maximal baer groups in translation planes and compatibility with homology groups

Geometriae Dedicata 59: 65-101, 1996. 65 © 1996 KIuwerAcademic Publishers. Printed in the Netherlands.

Maximal Baer Groups in Translation Planes and Compatibility with Homology Groups

MAURO BILIOTrI 1 and NORMAN L. JOHNSON 2 1Universit& di Lecce, Dipartimento di Matematiea, Via Arnesano, 73100 Lecce, Italy ZUniversity of Iowa, Department of Mathematics, Iowa City, Iowa 52242, U.S.A.

(Received: 13 September 1993)

Abstract. The translation planes with spreads in PG(3, q) that admit at least two Baer groups of order q - 1 are classified.

Mathematics Subject Classifications (1991): 51A40, 51A35, 51A10.

Key words: Baer homology groups, translation planes, Desarguesian planes.

1. In troduct ion

Consider the collineation groups in translation planes which may be generated by elements which fix subspaces of line size pointwise. The central collineations in such groups are homologies or elations. If a collineation fixes a line size space pointwise but does not fix a line pointwise then the collineation fixes a Baer subplane pointwise and is called a Baer collineation. If the net defined by the components of the Baer subplane is derivable then a Baer collineation becomes an elation or homology in the derived plane. For this reason, if the collineation fixes exactly one Baer subplane, the collineation is said to be a Baer-elation and otherwise a Baer-homology.

The collineation groups of a finite translation plane generated by the set of all elations has been determined in the 1960s by Hering and Ostrom (see [17, 36]). Concerning the groups generated by Baer-elations, Foulser [11] showed that the same groups appear in finite translation planes of characteristic p when generated by Baer p-groups (Baer-elations) where p is not 2 or 3.

Furthermore, when the plane has even order q2 and the spread is in PG(3, q), Johnson and Ostrom [33, 34] have a fairly complete theory of the groups generated by Baer involutions (Baer-elations). (Also, see Hering and Ho [18] for some generalizations.)

For arbitrary dimension, Dempwolff [7] considered the Baer 2-group case for finite translation planes of even characteristic where the Baer 2-group is 'large enough' and Jha and Johnson [22, 23] considered the situation when there are at least two large Baer p-groups for finite translation planes of characteristic p.

66 MAURO BILIOTTI AND NORMAN L. JOHNSON

There has considerable work on groups generated by affine homologies (see Ostrom [36-38], Dempwolff [8, 9], Johnson and Pomareda [31]) but here the situation is much more complex.

However, there is essentially nothing known regarding collineation groups generated by Baer-homology groups analogous to the studies of groups generated by homology groups.

In Jha and Johnson [22, 23], it was shown that if a translation plane of order pr admits at least two Baer p-groups of order > pr/2 then either the plane is a known plane of order 16 or the plane is Hall.

In this article, we ask the following question: If a translation plane of order pr admits at least two Baer groups of order dividing p~ - 1 and sufficiently large order, is the plane a Hall plane?

Given any translation plane of order qZ with spread in PG(3, q), there is an associated translation plane of order q4 with spread in PG(3, q2) which admits an elation group of order q2 and a Baer-homology group of order q + 1. Moreover, the elation group and Baer-homology group do not centralize each other so that there are actually 2q 2 Baer groups of order q + 1 in such planes. (The reader is referred to [28] or [20] for an explication of such translation planes.)

So, very little may be said in general about translation planes of order p~ that admit at least two Baer groups unless the Baer groups are of order > (p~/2 + 1). The maximal order of a Baer-homology group in a translation plane of order qe is q - 1. Translation planes of order q2 that admit at least two Baer-homology groups have been constructed by Johnson [29] following the general guidelines of the work of Jha [21 ] on generalized Hall planes. Actually, such planes (generalized Hall planes of type 2) admit q Baer groups of order q - 1. In such planes, the 'axes' for the Baer-homology groups lie in the same net defined by the components of any one of the Baer subplanes.

We are able to show that any translation plane of order q2 that admits two Baer-homology groups of order q - 1 is a generalized Hall plane (of type 1) