1 is the unique abelian maximalsubgroup of G.Let us find the numbers of elements of given order in G. If u is a generator of

A, then .ub3/2 D .ub/2 D ub2ub D ub2u�1 D b2 so o.ub/ D o.ub3/ D 4.Since all elements in Q � A have order 4, ˇ, ˇ3; aib, i D 1; 2; : : : ; 2n�1, are all2n�1 C 2 elements of order 4 in Q. Let Z D hzi < G be cyclic of order 4 such thatz 2 G � .A [ C/. Then QZ D G so Z \ Q D Z.Q/. We have z D uc, whereu 2 Q and hci D C . Then ˛ D z2 D u2c2 D u2˛ so u2 D 1, u D ˛ and z 2 C ,a contradiction. Thus, all cyclic subgroups of order 4 in G are contained in Q or C .There are two elements of order 4 in C . Thus, we obtained 2n�1 C 2C 2 D 2n�1 C 4

elements of order 4 so c2.G/ D 12.2n�1 C4/ D 2n�2 C2. By Exercise A (see formula

(C)), c1.G/ D 2n�1 C 3. Next, A1 contains exactly j�k.A1/j � j�k�1.A1/j D

2kC1 � 2k D 2k elements of orders 2k , 2 < k < n. Since

2n�1 C 3C 2n�1 C 4C .23 C C 2n�1/ D 2n C 7C 2n � 8

D 2nC1 � 1 D jG#j;

we conclude that c1.G/ D 2n�1 C 3, c2.G/ D 2n�2 C 2 and ck.G/ D 2 for k D

3; : : : ; n � 1.It remains to find five members of the set �1 � fQ;A1g, where A1 D ha; ci. We

have ˆ.G/ D G0 D ha2i D Ã1.A/. Since abc D ab D a�1 and .bc/2 D 1, we getD D ha; bci Š D2n . The group D contains exactly 2n�1 C 1 distinct involutions so


G�D contains exactly two involutions henceD is the unique dihedral member of theset �1. Thus,Q;A1 andD are all maximal subgroups of G containing A.Let D1 and D2 be nonabelian maximal subgroups of D; then D1;D2 2 �2 so

G=D1 Š E4 Š D2. Then Di (i D 1; 2) is contained in three maximal subgroupsD, Ui , Vi . By Proposition 13.18(b), exactly two among subgroups D, Ui , Vi are ofmaximal class, say D, Ui (i D 1; 2). Since G � D contains exactly two involutionsandDi < Ui , it follows that Ui Š SD2n . We denote Ui D SDi .Next, U D haci is another cyclic subgroup ofG of order 2n�1. Since cn�1.G/ D 2

and A G G, we get U G G. Set d D ac. Since G=U is abelian of type .2; 2/, exactlythree members of the set �1 containD, and one of them is A1. Thus,Q,D1, SD1 andSD2 are of maximal class. By Theorem 5.4, all other maximal subgroups of G are notof maximal class. We see thatQ andD1 are characteristic in G. It remains to find yettwo maximal subgroups of G. Set M D D1 � C and N D D2 � C ; then M ¤ N .Clearly,M Š N . Thus, �1 D fQ;D;SD1; SD2;M;N;A1g. Thus, we have provedthe following

Proposition A.16.1. (a) G D Q8 �C4 Š D8 �C4 if both these central products haveorder 24. All proper subgroups of G are metacyclic. The subgroup of G isomorphicwith Q8 is characteristic. G has exactly three subgroups Š D8, c1.G/ D 7.

(b) Let n > 3; then G D Q2n � C4 Š D2n � C4 Š SD2n � C4 provided jGj D 2nC1.Exactly five members of the set �1 have rank two, other two (namely,M and N ) haverank three. Two members of the set �1 that are isomorphic with Q2n and D2n , arecharacteristic in G. Exactly two members of the set �1 are isomorphic with SD2n .Next, c1.G/ D 2n�1 C 3, c2.G/ D 2n�2 C 2, ck.G/ D 2 .k D 3; : : : ; n � 1/.

Exercise 1. (a) Prove thatQ �C Š D �C Š SD �C , whereQ Š Q2m ,D Š D2m ,SD Š SD2m , C Š C2n , m > 3, n > 2, and all three central products have the sameorder 2mCn�1.

(b) Find ck.G/ for the group G from (a).

(c) List all maximal subgroups of the group G from (a).

Exercise 2. LetG D Q8 �Q8,H D D8 �D8, and F D Q8 �D8, jGj D jH j D jF j D

32. Let L 2 fG;H;F g. (a) Prove that G Š H but G 6Š F . (b) Find ck.L/, k D 1; 2.(c) Find the number of nonabelian subgroups of order 8 in L. (d) Find the number ofsubgroups of L isomorphic with Q8. (e) List all maximal subgroups of L.

Exercise 3. Find ck.G/, using the description of maximal subgroups ofG from Propo-sition A.16.1.

Exercise 4. LetG D M �N , whereM is a 2-group of maximal class and order 2n > 8

and N is nonabelian of order 8. Find the numbers of subgroups of maximal class andorders 8 and 2n in G.


Exercise 5. Let G be the central product with amalgamated centers of m copies of agroup of maximal class and order 2n, n > 3. (i) Find ck.G/ for all k 2 N. (ii) Describeall maximal abelian subgroups of G. (iii) Find the number of subgroups of maximalclass and order 2n in G.

Exercise 6. Let H D Q � C be a subgroup of a 2-group G such that Q Š Q2n ,C Š C4, andQ\C D Z.Q/. Suppose that G �H has no elements of order 4. Showthat H and Q are characteristic in G. (Hint. ��

2.H/ D hx 2 H j o.x/ D 4i D H

andQ is characteristic inH .)

Exercise 7. Prove that a 2-groupG D Q�C , whereQ is of maximal class,C is cyclicandQ \ C D Z.Q/, has no elementary abelian subgroups of order 8.

Exercise 8. Let G be a 2-group of Proposition A.16.1(b). Is it true that the number ofsemidihedral subgroups of order 2k , k 2 f4; : : : ; n � 1g, in G is even?

Exercise 9. Let G be a group of Proposition A.16.1(b) with n > 5. Prove that thenumber of subgroups of maximal class and order 2k in G equals 2n�kC2.

Solution. Let M < G be of maximal class and order 2k . Then C � M is of order2kC1 and CM D C � .CM \Q/, where CM \Q D Q2k . Next. CM has exactly 4subgroups of maximal class and order 2k so every subgroup ofQ of maximal class andorder 2k produces exactly three new subgroups of maximal class and order 2k in Gand these subgroups are not contained inQ. SinceQ contains exactly 2n�k subgroupsof maximal class and order 2k , the desired number is 2n�k C 3 2n�k D 2n�kC2.

Exercise 10. Let a 2-group G D QZ, where Q Š Q2n is maximal in G, Z Š C4 isnormal in G and CG.Z/ is abelian. Prove thatQ is a direct factor of G.

Solution. We proceed by induction on n. It suffices to show that Z.G/ Š E4. Supposethat n D 3. Then Z.G/ is of order 4 (Proposition 10.17). If Z.G/ is cyclic, thenZ.G/ D Z since Z is the only one cyclic subgroup of order 4 not contained inQ, andwe get a contradiction. We are done if n D 3. Now suppose that n > 3. Let nonabelianM < Q be maximal and set H D MZ. By induction, Z.MZ/ D R Š E4. Wehave CG.Z/ D AZ 2 �1, where A < Q is cyclic of index 2. In that case, AZ isabelian of type .2n�1; 2/. ThenH \AZ is noncyclic (indeed, Z is maximal cyclic inH ). Since H \ AZ is a maximal abelian subgroup of a nonabelian group H , we getR D Z.H/ � H \ AZ. Then CG.R/ � HAZ D HA D G, so Z.G/ D R Š E4.

3o. In this subsection we consider G D Q1 �Q2, where jGj D 25 and

Q1 D ha; b j a4 D 1; a2 D b2; ab D a�1i Š Q8;

Q2 D hc; d j c4 D 1; c2 D d2; cd D c�1i Š Q8:

Then G is extraspecial of order 25, and M1 D hc;Q1i, M2 D hd;Q1i, M3 D

hcd;Q1i 2 �1. For k D 1; 2, we have ck.G/ D ck.M1/C ck.M2/ C ck.M3/ � 2


ck.Q1/. By 1o, c1.Mi / D 7 and c2.Mi / D 4 for i D 1; 2; 3. Therefore, c1.G/ D

7C 7C 7� 2 D 19 and c2.G/ D 4C 4C 4� 2 3 D 6. Thus,Q1 andQ2 contain allcyclic subgroups of order 4 from G.All nonabelian subgroups of G of order 8 are contained in �2 and all cyclic sub-

groups of order 4 are normal inG. LetA < G be cyclic of order 4. Since CG.A/=A Š

E4, A is contained in exactly three abelian subgroups of order 8 so in 7 � 3 D 4 non-abelian subgroups of order 8. By �76, G contains exactly 20 nonabelian subgroups oforder 8. It follows from 1o that G has exactly two subgroups Š Q8, namely Q1 andQ2. By Lemma 64.1(q), G has no abelian subgroups of index 2.Let D be a dihedral subgroup (of order 8) in G. Since D0 D G0, we get D G G.

We have G D DCG.D/ (Lemma 4.3). Since Q1 and Q2 are the only quaternionsubgroups of G and they do not centralizeD, we get CG.D/ Š D8. Thus, Q8 �Q8 Š

D8 � D8.If A < G is cyclic of order 4, then CG.A/ Š Q8 � C4 .Š D8 � C4/. Since

G D D�CG.D/ and CG.D/ Š D8, it follows that G contains a subgroup isomorphicwith E8 (such subgroup is contained in D � Z, where Z is a noncentral subgroupof order 2 in CG.D/). If A and C are cyclic subgroups of order 4 in Q1 and Q2,respectively, then AC is abelian of type .4; 2/. In such a way we obtain all 9 abeliansubgroups of type .4; 2/ in G. Since all subgroups of order 8 are members of the set�2, there are in G exactly j�2j � 20� 9 D 6 elementary abelian subgroups of order 8.

4o. In this subsection G D Q � D, where jGj D 25, Q Š Q8 is as in 1o and

D D hc; d j c4 D d2 D 1; cd D c�1i Š D8. Then G is extraspecial.If e 2 D�hd i is an involution andM1 D Q�hci, M2 D Q�hd i,M3 D Q�hei,

then

c1.G/ D c1.M1/C c1.M2/C c1.M3/ � 2c1.Q/ D 7C 3C 3 � 2 1 D 11

so c2.G/ D jGj�c1.G/�1'.4/

D 10 since exp.G/ D 4. The number of nonabelian sub-groups of order 8 inG equals 20 (see �76). We have Q8 �Q8 6Š Q8 �D8 if both centralproducts are of order 32 (indeed, these groups have different numbers of involutions).If follows that if D < G is dihedral, then CG.D/ Š Q8 so the numbers of dihedraland quaternion subgroups in G are equal to 10.

Exercise 11. LetG D Q1 �Q2, whereQ1 andQ2 are generalized quaternion groups.Find ck.G/ for all k 2 N.

Exercise 12. Let G D M1 �M2, whereM1 Š M2m ,M2 Š M2n and jM1 \M2j D

min fjZ.M1/j; jZ.M2/jg. Find ck.G/ for all k 2 N.

Exercise 13. LetG D R�S , whereR and S are 2-groups of maximal class and orders2m, 2n, respectively, m > 3, n > 3, jGj D 2mCn�1. Find the numbers of nonlinearirreducible characters of each possible degree. Find k.G/.


Solution. We have jG W G0j D 24, G0 D ˆ.G/, Z.G/ D Z.R/ D Z.S/ is of order2 so G has a faithful irreducible character. The number of non-faithful irreduciblecharacters ofG equals the number of irreducible characters ofG=Z.G/ Š .R=Z.G//�.S=Z.G//, i.e., .2m�3 C 3/.2n�3 C 3/. Every faithful irreducible character of G is ofdegree 4 so the number of such characters equals jGj�jG=Z.G/j

42 D 2mCn�1�2mCn�2

42 D

2mCn�6. Note that k.G/ D jIrr.G/j.

Exercise 14. Let G D R � S , where R, S are 2-groups of maximal class. Find thenumber of nonabelian subgroups of order 8 in G.

Exercise 15. Let a 2-group G D QZ, where Q is a normal subgroup of maximalclass and Z is normal cyclic, Z \ Q D Z.Q/ and CG.Z/ is abelian. Is it true thatjZ.G/j > 2?

Exercise 16. Let G D HC be a p-group, where C D hci G G is cyclic of order p2

andH is a subgroup of maximal class and order ppC1, jG W H j D p. Find c1.G/.

Appendix 17

Alternate proofs of characterization theoremsof Miller and Janko on 2-groups,

and some related results

G.A. Miller [Mil9] has classified the minimal non-Dedekindian 2-groups. We offer,in Theorem A.17.1, another proof of Miller’s result. Then, in Theorem A.17.2, weclassify the 2-groups all of whose nonabelian maximal subgroups are of the formQ�

E, whereQ is generalized quaternion and exp.E/ � 2; our proof is based on Miller’stheorem. Corollary A.17.3 is a partial case of Theorem 90.1 classifying the 2-groupsall of whose minimal nonabelian subgroups have order 8.A nonabelian 2-group G is said to be generalized dihedral if it is nonabelian and

contains such subgroup A that all elements of the set G � A are involutions. ThenA is abelian, jG W Aj D 2, all subgroups of A are G-invariant, �1.A/ D Z.G/ andG=G0 is elementary abelian since �1.G/ D G (Burnside). It is easy to check thatA is the unique subgroup of G satisfying the above properties. Next, every minimalnonabelian subgroup of our group is isomorphic with D8. Nonabelian sections ofgeneralized dihedral groups are generalized dihedral.A p-group M � E is said to be an M�-group if M is of maximal class and E is

elementary abelian (we consider the group f1g as elementary abelian p-group). Theabove group is said to be an M�

3 -group if, in addition, jM j D p3. Nonabelian epi-morphic images ofM�-groups areM�-groups.Suppose that a p-group G D M � E, where M is nonabelian with cyclic center

and E > f1g, is elementary abelian. Suppose that nonabelian M1 < G has no directfactors of order p. We claim that thenM1 is isomorphic to a subgroup ofM . Indeed,it suffices to prove thatM1 \E D f1g. Assume, however, that X D M1 \E1 > f1g;then jX j D p since Z.M/ is cyclic. Since X � E, we getX 6� ˆ.G/ soG D X �G0

for some G0 < G. Then, by the modular law,M1 D X � .M1 \ G0/, contrary to thehypothesis. In particular, if jM1j D jM j, then G D M1 � E and M1 Š M . Next, ifM1 < G is minimal nonabelian, thenM1 is isomorphic to a subgroup ofM . In whatfollows, we use these facts freely.

Remarks. 1. Let G D M � C , where M is a p-group of maximal class and C D

hci is cyclic of order pn, n > 1. We claim that G contains a nonmetacyclic A1-subgroup H of order pnC2 with jH \M j D p2, unlessM is generalized quaternion.

A.17 Two theorems of Miller and Janko 493

Indeed, by Blackburn’s theorem (see Theorem 9.6),M contains a nonabelian subgroupD D hR; ai of order p3, where jRj D p2 and o.a/ � p2. Set u D ac; thenR \ hui D f1g, o.u/ D o.c/. We assert that L D hu;Ri D hui R is the desiredsubgroup. Indeed, jL0j D p since L is nonabelian, and d.L=L0/ D 2 so L is minimalnonabelian, by Lemma 65.2(a); we also have jLj D pnC2. If M is not generalizedquaternion, one can take from the start R Š Ep2; in that case, L is not metacyclicsince Ep3 Š R � �1.C / < L. (If M is generalized quaternion, then j�1.G/j D 4,so, by Lemma 65.1, all A1-subgroups of G are metacyclic.)

2. Suppose that a group G of order 2m > 24 is not of maximal class. LetH 2 �1 beof maximal class. Then the set �1 contains exactly four members, say H D H1, H2,H3, H4, of maximal class (Theorem 12.12(c)). Suppose that all nonabelian membersof the set �1 are M�-groups. We claim that then G itself is anM�-group. Assumethat our claim is false. Let Z < H be cyclic of index 2; then, since jH j � 16, Z (oforder � 8) is characteristic in H so normal in G. Next, G contains a normal abeliansubgroup R of type .2; 2/ (Lemma 1.4); then R\H D �1.Z/. Since A D RZ 2 �1

is not an M�-group, it must be abelian, by hypothesis, and jA W Zj D 2. Let F bea nonabelian maximal subgroup of H . Then RF 2 �1 since jRF j D jH j, and RFis anM�-group which is not of maximal class since jRF j � 16. It follows that R D

Z.RF /. Since R < A (otherwise, G D RA is of class 2 < 3 � cl.H/, by Fitting’slemma; see Introduction, Theorem 21), we get CG.R/ � A.RF / D AF D G soR D Z.G/. If L < R is of order 2 and L 6� H , then G D HL D H � L is anM�-group.

3. Let G be a p-group which is not of maximal class and A;H 2 �1, where A isabelian and H is of maximal class. Then jZ.G/j D p2 and G D HZ.G/. Indeed,G0 D H 0 has index p3 in G (Theorem 12.12(a)). Then jZ.G/j D 1

pjG=G0j D p2

(Lemma 64.1(q) = Lemma 1.1) so G D HZ.G/.

Theorem A.17.1 ([Mil9]). If G is a minimal non-Dedekindian 2-group, then G is ei-ther minimal nonabelian or isomorphic to Q16.

Proof. One may assume that G is not an A1-group (= minimal nonabelian) so jGj D

2m > 23. Let H D Q � E 2 �1, where Q Š Q8 and exp.E/ � 2. Suppose thatE D f1g; then m D 4. If CG.Q/ 6� Q, then G D QZ.G/ so Z.G/ is cyclic oforder 4 since G is not Dedekindian. Then G D Q � Z.G/ Š D8 � Z.G/ so G is notminimal non-Dedekindian since D8 is non-Dedekindian. Thus, CG.Q/ < Q so G isof maximal class (Proposition 10.17) hence G Š Q16. Next assume that jGj > 24 soE > f1g for all nonabelian H 2 �1. We have H 0 D Q0 G G and H=Q0 < G=Q0 iselementary abelian. Assume that G=Q0 is not an A1-group. Then there is in G=Q0 anonabelian maximal subgroup F=Q0 D .Q1=Q

0/ � .E1=Q0/. where Q1=Q

0 Š Q8

and exp.E1=Q0/ � 2. Then .Q1=Q

0/ \ .H=Q0/ is elementary abelian and maximalinQ1=Q

0, a contradiction. Thus, NG D G=Q0 is either abelian or minimal nonabelian.

(i) Let NG be minimal nonabelian; then jG0j D 4. Since exp. NH/ D 2 and j NG WNH j D 2, we get j NH j � 8 (Lemma 65.1). If NH Š E4, then j NGj D 8 so jG W G0j D 4.


In that case, G is of maximal class (Taussky’s theorem) so G Š Q16, contrary tom > 4. Now let NH Š E8. Since NG is generated by elements of order 4, it has twodistinct maximal subgroups NA and NB of exponent 4. Then A and B are abelian (if, forexample, A is nonabelian, then A0 D Q0 and exp.A=A0/ D 2, a contradiction). In thatcase, A \ B D Z.G/ so jG0j D 2 (Lemma 1.1), a contradiction.

(ii) Let NG be abelian; then G0 D H 0 D Q0 and G D Q � CG.Q/ (Lemma 4.3).If CG.Q/ has a cyclic subgroup L of order 4, then Q � L is not Dedekindian. Thus,exp.CG.Q// D 2 so CG.Q/ D Z.G/. If Z.G/ D Q0 � E1, then G D Q � E1 isDedekindian, a final contradiction.

A 2-group G is said to be a Q�-group if G D Q � E, where Q is generalizedquaternion and E is elementary abelian.

Theorem A.17.2. Suppose that all nonabelian maximal subgroups of a nonabelian2-group G are Q�-groups, jGj D 2m. Then G is either minimal nonabelian or aQ�-group.

Proof. Assume that G is neither minimal nonabelian nor of maximal class (if G is ofmaximal class, it is generalized quaternion). We also may assume, in view of LemmaA.17.1, thatm > 4. Then all proper nonabelian subgroups ofG areQ�-groups. Thereis a nonabelian H D Q � E 2 �1, where Q is generalized quaternion and E iselementary abelian. Suppose that E D f1g. Then, by Remark 2, G is a Q�-group.Next we assume that E > f1g for every choice of nonabelianH 2 �1.In view of Lemma A.17.1, one may assume thatH is chosen so that jQj > 23. Then

H 0 D Q0 is cyclic of order> 2 and normal inG. In that case,A D CG.�2.Q0// 2 �1

is abelian since exp.Z.A// > 2. Since E < A, we get CG.E/ � HA D G so thatE < Z.G/ (<, since Z.Q/ < Z.G/ and Z.Q/ 6� E). It follows from jG0j > 2 thatA is the unique abelian member of the set �1 (Lemma 65.2(c)). Take a nonabelianF 2 �1 � fHg (F exists, by Exercise 1.6(a)) and assume that E 6� F . Then there isX � E of order 2 such that X 6� F . In that case, G D F � X is aQ�-group, and weare done. Therefore, one may assume that E < ˆ.G/. Write NG D G=E. Note that ifL 2 �1 is nonabelian, then NL is either elementary abelian or anM�-group (generallyspeaking, E is not a direct factor of L). By the above, NG contains a maximal subgroupNH , which is generalized quaternion of order > 8. In view of Remark 2, the followingtwo possibilities for NG must be considered.

(i) Let NG be not of maximal class. Then NG D NQ � NC , where j NC j D 2 and NG hasa subgroup NH D NQ of maximal class and index 2 (Remark 2). Since E < Z.G/ andj NC j D 2, the subgroup C G G is abelian and C \Q � E \Q D f1g so G D Q C ,a semidirect product with kernel C . If F is a nonabelian maximal subgroup of Q(recall that jQj > 8), then F C 2 �1 is an Q�-group so FC D C � F and henceexp.C / D 2. Since Q is generated by its nonabelian maximal subgroups, we getG D Q � C so that G is aQ�-group.


(ii) Now let NG be of maximal class. In that case, d.G/ D 2 since E < ˆ.G/, andhence, by Schreier’s Theorem A.25.1, we get d.F / � 3 for all F 2 �1. It followsthat jEj D 2. Since E 6� G0 (otherwise, jG W G0j D 4 so, by Taussky’s theorem,G is of maximal class), we get E \ G0 D f1g; then G0 is cyclic of index 8 in G andG=G0 is abelian of type .4; 2/ since d.G/ D 2. Let A=G0 and B=G0 be two distinctcyclic subgroups of order 4 in G=G0. Since abelian epimorphic images of Q�-groupshave exponent 2, it follows that A and B are abelian maximal subgroups of G soA \ B D Z.G/. In that case, jG0j D 2 < jH 0j, a final contradiction.

Corollary A.17.3 (Janko; see Theorem 90.1). If all minimal nonabelian subgroups ofa nonabelian 2-group G are isomorphic to Q8, then G is aQ�-group.

Proof. We use induction on jGj. By induction, every proper nonabelian subgroup ofG is aQ�-group. Now the result follows from Theorem A.17.2.

A 2-group G is said to be a D�-group if G D D � E, where D is dihedral and Eis elementary abelian. (Of course,Q�- andD�-groups areM�-groups.)

Proposition A.17.4. Suppose that all nonabelian maximal subgroups of a nonabelian2-group G areD�-groups. Then one of the following holds:

(a) G is aD�-group.

(b) G is a generalized dihedral group of order 25 with abelian subgroup of type.4; 4/. The group G is special, d.G/ D 3.

Proof. Suppose that G is neither minimal nonabelian nor a D�-group. All minimalnonabelian subgroups of G are isomorphic to D8 so generated by involutions. Then,by Theorem 10.33, G D C A is a generalized dihedral group; here A is abelian ofindex 2 in G and all elements of the set G � A are involutions which invert A, andexp.A/ > 2. Since G is not dihedral, A is noncyclic. Let A2 � A be of type .4; 4/;then A1 D C A2 is neither dihedral nor a nontrivial direct product so A1 D A. Itfollows that A has no proper subgroups of type .4; 4/. If A D L�A0, where jLj D 2,then L is a direct factor of G; in that case G is a D�-group. Thus, assuming thatall invariants of A are greater 2, we conclude that A is abelian of type .4; 4/. Next,Z.G/ D �1.A/ � G0. By Taussky’s theorem, jG W G0j > 4, so Z.G/ D G0. It followsfrom �1.G/ D G that G0 D ˆ.G/ so G is special.

Lemma A.17.5. Suppose that all nonabelian maximal subgroups of a nonabelian 2-group G areM�

3 -groups and jGj D 2m. Then one of the following holds:


(b) G is of maximal class and order 16.

(c) The central product G D M � C , whereM is nonabelian of order 8 and C iscyclic of order 4, m D 4.


(d) G is generalized dihedral, m D 5, with abelian subgroup A of type .4; 4/ (as inProposition A.17.4(b)).

(e) G is anM�3 -group.

Proof. Groups (a–e) satisfy the hypothesis. One may assume thatG is neither minimalnonabelian nor of maximal class; then m > 3. In that case, all proper nonabeliansubgroups of G are M�

3 -groups so all A1-subgroups of G have the same order 8(Remark 1). Next, it follows from Proposition 10.17(a) that, if m D 4, then G is oneof groups of parts (a–c), (e). In what follows we also assume that m > 4 and G is notanM�

3 -group.Let M < G be an A1-subgroup; then jM j D 8. Let M < H 2 �1, where

H D M � E and exp.E/ D 2 since m > 4. Set D D hH 0 j H 2 �1i. ThenD � G0 \ �1.Z.G// .� ˆ.G// is elementary abelian and all maximal subgroupsof NG D G=D are abelian. It follows that NG is either elementary abelian or minimalnonabelian generated by involutions so of order 8 hence, in the second case, NG Š D8

(Lemma 65.2(c)).

(i) Assume that jDj D 2; then exp. NG/ D 2 since m > 4, so G0 D D and all A1-subgroups of G are normal. LetM < G be anA1-subgroup. ThenG D M �CG.M/

(Lemma 4.3). It is easily seen that, in view ofm > 4, exp.CG.M// D 2 so CG.M/ D

Z.G/. If Z.G/ D Z.M/ �E, when G D M �E is anM�3 -group.

In what follows we assume that jDj > 2. If U < G is nonabelian of order 2n, thend.U / D n � 1 since jˆ.U /j D 2.

(ii) Suppose that exp. NG/ D 2. Let M < G be minimal nonabelian; then H D

MCG.M/ 2 �1. It follows from jDj > 2 that there is anA1-subgroupM1 < G suchthatM 0

1 ¤ M 0. In view of Corollary A.17.3 and Proposition A.17.4, one may assumethat there are in G two nonisomorphic minimal nonabelian subgroups. Therefore, onemay assume from the start thatM Š Q8. ThenM\M1 D f1g so jMM1j D 26, by theproduct formula. Set U D hM;M1i; then d.U / � d.M/C d.M1/ D 4 < 6� 1 so, bythe previous paragraph, U D G. Using Remark 1, we get ŒM;M1� > f1g (otherwise,M � M1 contains an A1-subgroup of order 24). Therefore, one may assume that Mis not normal in U . Then some cyclic subgroup C1 < M1 does not normalize somecyclic subgroup C < M (of order 4). Since U1 D hC;C1i of order � 24 is generatedby two elements and 2 < 4 � 1, we get U1 D G, and we conclude that d.G/ D 2. Itfollows that G is minimal nonabelian (Lemma 65.2(a)), contrary to the assumption.Now we assume that NG Š D8. Since D < G0, we get jG W G0j D j NG W NG0j D 4 so

G is of maximal class (Taussky), a contradiction since jDj > 2 D exp.D/.

Theorem A.17.6. Suppose that all nonabelian maximal subgroups of a nonabelian2-group G areM�-groups. Then one of the following holds:


(b) The central product G D M �C is of order 16,M is nonabelian of order 8 andC is cyclic of order 4.


(c) G is generalized dihedral of order 25 with abelian subgroup A of type .4; 4/ asin Proposition A.17.4.

(d) G is anM�-group.

Proof. Groups (a–d) satisfy the hypothesis. The groupM �C , whereM is of maximalclass and C is cyclic of order > 2, satisfies the hypothesis if and only if M \ C D

Z.M/, jM j D 8 and jC j D 4.Assume that the theorem is false. One may assume, in view of Lemma A.17.5, that

G contains a maximal subgroup H D M � E, where M is of maximal class andorder > 8 and E is elementary abelian. Then M 0 D H 0 is cyclic of order � 4 andG-invariant. Let V D �2.H

0/; then V 6� Z.M/ so CG.V / is abelian of index 2 in Gsince the center of anyM�-group is elementary abelian. It follows from jG0j > 2 thatCG.V / D A is the unique abelian maximal subgroup of G (Lemma 65.2(c)). SinceE < A\H , it follows that E < Z.G/ (< since �1.M

0/ � Z.G/ and �1.M0/ 6� E).

Assume that E D f1g and G is not of maximal class. Then the set �1 has exactlythree members which are not of maximal class (Theorem 12.12(c)) so they are eitherabelian or M�-groups. If the set �1 has three distinct abelian members, we get jG W

Z.G/j D 4; then jM j D 8, jGj D 16 and G either anM�-group or as stated in (b). IfjGj > 16, then, by Remark 2, G must be anM�-group; then E > f1g, contrary to theassumption. Next we assume that E > f1g for every choice of H .By Lemma 65.2(c), H has only one abelian maximal subgroup, say A1, since

jH 0j > 2 so H \ A D A1. If E 6� ˆ.G/, then G D X � G0, where X � E isof order 2, G0 2 �1. However, by hypothesis, G0 is an M�-group so G is also anM�-group. Next we assume that E < ˆ.G/.Suppose that NG D G=E is not of maximal class. Since . NH Š/ NM < NG, we get

exp. NG/ � exp. NM/ D exp.M/ � 8. By Remark 2, we get NG D NM � NC wherej NC j D 2. Also, C G G is abelian and C \ M D E \ M D f1g so G D M C , asemidirect product with kernel C . If F is a nonabelian maximal subgroup ofM , thenF C is anM�-group, by hypothesis, so F GFC , and we conclude that FC D C �F

hence C is elementary abelian. Since M is generated by its two distinct nonabelianmaximal subgroups, we get G D M � C so that G is anM�-group.Next we assume that NG is of maximal class. In that case, d.G/ D 2 since E <

ˆ.G/, and hence, by Schreier’s Theorem A.25.1, we get d.F / � 3 for all F 2 �1.It follows that jEj D 2. Since E 6� G0 (otherwise, by Taussky’s theorem, G is ofmaximal class), we get E \ G0 D f1g and so G=G0 is abelian of type .4; 2/ sinced.G/ D 2. Let U=G0; V=G0 < G=G0 be distinct cyclic of order 4. Then A;B areabelian since exp.X=X 0/ D 2 for any M�-group X . We have A \ B D Z.G/ sojG0j D 2, m D 4, a final contradiction.

Theorem A.17.7. Suppose that all nonabelian maximal subgroups of a nonabelianp-group G, p > 2, are M�

3 -groups. Then either G is an M�3 -group or one of the

following holds:



(b) G is of maximal class and order p4.

(c) G D M �C is of order p4, whereM is nonabelian of order p3 and exponent pand C is cyclic of order p2.

(d) G is extraspecial of order p5 and exponent p.

(e) G is special of order p5, d.G/ D 3.

(f) G is special of order p6 and exponent p, d.G/ D 3.

(g) G is of order p5, jG0j D p3, Z.G/ < G0 is abelian of type .p; p/. If R < Z.G/is of order p, then cl.G=R/ D 3, G has no abelian subgroups of index p.

Proof. Suppose that the theorem holds for all groups of order < jGj. Then all propernonabelian sections of G areM�

3 -groups. Groups (a–d), (f) and also groups of expo-nent p from parts (e) and (g) satisfy the hypothesis. Set jGj D pm. One may assumethat G is not minimal nonabelian; then m > 3.Let M < G be minimal nonabelian; then jM j D p3. Assume that CG.M/ has

a cyclic subgroup C of order p2. Then jM � C j D p4, by Remark 1 and we getjM \ C j D p and Z.M � C/ D C so G D M � C , the group of part (c). IfCG.M/ < M , then G is of maximal class (Proposition 10.17). However, G hasno proper subgroups of maximal class and order p4, by hypothesis, so jGj D p4

(Theorems 9.5 and 9.6). In what follows we assume that m > 4.Let D be generated by derived subgroups of all nonabelian members of the set

�1; then D � G0 \ �1.Z.G// � ˆ.G/. If M < G is minimal nonabelian, thenM < H 2 �1, where H is an M�

3 -subgroup so that M0 D H 0 G G. In that case,

H=H 0 is elementary abelian. If a p-group G has a nonabelian maximal subgroup, ithas at least p such subgroups (Exercise 1.6(a)). It follows that all maximal subgroupsof G=D are abelian and at least p of them are elementary soG=D is either elementaryabelian or minimal nonabelian generated by elements of order p (in the last case,G=D is nonabelian of order p3 and exponent p by Lemma 65.1). By Lemma 65.1(u),jG0j � p3.

(i) Suppose that jDj D p. Then G=D is elementary abelian so G0 D D since m >

4. Let M < G be minimal nonabelian. Then, by Lemma 4.3, G D MCG.M/ andexp.CG.M// D p (Remark 1). If CG.M/ is abelian, then CG.M/ D Z.G/ D M 0�E

so G D M �E is anM�3 -group.

Now assume that CG.M/ is nonabelian (of exponent p). Let N � CG.M/ beminimal nonabelian. SinceM 0 D D D N 0, we getM \ N D Z.M/ D Z.N /. ThenMN is extraspecial so it is not anM�

3 -group, and hence G D MN is extraspecial oforder p5 and exponent p. In that case, G is as in (d).

(ii) Now let jDj > p. Then there are nonabelian H;F 2 �1 with H 0 ¤ F 0. Theset �1 has at most one abelian member since jG0j � jDj > p (Lemma 64.1(u)). ThenH=H 0 and F=F 0 are distinct elementary abelian so �1.G=D/ D G=D. Since p > 2


and cl.G=D/ � 2, we get exp.G=D/ D p. Therefore, if G=D is minimal nonabelian,then jG=Dj D p3 (Lemma 65.1). Next, exp.G/ � exp.D/ exp.G=D/ D p2.

(ii1) Assume that the quotient group G=D is minimal nonabelian (of order p3);then d.G/ D d.G=D/ D 2. Since jG0 W Dj D p, we get jDj D p2 and jG0j D p3

so jGj D jDjjG=Dj D p5. Let F and H be such as in the previous paragraph. ThenF D M � H 0 and H D M1 � F 0, where M and M1 are minimal nonabelian (notethat F 0H 0 � ˆ.G/ so H 0 < F and F 0 < H ). Since F=H 0 < G=H 0 is nonabelianof order p3 and d.G=H 0/ D 2, it follows from Proposition 10.17 that G=H 0 is ofmaximal class. Similarly, G=F 0 is of maximal class. If G has an abelian subgroup ofindex p, then p5 D jGj D pjG0jjZ.G/j D p6 (Lemma 1.1), a contradiction. Thus,all maximal subgroups of G are nonabelian and G is as in (g). It is easy to check thatif exp.G/ D p, then G satisfies the hypothesis (see the last paragraph of (i)).

(ii2) Now let G=D be elementary abelian; then G0 D D D ˆ.G/ and cl.G/ D 2.

Assume that exp.Z.G// > p. Let C � Z.G/ be cyclic of order p2; then C is notcontained in any nonabelianH 2 �1. IfH D M �E is as above, thenM �C is not asubgroup of anM�

3 -group (Remark 1), and we conclude that G D M �C is as in (c).

Now Let exp.Z.G// D p. Assume, in addition, that a subgroup X < Z.G/ of orderp is not contained in ˆ.G/. Then G D X � G0, where G0 2 �1, so G0 D M0 �E0,where E0 is elementary abelian andM0 is nonabelian of order p3. In that case, G D

M0 � .E0 �X/ is anM�3 -group.

Let, in what follows, Z.G/ � ˆ.G/; then Z.G/ D G0 D ˆ.G/ D D � Z.G/ soG is special. LetM < G be minimal nonabelian. ThenMˆ.G/=ˆ.G/ D MD=D Š

M=.M \D/ Š Ep2 .

If d.G/ D 2, then G is minimal nonabelian (Lemma 65.2(a)), a contradiction.

Suppose that d.G/ > 3. Then MD=D is contained in two distinct maximal sub-groups, say F=D and H=D, of G=D. SinceM is a direct factor in F and H (see theparagraph preceding Remark 1), we get NG.M/ � FH D G so M G G. It followsthat G D MCG.M/ since G0 D ˆ.G/ D D � CG.M/ and Sylow p-subgroups ofAut.M/ are nonabelian of order p3 and exponent p. Assume that CG.M/ has a mini-mal nonabelian subgroup N and suppose thatM \N D f1g. It follows from Remark1 that exp.M/ D p D exp.N /. Let T < M � N be the diagonal subgroup; thenT Š M is minimal nonabelian. However, T 6E M �N , contrary to what has just beensaid. Now letM \N D Z.M/; thenM �N is extraspecial so is not anM�

3 -subgroup,and we conclude that G D MN . Then jG0j D p < p2 � jDj, a contradiction. Thus,N does not exist so CG.M/ is elementary abelian (Remark 1). Since G D MCG.M/,we get CG.M/ D Z.G/. If Z.G/ D Z.M/ � E, then G D M � E; in that case,jG0j D p < jDj again, a contradiction.

Thus, d.G/ D 3. In that case, jGj D jG0jjG=G0j � p6. Suppose that jG0j D p3.ThenG0 D D D F 0 �H 0 �L0, where F;H andL are appropriate minimal nonabeliansubgroups of G. As above, exp.G=F 0H 0/ D exp.G=H 0L0/ D exp.G=L0F 0/ D p so,since F 0H 0 \H 0L0 \ L0F 0 D f1g, we conclude that exp.G/ D p and G is special.


Now letG be special of order p5 or p6, exp.G/ D p, jG0j D p2 or p3, respectively,and d.G/ D 3. If M < G is minimal nonabelian (in that case, jM j D p3), then theM�

3 -groupMG0 D M � E (here G0 D M 0 � E) is the unique maximal subgroup ofG containingM .

Appendix 18

Replacement theorems

Thompson’s replacement theorem (see Corollary A.18.3 below) receivedwide applica-tion in the study of arbitrary finite groups. Theorem A.18.4, which is due to Glauber-man, is a deep generalization of Thompson’s result. The proof of Theorem A.18.1presents some ideas of the proof of Theorem A.18.4 in a more easy form.

Theorem A.18.1 (Isaacs [Isa2]). Let G be a p-group and A < G abelian. Supposethat B < G is also abelian, A � NG.B/ and B 6� NG.A/. Then there exists anabelian subgroup A� < G such that

(a) jA�j D jAj,

(b) A \ B < A� \ B ,

(c) A� � NG.A/,

(d) exp.A�/ divides 2 exp.A/.

Proof. Without loss of generality, one may assume that G D AB; then B GG but A isnot normal in G so G is not Dedekindian and jG W Aj > p. We proceed by inductionon jGj. Let A < M < G, whereM 2 �1. ThenM D A.B \ M/ and B \M G G

since B;M G G. Suppose that B \M 6� NG.A/. By induction, applied to the triplefA;B \M;M g, there is an abelian subgroup A� � M satisfying (a), (c) and (d) andsuch that A\ .B \M/ < A� \ .B \M/, and so A\B < A� \B , and we are donein this case.Therefore, one may assume that B \ M � NG.A/. Then M D A.B \ M/ �

NG.A/, i.e., A GM so NG.A/ D M since A 6E G andM 2 �1. Let b 2 B be suchthat A ¤ Ab . Obviously, Ab is normal inM b D M so that H D AAb is normal inM . Since A and Ab are abelian, we get Z D A \ Ab � Z.H/ and cl.H/ � 2, byFitting’s lemma. By Exercise 1.18, exp.H/ divides 2 exp.A/. Since B is abelian andb 2 B , we get

Ab \ B D .A \ B/b D A \ B

and hence

Z \ B D .A \ Ab/ \ B D A \ .A \ B/ D A \ B:(1)

Setting A� D .H \ B/Z, we see that A� � H and A� is abelian since H \ B isabelian and Z � Z.H/. It remains to show that A� satisfies conditions (a–d).


By the modular law, H D A.H \ B/ and, since A < H , we have H \ B 6� A.Next,H \ B D .AAb/\ B � A \ B . Hence,

A� \ B D Œ.H \ B/Z� \ B � H \ B > A \ B

(> since A \ B � H \ B 6� A), and (b) follows.Since A� � M D NG.A/, i.e., (c) is true, it remains to prove (a). By (1), Z \B D

A \ B , andH \Z \ B D Z \ B D A \ B . We have

jA�j D j.H \ B/Zj DjH \ Bj jZj

jH \Z \ BjD

jH \ Bj jZj

jA \ Bj:

SinceH D AAb D A.H \ B/, we get

jAj2

jZjD

jAj2

jA \ AbjD jH j D

jAj jH \ Bj

jA \H \ BjD

jAj jH \ Bj

jA \ Bj:

Therefore, jAj D jZj�jH\BjjA\Bj

D jA�j, proving (a). Since exp.H/ divides 2 exp.A/ andexp.Z/ divides exp.A/, condition (d) also holds. The proof is complete.

Corollary A.18.2 ([Isa2]). Let G be a nonabelian p-group, p > 2, and let B G G beabelian. Suppose that x 2 CG.�n.B// has order � pn. Then ŒB; x� � �n.B/.

Proof. The subgroup A D h�n.B/; xi is abelian of exponent at most pn. Assume thatB 6� NG.A/. Then, by Theorem A.18.1, there exists an abelian subgroup A� < G

such that exp.A�/ � exp.A/ � pn and A� \ B > A \ B . However, A� \ B �

�n.B/ � A \ B , a contradiction. Thus B � NG.A/, and so ŒB; x� � ŒB;A� �

A \ B D �n.B/ (= since exp.A \ B/ � pn).

LetA.G/ be the set of all abelian subgroups of G of maximal order. If A 2 A.G/,then Z.G/ � CG.A/ D A. We did not assume, in Theorem A.18.1, that A 2 A.G/.

Theorem A.18.3 (Thompson’s replacement theorem [Tho3]). LetG be a p-group andB G G abelian. If A 2 A.G/ is such that B 6� NG.A/, then there exists A� 2 A.G/

such that A� \ B > A \ B , A� � NG.A/ and B � NG.A�/.

Proof. Let A� 2 A.G/ be such that jA� \ Bj is as large as possible. By TheoremA.18.1, the first two assertions hold. Assume that B does not normalize A�. Then,by Theorem A.18.1 again, there exist A�� 2 A.G/ such that A� \ B < A�� \ B ,contrary to the choice of A�.

Exercise 1. Let B be a subgroup of a p-group G and A 2 A.G/. Then B normalizesA if and only if ŒB;A;A� D f1g.

Solution. If ŒB;A;A� D f1g, then ŒB;A� � CG.A/ D A. If B � NG.A/, thenŒB;A;A� � ŒA;A� D f1g.

A.18 Replacement theorems 503

Definition 1. Let G be a p-group. The (characteristic) subgroup J.G/ D hA j A 2

A.G/i is called the Thompson subgroup of G (or J -subgroup of G).

If J.G/ � H < G, then J.H/ D J.G/ so J.G/ is a characteristic subgroup of anysubgroup of G containing J.G/. Clearly, Z.J.G// � A for all A 2 A.G/. Therefore,Z.J.G// D

T

A2A.G/ A. Next,A.J.G// D A.G/.

Exercise 2. Let G be a nonabelian p-group. If J.G/ is cyclic, then G is a 2-group ofmaximal class so jG W J.G/j D 2. (Hint. Let jJ.G/j D pn; then cn.G/ D 1 6 0

.mod p/. Use Theorems 1.10(b) and 1.17(b).)

Exercise 3. Let G be a p-group, T � G and A.G/ \ A.T / ¤ ¿. Then Z.J.G// �

Z.J.T //.

Theorem A.18.4 (Glauberman’s replacement theorem). Suppose that a p-group G,p > 2, has a normal subgroup B such that B 0 � Z.J.G// \ Z.B/. If A 2 A.G/ issuch that B does not normalizeA, then there isA� 2 A.G/ such that A\B < A�\B ,A� � NG.A/ and exp.A�/ divides exp.A/.

Proof. In view of Theorem A.18.3, one may assume that B is nonabelian so cl.B/ D

2. We proceed by induction on jGj. Set T D AB.� G/. Since A 2 A.T /, we haveJ.T / � J.G/ and Z.J.G// � Z.J.T // (Exercise 3). We also have B 0 � Z.J.T // \

Z.B/. Hence, if T < G, theorem follows by induction. We may, therefore, assumethat G D AB; then B GG, A 6E G so jG W Aj > p. We have

B 0 � Z.J.G// \ Z.B/ � A\ B � Z.G/:(1)

Let A < M 2 �1 and B1 D B \ M ; then J.M/ � J.G/, M D AB1 and B1 G G

since B;M G G. Next, we have B 01 � B 0 � Z.J.G// � Z.J.M// (Exercise 3) and

B 01 � Z.B1/. Therefore, B 0

1 � Z.J.M// \ Z.B1/. If B1 6� NM .A/, then the triplefA;B1;M g satisfies the hypothesis, and, there is A� 2 A.M/ such that

A \ B1 < A� \ B1; ŒA�; A;A� D f1g; exp.A�/ j exp.A/;

by induction. In that case, we have A.M/ � A.G/, and the theorem is true sinceA \ B1 D A \ .B \M/ D A \ B and A� \ B1 � A� \ B .Assume that B1 � NM .A/. We have ŒB;A� � ŒB;M� � B \ M D B1 so

ŒB;A;A� � ŒB1; A� � A, i.e., ŒB;A� normalizes A. Since B GG, we have ŒB;A;A� �

B\A. Consider the quotient group NG D G=B 0. Then NG D NA NB (bar convention!) andNA \ NB � Z. NG/ since NA and NB are abelian. Therefore, Œ NB; NA; NA� � NB \ NA � Z. NG/.Since B does not normalize A, there is an element b 2 B such that Ab ¤ A. SetD D Œb;A� D hŒb; a� j a 2 Ai. Since A;Ab GM , we have AAb E M so D � AAb .Then we have D � ŒB;A� � ŒB;M� � B \M D B1. We have AAb D DA. Indeed,

AAb D haab1 j a; a1 2 Ai D haa1Œa1; b� j a; a1 2 Ai � AD;


and since A;D � AAb, our claim follows. We have D 6� A (otherwise, AAb D

DA D A so Ab D A, which is not the case). Set

A� D D.A \ Ab/:

We will prove that A� has the required properties. We have A� D D.A \ Ab/ �

DA D AAb � M < G. Since cl.AAb/ � 2 (Fitting’s lemma) and p > 2, it followsthat exp.AAb/ D exp.A/, and so exp.A�/ divides exp.AAb/ D exp.A/.We have B 0 � Z.J.G// � A so B 0 � B\A; then B\AGB . Hence .A\B/b

�1

D

A \ B , and so A \ B D Ab \ B and A� D D.A \ Ab/ D D.A \ B/. Thus,

A \ B � A \ .A \ B/ D A \ .Ab \ B/ D .A \ Ab/\ B � A� \ B:

Since D D Œb;A� � B \ A� but D 6� A, we have A \ B < A� \ B . We haveA� � M D NG.A/ and so ŒA�; A;A� D f1g.It remains to prove that A� 2 A.G/. SetZ D A\Ab .� Z.AAb//. We have (since

A�A D D.A \ Ab/A D DA)

jA W Zj D jAb W Zj D jAAb W Aj D jDA W Aj D jA�A W Aj D jA� W .A� \ A/j:(2)

Next, Z D A \ Ab < D.A \ Ab/ D A�, and hence Z � A� \ A. Thus we get,by (2), jA�j jZj D jAj jA� \ Aj � jAj jZj so jA�j � jAj. If D is abelian, thenA� D DZ D D �Z is abelian, and we get A� 2 A.G/.It remains to show that D D Œb;A� is abelian, i.e., ŒŒb; u�; Œb; v�� D 1 for any

u; v 2 A. Setting w D Œb; v� and s D u�1, we have, by the Hall–Witt identity,

Œb; u;w�s Œs; w�1; b�w Œw; b�1; s�b D 1:

Since w 2 B.GG), we get

Œw; b�1; s�b 2 ŒB;B;A� D ŒB 0; A� � ŒZ.J.G//;A� D f1g:

The remaining factors of that formula are contained in ŒB;B� � Z.G/ (see (1)), hencethe above identity gives us

Œb; u;w� D Œs; w�1; b��1 D Œu�1; w�1; b��1:

Therefore, taking into account the inclusion Œu�1; w�1; b� 2 Z.G/, we get

Œb; u;w� D ŒŒu�1; w�1��1; b� D Œw�1; u�1; b�:

On the other hand, we have Œw�1; u�1� 2 Œb;A;A� since w�1 D Œb; v��1 � Œb;A�.Thus, in NG D G=B 0, Œw�1; u�1� corresponds to an element of the center (sinceŒ NB; NA; NA� � Z. NG/). So, we get

Œw�1; u�1� Œw; u� .mod B 0/:

A.18 Replacement theorems 505

Since B 0 � Z.B/, we have Œw�1; u�1; b� D Œw; u; b� for any b 2 B . By the above(recall that Œb; v� D w),

ŒŒb; u�; Œb; v�� D Œb; u;w� D Œw�1; u�1; b� D Œw; u; b� D Œb; v; u; b�:(2)

Applying the formula Œu; xy� D Œu; y�Œu; x�y to the equality Œb; uv� D Œb; vu� (uv D

vu since A is abelian), we obtain

Œb; uv� D Œb; v�Œb; u�Œb; u; v� D Œb; vu� D Œb; u�Œb; v�Œb; v; u�;

and so, since Œb; u�; Œb; v� 2 B ,

Œb; u; v� D ŒŒb; v�; Œb; u��Œb; v; u� Œb; v; u� .mod B 0/:

Since B 0 � Z.G/, we get Œb; u; v� D Œb; v; u�z with z 2 B 0 D Z.B/ so Œb; u; v; b� D

ŒŒb; v; u�z; b� D Œb; v; u; b�zŒz; b� D Œb; v; u; b�, which proves (see (2)) that

ŒŒb; u�; Œb; v�� D ŒŒb; v�; Œb; u��.D .ŒŒb; u�; Œb; v��/�1:

Since p > 2, we have ŒŒb; u�; Œb; v�� D 1. Thus, D is abelian, and the proof is com-plete.

Appendix 19

New proof of Ward’s theoremon quaternion-free 2-groups

Below we present a new proof of Ward’s theorem (see Theorem 56.1) which is due toZ. Bozikov. Recall that Ward’s theorem asserts that a nonabelian quaternion-free 2-group has a characteristic maximal subgroup. This proof is short but not so elementaryas given in �56.

Proof of Ward’s theorem. (i) First we assume thatG is a Q8-free 2-group which is alsoD8-free; thenG is modular (see ��43 and 73). We prove that thenG has a characteristicmaximal subgroup and use induction on jGj. By classification of nonabelian modular2-groups (see �73), there is AGG abelian and such thatG=A ¤ f1g is cyclic and thereis g 2 G and an integer s � 2 so that G D hA;gi and ag D a1C2s

for all a 2 A.We have f1g ¤ G0 < A. If G0 6� Z.G/, then CG.G

0/ � A, G=CG.G0/ is nontrivial

cyclic and CG.G0/ is a characteristic subgroup inG. In that caseG has a characteristic

maximal subgroup containing CG.G0/. Hence we may assume that G0 � Z.G/ and so

G is of class 2.Suppose that G0 is not elementary abelian. Then �1.G/ < G0 so G=�1.G

0/ isnonabelian of order < jGj. By induction, G=�1.G

0/ has a characteristic maximalsubgroup. Hence, we may assume that G0 is elementary abelian.For any x; y 2 G, Œx2; y� D Œx; y�2 D 1 since cl.G/ D 2, so ˆ.G/ D Ã1.G/ �

Z.G/. In particular, g2 2 Z.G/ and so the maximal subgroupM D hg2; Ai is abelian.IfM is the unique abelian maximal subgroup of G, thenM is characteristic in G, andwe are done. Therefore we may assume that G has another abelian maximal subgroupN ¤ M . We have M \ N D Z.G/ and jG W Z.G/j D 4. Since A 6� Z.G/,jA W .N \ A/j D 2 and therefore NA D G. Lemma 1.1 implies jG0j D 2. Also,we may assume that there is ˛ 2 Aut.G/ such that N D M˛ (otherwise, M ischaracteristic in G). Let h 2 N � A be such that N D hh;N \ Ai, and so G D

NA D hh;N \A;Ai D hh;Ai. Since h2 2 Z.G/, we haveM D hh2; Ai which is theunique abelian maximal subgroup of G containing A (recall that G=A is cyclic). Wehave h D gm for some m 2 M and so h acts the same way on A as the element g.Since h centralizesN \A, we have exp.N \A/ � 2s and all elements in A� .N \A/

are of order 2sC1, s � 2, which implies exp.N \A/ D 2s , in view jA W .N \A/j D 2.For any a 2 A � N , ah D a1C2s

and so G0 D ŒA; h� D hŒa; h� j a 2 Ai D ha2s

j

a 2 Ai D Ãs.A/. SinceM \N D Z.G/, we get .M \N/˛ D M \N . ButM˛ D N

A.19 New proof of Ward’s theorem on quaternion-free 2-groups 507

and so h.A �N/˛i D hA˛ �M i � N �M . Let a˛ D h0, where a 2 A �N so thato.h0/ D 2sC1 and hh0i covers N=.N \A/ since N=.N \A/ is cyclic. It follows thatN D .N \ A/hh0i and M \ N D .N \ A/hh2

0i and therefore exp.M \ N/ D 2s ,exp.N / D 2sC1, hh2s

0 i D G0 and Ãs.N / D G0. This implies Ãs.M/ D G0 andh2s

D h2s

0 . We compute (noting that Œh2; a2� D 1)

.ha/2 D haha D h2.ah/a D h2a1C2s

a D h2a2a2s

;

.ha/2s

D ..ha/2/2s�1

D .h2a2a2s

/2s�1

D h2s

a2s

D 1

(note that s � 2 and o.a/ D 2sC1). This implies that the third abelian maximal sub-groupK D .M \N/hhai is of exponent 2s . Since each abelian maximal subgroup ofG containsM \N D Z.G/, it follows that G has exactly three abelian maximal sub-groups M;N;K, where exp.M/ D exp.N / D 2sC1 and exp.K/ D 2s and thereforeK is characteristic in G, and we are done.

(ii) We assume now that G is a quaternion-free 2-group which is not D8-free. ByMain Theorem of �79, G is isomorphic to one of the groups of types (a), (b) or (c) ofthat theorem (these groups were named Wa-, Wb- and Wc-groups, respectively).

(ii)(a) Suppose that G is a Wa-group. Then G is a semidirect product G D hxi N ,where N is a maximal abelian normal subgroup of G with exp.N / > 2 and, if t is theinvolution in hxi, then every element in N is inverted by t .Since G=N > f1g is cyclic, we have �1.G/ D �1.hN; ti/ D N hti. Note that all

elements in Nt are involutions and each y 2 Nt inverts every element in N .IfN is not characteristic inG, then, since�1.G/ is characteristic inG and j�1.G/ W

N j D 2, there is ˛ 2 Aut.G/ such that �1.G/ D NN ˛ , where j�1.G/ W N ˛j D 2.Let u 2 N ˛ � N . Then u is an involution in Nt and so u inverts and centralizeseach element in N \ N ˛ . This implies that N \ N ˛ is elementary abelian. But thenN ˛ D .N \N ˛/hui is elementary abelian contrary to exp.N ˛/ D exp.N / > 2. ThusN is a characteristic subgroup of G with cyclicG=N > f1g. In that case, the maximalsubgroup U=N of G=N is characteristic in G=N ; then U is characteristic in G sinceN is, and we are done.

(ii)(b) Let G be a Wb-group. Then G D N hxi, where we may assume that N is amaximal normal elementary abelian subgroup of G and hxi is not normal in G.If �1.G/ D N , then, since G=N is cyclic > f1g, the maximal subgroup M of G

containing N is characteristic in G, and we are done.We may assume that�1.G/ > N . SinceG=N is cyclic, we have j�1.G/ W N j D 2.

If G ¤ �1.G/, then, since G=�1.G/ is cyclic > f1g, the maximal subgroupM of Gcontaining �1.G/ is characteristic in G, and we are done.Assume, in addition, that G D �1.G/ with jG W N j D 2 andN is not characteristic

in G (otherwise, we are done). There is ˛ 2 Aut.G/ such that G D NN ˛ , jN W

.N \ N ˛/j D 2, and N \ N ˛ D Z.G/ with jG W Z.G/j D 4. We note that N andN ˛ are elementary abelian maximal subgroups of G. On the other hand, each abelian


maximal subgroup of G contains N \ N ˛ D Z.G/, where jG W Z.G/j Š E4 and soG has exactly three abelian maximal subgroups. Let a be an element of order 4 inG � .N [ N ˛/ so thatM D .N \ N ˛/hai is the third abelian maximal subgroup ofG. Since exp.M/ D 4,M is characteristic in G, and we are done.

(ii)(c) Suppose that G is isomorphic to a Wc-group. Then G has an elementaryabelian normal subgroup N such thatG=N Š M2n , n � 4, and ifH=N D �1.G=N/,then H Š D8 � E2m and Z.H/ D N . Also, G=H is cyclic of order � 4. SinceH Š D8 � E2m , we have H D �1.G/. Therefore, ifM is the maximal subgroup ofG containing �1.G/, thenM is characteristic in G and we are done.

Appendix 20

Some remarks on automorphisms

In this section we prove some results on automorphisms of finite groups. ��32, 33, 34contain further information on automorphisms.

Theorem A.20.1. Let ' be an automorphism of a group G, N a '-invariant normalsubgroup of G. Then jCG=N .'/j � CG.'/j.

Proof. LetW be the semidirect product of G and h'i, compatible with action of ' onG; then N GW . By the Second Orthogonality Relation,

jCW .'/j DX

�2Irr.W /

j.'/j2 �X

�2Irr.W=N /

j.'/j2 D jCW=N .'/j:

By the modular law, CW .'/ D h'i CG.'/ and CW=N .'/ D h'i CG=N .'/ (bothproducts are semidirect), and now the result follows from the above long formula.

In the case where .jN j; o.'// D 1, the result may be strengthened.

Theorem A.20.2. Let G, ', N be as in Theorem A.20.1. If, in addition, .jN j; o.'// D

1, then CG=N .'/ D CG.'/N=N .

Corollary A.20.3. Let G be a group and ' an �.G/0-automorphism of G. Then (a)G D CG.'/ŒG; '� and (b) ŒG; '; '� D ŒG; '�.

Proof. (a) We know that ŒG:'�GhG;'i so ŒG:'�GG. As g' D gŒg; '�, ' acts triviallyon G=ŒG; '� so G=ŒG; '� D CG=ŒG;'�.'/ D CG.'/ŒG; '�=ŒG; '� (Theorem A.20.2),and hence G D CG.'/ŒG; '�, as was to be shown.

(b) By (a), ŒG; '� D ŒCG.'/ŒG; '�; '� D ŒŒG; '�; '� D ŒG; '; '�.

Corollary A.20.4 (= Corollary 6.5 (Fitting)). If a group G of Corollary A.20.3 isabelian, then G D CG.'/ � ŒG; '�.

Exercise 1. Let ˛ 2 Aut.G/, where G is a p-group and o.˛/ D p. If ˛ has exactly pfixed points then one of the following holds: (a) G is of order � pp and exponent p,(b) G is absolutely regular, (c) G is irregular of maximal class. (Hint. By Proposition1.8, hG;'i is of maximal class. Use Theorems 9.5 and 9.6.)


Exercise 2. Let p > 2, G D hai Š Cpn , n > 1 and ' 2 Aut.G/ be such that

' W a ! a1Cp. Then jCG.'pk

/j D pkC1 for k D 0; 1; : : : ; n� 1. Hence, the class ofthe natural extensionW of G by h'i, which is metacyclic, is n.

Exercise 3. Let G be a p-group and ' 2 Aut.G/#, and suppose that ' fixes all ele-ments of G of orders p and 4. Prove that o.'/ is a power of p. (Hint. Use Lemma10.8 and Frobenius’ normal p-complement theorem.)

Exercise 4. Let G D hai Š C2n , n > 2. Then Aut.G/ is an abelian group of type.2n�2; 2/ with two independent generators � W a ! a�1 and W a ! a5. The groupAut.G/ has exactly three involutions: � , 2n�3

W a ! a1C2n�1

and � D �2n�3

W

a ! a�1C2n�1

. Let W be the natural semidirect product of G and Aut.G/, theholomorph of G. Show that D D hG; �i Š D2nC1 , M D hG; 2n�3

i Š M2nC1 andS D hG; �i Š SD2nC1 . Check that cl.G/ D n. Moreover, construct the upper andlower central series of W . Find ck.W /.

Hint. The natural semidirect product H D �1.Aut.G// G has exactly three maxi-mal subgroups containing G, namely, D,M , S . Clearly, �1.H/ D �1.W / so

c1.W / D c1.H/ D c1.D/C c1.S/C c1.M/ � 2c1.G/

D .2n C 1/C .2n�1 C 1/C 3� 2 D 2n C 2n�1 C 3;

c2.W / D c2.D/C c2.S/C c2.M/ � 2c2.G/

D 1C .2n�1 C 1/C 2� 2 D 2n�1 C 2;

and, for k 2 f3; : : : ; ng, we have ck.W / D ck.D/ C ck.S/ C ck.M/ � 2ck.G/ D

1C 1C 2 � 2 D 2.

An automorphism ' 2 Aut.G/ is said to be fixed-point-free (or regular) if x' D x

for x 2 G implies x D 1. If G D F H is a Frobenius group with kernel H andcomplement F and f 2 F #, then h 7! hf is a fixed-point-free automorphism of H .

Proposition A.20.5. Let ' 2 Aut.G/ be fixed-point-free of order n.

(a) If x 2 G, then there exists exactly one y 2 G such that x D y�1y' .

(b) If x 2 G, then xx' : : : x'n�1

D 1.

(c) If o.'/ D 2, then x' D x�1 (in that case, G is abelian).

(d) If N GG is '-invariant, then the automorphism induced by ' on G=N , is fixed-point-free.

Proof. (a) If y�1y' D z�1z' for y; z 2 G, then zy�1 D .zy�1/' , so zy�1 D 1

since ' is fixed-point-free. Then y D z so jfy�1y' j y 2 Ggj D jGj. It follows thatx D y�1y' for exactly one y 2 G, proving (a).

(b) Let x 2 G. Then x D y�1y' for some y 2 G, by (a), and so

xx' : : : x'n�1

D y�1y'.y�1/'y'2

: : : .y�1/'n�1

y'n

D y�1y D 1:

A.20 Some remarks on automorphisms 511

(c) If o.'/ D 2, then, by (b), xx' D 1 so x' D x�1 for all x 2 G. If x; y 2 G,then x�1y�1 D x'y' D .xy/' D .xy/�1 D y�1x�1 so xy D yx, and G is abelian.

(d) Assume that .xN/' D xN for some x 2 G. Then x�1x' 2 N . As 'N isfixed-point-free, there exists y 2 N such that x�1x' D y�1y' . By (a), x D y 2 N ,so xN D N , i.e., ' induces a fixed-point-free automorphism on G=N .

Proposition A.20.6 ([Isa15, Lemma 3.2]). Let C D Z.G/, whereC is cyclic andG=Cis abelian. Then every automorphism of G, which is trivial on C and G=C , is inner.

Proof. Suppose that A D f� 2 Aut.G/ j ŒG; �� C and ŒC; �� D f1gg and note thatInn.G/ � A. Since jInn.G/j D jG=C j, it suffices to show that jAj � jG=C j.For each � 2 A, there is a well-defined map �� W G=C ! C defined by .Cg/�� D

Œg; ��, and we have �� 2 Hom.G=C;C / (check!). The map A ! Hom.G=C;C /defined by � ! �� is injective since if �� D �� , then Œg; �� D Œg; � for all g 2 G. Itfollows that g� D g� for all g and so � D . Thus, jAj � jHom.G=C;C /j. Since Cis cyclic, it follows easily from the fundamental theorem on abelian groups applied toG=C , that jHom.G=C;C /j � jG=C j, and the result follows.

Appendix 21

Isaacs’ examples

1o. Let b D ps�1p�1 D 1CpC Cps�1; then F�, the (cyclic) multiplicative group of

the field F D GF.ps/, has a unique subgroup C of order b. We define an action of Con P D Ap.s; �/, where � is the Frobenius automorphism of F (see �46), as follows:.x; y/c D .xc; ycpC1/ .c 2 C/. Taking into account that �.c/ D cp and writing.x; y/c D .xc; y�.c/c/, it is easy to check that this is in fact an action. Indeed,

Œ.x; y/.u; v/�c D .x C u; y C v C x�.u//c

D ..x C u/c; .y C v C x�.u//�.c/c/

D .xc; y�.c/c/.uc; v�.c/c/ D .x; y/c .u; v/c :

If .x; y/c D .x; y/, then c D 1, i.e., every element of C # induces a fixed-point-freeautomorphism of P . Therefore, C P D .C;P / is a Frobenius group with kernel Pand complement C . In what follows we retain the above notation.

We will construct a p-solvable group G such that b.Q/ D max f�.1/ j � 2 Irr.Q/gdoes not divide .1/ for all 2 Irr.G/, whereQ 2 Sylp.G/. This example disprovesone conjecture of the first author and is taken from Isaacs’ note (unpublished).Let P D Ap.s; �/. In what follows, we assume that s D p > 2; then P D

Ap.p; �/ and o.�/ D p. We define an action of S D h�i on P as follows: .x; y/� D

.�.x/; �.y//. Set G D .S C/ P . Next, .x; y/� D .x; y/ if and only if x; y 2 F0,the prime subfield of F, and so jCP .�/j D jF0j2 D p2, jCZ.P /.�/j D jF0j D p.Thus, � fixes exactly p classes of Z.P / and p classes of P=P 0 D P=Z.P /. ByBrauer’s permutation lemma, � fixes exactly p linear characters of both groups Z.P /and P . If � 2 Lin.Z.P // � f1Z.P /g is one of them and H D ker.�/, then � fixes alllinear characters of Z.P / with kernel H . It follows that � fixes exactly one maximalsubgroup of Z.P /. We choose � so that it raises every element of F in power p (i.e., �is a Frobenius automorphism of F). If c 2 C and �.c/ D c, then cp�1 D 1 so c D 1

since .jC j; p � 1/ D .b; p � 1/ D .1C p C C pp�1; p � 1/ D 1 in view of

1CpC Cpp�1 D pC .p� 1/C .p2 � 1/C C .pp�1 � 1/ 1 .mod p� 1/

(recall that p > 2). Therefore, S C D .S ; C / is a Frobenius group with kernel Cand complement S , jS C j D pb. The group S C has exactly jC j D b subgroupsof order p, and all these subgroups are conjugate in S C (Sylow). Then, if T is a

A.21 Isaacs’ examples 513

subgroup of order p in S C D T C , then T fixes exactly one maximal subgroupHT

of Z.P /. Since b is the number of maximal subgroups of Z.P /, there is a one-to-onecorrespondence T $ HT , where T is a subgroup of order p in S C and HT is amaximal subgroup of Z.P / fixed by T .By the previous paragraph, S fixes exactlyp2�p nonlinear irreducible characters of

P (these characters are irreducible constituents of the induced character ˛P , where ˛runs over all nonprincipal linear characters of Z.P /=HS andHS is a unique maximalsubgroup of Z.P / fixed by S). It follows that ˛S �P is the sum of p2 irreducible char-acters of SP of degree p.p�1/=2. In particular, the inertia subgroup IG.˛/ D S P ,and so all irreducible constituents of ˛G have degree bp.p�1/=2. If T < S C is of or-der p, then, ifHT is the maximal subgroup of Z.P / fixed by T and ˇ is a nonprincipalirreducible character of Z.P /=HT , then IG.ˇ/ D T P . If 2 Irr1.G=Z.P //, then.1/ divides jG W P j D bp, by Ito’s theorem on degrees (see Introduction, Theorem17). Therefore, the p-part of the degree of every irreducible character of G does notexceed p.p�1/=2.However, S P 2 Sylp.G/ has an irreducible character of degree p

.pC1/=2. This isbecause S definitely does not stabilize all nonlinear irreducible characters of P . Thisproves, that b.P / D max f .1/ j 2 Irr1.P /g − .1/ for all 2 Irr.G/. We do notknow if there exists an analogous example for p D 2.Isaacs [Isa4] proved that for every set S of powers of a prime p such that p0 D

1 2 S , there exists a p-group P such that cd.P / D S (see Theorem A.21.2, below).Next, he showed (in the letter to the author) that for each p-group P there exists agroup G with cyclic Sylow p-subgroup C such that cd.G/ D cd.P /. We will provethis result. Moreover, we show that, for every set S of powers of a prime p containing1 D p0 and with maximum member pa, there exists a group G having a cyclic Sylowp-subgroup C of order pa such that cd.G/ D S . Let a prime q ¤ p. For eachmember pe 2 S , let Ve be an elementary abelian q-group on which a cyclic group oforder pe acts faithfully and irreducibly. Let C act on Ve with kernel of order pa�e

(in that case, C acts on Ve irreducibly), and let W be the direct product of the groupsVe for pe 2 S . so that C acts on W . Let G be the semidirect product of W withC . We claim that cd.G/ D S . Obviously, it is enough to prove that pa 2 cd.G/; byIto’s theorem (Introduction, Theorem 17), the set cd.G/ contains only powers of p notexceeding jC j D pa. Let W D Ve1

� � Ves, where S D f1; pe1 ; : : : ; pes D pag.

Let xi 2 V #ei, i D 1; : : : ; s x D x1 : : : xs . Then the normal closure of hxi in G is

W , the socle of G. Therefore, Irr.G/ has a faithful character (Gaschutz; see [BZ,Chapter 9]). We claim that .1/ D pes D pa. Note that C Va is a Frobenius group.Since Va 6� ker./, it follows that the restriction Va

has a non-principal constituent,and so .1/ � pa in view of cd.C Va/ D f1; pag. Since .1/ divides pa, by Ito’stheorem, we get .1/ D pa.It is not surprising that the subgroup W in the previous example is abelian. Indeed,

if the degrees of all irreducible characters of G are powers of a fixed prime p, then Ghas an abelian normal p-complement (Ito).


2o. Let S be the set of powers of p such that 1 2 S . We will prove that there existsp-groups G such that cd.G/ D S . Moreover, Isaacs [Isa4] has showed that thereexists a p-group G of class � 2 such that cd.G/ D S .Let U be an abelian group which acts on an abelian group A. Let QA D Lin.A/

be the group of linear characters of A. As we know (see [BZ, �1.9]), there existsa natural isomorphism of QA onto A. If ˛ 2 QA, u 2 U and a 2 A, then, setting˛u.a/ D ˛.uau�1/, we define the action of U on QA.

Lemma A.21.1 ([Isa4]). Let U be an abelian group acting on an abelian group A andS the set of the sizes of the U -orbits in this action. Write G D U QA, where thesemidirect product is constructed with respect to the action of U on QA, induced by thegiven action of U on A. Then cd.G/ D S . Furthermore, if ŒA;U;U � D f1g, thencl.G/ � 2.

Proof. Let � 2 Lin. QA/ and T the stabilizer of � inU . Since the cyclic group QA= ker.�/is centralized by T , which is abelian, it follows that QAT= ker.�/ is abelian and thus

every 2 Irr.� QAT / is linear. Obviously, QAT D IG.�/, the inertia subgroup of �in G, and thus every 2 Irr.�G/ has degree jG W QAT j D jU W T j. Therefore,cd.G/ D fjU W T j j T is a stabilizer in U of some � 2 Lin. QA/g. However, thereis a natural correspondence between Lin. QA/ and A and this defines a permutationisomorphism of the actions of U on A and Lin. QA/, respectively, and we conclude thatcd.G/ D S .Now suppose that ŒA;U;U � D f1g. If ˛ 2 QA, we have for u 2 U and a 2 A that

Œ˛; u�.a/ D .˛�1˛u/.a/ D ˛.a�1/˛.uau�1/ D ˛.Œa; u�1�/:

Therefore, if v 2 U , we get

Œ˛; u; v�.a/ D Œ˛; u�.Œa; v�1�/ D ˛.Œa; v�1; u�1�/ D ˛.1/ D 1;

and so Œ QA;U;U � D f1g. Since QA is abelian and Œ QA;U � � QA, this yields CG.Œ QA;U �/ �QAU D G, i.e., Œ QA;U � � Z.G/. Since G=Œ QA;U � D QAU=Œ QA;U � is abelian, it followsthat cl.G/ � 2.

Now we are ready to prove the main result of this subsection.

Theorem A.21.2 ([Isa4]). Let p be a prime and 0 D e0 < e1 < < em integers.Then there exists a p-group that is generated by elements of orderp and has nilpotenceclass � 2 such that cd.G/ D fpei j 0 � i � mg.

Proof. Let u1; : : : ; uembe generators of U Š Epem . Let A be an elementary abelian

p-group with basis

a1; z1;1; : : : ; z1;e1; a2; z2;1; : : : ; z2;e2

; : : : ; am; zm;1; : : : ; zm;em;

and define an action of U on A as follows.

A.21 Isaacs’ examples 515

Put .zi;/u� D zi; for all i; ; � and .ai /

u� D ai if � > ei and .ai /u� D aizi; if

� � ei . Since the automorphisms of A defined this way all have order p and commutepairwise, this does define an action of U on A.Let us compute the sizes of the U -orbits of this action. Write Z D hzi; j �

ei i � A and take a 2 A. Then Z � Z.G/, where G D U A is the natural semidirectproduct with kernel A. Assume that a 2 A�Z. There exists, then, a unique subscripti such that a D bcz, where b 2 haj j j < ii, 1 ¤ c 2 hai i, z 2 Z.Suppose that u 2 U . If (the expression of) u involves the generator u with � ei ,

the exponents of zi; in (expressions of) a and au will not be equal, and u does notcentralize a. Thus CU .a/ � hu j � > ei i. Since the reverse inclusion is obvious, weget CG.a/ D hu j � > ei i.We now have jU W CU .a/j D pei and we see that the orbit sizes of the action of U

on A are precisely the numbers pei for 0 � i � m. Since ŒA;U � � Z � Z.G/, wehave ŒA;U;U � D f1g and the result follows by Lemma A.21.1.

Appendix 22

Minimal nonnilpotent groups

A group G is said to be minimal nonnilpotent if it is not nilpotent but all its propersubgroups are nilpotent. In this section we present information on the structure ofminimal nonnilpotent groups. The results of Theorem A.22.1 are due to O.Y. Schmidt[Sch2] and Y.A. Golfand [Gol]; L. Redei [Red] has classified such groups.

Theorem A.22.1. Let G be a minimal nonnilpotent group. Then jGj D paqˇ , wherep and q are distinct primes. Let P 2 Sylp.G/, Q 2 Sylq.G/. Let b be the order of q.mod p/.

(a) One of the subgroups P ,Q (sayQ) coincides with G0.

(b) IfQ is abelian, thenQ \ Z.G/ D f1g andQ Š Eqb .

(c) P is cyclic and jP W .P \ Z.G//j D p.

(d) If Q is nonabelian, then it is special and jQ=Q0j D qb , Q \ Z.G/ D Z.Q/.The number b is even.

Proof. Let us show by induction on jGj that G is solvable. Assuming that G is aminimal counterexample, we conclude that G is nonabelian simple.LetH and F be different maximal subgroups of G whose intersectionD D H \F

is as large as possible. Suppose that D > f1g. Since H;F are nilpotent, it followsthat NH .D/ > D, NF .D/ > D, and therefore NG.D/ — H and NG.D/ — F .Let NG.D/ � R, where R is maximal in G. Obviously, H ¤ R ¤ F . HoweverjH \Rj > jNH .D/j > jDj, contrary to the choice ofH and F . HenceD D f1g.By Sylow’s theorem, there exist in G two maximal subgroups H , F of different

orders. Then

jG#j D jGj � 1 �X

x2G

.H #/x C .F #/x j

D jG W H j jH #j C jG W F j jF #j D 2jGj � jG W H j � jG W F j;

i.e., jG W H j C jG W F j > jGj, which is impossible. Hence, G is solvable.Let H G G be of prime index p; then H D F.G/, the Fitting subgroup of G, and

H is the only normal subgroup of prime index in G. Let Q be a p0-Hall subgroupof H ; then Q is a normal p-complement in G. Let P 2 Sylp.G/, jP j D pa. ThenG=Q Š P , and G=Q contains only one subgroup of index p. Therefore P Š C.pa/.If P0 D P \H , then CG.P0/ � hP;Qi D G and P0 � Z.G/ so P0 D P \ Z.G/.

A.22 Minimal nonnilpotent groups 517

Assume that the p0-Hall subgroup Q of G is not primary. Then Q D Q1 � Q2,where f1g < Q1 2 Syl.G/ and Q2 > f1g is a Hall subgroup of G. By assumption,PQ1 < G, PQ2 < G; therefore, PQ1 D P � Q1, PQ2 D P � Q2, and soCG.P / D G and G D P � Q is nilpotent, a contradiction. Thus, Q 2 Sylq.G/,where q ¤ p is a prime. Since G=Q Š P is abelian, G0 � Q. Since G contains onlyone normal subgroup of prime index, we getQ D G0.Let Q0 G G be such that Q0 < Q and P;P1 2 Sylp.G/ different. Since we have

PQ0; P1Q0 < G, it follows that CG.Q0/ � hP;P1i D G (the subgroup hP;P1i,which contains two different Sylow p-subgroups, is not nilpotent), and soQ0 � Z.G/.

(i) Suppose that Q is abelian. Since G has no nontrivial direct factors, it followsfrom from Corollary 6.5 that Q \ Z.G/ D f1g, and so, by the result of the previousparagraph,Q is a minimal normal subgroup of G so elementary abelian.Let jQj D qˇ and let b be the order of q .mod p/, P D hxi; then xp 2 Z.G/.

If v0 2 Q# and v D v0vx0 : : : v

xp�1

0 , then vx D v. Since x normalizes Q1 D

hv0; vx0 ; : : : ; v

xp�1

0 i, it follows that Q1 G G; therefore, by the previous paragraph,Q1 D Q and ˇ � p. The equality vx D v implies that v D 1, since v 2 Z.G/\Q D

f1g. Therefore, ˇ < p. Next, P does not normalize nontrivial subgroups of Q so thenumber s of subgroups of order q inQ is divisible by p. Let ˇ D kb C t , 0 � t < b.Assume that t > 0; then b > 1 so p − q � 1. We have,

s DqkbCt � 1

q � 1D qt

qkb � 1

q � 1Cqt � 1

q � 1

so p divides qt �1q�1

, a contradiction since 0 < t < b. Thus, t D 0 so ˇ D kb.

Assume that k > 1. The number of subgroups of order qb inQ is

d D.qkb � 1/ : : : .qkb � qb�1/

.qb � 1/ : : : .qb � qb�1/:

Since kb�i 6 0 .mod b/ for i 2 f1; : : : ; b�1g, we get qkb �qi D qi .qkb�i �1/ 6 0

.mod p/. Since P does not normalize subgroups of Q of order qb , it follows that p

divides d so that p divides qkb �1

qb�1D qb.k�1/ C C qb C 1 k .mod p/ hence

k � p. But then ˇ > p, a contradiction. Thus, k D 1, ˇ D b, jQj D qb .

(ii) Now suppose that Q is nonabelian. In that case, as we have proved, all G-invariant subgroups, properly contained inQ, lie in Z.G/\Q D Z.Q/. In particular,ˆ.Q/ � Z.Q/. By (i), applied to G=Q0, we must haveQ0 D Z.G/, and we concludethat Q0 D ˆ.Q/ D Z.Q/, i.e., Q is special. Let L < Z.Q/ be maximal. Then bywhat has just been proved,Q=L is extraspecial of order qbC1 so b is even.

If G is the group of Theorem A.22.1 and jQ \ Z.G/j D qc , then G is called anS.pa; qb; qc/-group, where jP j D pa, jQj D qˇ D qbCc . As we saw, if c > 0, thenb is even. Y.A. Golfand [Gol] has shown that c � b=2, but we omit the proof.Set mc.G/ D k.G/

jGjand f.G/ D 1

jGj

P

�2Irr.G/ .1/.


Theorem A.22.2 ([BZ, Chapter 11]). Let G D PQ be an S.pa; qb ; qc/-group, P 2

Sylp.G/, Q D G0 2 Sylq.G/ and 2 Irr1.G/.

(a) IfQ\Z.G/ � ker./, then .1/ D p; ifQ\Z.G/ 6� ker./, then .1/ D qb=2.

(b) mc.G/ D q2Cp2qc�1

p2qbCc , f.G/ D qbCpqcCb=2Cp�pqb=2�1

pqbCc .

(c) If H � Z.G/, then f.G=H/ � f.G/. If c > 0 and f1g < H < Z.Q/, thenf.G=H/ D f.G/ if and only if p D 1C qb=2, q D 2, f.G/ D 2�.b=2/.

(d) If p > q and f.G/ � 1q, then p D 3, q D 2, G=.P \ Z.G// 2 fA4;SL.2; 3/g,

f.G/ D 12.

Appendix 23

Groups all of whose noncentralconjugacy classes have the same size

In this section we prove some results from [Ito2]. For x 2 G, we write iG.x/ D jG W

CG.x/j and call that number the index of x in G. Recall that G is said to be p-closed(p-nilpotent) if Sylow p-subgroup is normal in G (G has a normal p-complement).Recall that �.G/ is the set of all prime divisors of jGj.

Theorem A.23.1. Let p; q 2 �.G/ be distinct and pq − iG.x/ for all x 2 G. ThenG is either p- or q-nilpotent. If G is p-nilpotent, then NG.P / D CG.P / for P 2

Sylp.G/.

Proof. Take P 2 Sylp.G/, Q 2 Sylq.G/ and let x 2 G with p − iG.x/. ThenP y � CG.x/ for some y 2 G so x 2 CG.P

y/, i.e., any element of G whose indexis prime to p, is contained in at least one conjugate subgroup of CG.P /. Similarly,any element of G whose index is prime to q, is contained in at least one conjugatesubgroup of CG.Q/. Therefore, by hypothesis, we obtain the following inequality:

jG W NG.CG.P //j.jCG.P /j � 1/C jG W NG.CG.Q//j.jCG.Q/j � 1/C 1 � jGj:

Dividing the both sides by jGj, we obtain

jCG.P /j

jNG.CG.P //jC

jCG.Q/j

jNG.CG.Q//j

� 1C�

jNG.CG.P //j�1 C jNG.CG.Q//j

�1 � jGj�1�

> 1:

At least one of two summands on the left hand side equals 1. Let, for definiteness,jCG.P /j

jNG.CG.P //jD 1, or, what is the same, CG.P / D NG.CG.P //. Since CG.P / E

NG.P /, we get CG.P / D NG.CG.P // � NG.P /.� CG.P // so NG.P / D CG.P /.By Burnside’s normal p-complement theorem, G is p-nilpotent.

Exercise 1. Study the structure of G such that, for some non-conjugate A;B < G, wehave

S

x2G.A [ B/x D G.

Definition 1. The centralizer F of an element x 2 G � Z.G/ is said to be free if thereare no elements y; z 2 G � Z.G/ such that CG.y/ < F and F < CG.z/.


Proposition A.23.2. Let F be a free centralizer in G. Then F D P � A, whereP 2 Sylp.F / and A is abelian. If F contains two elements x; y 2 G such that�.o.x// D fpg, �.o.y// D fqg, p ¤ q, and CG.x/ D F D CG.y/, then F isabelian.

Proof. There are p 2 �.F / and a p-element x 2 G such that F D CG.x/ since Fis free. Let y 2 F with �.o.y// D fqg, where q ¤ p. Then hxyi D hxi � hyi soCG.xy/ � CG.x/ D F . Since F is free, we get CG.xy/ D F so y D x�1.xy/ 2

Z.F /, and we conclude that F D P � A, where P 2 Sylp.F / and A is abelian(here we use Burnside’s normal p-complement theorem). Assume, in addition, thaty 62 Z.G/. Then CG.y/ D F and F D Q�B , whereQ 2 Sylq.F / and B is abelian,by the above. SinceQ � A soQ is abelian, we conclude that F is abelian.

Exercise 2. Let H G G be a �-Hall subgroup. Suppose that iG.x/ is a �-number forall x 2 H . Prove that then G D F �H .

Solution. Let F be a � 0-Hall subgroup of G (Schur–Zassenhaus). If x 2 H , thenF y � CG.x/ for some y 2 H so x 2 CG.F

y/ D CG.F /y . Let x1; : : : ; xn 2 H be

representatives of the set of allH -classes such that xi 2 CG.F /, i D 1; : : : ; n, and setD D fx1; : : : ; xng; thenD D H (Burnside). It follows that G D F �D D F �H .

We say that a group G has an abelian partition with kernelK, if G is a set-theoreticunion of some abelian subgroups each pair of which has intersection K; then K �

Z.G/.

Definition 2. A p-group G is said to be of type (F) if, for every x 2 G � Z.G/, thecentralizer CG.x/ is free.

Exercise 3. If the centralizers of all noncentral elements of G are abelian, then G hasan abelian partition with kernel Z.G/ and G is of type (F).

Exercise 4. Let x 2 N GG. Prove that iN .x/ divides iG.x/. (Hint. LetK be aG-classcontaining x. Then Inn.G/ permutes transitively the set of N -classes contained inK.)

Let cs.G/ be the set of class sizes of G. If cs.G/ D f1; ng with n > 1, then G is oftype (F).

Lemma A.23.3. Suppose that cs.G/ D f1; ng and G is nonnilpotent. Then F D

CG.x/ is abelian for all x 2 G � Z.G/.

Proof. Suppose that F is a p-subgroup for some p 2 �.G/. Then all p0-elementsof G lie in Z.G/ so G is nilpotent, a contradiction. Thus, j�.F /j > 1. By LemmaA.23.2, F D P � A, where F 2 Sylp.F / and A > f1g is abelian. Assume that P isnonabelian so one may assume that �.o.x// D fpg. By Lemma A.23.2, A � Z.G/.Since G is nonnilpotent, there is y 2 G � Z.G/ with �.o.y// D fqg, q ¤ p. SetF1 D CG.y/; then jF1j D jF j and F1 D Q�A1, whereQ 2 Sylq.F1/ andA1 > f1g

is abelian. But A � Z.G/ < F1 so A1 D A since jA1j D jAj. It follows that P � A1

so P is abelian, contrary to the assumption.

A.23 Groups all of whose noncentral conjugacy classes have the same size 521

TheoremA.23.4. LetG be a group with cs.G/ D f1; ng. ThenG D P �A is nilpotentwith P 2 Sylp.G/, A � Z.G/ and �.n/ D fpg.

Proof. Suppose that G is nonnilpotent. Then, by Lemma A.23.3, centralizers of non-central elements of G are abelian so G admits an abelian partition † D fCG.xi /g

miD1

with kernel Z.G/ all of whose components are centralizers of noncentral elements ofG and therefore have the same order 1

njGj (Exercise 1). In that case, G=Z.G/ has

a nontrivial abelian partition with kernel f1g all of whose components have the sameorder 1

njG W Z.G/j (i.e.,G=Z.G/ is an equally partitioned group). By Isaacs’ Theorem

68.4, there is a prime p such that exp.G=Z.G// D p; then G is nilpotent, i.e., G isnot a counterexample. Clearly, P 2 Sylp.G/ is a unique nonabelian Sylow subgroupof G. Since G D PZ.G/, we get �.n/ D fpg.

Proposition A.23.5. Let G be a nonabelian p-group of type (F). Then there exists anabelian A GG such that Z.G/ � A and exp.G=A/ D p.

Proof. This follows from Exercise 7.22.

A nonabelian p-group G in which iG.x/ D pe for all x 2 G � Z.G/ is called anIto p-group. Since an Ito p-group is of type (F), we deduce from Proposition A.23.5the following

Corollary A.23.6. If G is an Ito p-group, then exp.G=A/ D p for a suitable abelianA GG. In particular, if p D 2, then G=A is elementary abelian.1

Let G be an Ito p-group. It follows from Ito’s proof of Proposition A.23.5 (seealso the solution of Exercise 7.21) that, if x 2 Z2.G/ � Z.G/, then xp 2 Z.G/ soŒx; y�p D Œxp; y� D 1 for all y 2 G hence fŒx; y� j y 2 Gg D L.x/ � Z.G/ andexp.L.x// D p so L.x/ is elementary abelian. As we know, the map y 7! Œx; y�

is the isomorphism G=CG.x/ Š L.x/ so jL.x/j D pe , where pe D iG.x/. SinceL.x/ Š Epe , we conclude that ˆ.G/ � CG.x/. Thus, we have the following

Proposition A.23.7. Let G be an Ito p-group with iG.x/ D pe for all x 2 G � Z.G/.Then d.G/ � e and j�1.Z.G//j � pe.

Let Z.G/ be cyclic. Then j�1.Z.G//j D p so, since Epe Š L.x/ � Z.G/, we gete D 1. In that case, by [Kno1] (see also Exercise 2.7), jG0j D p so cl.G/ D 2.Ishikawa [Ish] has proved that if G is a group of Theorem A.23.5, then cl.G/ � 3.

1As A one can take a subgroupM.G/ D hx 2 G j CG.x/ D CG.xp/i (see Exercise 7.22).

Appendix 24

On modular 2-groups

A p-group G is modular if every two subgroups of G are permutable (for arbitrarygroups the definition is another). Sections of modular p-groups are modular. In thissection we offer a generalization of a part of Iwasawa’s theorem on modular 2-groups[Iwa] (see also �73).Let H2 D ha; b j a4 D b4 D 1; ab D a�1i be the unique nonabelian metacyclic

group of order 16 and exponent 4. The groupH2 has exactly three (central) involutionsa2, b2 and a2b2, the quotient group H2 D H2=ha

2b2i Š Q8. Indeed, the groupH2 D hNa; Nbi has three distinct cyclic subgroups of index 2, namely h Nai, h Nbi, h Na Nbi

of order 4, and our claim follows. Since exp.H2/ D 4, a2b2 is not a square. Sinceha2i D H 0

2, it follows that ha2i, hb2i and ha2b2i are characteristic in H2. Note thatH2 is nonmodular sinceH2=hb

2i Š D8 is nonmodular.We define the 2-group F as follows:

(F1) F D hQ;Zi, whereQ Š Q8, Z Š C8, jQ \Zj D 4.

(F2) jF j D 25.

(F3) CF .Q/ Š E4.

(F4) �2.F / D QCF .Q/.

We have NF .Q\Z/ � hQ;Zi D F soQ\ZGF . By (F1) and (F2),QZ ¤ ZQ

so F is not modular, and F =.Q \ Z/ Š D8. By (F3) and (F4), �2.F / Š Q � C2

hence c1.F / D 3, c2.F / D 6 so c3.F / D jF j�j�2.F /j'.8/

D 4 and F is not of maximalclass. By (F1) and (F2), Z 6E F . Let U=.Q \ Z/ be the cyclic subgroup of order4 in F =.Q \ Z/. Then U is nonabelian (otherwise, CF .Q \ Z/ � UZ D F soQ \ Z � Z.F /, a contradiction since jQ W .Q \ Z/j D 2 and Q is nonabelian).If exp.U / D 4, then U Š H2 is minimal nonabelian so c2.U / D 6 D c2.F /,c1.U / D 3 D c1.F / so U D �2.F /, a contradiction since �2.F / .> Q/ is notminimal nonabelian. Thus, exp.U / D 8 so U Š M24 .The group F is uniquely determined. This follows from �52. Let G be a 2-group

of order > 24 with�2.G/ D Q8 �C2. Then jGj D 25 and G is uniquely determined:G D hw;yi, where

w4 D y8 D 1; Œw; y� D t; y2 D tv; w2 D v2 D u;

v2 D t2 D Œt; w� D 1; Œt; y� D Œv;w� D u; vy D v�1:

Here Z.G/ D hui is of order 2, CG.t/ D hti � hv;wi, where hv;wi Š Q8.

A.24 On modular 2-groups 523

In Theorem A.24.1 we offer another approach to the proof of the following assertionfrom [Iwa]: If a modular 2-group has a section Š Q8, it is Dedekindian. In fact,Theorem A.24.1 is a generalization of the above assertion since in that theorem we donot assume that G is modular.

Theorem A.24.1. A 2-group G is Dedekindian if it satisfies the following conditions:

(i) Every two subgroups of G of order � 4 are permutable.

(ii) G has a section isomorphic to Q8.

(iii) G is H2-free.

(iv) G has no subgroups isomorphic to F .

We first prove the following

Lemma A.24.2. Suppose that a 2-group G is H2-free and such that G=N Š Q8 forsome N GG. Then G D Q N withQ \N D f1g andQ Š Q8.

Proof. We use induction on jGj. One may assume that jGj > 23. First let jN j D 2.SinceG is not of maximal class, it has no cyclic subgroups of index 2 (Lemma 64.1(t)).IfN 6� ˆ.G/, we getG D N �Q, and we are done. Now letN < ˆ.G/. In that case,all maximal subgroups of G are abelian of type .4; 2/ soG is minimal nonabelian, and�1.G/ Š E4. Then, by Lemma 65.1, G is metacyclic of exponent 4 so G Š H2,contrary to the hypothesis.Now suppose that jN j > 2. Let L be a G-invariant subgroup of index 2 in N . By

the previous paragraph, G=L D .Q=L/ � .N=L/, where Q=L Š Q8. By inductionapplied toQ, we haveQ D Q1 L, whereQ1 \ L D f1g andQ1 Š Q8. We have

Q1 \ N D .Q1 \Q/\N D Q1 \ .Q \N/ D Q1 \ L D f1g

so G D Q1 N is a semidirect product.

Proof of Theorem A.24.1. We use induction on jGj. The hypothesis is inherited bysubgroups. All subgroups ofG of exponent� 4 are modular, by (iii), so exp.�1.G//D

2. By hypothesis, there is a chain L E K � G such that K=L Š Q8. Since K isH2-free, we get K D Q L, whereQ \ L D f1g andQ Š Q8 (Lemma A.24.2). SetQ D ha; b j a4 D b4 D 1; ab D a�1i, A D hai, B D hbi. Let Q � H 2 �1. Byinduction, H D Q � E, where exp.E/ � 2. Let z 2 G �H , where o.z/ is as smallas possible. Set Z D hzi. We get exp.H/ � 4 so jZj � 2 exp.H/ D 8.Let jZj D 2. Then ZA D AZ and BZ D ZB are abelian of type .4; 2/ since

G has no subgroups Š D8. If y 2 E, then hz; yi Š E4 since hzihyi < G. ThenG D H � Z D .Q � E/ �Z D Q � .E �Z/ is Dedekindian.Let jZj D 4. Then Z is permutable with all subgroups ofH , by (i). SetM D QZ;

then M � G. Assume that Q \ Z D Z.Q/; then jM j D 24. Since a 2-group ofmaximal class and order � 24 has two non-permutable subgroups of order 4,M is not


of maximal class. Therefore, by Lemma 64.1(i) (= Proposition 1.17),M D Q�Z.M/.Next, Z.M/ Š E4 (otherwise, �1.M/ D M is not elementary abelian, by Appendix16). It follows that M D Q � L, where jLj D 2. Then L 6� H and jLj D 2 <

jZj, contrary to the choice of z. Thus, Q \ Z D f1g. By the previous paragraph,�1.Z/ < Z.M/. If D D Z.Q/ � �1.Z/ .Š E4/, then D G M and M=D Š E8 soexp.M/ D 4 andM is modular, by (i). Then, by the previous paragraph,M=�1.Z/ D

.Q�1.Z/=�1.Z// � .Z=�1.Z// so Z GM . Since A Z 6Š H2, it follows that AZis abelian of type .4; 4/. Similarly, B Z Š A � Z. It follows that Z centralizesAB D Q so M D Q � Z. Let T D ha2z2i. Then M=T Š Q8 � C4 is of order 16so Q8 � C4 Š D8 � C4 (Appendix 16). The inverse image of D8 of exponent 4 is notmodular since D8 is not modular, a contradiction.It remains to consider the case jZj D 8. Then H \ Z is cyclic of order 4. Since

Ã1.H/ D Z.Q/, we get Z.Q/ < Z \H < Z. Since CH .Q/ D Z.H/ is elementaryabelian, we may assume that Z \ H does not centralize a subgroup A of order 4in Q. Then A.H \ Z/ D .H \ Z/A and �1.H \ Z/ D �1.A/ .D Z.Q// soA.H \Z/ D Q1 Š Q8. Since every nonabelian subgroup of order 8 is a direct factorof H , one may assume from the start that jQ \ Zj D 4. Set D D hQ;Zi. Assumethat jDj D 24; then D D QZ D ZQ has the cyclic subgroup Z of index 2 so Dis of maximal class (Lemma 64.1(t)). Since D has two non-permutable subgroups oforder 4, we get a contradiction. Thus, jDj > 24 so Z does not normalize Q. ThenD=.Q \Z/ Š D2n for n > 2 sinceD=.Q \ Z/ is generated by two non-permutablesubgroups Q=.Q \ Z/ and Z=.Q \ Z/ of order 2. Since exp.H=.Q \ Z// D 2,we get exp.G=.Q \ Z// � 4 so n D 3. We conclude that D=.Q \ Z/ Š D23

so jDj D jQ \ ZjjD=.Q \ Z/j D 25. As above, D is not of maximal class soCD.Q/ 6� Q (Lemma 64.1(i)). Since QCD.Q/ is modular, CD.Q/ Š E4 (Appendix16). In that case,QCD.Q/ D Q � L, where jLj D 2. Clearly, QCD.Q/ D D \H .Take y 2 D�QCD.Q/ and set Y D hyi. Since y 62 H , we get jY j D 8, by the choiceof z. Then c3.D/ D 4, c2.D/ D c2.QCD.Q// D 6, c1.D/ D c1.QCD.Q// D 3. Itfollows that D Š F , contrary to the hypothesis. Thus, G has no cyclic subgroups oforder 8. The proof is complete.

Thus, the classification of modular 2-groups G is reduced to the case where G isquaternion-free. If G of Theorem A.24.1 is a minimal counterexample, it is minimalnon-Dedekindian so the result follows from Theorem A.17.3. However, the proof ofTheorem A.24.1 presents independent interest.

Corollary A.24.3. A 2-group G is Dedekindian if it satisfies the following conditions:(i) Whenever U;V � G with jU j � 4, jV j � 8, then U and V are permutable. (ii) Q8

is involved in G. (iii) G is H2-free.

Indeed, G has no subgroups Š F so G is Dedekindian, by Theorem A.24.1.

Corollary A.24.4 ([Iwa]). Suppose that a 2-group G is modular. If Q8 is involved inG, then G is Dedekindian.

A.24 On modular 2-groups 525

Indeed, H2 and F are nonmodular since they have epimorphic image Š D8. Nowthe result follows from Theorem A.24.1.

Corollary A.24.5. Let a non-Dedekindian 2-group G be H2-free and let �2.G/ < G

be nonabelian Dedekindian. Then G has a subgroup Š F .

Indeed, if G has no subgroups isomorphic to F , it satisfies the hypothesis of Theo-rem 24.1 (see Theorem 1.20) so G is Dedekindian.Now we state an analog of Lemma A.24.2 for p > 2. Suppose that a p-group G,

p > 2, has no sections of order p4 which are either of maximal class or nonmetacyclicminimal nonabelian of order p4. LetNGG be such thatG=N Š S.p3/, the nonabeliangroup of order p3 and exponent p. Then G D S N , where S \ N D f1g. Assumethat our assertion is false. As in the proof of Lemma A.24.2, one may confine to thecase jN j D p. Then d.G/ D 2. SinceG is not of maximal class, we get jZ.G/j D p2.In that case, however, G is a nonmetacyclic A1-group, a contradiction.

Appendix 25

Schreier’s inequality for p-groups

If d.G/ is the minimal number of generators of (not necessarily prime power) groupG,H < G, then the following inequality due to O. Schreier holds:

d.H/ � 1C .d.G/ � 1/jG W H j:(1)

Consider the regular wreath productG0 D Epd�1 wrCn, wherep 2 �.n/, n; d > 1.Let H0 be the base subgroup of G0; then H0 Š Ep.d�1/n. Let Cn D hxi; then G0 D

hx;H0i. Define the group G D hy;H0i as follows: hyi \ H0 D f1g, o.y/ D pn,yn 2 Z.G/, hy D hx for all h 2 H0. Set H D hyn;H0i; then H D hyni � H0 Š

Ep1C.d�1/n. Since yn � ˆ.hyi/ \ Z.G/, we get hyni � ˆ.G/. Since G=hyni Š G0,we have d.G/ D d.G0/. Since G0 D hx;Epd�1i, we get d.G/ � d . Since d.H/ D

1C .d � 1/n, we get d.G/ D d , by (1), so estimate (1) is attained. We offer anotherproof of (1) for p-groups.

Theorem A.25.1. Let H be a subgroup of index pm in a p-group G. Then d.H/ �

1C pm.d.G/ � 1/.

Proof. We use induction on m.

(i) Let m D 1. Then ˆ.H/ G G and ˆ.H/ � ˆ.G/. Hence d.G=ˆ.H// D d.G/,d.H=ˆ.H// D d.H/ so, without loss of generality, one may assume that ˆ.H/ D

f1g. ThenH Š Epd.H/ .Take x 2 G �H ; then G D hxiH , xp 2 Z.G/ \ˆ.G/. Setting NG D G=hxpi, we

get d. NG/ D d.G/. It follows from �1. NG/ D NG that NG0 D ˆ. NG/.Let, in addition, d.G/ D 2. Then j NG W NG0j D p2 so d.G/ D d. NG/ D 2 and NG is

either abelian of order p2 or of maximal class. In the first case, jGj D p3 so (1) is true.If NG is of maximal class, the d. NH/ � p (Theorems 9.5 and 9.6) so d.H/ � pC1, and(1) holds again. (If, in addition, o.x/ D p, then jGj � ppC1 so d.H/ � p.)Next suppose that d D d.G/ > 2. Then G D hx; x2; : : : ; xd i where x2; : : : ; xd 2

H . Set Ai D hx; xi i (i D 2; : : : ; d ). Since d.Ai/ D 2 and H \ Ai is elementaryabelian of index p in Ai , we have jAi j � p2Cp, i D 2; : : : ; d , by the previousparagraph. Since NG.H \ Ai/ � HAi D G, H \ Ai is normal in G. Next, A2.H \

A3/ : : : .H \ Ad / � hx; x2; : : : ; xd i D G and the factors A2;H \ A3; : : : ;H \ Ad

are pairwise permutable. Now, for i D 2; : : : ; d , we get

jAi=hxpij � p1Cp; j.H \ Ai /=hx

pij � pp ; j NGj D jG=hxpij � p1C.d�1/p;

A.25 Schreier’s inequality for p-groups 527

and

jGj � pj NGj � p2C.d�1/p; jH j � p1C.d�1/p; d.H/ � 1C .d � 1/p;

proving (1) for G andH in the casem D 1.1

(ii) Let m > 1. Take in G a maximal subgroup F containing H . By induction,d.H/ � 1C pm�1.d.F / � 1/. By (1), d.F / � 1 � p.d.G/ � 1/ so

d.H/ � 1C pm�1 p.d.G/ � 1/ D 1C pm.d.G/ � 1/;

and (1) is also proved for m > 1.

Definition. A p-group G is said to be generalized regular if exp.�1.H// D p forevery nontrivial sectionH of G.

Generalized regular p-groups coincide with P2-groups (see �11). The group Q8 isgeneralized regular but irregular.

Theorem A.25.2. LetH be a maximal subgroup of a p-group G.

(a) If exp.G/ D p, then d.H/ � .p � 1/.d.G/ � 1/.

(b) If G is generalized regular, then d.H/ � 1C .p � 1/.d.G/ � 1/.

Proof. We may assume that G is nonabelian and, as before, ˆ.H/ D f1g; then H Š

Epd.H/ . Set d D d.G/.

(a) Let exp.G/ D p; then ˆ.G/ D G0 so jG W G0j D pd and, by Lemma 64.1(q),jZ.G/j D pd�1.

(i) Let d D 2. Then G is of maximal class so jGj � pp (Theorem 9.5) andjH j � pp�1, d.H/ � p � 1 D .d � 1/.p � 1/.

(ii) Let d > 2. Take x D x1 2 G � H ; then G D hx1; x2; : : : ; xd i, wherex2; : : : ; xd 2 H . For i D 2; : : : ; d , consider the subgroup Ai D hx1; xi i; then, by (i),jAi j � pp . We have NG.H \ Ai/ � HAi D G so H \ Ai is G-invariant of order� pp�1 for i > 1. It follows from

A2.H \ A3/ : : : .H \ Ad / � hx1; x2; : : : ; xd i D G

that jGj � p1C.p�1/.d�1/, jH j � p.p�1/.d�1/ so d.H/ � .d � 1/.p � 1/.

(b) Let H 2 �1 and, as above, ˆ.H/ D f1g, i.e., H is elementary abelian. Ifx D x1 2 G �H , then xp 2 Z.G/, G D hxi H . As above, G D hx1; x2; : : : ; xd i,where x2; : : : ; xd 2 H , x1 D x. Let G0 D G=hxpi, H0 D H=hxpi. If yi D xi hx

pi,then G0 D hy1; : : : ; yd i, y2; : : : ; yd 2 H0. Since G0 is generalized regular and�1.G0/ D G0, we have exp.G0/ D p. By (a), d.H0/ � .d � 1/.p � 1/. Hence,d.H/ � 1C d.H0/ � 1C .d � 1/.p � 1/, completing the proof of (b).

1If o.x/ D p, then, as we know, jAi j � ppC1 for all i D 2; : : : ; d so jGj � p1Cp.d�1/ andd.H/ � p.d � 1/. Thus, if d.H/ D 1C p.d � 1/, then xp ¤ 1 for x 2 G �H or, what is the same,�1.G/ D H .


It is possible to show that estimates of Theorem A.25.2 are best possible.Let G be a generalized regular p-group and letH < G be of index pm > p. Using

Theorem A.25.2(b) and induction on m, it is easy to show that d.H/ � 1 C .p �

1/m.d.G/ � 1/. In particular, if p D 2, then d.H/ � d.G/.

Appendix 26

p-groups all of whose nonabelian maximalsubgroups are either absolutely regular

or of maximal class

In this section we prove the following

Theorem A.26.1. Let a nonabelian p-group G be neither minimal nonabelian norabsolutely regular, p > 2 and jGj > ppC1. If all nonabelian maximal subgroups ofG are either absolutely regular or of maximal class, then one of the following holds:

(a) G is of maximal class.

(b) G D B � C where B is absolutely regular, jC j D p, j�1.G/j D pp , �1.G/ �

Z.G/, d.G=�1.G// D 2. All maximal subgroups of B containing �1.B/, areabelian.

(c) G is an Lp-group (see ��17, 18), jG W CG.�1.G//j D p.

Groups (a)–(c) satisfy the hypothesis.

Proof. The last assertion is checked easily as will be clear from the proof. It remainsto show that if G satisfies the hypothesis, it is one of groups (a)–(c).If G is of maximal class (of order > ppC1), then all maximal subgroups of G

are either absolutely regular or of maximal class (Theorem 9.6) so G satisfies thehypothesis. In what follows we assume that G is not of maximal class.

Suppose that exp.G/ D p. Then G has no absolutely regular maximal subgroups.By Theorem 9.5, G has no subgroups of maximal class and index p. Since G is notminimal nonabelian, we get a contradiction. Thus, exp.G/ > p.Let G be regular. Then, by Theorem 9.5, G has no subgroups of maximal class

and index p. Assume that j�1.G/j > pp . Then the set �1 has no absolutely regu-lar members. Since G is not minimal nonabelian. we get a contradiction. Now letj�1.G/j D pp. Then all maximal subgroups of G, containing �1.G/, are abelian(Theorem 9.5). If G=�1.G/ is cyclic, then G is an Lp-group. Now assume thatG=�1.G/ is noncyclic. Then there are in G two distinct maximal subgroups A andB that contain �1.G/; moreover, d.G=�1.G// D 2 (Exercise 1.6(a)). In that case,�1.G/ � A \ B D Z.G/ so jG0j D p. If �1.G/ � ˆ.G/, then d.G/ D 2 so G isminimal nonabelian, contrary to the hypothesis. Otherwise, there is X < �1.G/ of


order p such that G D X �M , whereM 2 �1 is absolutely regular. In that case, Gis as in (b).Next we assume that G is irregular. Since G is not of maximal class, it contains a

normal subgroup R of order pp and exponent p.

(i) Suppose that jGj > ppC2. Then all maximal subgroups of G containing Rare neither absolutely regular nor of maximal class (Lemma 64.1(f)). Therefore, ifR < A, where A is maximal in G, then A is abelian. Assume that R < �1.G/. Letx 2 G � R be of order p; then L D hx;Ri is elementary abelian of order ppC1.Consideration of intersection of a maximal subgroup, say H , with L shows that His neither of maximal class nor absolutely regular. Then all maximal subgroups of Gare abelian, a contradiction since G is not minimal nonabelian. Thus, R D �1.G/.Therefore, if G=R is cyclic, thenG is an Lp-group so it is as in (c). Suppose that G=Ris noncyclic. Since all maximal subgroups of G, containing R, are abelian, it followsthat R � Z.G/ and jG W Z.G/j D p2 so cl.G/ D 2 and G is regular, contrary to theassumption.

(ii) Let jGj D ppC2.

(ii1) Suppose that G=R is cyclic (of order p2). LetD < R be G-invariant of indexp2 and C D CG.R=D/; then jG W C j � p. Since cl.G/ � p > 2, it follows thatC 2 �1 is not of maximal class so abelian. If �1.G/ D R, then G is an Lp-group.Assume that j�1.G/j D ppC1; then �1.G/ is regular so elementary abelian. In thatcase, all maximal subgroups of G are abelian, a contradiction.

Now let G=R Š Ep2 .

(ii2) Suppose that allM < G such that R < M , are abelian. Then R D Z.G/ andcl.G/ D 2 so G is regular, contrary to the assumption.

(ii3) Now suppose that there is nonabelian M < G such that R < M . Then Mis of maximal class so the number of subgroups of maximal class and index p in Gis exactly p2 (Theorem 12.12(c)). Since d.G/ D 3 and G has no absolutely regularmaximal subgroups (Theorem 12.12(b)), the number of abelian subgroups of index pinG is exactly pC 1. In that case, cl.G/ D 2 soG is regular, a final contradiction.

Since absolutely regular 2-groups are cyclic, similar result for p D 2 gives thefollowing

Exercise 1. If all nonabelian maximal subgroups of a nonabelian 2-group G are ofmaximal class, then one of the following holds:


(b) G is of maximal class.

(c) G D DZ.G/ is of order 16, where G is nonabelian of order 8.

Exercise 2. Classify the nonabelian p-groups of order ppC1, p > 2, all of whosenonabelian maximal subgroups are either absolutely regular or of maximal class.

Research problems and themes II

This is the second part of the list written by the first author.

701. Study the irregular p-groups, p > 2, all of whose metabelian subgroups areregular.

702. Classify the minimal non-quaternion-free 2-groups. (For a solution, see �80.)

703. Study the p-groups G such that all elements of the setG�G0 (G� Ã1.G/) havethe same order > p and < exp.G/.

704. Describe the p-groups with cyclic subgroup of index p3 acting as in �74.

705. Classify the U3-groups (see �64).

706. Classify the p-groups G such that for every (i) H < G there exists R G G withG=R Š H , (ii) R GG there existsH � G such that G=R Š H .

707. Construct a p-group G D XY (X;Y < G) such that CY .x/ D f1g for all x 2 X .

708. Set ˛.G/ D max fd.A/ j A � G; A0 D f1gg, .G/ D max fd.H/ j H � Gg.Produce a good estimate of .G/ in terms of ˛.G/.

709. Does there exist special p-groups all of whose maximal subgroups are special?If so, classify such groups.

710. Let jˆ.G/j D pn. Study the structure of G D Gn provided cl.G/ D nC 1. Findthe minimal n such that all Gn (p > 2) are irregular.

711. Does there exist p-groups of order> p2 and exponent> p, all of whose maximalsubgroups are generated by elements of order p? If so, study their structure.

712. Study the p-groups G of exponent p such that jZp.G/j D pp .

713. Classify the p-groups which are lattice isomorphic to p-groups with abeliansubgroup of index p.

714. Is it true that a p-group with a subgroup of order ppC2 and exponent p has anormal subgroup of order ppC1 and exponent p?

715. Study the p-groups G such that Aut.G/ is an extraspecial p-group.

716. Classify the p-groups G such that G=R is special for all R � Z.G/ of order p.


717. Let ˆ.G/ < N < G. Study the structures of N and G provided all maximalsubgroups of G not containing N , are metacyclic.

718. Let M 2 �1. Classify the p-groups G all of whose nonabelian L 2 �1 � fM g

are minimal nonabelian.

719. Study the p-groups G such that for every x 2 ˆ.G/ there is a minimal basisa1; : : : ; ad with x 2 ha1i.

720. Is it true that the derived length of a p-group G is bounded provided thatexp.Aut.G//p D p? Study the p-groups G satisfying the last condition.

721. Let G be group of order pm with exactly pm�2 minimal nonabelian subgroups.Is it true that jG0j is bounded?

722. (Old problem) Study the p-groups with elementary abelian Aut.G/.

723. Study the pairs H < G of p-groups, if they exist, with cd.G/ D cd.H/ andjIrr.G/j D jIrr.H/j.

724. Classify the p-groups with self centralizing abelian subgroup A of order p3. (Forp D 2 and noncyclic A, see ��51, 77.)

725. Study the p-groups all of whose maximal subgroups are either metacyclic or havederived subgroup of order � p.

726. Classify the p-groups G such that Aut.G/ 2 f†pn;UT.n; p/g for some n.

727. Does there exist a 2-groupG such thatˆ.G/ is isomorphic to a Sylow 2-subgroupof some Suzuki simple group Sz.2m/? If so, find all such m.

728. (i) Study the p-groups G, p > 2, such that G=Ã1.G/ is extraspecial. (ii) Doesthere exist among groups satisfying (i), irregular p-groups?

729. Study the p-groups G with a minimal basis a1; : : : ; ad such that every elementofG can be presented in the unique way in the form a1

1 : : : ad

dwith 0 � i < o.ai /,

i D 1; : : : ; d .

730. Classify the p-groups whose nonabelian subgroups of index p2 are metacyclic.

731. Classify the p-groups all of whose two-generator subgroups are metacyclic.

732. Study the p-groups G with equal numbers of principal series in G and G=G0.

733. (Passman [Pas]) Classify the 2-groups all of whose nonnormal subgroups arecyclic. (This problem is solved in �16.)

734. Classify the p-groups G such that, whenever nonabelian H 2 �1, then H D

EZ.H/, where E is extraspecial.

735. (Isaacs–Slattery) Given p > 2, does there exist a p-group G of class p C 1 withcd.G/ D f1; p2; p4g?

Research problems and themes II 533

736. Classify the p-groups G such that, whenever H � G is nonabelian, thenCG.H/ D Z.H/.

737. Classify the p-groups G all of whose maximal cyclic subgroups coincide withtheir centralizers.

738. For definition of an An-group, see ��65, 71, 72. There exists ˛.n/ such thatjG0j � p˛.n/ for all An-groups G, and for someAn-groupH we have jH 0j D p˛.n/.Give good estimate for ˛.n/. Even ˛.4/ is not known (it is known only that ˛.4/ � 6;see �72). Is it true that ˛.nC1/ > ˛.n/ for all n? Janko conjectured that ˛.n/ � nC1.

739. (i) Study the An-groups G satisfying jG0j D p˛.n/. (ii) Is it true that ˛.4/ D 6?(See #738.)

740. Study the p-groups G such that G=Ã2.G/ is special.

741. Classify the p-groups G with jG W Z./j � p5 for all 2 Irr1.G/. (HereZ./ D hx 2 G j j.x/j D .1/i.)

742. Classify the p-groups in which every nonabelian two-generator subgroup is eithermetacyclic or minimal nonabelian. (See #699.)

743. Study the 2-groups G such that every proper nonabelian subgroup of G has asection (i) Š Q8, (ii) Š D8.

744. Study the p-groups G withH 0 D G0 for allH 2 �1. (As Mann proved [Man12],these groups coincide with groups without irreducible characters of degree p.)

745. Classify the An-groups G, n > 1 (see #738), with cyclic G0 of order pn�1 (see�72).

746. Study the 2-groups G satisfying (i)�1.G/ Š D2n �C4 (for n D 3, see Theorem43.9), (ii)�1.G/ D D2n � E2m .

747. Let a p-group G be anAn-group. Is it true that exp.G0/ � pn.

748. Let G be a p-group of maximal class and max f.1/ j 2 Irr.G/g D pbg. Isit true that (i) cd.G/ D f1; p; p2; : : : ; pbg? (ii) all faithful irreducible characters havethe same degree pb?

749. Classify the p-groups G of coclass 2 without subgroups of maximal class andindex p. (For such G, either jG W G0j D p2, and then p > 2, or G=K3.G/ is minimalnonabelian of order p4.)

750. Study the nonabelian p-groups G such that H 0 D ˆ.H/ for all nonabelianH � G. Moreover, study the nonabelian p-groups G such that, whenever H � G isnonabelian, thenH=H 0 is homocyclic.

751. Study the p-groups G with Z.H/ D Z.G/ for anyA1-subgroup H < G.


752. Study the p-groups all of whose maximal subgroups have trivial Schur multipli-ers.

753. (Ito) Let a p-group G have a faithful irreducible character of degree p2. Studythe abelian subgroups of G.

754. Study the p-groups with an automorphism ¤ id leaving unchanged all maximalabelian subgroups.

755. (Old problem) Classify the 2-groups containing exactly three involutions. (Thisis solved in �83.)

756. Describe the group of all automorphisms � of the abelian p-group G such that��1.G/ D id.

757. Study the p-groups G with CG.x/ � H for all A1-subgroups H < G andx 2 H � Z.G/. (For a solution, see Theorem 92.1.)

758. Classify the p-groups all of whose maximal subgroups are of the formM � E,whereM is metacyclic and E is abelian.

759. Is it true that if G is irregular of order pm with c1.G/ D 1 C p C C pm�2,then either exp.G/ D p2 or jG W �1.G/j � p?

760. Classify the p-groups all of whose nonabelian maximal subgroups are either class2 or of maximal class.

761. Find the greatest s 2 N such that, wheneverG is an irregular Ls-group (see �17),then Ã1.G/ is cyclic. (I think that s D p.)

762. Classify the p-groups G containing an element t of order p such that CG.t/ D

hti �M , whereM Š Mpn .

763. Let G be a p-group and M < G be of maximal class with CG.M/ < M .Estimate ranks of abelian subgroups, abelian normal subgroups, subgroups, normalsubgroups of G in terms ofM .

764. Study the p-groups G containing an extraspecial subgroup E such that CG.E/ <

E.

765. Classify the nonabelian p-groups such that the orders of elements of any theirminimal basis are pairwise distinct. (The group SD2n , n > 3, satisfies the aboveproperty.)

766. (i) (Blackburn) Study the p-groups G, p > 2, with jG=K3.G/j D p3. (ii)Moreover, classify the p-groups G such that G=K3.G/ is extraspecial (minimal non-abelian).

767. Classify the two-generator p-groups, p > 2, all of whose maximal subgroups aretwo-generator. (See �70 and Theorem 71.7.)


768. Classify the p-groups all of whose nonnormal subgroups are abelian.

769. Study the p-groups G such that Z.H/ is either cyclic orŠ Ep2 for all nonabelianH < G.

770. Construct, for each p > 2, a p-group G all of whose maximal regular subgroupsare not normal.

771. Study the p-groups G with d.G/ < d.H/ for H 2 �1.

772. Study the p-groups G such that A \ B is cyclic for all distinct A1-subgroupsA;B < G.

773. Let G GW , whereW is a 2-group and let all W -invariant maximal subgroups ofG be metacyclic (two-generator). Study the structure of G.

774. Does there exist a p-group H such that H Š G0 for some p-group G butH 6Š ˆ.W / for all p-groups W ?

775. Does there exist a p-group H such that H Š ˆ.G/ for some p-group G butH 6Š W 0 for all p-groups W ?

776. (Ito–Ohara) Classify the nonmetacyclic 2-groups G D AB , where A and B arecyclic. (For a solution, see �87.)

777. Study the p-groups containing only one non-two-generator maximal subgroup.

778. Classify the p-groups all of whose nonabelian maximal subgroups are eitherminimal nonabelian or metacyclic.

779. Study the p-groups with metacyclicG=Z.G/. (It follows from Theorem 36.1 thatif, in addition, Z.G/ < G0, then G is metacyclic.)

780. Study the p-groups G such that all nonidentity cyclic direct factors of maximalabelian subgroups of G are maximal cyclic subgroups of G.

781. Let H2 D ha; b j a4 D b4 D 1; ab D a�1i. Study the 2-groups G such that (i)G is H2-free, (ii) CG.H/ D Z.H/ for someH Š H2.

782. Study the p-groups all of whose normal non-characteristic subgroups have thesame order.

783. Classify the nonabelian p-groups covered by nonabelian subgroups of order p3.

784. Classify the p-groups G without special sections.

785. Study the p-groups G such that NG.H/ � HG (HG � NG.G/) for all nonnor-malH < G.

786. (Old problem) Classify the p-groups G with ˆ.G/ Š Ep2 . (By Proposition 4.9,if p D 2, then ˆ.G/ � Z.G/.)


787. Classify the p-groups all of whose nonnormal subgroups have exponent p (ele-mentary abelian).

788. Study the p-groups in which every two noncommuting elements generate a p-group of maximal class. (A solution in case p D 2 follows from Theorem 90.1.)

789. Study the p-groups, p > 2, all of whose irregular sections are two-generator.

790. Study the 2-groups G such that CG.t/=hti is of maximal class for an involutiont 2 G.

791. Classify the p-groups all of whose nonabelian maximal subgroups are two-generator. (See �70.)

792. Let ı.G/ be the minimal degree of faithful representation of a group G by per-mutations (the degree of G). For p-groups A;B , find ı.A � B/, where A Š B andA \ B D Z.A/.

793. Classify the p-groups G, p > 2, such that, whenever A < H � G, where A is amaximal abelian subgroup of G, thenH is irregular.

794. Study the p-groups all of whose maximal regular subgroups have the same order.

795. Study the p-groups all of whose subgroups of fixed order pr , r > 3, are two-generator.

796. LetG be a group of order pm with exactly pm�3 minimal nonabelian subgroups.Is it true that the set �1 has an abelian member?

797. Study the p-groups G such that Z.G=HG/ is cyclic for every nonnormalH < G.

798. Classify the 2-groups with exactly two involutions which are squares.

799. Classify the p-groups with minimal nonabelian G=Z.G/.

800. Does there exist a p-group such that every its maximal subgroup is a directproduct of nonabelian groups of order p3?

801. Classify the p-groups all of whoseA1-subgroups are of order � p4.

802. Estimate the order of a p-group G D ��2.G/ in terms of c2.G/. (See �64.)

803. LetG be a p-group with j�1.G/j D pn. Does there exist a constant c D c.p; n/

such that jG=Ã1.G/j � pc? Moreover, if �1.G/ is of order pn and exponent p, doesthere exist a constant f D f .p/ such that jG=Ã1.G/j � pf n? (See the Remark in�15 where this problem is solved by Mann for p > 2. See also [Man5, II].)

804. Does there exist an An-group G with metacyclic G0 and jG0j D p˛.n/ (see#738)? If so, classify such groups.

805. Study the p-groups G such that AG is minimal nonabelian for all nonnormalabelian A < G.


806. Find a necessary and sufficient condition for a p-group to have a p-admissibleHall chain (see ��24, 88).

807. Classify the p-groups G such that jCG.Z/=Zj D jAut.Z/jp for all cyclic Z <

G.

808. Study the p-groups G containing H 2 �1 such that each maximal cyclic sub-group of H is a maximal cyclic subgroup of G.

809. Does there exist an A4-group G with nonabelian G0? (See �72.)

810. Study the p-groups G such that, whenever A < G is not normal and x 2 G �

NG.A/, then A \ Ax D AG .

811. Study the p-groups in which normalizers of all subgroups are two-generator.

812. Classify the p-groups G such that jNG.Z/ W Zj D p for all maximal cyclicZ < G.

813. Describe the one-stepped maximal subgroups of the following groups: †pn 2

Sylp.Spn/ and UT.n; p/ 2 Sylp.GL.n; p// (see �64).

814. Is it true that G Š †p2 if sk.G/ D sk.†p2/ for k D 1; 2?

815. Let A 2 f†pn;UT.m;pn/g (see #813). Classify the p-groups G with ck.G/ D

ck.A/ for all k.

816. Let 1 < k < m and let G be a group of order pm satisfying sk.G/ � 'm�1;k ,where 'm;k D sk.Epm/. Study the structure of G. (See Theorem 5.17.)

817. Classify the p-groups G in which the intersection of all A1-subgroups is notcontained in Z.G/.

818. Suppose that the lattices of normal subgroups of p-groups G andH are isomor-phic. Describe the structure of H if G Š †pn .

819. Classify the p-groups in which the normalizer of every nonnormal abelian sub-group is either abelian or anA1-subgroup.

820. Study the p-groups G such that ˆ.A/ � Z.G/ for all minimal nonabelian sub-groups A < G but ˆ.G/ 6� Z.G/.

821. Improve the estimate given in Theorem 15.4.

822. (Roitman) Let A G G be p-groups, let jA W CA.g/j � pp for all g 2 G � A.Study the structure and embedding of A in G. (See Appendix 7.)

823. Study the p-groups G in which eachM 2 �1 equals CG.x/ for some x 2 G.

824. Classify the metacyclic p-groups all of whose maximal subgroups are character-istic.


825. Let jGj D pm, let H < G be of order ph and 1 < k < m � h. SetM D fF <

G j H < F; jF j D phCkg. Is it true that if jMj D 'm�h;k , thenH GG?

826. Classify the p-groups all of whose minimal nonabelian subgroups have cyclicsubgroups of index p.

827. Study the p-groups G D ha; b; ci such that ha; bi; hb; ci; ha; ci 2 �1.

828. Study the p-groups G covered by normal abelian subgroups.

829. Study the p-groups with exactly pC 1 subgroups of order pp and exponent p indetail. (See Theorem 13.23.)

830. Is it true that jG W Z.G/j is bounded provided jG W NG.H/j � p for allH < G?(Reported by Mann: According to results of Mann and Vaughan-Lee, in that casejG=Z.G/j � p6.)

831. Find ˛1.H � C/ in terms of H , where C is cyclic. H \ C D Z.H/ of order p.

832. Classify the p-groups G with ˛1.G/ < j�1j. (For exp.G/ D p, see �76.)

833. Estimate jG0j is terms of ˛1.G/.

834. Study the p-groups, p > 2, all of whose sections of exponent p are abelian.

835. Study the p-groups G of class c > p such that Zc�1.G/ is of exponent p.

836. Is it true that for each p-group H there exists a p-group G such thatH Š H1 �

ˆ.G/ and �1.ˆ.G// D ˆ.G/?

837. (Kazarin) For each abelian p-group G we can determine the parameter t .G/ as aminimal n such that G is a subgroup of GL.n; p/. How to calculate t .G/?

838. (Kazarin) Let R be an associative nilalgebra over the field of characteristic p.Determine the “circle” operation as follows (N. Jacobson): a ı b D ab C a C b andobtain the p-group R�, the adjoint group of R. What is the structure of the adjointgroup of a finite nilpotent algebra?

839. (Kazarin) Is it true that if the minimal number of generators of the adjoint groupR� is bounded (R is a nilpotent algebra over the field of characteristic p), then jRj isalso bounded? This problem is connected with Golod’s examples of infinite finitelygenerated groups.

840. Let G D ES.m;p/ be an extraspecial group of order p2mC1. Find the numberof extraspecial subgroups of given order in G. Find sn.G/ for all n.

841. Classify the 2-groups G with jCG.B/j D 4 for a nonabelian B < G, jBj D 8.

842. Study the p-groups G with jZ.M/j � p2 for allM 2 �1.

843. Study the p-groups G with jM W Z.M/j � p2 for allM < G with jG W M j D

p2.


844. LetN > f1g be a subgroup of a p-groupG. Study the structure ofG if, wheneverx 2 G �N , then cl.hN;xi/ D cl.N /.

845. Study the p-groups G such that Sylow p-subgroups of Aut.G/ areA1-groups.

846. Classify the p-groups in which any two distinct A1-subgroups have cyclic inter-section.

847. Study the p-groups G such that, for some N GG and all abelianA < G, we havejAN W N j � p.

848. Find all minimal nonabelian p-groups with trivial Schur multiplier.

849. Let A be a maximal abelian normal subgroup of a p-group G. Suppose that forevery a 2 A � Z.G/ there is x 2 G such that ha; xi is minimal nonabelian. Study thestructure of G.

850. Classify the non-Dedekindian p-groups G such that, whenever F < G is non-normal, then jG W F Aut.G/j � p.

851. Study the p-groups G such that exp.HG/ D exp.H/ for all minimal nonabelianH � G.

852. Study the p-groups G in which all A1-subgroups are isomorphic. (For the case,where, in addition, allA1-subgroups have exponent 4, see �57.)

853. Classify the p-groups G such that CG.CG.H// D H for all nonabelian (minimalnonabelian) H � G.

854. Classify the p-groups G in which every nonnormal subgroup is either abelian orminimal nonabelian.

855. Classify the pairs H < G of p-groups with ˇ1.G;H/ 2 fp � 1; p; p C 1g (see�76).

856. Study the p-groups, p > 2, with exactly one minimal irregular subgroup. More-over, study the p-groups G with exactly one irregular member of the set �1.

857. Find the least upper bound of orders of A1-subgroups of †pn and UT.n; p/.

858. Classify the p-groups G such that ker./ � Z.G/ for all 2 Irr1.G/.

859. Study the p-groups which are generated by centers of their A1-subgroups.

860. Classify the p-groups covered byA1-subgroups.

861. Classify the p-groups all of whose maximal subgroups, but one, are either abelianor minimal nonabelian.

862. Study the p-groups in which every subgroup is generated by its elements of max-imal order. (Regular p-groups satisfy this property, however, p-groups of maximalclass and exponent > p2 does not satisfy).


863. Classify the p-groups G with two maximal subgroups which are special.

864. Let i 2 f1; : : : ; d.G/� 1g. Is it true that the number of subgroups H of maximalclass and such that H 2 �i , is divisible by p? (This is true if i D 1, by Theorem12.12(c)).

865. Does there exist a p-group H of maximal class, of order pp and exponent p > 3such that for every irregular p-group G of maximal class, we have G=Ã1.G/ 6Š H?

866. Classify the 2-groups G all of whose proper nonabelian subgroups are of theform H2 � E, where H2 D ha; b j a4 D b4 D 1; ab D a�1i and E is elementaryabelian. (The solution follows from Theorem 57.3.)

867. Study the p-groups G such that p2 − jAut.G/=Inn.G/j.

868. Study the p-groups all of whoseA1-subgroups have the same exponent p2.

869. Study the p-groups G with jNG.F /j D jNG.H/j for any F;H 6E G.

870. Classify the special p-groups possessing an abelian maximal subgroup.

871. Study the groups G of exponent p having exactly one normal subgroup of indexpi for i D 2; : : : ; p � 1.

872. Study the p-groups such that the normal closure of each their abelian subgroupis either abelian or an A1-subgroup.

873. Study the irregular p-groups, p > 2, such that the normal closure of each theirnonnormal cyclic subgroup is regular.

874. Study the p-groups all of whose metacyclic subgroups are abelian.

875. Study the p-groups G such that their holomorph is nilpotent of the same class asG.

876. Study the p-groups G with NG.A/ D ACG.A/ for all nonabelian A < G.

877. Study the p-groups G with cl.BG/ D cl.B/ for all nonabelian B � G.

878. Construct a p-group G such ˆ.G/ D M �N andM;N of maximal class.

879. Study the normal and power structures of an irregular p-group, p > 3, which isa product of two metacyclic subgroups.

880. Let G D †pn . Describe allH < G such that d.H/ D pn�1.

881. Study the p-groups G with Aut.G/ Š Aut.A/ for some abelian p-group A.

882. Is it true that for each p-group G with trivial multiplier there is a p-group Wwith d.W / D d.G/ and W=N Š G for some f1g < N GW ?

883. Study the p-groups G with abelian��n.G/ of type .p

n; p; : : : ; p/, n > 1.


884. Study the 2-groups G with CG.t/ Š C2n � E2m , n > 1, for some involutiont 2 G.

885. Study the 2-groups G with CG.t/ D M � E, whereM is of maximal class andE is elementary abelian. (For jEj D 2, see ��49, 51.)

886. Study the 2-groups G D C1 : : : Cd , where C1; : : : ; Cd are cyclic not necessarilypairwise permutable and d D d.G/.

887. Let G be an abelian 2-group of exponent � 4 and d.G/ D n. Find the maximumof ranks of subgroups of Aut.G/. The same question for arbitrary abelian p-group.

888. Classify the 2-groups G with G=Z of maximal class for some cyclic Z GG.

889. Classify the p-groups G such that all maximal subgroups of its A1-subgroupsare maximal abelian subgroups of G. (For a solution, see �92.)

890. Classify the p-groups which are not generated by noncyclic subgroups of indexp3.

891. Classify the metacyclic p-groups with trivial Schur multiplier.

892. Does there exist an An-group, with noncyclic derived subgroup of exponent> pn�1? If so, classify such groups.

893. Describe the set f˛1.G/ j G is anA3-group g.

894. Give an upper estimate of the number jG W Z.G/j, where G runs over the set ofAn-groups.

895. Classify the p-groups G with ˛2.G/ D 1 (˛n.G/ is the number ofAn-subgroupsin a p-group G).

896. Classify the 2-groups all of whose A1-subgroups are isomorphic to (i) M2n orQ8, (ii) M2n or D8, (iii) M2n or Q8 or D8, (iv) M2n or H2 D ha; b j a4 D b4 D

1; ab D a3i.

897. Study the p-groups G such that �1.A/ � Z.G/ for all minimal nonabeliansubgroups A < G but �1.G/ 6� Z.G/.

898. Classify the nonabelian groups G of exponent p such, that, whenever A;B � G

are minimal nonabelian, then A \ B > f1g.

899. Study the 3-groups, all of whose minimal nonabelian subgroups are maximalregular.

900. Classify the 2-groups G such that �2.G/ is abelian of type .4; 2; : : : ; 2/.

901. Describe the structure of Aut.G/, whereG D ES.m;p/�Cpn (G D ES.m;p/�Cpn is of order p2mCn); here ES.m;p/ is extraspecial of order p2mC1.

902. Study the p-groups G such that Œ�i.G/;Ãi .G/� D f1g for all i .


903. Find ˛1.A�C/, whereA is a minimal nonabelian p-group and C Š Cpn , n > 1.

904. Study the p-groups G such that, whenever A;B < G are distinct minimal non-abelian, then jA \ Bj � p2. (Such G contains anA1-subgroup of order p3.)

905. Classify the irregular p-groups of order pm with maximal possible number ofsolutions of equation xp D 1. (For p D 2, see Exercise 10.20.)

906. Classify the p-groups having a representation group of exponent p. (Accordingto D.L. Johnson, the multipliers of noncyclic groups of exponent p are nontrivial.)

907. Let H2 D ha; b j a4 D b4 D 1; ab D a�1i. Classify minimal non-H -free2-groups. (See ��78, 80.)

908. Describe the representation groups of extraspecial p-groups.

909. Study the structures of representation groups of homocyclic p-groups.

910. Classify the p-groups whose A2-subgroups H satisfy (i) ˛1.H/ D p, (ii)˛1.H/ D p C 1, (iii) ˛1.H/ � p C 1.

911. Classify the A1-groups (A2-groups), isomorphic to G-invariant subgroups ofˆ-subgroup of a p-group G. (The group H2 from #907 does not satisfy the abovecondition.)

912. Study the p-groups in which the centralizer of every noncentral element is (i) anA1-subgroup; (ii) either abelian or anA1-subgroup, (iii) of class � 2.

913. Classify the p-groups in which the normalizer of every nonnormal cyclic sub-group is such as in #912.

914. Study the 2-groups G satisfying G0 D Ã3.G/.

915. Study the p-groups G of order pm and classm�2 satisfying one of the followingconditions: (i) d.G/ D 2 and jG W G0j D p3, (ii) jZ.G/j D p2.

916. (i) Does there exist a p-group all of whose maximal subgroups are direct productsof extraspecial (special) groups? If so, classify such groups. (ii) Study the p-groupsall of whose nonabelian maximal subgroups are special.

917. Classify the p-groups all of whose irreducible characters are induced fromabelian subgroups (fromA1-subgroups).

918. Study the p-groups G such that, whenever A < G, then G contains a normalsubgroup B Š A.

919. Given n > 3, does there exist a 2-group, which is not of maximal class, containingexactly one proper subgroup isomorphic with D2n (Q2n , SD2n)?

920. Classify the p-groups all of whose A1-subgroups are of order p3. (See also#115. For p D 2, this problem is solved in �90.)


921. LetG,G0 be p-groups of the same order containing, for each n, the same numberof conjugacy classes of size pn. Is it true that G0 is special if G is?

922. Given a group �, find a group G such that � is a representation group of G orprove that such G does not exist.

923. Suppose that a p-groupG contains a subgroupH D M�C , whereM is irregularof maximal class and jC j D p. Describe the structure of G if it has only one subgroupŠ H .

924. Study the p-groups G such that, whenever its nonabelian epimorphic image Hhas cyclic center, then jZ.H/j D p.

925. Find the representation groups of M � N , where M and N are 2-groups ofmaximal class.

926. Classify the 2-groups G such that whenever H 2 �1 is nonabelian, then H D

M �E, whereM is minimal nonabelian and E is abelian.

927. Study the representation groups of p-groups of maximal class, p > 2.

928. (I. D. Macdonald) Let n > 2. Classify the p-groups G of class 2n all of whosemaximal subgroups are of class n.

929. Given a nonabelian group G of exponent p, does there exist a p-group W suchthatW=S Š G, where S is the socle of W ?

930. Does there exist a group of exponent p and order > pp such that it has only onenormal subgroup of order pi , i D 1; 2; : : : ; p?

931. Given k 2 N, let a p-groupG be such that, for every abelian groupA of order pk ,there is A1 � G isomorphic with A. Give a realistic upper bounds of pnk D min jGj.

932. Classify the non-p-abelian p-groups all of whose proper subgroups (sections)are p-abelian.

933. Classify the p-groups G satisfying the following condition. Whenever H < G

is nonnormal, there exists exactly one maximal subgroup of G containing H . (Thisproblem was solved in �84.)

934. Does there exist a p-group with two representation groups of different classes?

935. Study the p-groups having the same class as all their representation groups.

936. Let p > 3, let G be a p-group of maximal class and order > ppC1 andH < G,H 6� G1. Is it true that if jH j D pp�2 and p is large, thenHG > f1g?

937. Study the p-groups all of whose noncyclic (nonabelian) subgroups of the sameorder are isomorphic.

938. Study the 2-groups G D hx1; x2; : : : ; xni, where x1; x2; : : : ; xn are involutionssuch that jhxi ; xj ij � 8 for all i ¤ j .


939. Find all n 2 N such that there exists anAn-group G with ˛1.G/ � p3.

940. Find all k such that the groups †pn and UT.n; p/ haveHk-chains. (See �88.)

941. Classify the powerful p-groups with abelian subgroup of index p.

942. Study the regular p-groups all of whose proper subgroups are powerful.

943. Study the regular p-groups G all of whose proper subgroups containing G0

(ˆ.G/), are powerful.

944. Classify the p-groups all of whose powerful sections are abelian.

945. Classify the p-groups G such that ˛1.H/ D pd.H/�1 for all H � G which areneither abelian nor minimal nonabelian.

946. Classify the p-groups all of whoseA1-subgroups are characteristic.

947. Classify the p-groups in which any two nonnormal abelian subgroups of the sameorder are conjugate.

948. Find all n 2 N such that there exists a nonabelian p-group G all of whosesubgroups of order (index) pn are isomorphic.

949. Find ˛m;e D max f˛1.G/ j jGj D pm; exp.G/ D peg. Is it true that ˛m;1 �

˛m;e?

950. Classify the p-groups in which any two nonnormal subgroups of the same orderare contained in the same number of maximal subgroups.

951. Classify the p-groups G admitting an automorphism ˛ of order p such thatCG.˛/ is cyclic. (See �48 and [Bla13]).

952. Classify the 2-groups G possessing an automorphism ˛ of order 2 such thatCG.˛/ is of maximal class. (See �49.)

953. Classify the p-groups all of whoseA1-subgroups have normal complements.

954. Let a p-group G D AwrB be a standard wreath product. Study the structure ofG=Z.G/.

955. Study the pairs H < G of p-groups such that jG W H j D p, exp.H/ > p andCH .x/ is of exponent p for all x 2 G �H .

956. Study the p-groups G such that G=G00 is special.

957. Study the p-groups whose ˆ-subgroup is special. (The Burnside group B.4; 2/of order 212 satisfies the above condition; see �60.)

958. Classify the 2-groups with exactly eight A1-subgroups. Moreover, classify thep-groups G with ˛1.G/ D .p2 C p C 1/C 1. (See �76.)

959. Classify the p-groups of maximal class, p > 2, whose representation groups arealso of maximal class.


960. Classify the representation groups of Suzuki 2-groups A.m; �/ (see �46).

961. Describe the representation groups of abelian p-groups.

962. Study the p-groups all of whose subgroups (nonabelian subgroups) of index p2

are isomorphic.

963. Classify the p-groups G such that G has a representation group with cycliccenter.

964. Classify the minimalM24-free 2-groups.

965. Classify (i) the minimal non-F -free 2-groups, where F 2 fD8 � C2;Q8 � C2;

D8 � C4g.

966. Classify the 2-groups containing a nonabelian subgroup of order 8, all of whosenonabelian subgroups of order 16 are of the formM �C , where C is cyclic of order 4.

967. Study the p-groups G with an element x of order p which is contained in onlyone maximal subgroup of G.

968. (Zhmud) Study the p-groups G such that, whenever x; y 2 G have the samenormal closures, then they are conjugate.

969. Find the least upper bound of ranks of Ep4-free p-groups.

970. Given a 2-group G and n > 3, let sdn.G/ be the number of proper subgroupsof G isomorphic with SD2n . Which members of the set f1; 2; 3g may be values of thefunction sdn.�/?

971. Study the p-groups G such that jH W H 0j � p3 for all nonabelian H � G.

972. Study the pairs H < G of 2-groups such that jG W H j D 2 and G � H is theunion of at most four conjugacy classes.

973. Letpn be the minimal order ofA1-subgroups of a p-groupG. Study the structureof G if the number of A1-subgroups of order pn in G is � p C 1.

974. Classify the groups G of exponent p such that jG=K3.G/j D p3.

975. Study the structure of ˆ-subgroups of quaternion-free 2-groups.

976. Suppose thatN is a properG-invariant subgroup ofˆ.G/, whereG is a p-group.Does there exist a p-group W such that N Š ˆ.W /?

977. (Mann) Classify the p-groups with exactly p2 � p minimal characters (see Ap-pendix 10).

978. (Mann) Classify the p-groups with exactly p2 � 1 minimal characters.

979. Classify the p-groups G such that G has a subgroup H of index pn withHG D

f1g and G contains a subgroup E with d.E/ D 1C p C C pn�1.


980. Let G be a p-group such that f1g < Hp.G/ < G. Is it true that the class ofG=Hp.G/ is bounded?

981. Classify the two-generator 2-groups containing exactly one maximal subgroupthat is not two-generator. (See [BJ2].)

982. Study the 2-groups such that�2.G/ is extraspecial (special).

983. Classify the p-groups which are not generated by minimal nonmetacyclic sub-groups.

984. Classify the p-groups G with nonabelian derived subgroup (ˆ-subgroup) suchthat all proper subgroups of G have abelian derived subgroups (ˆ-subgroups).

985. Classify the p-groups G with non-absolutely regular derived subgroup (ˆ-sub-group) such that all proper subgroups of G have absolutely regular derived subgroups(ˆ-subgroups).

986. Classify the p-groups G with nonmetacyclic derived subgroup (ˆ-subgroup)such that all proper subgroups ofG have metacyclic derived subgroups (ˆ-subgroups).

987. Study the p-groups all of whose sections of exponent � p2 are abelian.

988. Study the p-groups G such that ˆ.G/ D E � Cpn , where E is extraspecial.

989. Study the p-groups all of whose maximal nonnormal subgroups have cyclicintersection.

990. Study the p-groups all of whose Thompson critical subgroups are special (see�14).

991. Does there exist a nonmetacyclic 2-group with exactly one proper subgroup iso-morphic withH2 D ha; b j a4 D b4 D 1; ab D a�1i. If so, classify all such G.

992. Study the 2-groups G in which the centralizer of every element of order 4 isabelian.

993. Study the p-groups G such that ˆ.G/ and G0 are special and G0 < ˆ.G/.

994. Study the 2-groups G such that ˆ.G/ is nonabelian of order 24. (For a solution,see Theorem 85.1.)

995. Classify p-groups all of whose nonnormal subgroups are either abelian or ex-traspecial.

996. Study the p-groups G with n.G/ D jIrr1.G/j D kG.G0/ (here kG.G

0/ is thenumber of G-classes contained in .G0/#). (If G is a normal Sylow 2-subgroup ofminimal nonnilpotent subgroup of order 5 26, then kG.G

0/ D 3 D n.G/.)

997. Study the 2-groups G containing an abelian subgroup A of type .4; 4/ (of type.4; 2; 2/) with CG.A/ D A.


998. Study the p-groups G such that (i) jNG.H/ W H j D p for all A1-subgroupsH < G, (ii) NG.H/ is anA2-group for everyA1-subgroupH < G. (iii) Classify the2-groups G containing a subgroup H2 D ha; b j a4 D b4 D 1; ab D a3i and suchthat jNG.H2/ W H j D 2.

999. Classify the 2-groups G, containing the subgroup A D D8 � C4 such that (i)CG.A/ D Z.A/, (ii) jNG.A/ W Aj D 2.

1000. Let a p-group G D AwrB , where the active factor B is a regular permutationgroup. Find the structure of the multiplier of G in terms of A and B .

1001. Study the structure of the p-groups with exactly two class sizes (see [Ish]).

1002. Study the p-groups G such that a Sylow p-subgroup of Aut.G/ is isomorphicto (i) †pn , (ii) UT.n; p/.

1003. (Old problem) Classify the p-groups G such that jGj − jAut.G/j.

1004. (Old problem) Study the p-groups G such that jAut.G/ W Inn.G/jp D p.

1005. Let k > 1 be fixed. Study the p-groups G such that Aut.G/ acts on the set ofall subgroups of G of order pk transitively.

1006. Study the p-groups G such that A=AG is cyclic for all A � G.

1007. Study the p-groups G with jCG.H/ W Z.H/j � p for all nonabelianH � G.

1008. Compute the groups of central automorphisms of †pn and UT.n; p/.

1009. Study the p-groups G such that Aut.G/ acts transitively on the set of all A1-subgroups of G.

1010. Study the p-groups G such that all p-automorphisms of G fix all G-classes.

1011. Study the p-groups G, p > 2, with jG=G00j D p5.

1012. Does there exist a p-groupG of exponent > p > 2, such that all elements of theset G �ˆ.G/ have the same order p? (According to Khukhro, such 7-groups exist.)

1013. Find the set fd.A/g, where A runs through all maximal abelian subgroups of†pn (UT.n; p/).

1014. Study special quotient groups of representation groups of Epn .

1015. Classify the p-groups all of whose nonnormal subgroups are metacyclic.

1016. Classify the p-groups all of whose two distinct A1-subgroups generate an A2-subgroup.

1017. Study the p-groups G such that, for all minimal nonabelian A < G, (i)CG.A/ D Z.G/, (ii) CG.A/ D Z.A/.

1018. Study the p-groups G such that, whenever F;H < G and FG D HG , then FandH are conjugate in G.


1019. Study the p-groups G of class 2 with k.G/ D jZ.G/j C k.G=Z.G// � 1.

1020. Classify the p-groups G such that Aut.G/ is a p-group which is (i) metacyclic,(iii) of maximal class, (iv) extraspecial.

1021. Classify the p-groups G with CG.A/ ¤ CG.B/ if A;B < G and jAj ¤ jBj.

1022. Study the structure of the ˆ-subgroup of a p-group G provided ˆ.G/ is non-abelian and jˆ.G/ W ˆ.G/0j D p3.

1023. Study the structure of the ˆ-subgroup of a p-group G provided ˆ.G/ is non-abelian and jZ.ˆ.G//j D p2.

1024. Is it true that limjcd.G/j!1.jcd.G/j � dl.G// D 1, where G runs over allp-groups and dl.G/ is the derived length of G, cd.G/ D f.1/ j 2 Irr.G/g?

1025. Study the structure of the ˆ-subgroup of a p-group G provided ˆ.G/ containsa subgroup B of order p3 such that Cˆ.G/.B/ D B .

1026. Classify the 2-groups such that, whenever A;B 2 �1 are distinct, then A\B ismetacyclic.

1027. LetN GG be the union of kG.N / conjugacy classes ofG. Classify the p-groupsG satisfying k.G/ � kG.N /C k.G=N/C p � 2.

1028. Study the p-groups all of whose Thompson critical subgroups have cyclic cen-ters.

1029. Classify the irregular p-groups with exactly one k-admissible Hall chain (see�88).

1030. Classify the p-groups G with�1.H/ D H for all nonmetacyclicH � G. (Forsolution, see [Ber33] and [BozJ4].)

1031. Does there exist p-groups G such that, whenever A;B 2 �1 are distinct, thenA \ B is anA1-subgroup? If so, classify these groups.

1032. Study the structure of a p-group G if CG.x/ � ˆ.G/ for all x 2 ˆ.G/�Z.G/.

1033. Describe the p-groups G with j� W Z.�/j < jGj, where � is a representationgroup of G.

1034. Study the p-groups covered by nonabelian metacyclic subgroups.

1035. Study the p-groups all of whose two-generator subgroups are of order � p4.

1036. Classify the p-groups all of whose normal nonabelian subgroups have index� p2.

1037. Let G 2 f†pn/;UT.m;pn//g. Describe M.G/, where M.G/ D hx 2 G j

CG.xp/ D CG.x/i is the Mann subgroup of G.


1038. Let n < m. Classify the groups G of order pm with sn.G/ < sn.Epm/ and suchthat there does not exist a group H of order pm with sn.G/ < sn.H/ < sn.Epm/.

1039. Let S be a p-group. Does there exist a p-group G such that S is isomorphic toa direct factor of ˆ.G/? The same problem for Ã1.G/ and G0 instead of ˆ.G/.

1040. Let J.G/ be the subgroup generated by all abelian subgroups of maximal orderin a p-group G. Study the p-groups G such that J.G/ � G0.

1041. Does there exist 2-groups containing exactly one proper U2-subgroup (see �67)of given order?

1042. Study the p-groups G, p > 2, with G=G00 of maximal class.

1043. (Old problem) Estimate the number of p-groups of maximal class and givenorder.

1044. Let G be a p-group and M.G/ its Mann subgroup. Classify the p-groups Gsuch that the cyclic subgroup M.G/ > Z.G/.

1045. Let all members of the set �1 be irregular p-groups. Find the least upper boundfor jGj.

1046. Study the irregular p-groups with exactly two nontrivial characteristic sub-groups.

1047. LetM1.G/ be the subgroup generated by all cyclic subgroups C < G such thatNG.Ã1.C // D NG.C /. Study the structure of M1.G/.

1048. Study the p-groups all of whose cyclic subgroups of order > p are characteris-tic.

1049. Study the p-groups having only one Thompson critical subgroup (see �14).

1050. Find M.G/, where p-group G D AwrB is the standard wreath product (hereM.G/ is the Mann subgroup of G).

1051. Let G D A B be a semidirect product with kernel B . Consider the followingsituations: (i) M.G/ D M.B/, (ii) M.G/ D M.A/ � M.B/.

1052. Study the p-groups withM.G/ > Z.G/ andM.H/ D Z.H/ for allH < G.

1053. Study the p-groups G with CG.x/ < CG.xp/ for all x 2 G �ˆ.G/.

1054. Let G D �1.G/ be a p-group of class > 2. Describe the structure of Gprovided jhx; yij � p3 for all elements x; y 2 G of order p.

1055. Study the p-groups G such that hx; yi is either regular or of maximal class forall x; y 2 G.

1056. Study the p-groups in which any two distinct maximal abelian subgroups havedistinct orders (exponents).


1057. Study the 2-groups G of exponent > 4 containing a maximal subgroup H suchthat all minimal nonabelian subgroups of G not contained in H have order 8. (See�90.)

1058. Classify the 2-groups G containing an involution t such that CG.t/=hti has acyclic subgroup of index 2.

1059. Study the 2-groups G satisfying �1.G/ D ES.m; 2/ � E2n , where ES.m; 2/ isan extraspecial group of order p1C2m.

1060. Study the p-groups G with an involution t such that CG.t/ Š ES.m:2/ � C2.

1061. Describe the non-Dedekindian p-groups G such that, whenever H 6E G, thenG=HG has exactly one subgroup of order p.

1062. Let A < G be a maximal abelian subgroup of a p-group G, jG W Aj > p. Studythe structure of G if, whenever A < H 2 �1, then jH 0j D p.

1063. Let �1 be a representation group of G, �2 a representation group of �1, andso on. Such a series we call a �-series of G. Is it true that each p-group has a finite�-series? (We do not assert that all �-series are finite.)

1064. Construct all �-series for Epk , p > 2, k 2 f2; 3g.

1065. Construct all �-series for nonabelian metacyclic p-groups. (By Theorem 47.4,all members of these series are metacyclic.)

1066. Construct all �-series for groups (i) Ap.m; �/ (see �46), (ii) for nonabelianSylow subgroups of minimal nonnilpotent groups, (ii) for all minimal nonabelian p-groups.

1067. Construct all �-series for 2-groupsM � C2 (whereM is of maximal class).

1068. Let G be a p-group, p > 2. Study the structure of ˆ.G/ if it is irregular oforder p2Cp .

1069. Construct all �-series for all 14 groups of order 24.

1070. Given a p-group G, let G0 be a group containing a normal subgroup N of orderp such that G0=N Š G and N � ˆ.G0/. Such G0 we call a ˆ-extension of G if itexists. Study the p-groups which have no nontrivial ˆ-extensions.

1071. Classify the p-groups G of exponent pe such that the subgroup ��e .G/ D hx 2

G j o.x/ D pei is abelian.

1072. Classify the 2-groups all of whose subgroups of order 25 are two-generator.

1073. Study the p-groups G such that CG.x/ is anA1-subgroup of order p4 for somex 2 G of order p.

1074. Classify the p-groups G with Ã1.G/0 > f1g and Ã1.H/

0 D f1g for allH < G.


1075. Study the p-groups with abelian Thompson critical subgroups for all H < G

but nonabelian Thompson critical subgroups for G (see �14).

1076. Study the p-groups G such that (i) M.G/ D ˆ.G/, (ii) M.G/ D Ã1.G/.

1077. For all k, find the minimal m D mk;p such that each group of order pm has an

abelian subgroup of order pk .

1078. Study the pairs H < G of p-groups such that H is a nonabelian normal sub-group of G and all characters in Irr1.H/ are not G-invariant.

1079. Let N be the set of all n 2 N such that there does not exist a 2-group G withjIrr1.G/j D n. Is it true that the setN is nonempty and finite?

1080. Give a realistic estimate of jG0j, where G is a 2-group with jIrr1.G/j � n.

1081. Does there exist a p-group H with G=M.G/ 6Š H for all p-groups G?

1082. Study the pairs N < G of p-groups such that all elements of the coset xN havethe same order for all x 2 G �N .

1083. Study the p-groups, all of whose maximal subgroups are nontrivial centralproducts.

1084. Classify the 2-groups G such that G � Z.G/ is the union of � 12 conjugacyclasses.

1085. Is it true that ˛1.G/ � ˛1.G=N/ for f1g < N GG?

1086. Classify the p-groups G D �1.G/ of order pn with minimal possible c1.G/.

1087. Classify the special p-groups G with jZ.G/j D p2.

1088. Classify the p-groups, p > 2, all of whose nonabelian maximal subgroups areof the formM �E, whereM is of maximal class and E is elementary abelian.

1089. Let G be an abelian p-group. Find the class of a Sylow p-subgroup (i) ofAut.G/, (ii) of the holomorph of G.

1090. Does there exist a p-group G of arbitrary large order with dl.P / D dl.G/,where P is a Sylow p-subgroup of the holomorph of G?

1091. Classify the 2-groups G with jCG.˛/j D 2 for ˛ 2 Aut.G/ of order 4. (See�77.)

1092. Suppose that M < G is metacyclic and, whenever M < N � G, whenexp.N / > exp.M/. Study the structure of G.

1093. Classify the 2-groups containing exactly three subgroups Š D8. (There is onlyone 2-group containing exactly one proper subgroup Š D8, namely SD16.)

1094. Classify the 2-groups containing exactly one proper subgroup Š Q2n .

1095. Let � be a representation group of Epn . Find all possible values of jAut.�/j.


1096. Let k > 3 be fixed. Study the 2-groups all of whose normal subgroups of order2k are two-generator.

1097. Classify the p-groups G such that there exists a pair R GH such that jRj D p,H=R Š G and n.H/ � n.G/ D p � 1. (Here n.G/ D jIrr1.G/j.)

1098. Classify the groups G of order 22nC1 such that there exists a pair R GH withjRj D 2, H=R Š G and n.H/ � n.G/ D 2.

1099. Study the pairs of p-groups N GG such that jN j D p3 and n.G/� n.G=N/ D

p2 C p � 2.

1100. Classify the special p-groups G all of whose maximal abelian subgroups haveorder pjG0j, p > 2.

1101. Classify the p-groups G with U \V GG for any distinct U;V < G of the sameorder.

1102. Study the p-groups G D A M.G/ with A \ M.G/ D f1g (here M.G/ is theMann subgroup of G).

1103. Study the p-groups G withM.A/ � M.G/ for all nonabelian A < G.

1104. Classify the 2-groups containing a normal subgroup of maximal class and in-dex 4.

1105. Suppose that H 2 �1 and all cyclic subgroups that are not contained in H , arenormal in G. Study the structure of G.

1106. Suppose that G is a p-group with jÃi�1.G/ W Ã1.Ãi�1.G//j � pp for somei 2 N. Is it true that then jÃi�1.G/ W Ãi.G/j � pp?

1107. Classify the p-groups G such that jCG.H/j � p2 for all nonabelianH � G.

1108. Classify the p-groups having only one .p� 1/-admissible Hall chain (see ��24,88).

1109. Let G run over all groups of exponent pe , e > 2. Does there exist c D c.e/

such that exp.Ãc.G// < pe for all such G? (For definition, see �23. For p > 2, see[Wil2].)

1110. For G 2 f†pn;UT.n; pk/g, study the structures of the following quotientgroups: Ãn�1.G/, Ãn�1.G/, G=Ã2.G/ and G=Ã2.G/.

1111. Find �.Aut.G// for G Š UT.n; pk/.

1112. (N. Ito) Let B0 D B.d; pe/ be the maximal finite group of exponent pe with dgenerators, d > 1, p > 2. Set G D B0=K3.B0/. (i) Find cd.G/ and the number ofcharacters of each degree n in Irr.G/. (For e D 1, see [IM].) (ii) Find the class sizesvector of G.


1113. LetG be a p-group. A series E W f1g D E0 < E1 < < En D G is said to bean E-series of G of length l.E/ D n, if EiC1=Ei is a normal subgroup of exponent pand maximal order in G=Ei , i D 0; 1; : : : ; n� 1. Does there exist a p-group G whichhas two E-series of different lengths?

1114. Study the 2-groups G with abelian G=Ã1.Ã1.G//.

1115. Describe all A1-subgroups of a p-group G D M � C (M D M � C withM \ C D �1.C /), whereM is minimal nonabelian and C is cyclic.

1116. Study the p-groups G such that Ã1.G/ is powerful.

1117. Study the p-groups G, p > 2, such that jˆ.G/ W Ã1.G/j D p.

1118. Define the series G D ˆ0.G/ > ˆ1.G/ > ˆ2.G/ > > ˆn.G/ > asfollows. ˆ0.G/ D G, ˆ1.G/ D ˆ.G/, ˆiC1.G/ D ˆ.ˆi .G//. Study the p-groupsG such that the above series is lower central.

1119. Given a p-group G, let Ei .G/ be least normal subgroups of G such thatG=Ei .G/ is generated by elements of order p, i D 1; 2. Does there exist a p-group Gsuch that jE1.G/j ¤ jE2.G/j.

1120. Study the p-groups G with H 0 D Ã1.H/ for all nonabelian H < G butG0 ¤ Ã1.G/.

1121. Study the p-groups G with G=M.G/ 2 fMpn;Cpn � Cpg.

1122. Classify the p-groups G such thatH=HG is cyclic for allH < G.

1123. LetG be a nonabelian p-group, d.G/ D 3. Study the structure ofG ifAˆ.G/ 2

�1 (AG0 2 �1) for allA1-subgroups A < G.

1124. Classify the 2-groups, all of whose nonabelian maximal subgroups are of theform T � C , where T has a cyclic subgroup of index 2 and C is cyclic.

1125. Given a p-groupG, is it true that there exists a p-groupH such that (i)ˆ.H/ Š

Ã1.G/, (ii) Ã1.H/ Š ˆ.G/, (iii)H 0 Š ˆ.G/, (iv) ˆ.H/ Š G0?

1126. Let the Burnside 2-group G D B.2m; n/, m;n > 1. Study the structures ofsubgroups �1.G/, �2.G/ and ��

2.G/, ˆ.G/, G0. Find k.G/, cd.G/ and cl.G/. (For

m D 2 D n, see �60.)

1127. Does there exist, for each e > 1, a p-group G of exponent pe such thatexp.Ãe�1.G// D p and Ãe�1.G/ < Ãe�1.G/?

1128. Study the regular p-groups, p > 2, which are also powerful.

1129. Study the 2-groups G with ˆ.G/ elementary abelian and squares constitute abasis of ˆ.G/.

1130. Classify the 2-groups all of whose nonabelian maximal subgroups are of theform T �E, where T Š M2n and E is elementary abelian.


1131. Does there exist anA4-group G, p > 2, such that ˛1.G/ D 1C .p2 CpC 1/?

1132. Let a p-groupG D A1� �An, whereA1; : : : ; An are minimal nonabelian andAi \ Aj D Z.Ai / D Z.Aj / for all i ¤ j . Describe (i) maximal abelian subgroups ofG, (ii) degrees of irreducible characters of G and the number of irreducible charactersof each degree.

1133. Study the powerful p-groups all of whose maximal subgroups are powerful.

1134. Classify the p-groups G in which any nonnormal subgroup is contained inexactly one maximal subgroup of G. (This problem was solved in �28.)

1135. Describe the structures of ˆ-subgroups having derived subgroup of order p.

1136. Let f1g < H < G. Suppose that, wheneverK GH , thenH \KG D K. Studythe embedding of H in G and the structure ofH=HG .

1137. Study the p-groups G satisfying �i .H/ D Ãe�i.H/, i D 1; : : : ; e � 1, for allits sectionsH of exponent pe , where pe 2 fp; : : : ; exp.G/g.

1138. Let G be an irregular p-group such that the indices of the chain C W G >

Ã1.G/ > Ã2.G/ > are � pp and suppose that the length of C is n. Estimate thelength of the chain G > Ã1.G/ > Ã2.G/ > (see �24).

1139. Study the p-groups G of exponent pe such that the length of the chain G >

Ã1.G/ > Ã2.G/ > equals e C 1.

1140. Give an algorithm for construction of a k-admissible Hall chain for an arbitraryabelian p-group (see �24).

1141. Find the number of k-admissible Hall chains in a homocyclic p-group.

1142. Classify the p-groups, all of whose A1-subgroups have the same order pn.Consider case n D 3 in detail,

1143. LetG be a special group with d.G/ D d . Suppose that G=M is extraspecial forall maximal subgroupsM of G0. Estimate jG0j.

1144. LetG be a special group with d.G/ D d . Suppose that G=M is extraspecial forall maximal subgroupsM of G0. Estimate jG0j.

1145. Classify the p-groups of maximal class and order ppC1 such that j�1.G/j D

pp. (As Mann showed, the set of such G is nonempty.)

1146. Classify the p-groups G withM \H GG for allM 6� H 2 �1.

1147. Study the p-groups G with ep.G/ D 2p C 1, where ep.G/ is the number ofsubgroups of order pp and exponent p in G.

1148. Let 1 < k < p. Does there exist a p-group G of order ppC1 with ep.G/ D

k? (As Mann [Man5] has showed, there is irregular G of order ppC1, p > 2, withep.G/ D 1.)


1149. LetM 2 �1. Suppose that jG W HG j D p for all subgroups (cyclic subgroups)H < G not contained inM . Study the structure of G.

1150. Classify the subgroups of class 2 of maximal order in †pm and UT.n; p/.

1151. Find the orders of Aut.†pn/ and Aut.UT.n; p//.

1152. Find the character degrees vectors for †pn and UT.n; p/.

1153. Study the p-groups G such that, whenever ; 2 Irr.G/ with .1/.1/ �

b.G/ D max f�.1/ j � 2 Irr.G/g, then 2 Irr.G/.

1154. Study the p-groups without subgroups of maximal class.

1155. Find for G 2 f†pn; UT .n; p/g such number kn that G has an irregular sub-group of order pkn but all subgroups of G of order < pkn are regular. Find maxfjRj j

R < G is regular g.

1156. Study the 2-groups G with (i) G=M.G/ of maximal class, (ii) cyclic G=M.G/

(hereM.G/ is the Mann subgroup of G).

1157. Does there exist, for each n > 1, a p-group G with abelianÃn.G/ and irregularÃn.G/?

1158. Does there exist a p-group G such that, for some n 2 N, the subgroups Ãn.G/

and Ãn.G/ are abelian but different?

1159. Let G be a p-group, P is a Sylow p-subgroup of the holomorph of G. Studythe structure of G provided jG W ŒG;P �j D p.

1160. Study the p-groups whose ˆ-subgroups are irregular of order ppC2.

1161. Does there exist a p-group G such that any two different maximal subgroups ofexponent p in G have distinct orders?

1162. Classify the 2-groups all of whose minimal nonabelian subgroups are isomor-phic to (i) H2 D ha; b j a4 D b4 D 1; ab D a3i or Q8, (ii) H2 or D8, (iii) H2 or Q8

or D8. (Problem (i) was solved by Janko; see �92. If the intersection of all nonnormalsubgroups of G is > f1g, then G satisfies (i). Janko also deduced the last result from(i); see �92.)

1163. Study the p-groups G with special�1.G/ (�2.G/).

1164. Classify the p-groups G such that jG W HG j D p for all nonnormal H < G.(For a solution, see �62.)

1165. Study the p-groups, p > 2, in which the number of subgroups of maximal classand order pp is not a multiple of p.

1166. Does there exist a P -group G (see �11) with non-powerful Ã1.G/?


1167. Study the structure of a p-group G D �1.G/ if it has no subgroups of orderppC1 and exponent p.

1168. Classify the 2-groups containing a U2-subgroup of index 2 (see �67).

1169. Estimate an D max f˛1.G/ j jGj D png. Is it true that if n large andG of orderpn is such that ˛1.G/ D an, then exp.G/ D p?

1170. Study the p-groups G such that, in all sections of G, upper and lower centralseries coincide.

1171. Find exp.Aut.G//, where G is abelian of given type.

1172. Study the p-groups G with extraspecial (special) NG.H/ for some H < G.

1173. Describe abelian subgroups in the holomorph of a 2-group of maximal class.

1174. LetG be an abelian group of type .pn; p; : : : ; p/, n > 1. Describe the structureof a Sylow p-subgroup of Aut.G/.

1175. Classify the 2-groups G with nonabelian �1.G/ of order 24.

1176. Study the p-groups all of whose maximal regular subgroups are isomorphic.

1177. Let R < G be abelian of type .p; p/. Study the structure of G provided CG.R/

is metacyclic.

1178. Study the p-groups generated by normal subgroups of order p3.

1179. Study the p-groups generated by normal A1-subgroups.

1180. Study the p-groupsG such that Aut.G/ acts transitively on the set of elementaryabelian subgroups of G of order p2.

1181. Classify the p-groups all of whoseA2-subgroups are metacyclic.

1182. Does there exist a p-admissible Hall chain in †pn 2 Sylp.Spn/, UT.n; p/ 2

Sylp.GL.n; p//?

1183. Study the p-groups all of whose proper nonabelian epimorphic images are spe-cial.

1184. Study the p-groups in which any two non-conjugate maximal abelian subgroupshave distinct orders.

1185. Study the p-groups that have maximal regular subgroup of order ppC1.

1186. Study the p-groups G such that HG is irregular of maximal class for somenonnormal absolutely regularH < G.

1187. Classify the p-groups, p > 2, with exactly one 2-admissible Hall chain.

1188. Find max fm j exp.G/p D exp.Epmg, where G runs over all nonabelian groupsof order pm and exponent p.


1189. Study the p-groups of exponent pe all of whose metacyclic subgroups haveorder � peC1.

1190. Study the p-groups G with cyclicM.G/=Z.G/ > f1g (hereM.G/ is the Mannsubgroup of G).

1191. Study the nonabelian p-groups G D �1.G/, p > 2, in which every twononcommuting elements of order p generate a p-group of maximal class.

1192. Let P be a Sylow p-subgroup of the holomorph of Cpn . Study the structure ofAut.P /.

1193. Study the structure of Aut.P /, where P is a Sylow p-subgroup of the holo-morph of a p-group with cyclic subgroup of index p.

1194. Study the p-groups G withM.G/ D G0.

1195. Study the p-groups G such that, wheneverH 6E G is of order pp and exponentp, thenHG is of maximal class.

1196. Study the p-groups of exponent pe > p, all of whose cyclic subgroups of orderpe are normal.

1197. Find ck.G/ for all k, where G 2 f†pn;UT.n:p/g.

1198. Given e > 2, does there exist a p-group G of exponent pe such that the quotientgroup Ãi.G/=ÃiC1.G/ is irregular for i D 1; : : : ; e � 1?

1199. Classify the p-groups, p > 2, all of whose nonabelian two-generator subgroupsare absolutely regular.

1200. Let G be of exponent pe > p2 and suppose that Ãe.G/ D f1g. Is it true thatthen Ãi.G/ D Ãi.G/ for all i?

1201. Does there exist a constant C such that, for each e � C and each p-group G ofexponent pe , the subgroup Ãn.G/ > f1g is regular? (The answer is ‘no’, by [Wil2].)

1202. Classify the p-groups in which any two noncommuting elements generate eitherA1-subgroup or subgroup of maximal class.

1203. Study the nonabelian p-groups all of whose nonabelian epimorphic images havecyclic centers.

1204. Study the p-groups G in which each characteristic subgroup equals �i .G/ \

Ãj .G/ for some i; j 2 N.

1205. Classify the p-groups possessing a subgroup Z of order p which is containedin only one abelian subgroup of type .p; p/.

1206. Classify the 2-groups G withH < G such that NG.H/ is a U2-group.

1207. Classify the p-groups G possessing a subgroup A of order p2 such that that thecentralizer of A in G has order p3. (For p D 2, see ��51, 77.)


1208. Classify the 2-groups G possessing a subgroup H of order 8 such that that thenormalizer of H in G has order 16.

1209. Let G be a two-generator p-group. Suppose that exactly p maximal subgroupsF of G satisfy jZ.F /j D p and exactly p maximal subgroups H of G satisfy jH W

H 0j D p2. Is it true that G is of maximal class. (The answer is ‘yes’ for p D 2, byTaussky’s theorem and Theorem 5.4.)

1210. Given k, let a p-group G contains all types of groups of order pk . Give arealistic upper bound of jGj. (See also #931.)

1211. Study the p-groups of class > 3 covered (generated) by normal subgroups ofclass � 2.

1212. Classify the p-groups G with metacyclic NG.M/ for someM < G.

1213. Let �1.G/ be the number of conjugate classes of A1-subgroups of G. Classifythe p-groups G with �1.G/ 2 fp � 1; p; p C 1g (see Proposition 76.15).

1214. Let a p-group G D A B be a semidirect product with kernel B , the subgroupsA and B are abelian. Estimate cl.G/ in terms of A, B and action of A on B .

1215. Study the p-groups G such that d.G/ < d.H/ for all H 2 �1. (Almost allnonmetacyclicA1-groups satisfy this condition.)

1216. Study the p-groups G with ker./\ ker./ D f1g provided ; 2 Irr1.G/ and.1/ ¤ .1/.

1217. Study the nonabelian p-groups G all of whose characters from Irr1.G/ havekernels of the same order.

1218. (Old problem) Classify the p-groups with cyclic derived subgroup.

1219. Find max fjG0jg, where G runs over all An-groups with (i) abelian G0, (ii)metacyclic G0.

1220. Study the p-groups whose cyclic subgroups are characteristic in their centraliz-ers.

1221. Study the p-groups G with NG.H/ D HCG.H/ for allA1-subgroupsH < G.

1222. Let G be a 2-group of order 22.eC1/C1 and exponent 2e, e > 1, such thatjÃe�1.G/j D 25. Study the structure of Ãe�1.G/ provides it is nonabelian.

1223. Let G be a p-group of order pp.eC1/ and exponent pe , p > 2, e > 1, such thatjÃe�1.G/j D p2p . Study the structure of Ãe�1.G/ provided it is irregular (see �23.)

1224. Study the p-groups G containing only one maximal subgroup with center oforder > p.

1225. Let R � ˆ.G/ be a G-invariant nonabelian subgroup of order � p6. Describethe structure of R.


1226. Let R � ˆ.ˆ.G// be a G-invariant nonabelian subgroup of order p5 or p6.Describe the structure of R.

1227. Does there exist a nonabelian p-group G that is generated by kernels of itsnonlinear irreducible characters. If answer is ‘yes’, classify such groups.

1228. Study the finite 2-groups generated by two elements x and y of order 2 and 4,respectively. Describe the structures of ˆ-subgroups of such groups.

1229. Describe the structures of G0, where G is as in #1228.

1230. Suppose that G is a nonabelian p-group of exponent pe such that ��e .G/ D

hx 2 G j o.x/ D pei is abelian. Is it true that the dl.G/ is bounded?

1231. A group G is said to be a Q-group if, whenever a; b 2 G generate the samesubgroup, then a and b are conjugate in G. Classify theQ-groups of exponent � 8.

1232. Study the p-groups G with Autc.G/ D Z.Aut.G//. (Here Autc.G/ is the groupof central automorphisms of G.)

1233. Study the p-groups G such that, for everyH < G of index at most p2, one hasZ.H/ � Z.G/.

1234. Study the p-groups in which the centralizer of each noncentral element of com-posite order has cyclic subgroup of index p.

1235. Study the p-groups G containing only one maximal subgroup, say A, such thatjA W A0j > p2. (If p D 2, then G is of maximal class by Taussky’s theorem andTheorem 5.4.)

1236. (This problem was inspired byMann’s paper [Man32] on skew 2-groups.) LetGbe a 2-group and t .G/ the number of involutions inG. Study the 2-groupsG such that,wheneverN is a nonidentity G-invariant subgroup in ˆ.G/, then (i) t .G/ < t.G=N/,(ii) t .G/ > t.G=N/.

1237. Does there exist a p-group G of order ppC2, p > 2, which is not of maximalclass and such that G=N is of maximal class for each minimal normal subgroup N ofG.

1238. Study the irregular p-groups G such that Kp.G/ D Ã1.G/ and G=Ã1.G/ is ofmaximal class.

1239. Classify the p-groups G such that ı.H/ � 1p jH j for allH < G. Here ı.G/ is

the minimal degree of representation of G by permutations.

1240. Let N.X/ be the set of nonnormal subgroups of a p-group X . Let G and Hbe non-Dedekindian p-groups. Suppose that there is a 1 � 1-correspondence betweenN.G/ andN.H/ such that corresponding subgroups and their normalizers are isomor-phic. Is it true that G andH are not necessarily isomorphic?

1241. Classify the p-groups G in which all members of the set �2 areA1-groups.


1242. Classify the p-groups all of whose maximal subgroups are either abelian orspecial.

1243. Study the 2-groups G without normal subgroup Š E24 , but all of whose maxi-mal subgroups have normal subgroups Š E24 .

1244. Study the structure of a p-group G if for any A1-subgroup A < G we havejG W AG j D p.

1245. Study the p-groups G such that, whenever A;B < G are distinct maximalabelian, then (i) NG.A \ B/ D hA;Bi; (ii) A \ B GG.

1246. Study the p-groups G such that, whenever U;V are distinct A1-subgroups ofG, then U \ V is maximal either in U or in V .

1247. Let d > k > 2 be fixed. Find the minimal number of nonabelian members inthe set �k , where G is a nonabelian p-group with d.G/ D d . (See �76.)

1248. Study the p-groups G such that, whenever A < G is minimal nonabelian, thenthere is in G only one subgroup of order pjAj containing A.

1249. Let G be a p-group with c1.G/ D 1C p C C pp�1 C kpp , where k > 1.Let ep.G/ be the number of subgroups of order pp and exponent p in G. Estimateep.G/ is terms of k.

1250. Study the irregular p-groups G with cl.H/ � 12.p C 1/ for allH < G.

1251. Study the p-groups without normal cyclic subgroups of order p2.1

1252. Study the p-groups G such that for any nonabelian H 2 �1, all A1-subgroupsinH are conjugate in G. (In that case,H has a G-invariant abelian subgroup of indexp, by Theorem 10.28.)

1253. A p-group G is said to be generalized metacyclic with respect to its cyclicsubgroup C if there exists only one maximal chain of subgroups connecting C andG. Classify the p-groups which are generalized metacyclic with respect to all theirmaximal cyclic subgroups. (Such group must be two-generator.)

1254. Study the p-groups with only one abelian subgroup of type .pn; p/, n > 1.

1255. Study the p-groupsG such that all members of the set�1[�2 are two-generator.(See �70.)

1The group †pn 2 Sylp.Spn/, n > 1, has no normal cyclic subgroups of order p2, unless pn D 4.

Indeed, assume that pn > 4 and let L be a normal cyclic subgroup of order p2 in Gn D †pn . Since G2

is of maximal class, it has a normal abelian subgroup of type .p; p/ so it has no normal cyclic subgroupsof order p2, hence n > 2. Let B D H1 � �Hp , where Hi Š †pn�1 is the i th coordinate subgroupof the base B of the wreath productGn D Gn�1 wr Cp . Then�1.L/ D Z.G/ and L\Hi D f1g for alli . Assume that L < B ; then CG.L/ � H1 � �Hp D B , a contradiction, since Z.B/ is elementaryabelian. Thus, L 6� B . Since Z.Hi / centralizes L (consider the centralizer of L in semidirect productHi L and take into account that Z.Hi / is of order p) so CG.Z.Hi // � BL D G, we get Z.Hi / � Z.G/for all i , a contradiction since jZ.G/j D p.


1256. Classify the p-groups G such that all members of the set �2 are metacyclic.

1257. Study the p-groups G such that all members of the set �2 are centralizers ofappropriate elements.

1258. Estimate the number max jfk.G/ j jGj D pngj (see [BI]).

1259. Study the p-groups G with c1.G/ D 1 C p C C pk and exactly k C 1

conjugate classes of subgroups of order p.

1260. Study the 2-groups with exactly two conjugate classes of four-subgroups. (Thisproblem was solved by Janko, according to his letter at 16/05/07.)

1261. Study the p-groups all of whose nonnormal subgroups of the same order areconjugate. (For a solution, see �58.)

1262. Does there exist an universal constantC D C.p/ such that, wheneverG=�C .G/

is regular for a p-group G, then G is also regular?

1263. Study the nonabelian p-groups all of whose nonabelian subgroups have centersof orders at most p2.

1264. Is it true that f˛1.G/ j G is a nonabelian 2-group g D N? (See �76.)

1265. Is it true that for each n there exists k with ˛1.G/ 62 fk C 1; : : : ; k C ng for allp-groups G?

1266. Classify the p-groups G with minimal nonabelian Aut.G/.

1267. Study the p-groups G such that all members of the set �2 are special.

1268. Study the p-groups, in which each 4-fold commutator Œx; y; x; y� equals 1 forall x; y 2 G.

1269. Study the p-groups G such that CG.x/ is absolutely regular for all x 2 G �

Z.G/.

1270. Study the p-groups without subgroups of order ppC1 and exponent p thatcontain a subgroup Š Epp .

1271. Classify the p-groups G of exponent > p such that �1.ˆ.G// D ˆ.G/ but�1.ˆ.H// < ˆ.H/ for allH 2 �1.

1272. Study the p-groups G of exponent pe , e > 1, with c1.G/ >Pe

iD2 ci .G/.

1273. Classify the pairsHGG of p-groups such that there exists only one 1-admissiblechain inH (see �88).

1274. Study the p-groups G such that whenever H is a nonnormal subgroup of G,then exp.NG.H// D exp.H/.

1275. Let a p-group G be neither abelian nor anA1-group. Study the structure of thegroup A D f˛ 2 Aut.G/ j M˛ D M for all nonabelian subgroupsM < Gg.


1276. Study the p-groups G containing a minimal nonabelian subgroup H such thatH is the unique minimal nonabelian subgroup of its order in G.

1277. Classify the groups that are not generated by their noncyclic subgroups of in-dex p4.

1278. (Old problem) ClassifyA3-groups.

1279. Let � be the representation group of the restricted Burnside group B.d; p/.Find the maximum of ranks of proper subgroups (of abelian subgroups) of �.

1280. Classify the p-groups covered by normal extraspecial (special) subgroups.

1281. Classify the p-groups that are not generated by minimal nonmetacyclic sub-groups. In particular, classify the p-groups G such that all subgroups of G not con-tained in a fixedH 2 �1, are metacyclic.

1282. Study the p-groups containing exactly one subgroup of order p.p�1/kC2 andexponent � pk .

1283. Let G be a p-group of maximal class and order > ppC1 and t D jfH 2 �1 j

ep.H/ > 0gj. Find all possible values of t . (Here ep.G/ is the number of subgroupsof order pp and exponent p in G.)

1284. Study An-groups G, n > 2, with ˛n�1.G/ D 1.

1285. Suppose that a p-group G is neither abelian nor an A1-group. Classify thep-groups G such that ˛1.G/ D pd.G/�1 (see �76, Theorem B).

1286. Classify the 2-groups with exactly one subgroup of order 2which is not maximalcyclic.

1287. Study the p-groupsG such thatU D Z.CG.U // for all cyclic subgroupsU < G

of orders > p.

1288. Classify the p-groups in which the number ofA1-subgroups of index p is > p.

1289. Study the p-groups G with jNG.A/j ¤ jNG.B/j for all non-conjugate nonnor-mal subgroups A and B of the same order.

1290. Study the p-groups in which any abelian subgroup is contained in a two-generator subgroup.

1291. Classify the p-groups of class > 2 in which any two subgroups of the sameorder have the same class.

1292. Study a pair H G G of p-groups such that Zp.G/ is regular but Zp.H/ isirregular.

1293. Study the p-groups G with cl.NG.A// D cl.A/ for all nonnormal nonabelianA < G.


1294. Study the p-groups G with (a) ˆ.NG.A// D ˆ.A/, (b) NG.A/0 D A0, (c)

Ã1.NG.A// D Ã1.A/ for all A 6E G.

1295. Study the structure of the subgroupT

H NG.H/, where H runs over the set ofsubgroups of G of class 2.

1296. Study the p-groups all of whose two-generator subgroups are P -groups. (See�11.)

1297. Study the irregular p-groups with regular normalizers of nonnormal subgroups.

1298. Study the irregular p-groups with regular centralizers of noncentral elements.

1299. Let G be a p-group of order pnC2 with jG0j D p2. Find all possible degreesvectors of G.

1300. Find the degrees vector of Pd;n, a Sylow p-subgroup of Aut.Hd;n/, whereH D Hd;n is homocyclic with d.H/ D d and exp.H/ D pn.

1301. Let D1.pe/ be the set of all possible values of k.G/ provided G runs over all

groups of order pe and exponent p. Find the minimal value of e (or prove that it doesnot exist) such that limp!1D1.p

e/ D 1.

1302. Compute d.A/ and exp.A/, where A 2 Sylp.Aut.G// and G is abelian of giventype.

1303. Study the p-groups G, p > 2, such that Hp.G/ < G is special (minimalnonabelian).

1304. Study the p-groups G all of whose subgroups of index p2 that are not membersof the set �2, are not G-invariant.

1305. Study the minimal non-p-abelian p-groups.

1306. Study the p-groups G such that Inn.G/ is not characteristic in Aut.G/.

1307. Classify the p-groups lattice isomorphic withA2-groups.

1308. Study the p-groups G all of whoseA1-subgroups H satisfy jH=HG j � p.

1309. Study the 2-groups all of whose minimal nonmetacyclic sections areŠ E8.

1310. Study the nonmetacyclic 2-groups, all of whose minimal nonmetacyclic sub-groups are isomorphic.

1311. Study the p-groups G such that whenever ; 2 Irr.G/ have the same degree,then all nonlinear irreducible constituents of their product have the same degree.

1312. Study the p-groups G such that, for each 2 Irr1.G/, the set Lin.G/ containsall irreducible characters of G of degree .1/.

1313. (Old problem) Characterize the p-groups ˆ such that there exists a p-group Gwith ˆ.G/ Š ˆ.


1314. Give a condition sufficient for the set R of elements x 2 G such that hx; yi isregular for all y 2 G.

1315. Let G be a p-group such that d.Aut.G// D 2. Estimate d.G/. Is it possible toestimate d.G/ is terms of d.Aut.G//?

1316. Classify the p-groups such that Aut.G/ acts transitively on every set of nonnor-mal subgroups of G of the same order.

1317. Find all p-groups H with �1.H/ D H and such that there is no p-group G oforder > jH j satisfying �1.G/ Š H?

1318. (Janko) Let a nonabelian 2-groupG possess a cyclic subgroup L of order 4 suchthat CG.L/ is abelian of type .4; 2/. Is it true that G is of coclass 2?

1319. Study the 2-groups G containing an element t such that CG.t/ is (i) abelian oftype .4; 4/, (ii) metacyclic. (See ��48, 49.)

1320. Study the p-groups G such that Ã1.G/ is nonabelian of order p4.

1321. Study the 2-groups all of whose A1-subgroups are isomorphic with M2n . (Forn D 4, see Theorem 57.6.)

1322. Describe Aut.G/ for allA2-groups.

1323. Study the 2-groups G with CG.x/ Š A.m; �/ for some x 2 G, o.x/ D 2 (see�46).

1324. Describe the automorphism groups of nonabelian Sylow subgroups of all mini-mal nonnilpotent groups.

1325. Suppose that G is a p-group, p > 2, which is neither abelian nor an A1-groupand ˛ an automorphism of G of order 2. Is it true that G contains an ˛-invariantA1-subgroup? (For more general result, see Lemma 31.4(c).)

1326. Given a nonabelian p-group H with center of order > p and k 2 N, does thereexist a p-group G of order pk jH j such that CG.x/ Š H for some x 2 G?

1327. Let ˛ 2 Aut.G/ be of order 2, where G is a nonabelian p-group, p > 2. Studythe structure of G if it has no ˛-invariant nonnormal subgroup.

1328. (Old problem) Study the irregular p-groups all of whose proper subgroups areregular.

1329. Study the p-groups without minimal nonabelian epimorphic images.

1330. Find max fd.G/g, where G runs over all 2-groups with �1.G/ Š E2n , n isfixed.

1331. Let G be a p-group, p > 2, with j�1.G/jj D pn and exp.�1.G// D p. Is ittrue that jG=Ã1.G/j is bounded?


1332. Classify the groups of exponent p whose Schur multipliers have order p. (Ac-cording to D.L. Johnson, all noncyclic groups of exponent p have nontrivial Schurmultipliers.)

1333. LetG be a p-group of maximal class and order pm > p3. Find lower and upperestimates for ˇ1.G;G1/. (See �76.)

1334. Study the p-groups G such that exp.NG.H// D exp.H/ for all nonnormal(nonnormal abelian)H < G.

1335. Let A;B 2 �1 be distinct. Suppose that jA0j D pn D jB 0j and jG0j D p2nC1.Study the structure of G0. Is n bounded?

1336. Let M be a p-group of maximal class and order pn > ppC1 with ep.M/ D

pn�p . Does there exist a p-group G of maximal class and order pjM j that contains asubgroup isomorphic toM ?

1337. Classify the metacyclic p-groupsM such that there does not exist a metacyclicp-group containing a maximal subgroup isomorphic toM . (IfM Š SD2n , then theredoes not exist a metacyclic 2-group containing a maximal subgroup isomorphic toM .)

1338. Let Rd be the set of all representation groups of Epd . Find (i) the orders ofnormal subgroups N of G such that G=N is extraspecial of order p2nC1 (G 2 Rd ),(iii) jAut.G/j for all G 2 Rd .

1339. LetH D H.d;pe/ be a homocyclic group of rank d > 1 and exponent pe > p,let RH be the set of all representation groups of H . Consider for G 2 RH the samequestions as in #1338.

1340. Let G D UT.n; p/ 2 Sylp.GL.n; p//. Study the p-groups which are latticeisomorphic with G.

1341. Classify the p-groups G of exponent > p2 such that ��3.G/ is an Lp-group.

1342. Classify the p-groups G with NG.A \ B/ � hA;Bi for any two nonincidentA;B < G.

1343. Study the 2-groups G containing an abelian subgroup of type .4; 2/ such thatNG.A/ is abelian of type .4; 4/.

1344. Let Rd be the set of representation groups of Epd , d > 1. Find f˛1.G/ j G 2

Rd g.

1345. Let G and G0 be lattice isomorphic p-groups. Suppose that there is in G anormal subgroup of order pn and exponent p. Is it true that G0 contains a normalsubgroup of order pn and exponent p?

1346. Find a realistic upper bound for ˛1.G/, where jGj D pm and exp.G/ D p.

1347. Classify the nonabelian groups of exponent p all of whose two-generator sub-groups have order p3 (� p4).


1348. Let 3 < n < m and let G be a group of order pm and exponent p with atwo-generator subgroup of order pn. Study the structure of G if it has at most p2

two-generator subgroups of order pn.

1349. (Isaacs) Let a q-group Q act on a 2-group G so that Q-orbits have pairwisedistinct sizes. Describe the structure of G. (See Appendix 3.)

1350. Study the p-groups G such that all its A2-subgroups have exponent < exp.G/.

1351. Classify the p-groups with homocyclic maximal subgroup.

1352. Classify the p-groups G such that NG.H/ E G for allH � G.

1353. Let k2.G/ denote the number of conjugacy classes of abelian subgroups of type.p; p/ in a p-group G. Classify the p-groups G with k2.G/ D 3.

1354. Study the p-groups in which all maximal subgroups of exponent p are maximalregular.

1355. Let a p-groupG D E0E1, whereE0 andE1 are elementary abelian,E0\E1 D

f1g. Find min fc1.G/g in terms of E0 and E1.

1356. (This problem was inspired by Mann’s letter in June 2006.) Present a two-generator p-group G, p > 2, with irregular (i)ˆ.G/, (ii) Ã1.G/, (iii) Ã2.G/, (iv) G0,(v) G00.

1357. Study the structure of G D AB , where cl.A/; cl.B/ � 2 and AGBG D f1g.

1358. Study theAn-groups G with d.G/ D nC 1.

1359. Classify the nonabelian p-groups G such that any two its subgroups of the sameindex pk (k D 1; 2) are isomorphic.

1360. Let A < G be p-groups with jAj D p.p�1/kC2, exp.A/ D pk > p. Supposethat A < A1 � G implies exp.A1/ > p

k . Study the structure of G. (See �24.)

1361. Study the p-groups G such that for every minimal basis fa1; : : : ; ad g we haveQd

iD1 o.ai / D jGj.

1362. Classify the p-groups R of order pp , p > 2, such that there exists an irregularp-group G of maximal class such that: (i) R Š R1 < G, however (ii) R1 6� G1.

1363. (Ito) Let n > 1 be odd. Is the groupG D ha; b j a4n D 1; b2 D a2n; ab D a�1i

Hadamard (see #247)?

1364. Study the 2-groups G such that �2.G/ is an U2-group (see ��18, 67).

1365. Study the 2-groups G such that �2.G/ Š D2m � D2n .

1366. Let a p-group G be irregular but not of maximal class. Then c2.G/ kpp�1

.mod pp/ (Theorem 13.2(b)). Find all possible values of k.


1367. Study the p-groups G such that whenever Z is a nonnormal maximal cyclicsubgroup of G then G=ZG is cyclic.

1368. Let G be a representation group of an elementary abelian p-group of rank d .Find the class of a Sylow p-subgroup of Aut.G/.

1369. Let Rd be the set of all representation groups of Epd . Find jAj, where A is thegroup of all automorphisms of G 2 Rd fixing all elements of Z.G/.

1370. Let A D h 2 Aut.G/ j M D M for all A1-subgroupsM < Gi. Study thestructure of A.

1371. Study the p-groups G containing a special subgroup E such that, wheneverE < M � G, then exp.M/ > exp.E/. (See �83.)

1372. Let G D B.4; n/ be the free group of exponent 4 and rank n. (i) Findmax fd.M/ j M < Gg. (ii) (Old problem) Compute jGj. (iii) Describe the lowerand upper central series of G. (iv) Describe the members of the derived series of G.(v) Find max fd.A/ j A < G;A0 D f1gg. (vi) Find c1.G/. (vii) Does there exist a2-admissible Hall chain in G? Find the number of principal series in B.4; 2/. (viii)Find cd.G/ and the number of irreducible characters of G of given degree. (See �60.)

1373. Let H be a p-group. Find a p-group G of minimal possible order such that His isomorphic to a subgroup of ˆ.G/.

1374. Find the number of maximal series in the abelian p-group of given type.

1375. Let G be a two-generator nonmetacyclic p-group of order > p4, p > 2. Is ittrue that the number of two-generator members of the set �1 is a multiple of p?

1376. Classify the p-groups G such that �2.G/ D E � E1, where E is extraspecialand E1 is elementary abelian.

1377. For groups of order p4 describe (i) automorphism groups, (ii) holomorphs, (iii)representation groups.

1378. Study the irregular p-groups, p > 2), all of whose regular subgroups are ofclass � 2. (In that case, p D 3.)

1379. Study the p-groups all of whose cyclic subgroups of some fixed order are con-jugate. (For a solution, for order 4, see Theorem 89.8.)

1380. Suppose that a p-group G D �e.G/ is of order ppe. Describe the structure ofG provided exp.G/ > pe.

1381. Does there exist a special p-group such that all its maximal subgroups arecharacteristic?

1382. Let G be a p-group of exponent > p and R < G a fixed subgroup of orderpp and exponent p. Suppose that if Z < G is arbitrary cyclic of order > p, thenR \Z D f1g. Describe the structures of R and G.


1383. Let A be an A1-subgroup of a p-group G such that there is in G only onesubgroup of order pjAj containing A. Describe the structure of G.

1384. Study the irregular p-groups G such that exp.�1.ZG// D p for all cyclic

Z < G.

1385. Describe maximal subgroups of standard wreath product G D A � B , where Aand B are irregular p-groups of maximal class.

1386. Study the structure of an irregular p-group G such that, whenever H < G ismaximal abelian andH < F < G with jF W H j D p, then exp.F / > exp.H/.

1387. Study the noncyclic p-groups G possessing only one normal subgroup of orderpi for i D 1; : : : ; p. Is it true that G is irregular?

1388. Study the p-groups G such that (i) j��2.G/j D ppC1, (ii)��

2.G/ is irregular oforder ppC2.

1389. Determine the structure of the Schur multiplier and representation group ofUT.n; p/.

1390. Let � be a representation group of Epn . Determine the structure of the Schurmultiplier of �.

1391. Does there exist a p-group G of exponent > p satisfying Hp.G/ D Ã1.G/?

1392. Does there exist a p-group G such that A \ Z.G/ D f1g for all minimalnonabelian A < G?

1393. Let G be an irregular p-group satisfying�1.G/ D G. Study the structure of Gprovided all maximal subgroups of exponent p in G have the same order pp .

1394. Study the p-groups G such that, whenever H < G is nonnormal of exponentp, then exp.�1.NG.H/// D p but exp.�1.G// > G.

1395. Classify the 2-groups G such that ��3.G/ is a U2-group. (Note that if n > 3,

then ��n.G/ is not an U2-group.)

1396. Study the p-groups G such that Œ�i.G/;Ãi .G/� D f1g for all i .

1397. Study the nonabelian p-groups G such that jG= ker./j D p.1/2 for all 2

Irr1.G/. For partial case of this problem, see #1.

1398. Find a sufficient condition for existence of 3-admissible Hall chains in a 2-group(see �88).

1399. Let G be a p-group and N � ˆ.ˆ.G// be G-invariant. Describe the structureof N if it is nonabelian of order � p6.

1400. LetG be a p-group such thatˆ.G/ D E�E1, whereE andE1 are extraspecial.Describe the structure of G.

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Author index

AAn, L.-J., �65

BBerkovich, Y., ��47, 48, 55, 58, 59,

62–65, 69, 72, 76, 88,Appendices 16, 17, 21, 24, 27Blackburn, N., ��66, 69Bozikov, Z., ��52, 85, Appendix 19Burnside, W., ��60, 61

CCepulic, V., �65Cossey, J., Appendix 26

GGlauberman, G., Appendix 18Golfand, Y.A., Appendix 22

HHall, P., �88

IIsaacs, I.M., ��47, 68, Appendices 18, 20,

21Ito N., �64Iwasawa, K., ��73, 79

JJanko, Z., ��48–56, 60-63, 64–67, 70–87,

89–92, Appendix 17, Problems

KKazarin, L. S., ��58, 65, 71

LLi Shirong, �63

MMann, A., ��64, 65, 76Miller, G. A., Appendix 17

NNapolitani, F., �73

PPyliavska, O., �65

RRedei, L., �65, Appendix 22

SSagirov, I. A, �46Sanov, I. N., �60Schmidt, O. Y., �59, Appendix 22Schreier, O., Appendix 25Sheriev, V.A., �65Suzuki, M., �48

TThompson, J. G., Appendix 18

Tobin, S. J., �60

VVishnevetsky, A., �46

XXu, M.Y., �65

WWard, H.N., �56, Appendix 19Wilkens, B., �79

ZZhang, Q., �65

Subject index

Aabelian subgroups of small index, �76absolutely regular p-groups, �88˛n.G/, �76An-groups, �72A2-groups, properties, defining relations

of ��65, 71An-groups, n D 3; 4, derived subgroup of,

�72An-groups with cyclic derived subgroup

of order pn, �72An-groups with cyclic derived subgroup

of order pn�1, p > 2, �72artifical square, �56automorphism group, Appendix 20

Bˇ1.G;H/, �76N1.G;H/, �76Blackburn’s theorems, ��66, 69Burnside group B.4; 2/ of order 212,

structure of, �60

Ccentralizer of involution, ��48, 49central products of some 2-groups,

Appendix 16central products of 2-groups of maximal

class with cyclic subgroup oforder 4, Appendix 16

characteristic maximal subgroup inquaternion-free 2-groups, �56

characterization of groups of exponent p,�68

characterizations of p-groups of maximalclass, �69

characters od Suzuki p-groups Ap.m; �/,�46

classification ofA2-groups, �71classification of modular p-groups, �73

DDedekindian groups, �73derived subgroups ofAn-groups,

n D 2; 3; 4, �72dominating chain, �88

Eelements of order � 4 in 2-groups, �75estimate of j�k.G/j and j��

k.G/ in terms

of ck.G/, �64extraspecial p-groups, �83

Ffree centralizer, Appendix 23

Ggenerators of p-groups, the number of,

Schreier’s theorem, Appendix25

�1.H/, �76�H

1 , �76group of type (F), Appendix 23groups all of whose nonabelian maximal

subgroups are either absolutelyregular or of maximal class,Appendix 26

groups all of whose subgroups of indexp2 are abelian (= A2-groups),��65, 71

groups G of exponent p and order pm

satisfying ˛1.G/ D pm�3, �76groups without normal elementary abelian

subgroups of order p3, ��50, 69

HHall chains (p-admissible and

k-admissible = Hk-chains), �88

IIsaacs’ examples, Appendix 21

Subject index 595

Ito k-series, �64Iwasawa’s theorem on modular p-groups,

new proof due to Janko, �73,Appendix 24

JJanko’s theorems on 2-groups with small

centralizer of an involution,��48, 49

Janko’s theorem on 2-groups withoutnormal elementary abeliansubgroup of order 8, �50

Janko’s theorem on 2-groups withselfcentralizing elementaryabelian subgroup of order 8, �51

Janko’s theorem on 2-groups G withj�2.G/j D 24, �52

Janko’s theorems on 2-groups G withcn.G/ D 4, n > 1, ��53, 54

Janko’s theorem on 2-groups G in whichthe subgroup generated by allelements of order 4, is of order4, �55

K�1.G/, �76�1.M/, �76k-stepped p-groups, �64

Mmaximal abelian subgroups, ��91, 92maximal centralizers of p-groups,

Appendix 23M3-groups, �76.metacyclicAn-groups with derived

subgroup of order pn, ��65metacyclic groups, ��65, 66, 69metacyclic p-groups with derived

subgroup of order pn areAn-groups, �65

minimal nonabelian subgroups, thenumber of, ��65, 76, 90, 92

minimal nonmetacyclic p-groups, ��66,69

minimal nonmodular p-groups, �78minimal nonnilpotent groups, Appendix

22minimal non-quaternion-free 2-groups,

�80

modular p-groups, Iwasawa’sclassification of, �73

Nnonmodular quaternion-free 2-groups,

structure of, �79number of cyclic subgroups of given order

in some central and wreathproducts, Appendix 16

O��

n.G/, �55one-stepped p-groups, �64

PpairsH < G with ˇ1.G;H/ D p�1, �76pairsH < G with ˇ1.G;H/ D p, �76p-groups all of whose nonnormal

subgroups are conjugate, �59p-groups all of whose cyclic subgroups of

composite orders are normal,�63

p-groups generated by elements of givenorder, �64

p-groups G satisfying ˛1.G/ � pd.G/�3,�76

p-groups G with 1 < ˛1.G/ < p2 are

A2-groups, �76p-groups G with ˛1.G/ D p2, �76p-groups G with ˛1.G/ D p2 C p C 1,

�76p-groups G with d.G/ > 2 all of whose

maximal subgroups aretwo-generator, �70

p-groups with abelian subgroups of indexp2, ��65, 71

p-groups with abelian n-th maximalsubgroups, n D 3; 4, �72

p-groups with a cyclic subgroup of indexp2, Janko’s proof, �74

p-groups with extraspecial�1.G/ or��

2.G/, �83p-groups G, p > 2, with c1.G/ D p C 1,

�69p-groups G with metacyclic ��

2.G/, �86p-groups G with �1.G/ D p � 1, �76p-groups G with �1.G/ D p, �76p-groups without normal subgroup

Š Ep3 , p > 2, �69


p-groups, p > 2, without normalsubgroup of order p3 andexponent p, �69

p-groups with a uniqueness condition fornonnormal subgroups, �84

p-groups with few minimal nonabeliansubgroups, �76

p-groups with large normal closures ofnonnormal cyclic subgroups,�62

p-groups G with j��2.G/j D ppC1, �55

p-groups G with��2.G/ extraspecial, �83

p-groups G with��2.G/metacyclic, �86

p-groups G with��n.G/metacyclic,

n > 2, �86

Qquaternion-free 2-groups, structure of,

Janko–Wilkens theorem, �79

SSchreier’s inequality for p-groups,

Appendix 25subgroup structure ofA2-groups, �65

Ttwo-generator modular p-groups are

metacyclic, �732-groups all of whose minimal normal

subgroups are of order 8, �902-groups all of whose minimal normal

subgroups are isomorphic andhave exponent 4, �57

2-groups in which the centralizer of aninvolution is abelian of type.2; 2m/, �48

2-groups in which the centralizer of aninvolution is C2 � Q2m , �49

2-groups G with an involution t such thatCG.t/ Š C2 �M , whereM isof maximal class, �51

2-groups with exactly four cyclicsubgroups of order 4, �54

2-groups with exactly four cyclicsubgroups of order 2n, n > 2,�55

2-groups with exactly one nonmetacyclicmaximal subgroup, �87

2-groups with exactly three involutions,�82

2-groups with nonabelian Frattinisubgroup of order 16, �85

2-groups G with j�2.G/j D 16, �522-groups G with j�3.G/j D 25, �522-groups with 7 and 11 involutions, �642-groups G with j��

2.G/j D jhx 2j

o.x/ D 4ij D 16, �552-groups G with��

2.G/ D Q2n � C2, �752-groups without elementary abelian

subgroups of order 8, �492-groups without normal elementary

abelian subgroup of order 8, �502-groups with selfcentralizing abelian

subgroup of type .4; 2/, �772-groups with selfcentralizing elementary

abelian subgroup of order 8, �51

Uunitriangular group, ProblemsU2-groups, determination, defining

relations of, ��64, 67

WWard’s theorem on quaternion-free

2-groups, �56, Appendix 17Wilkens 2-groups (Wa-, Wb- and

Wc-groups), �79

groups 2 groups janko vol 2

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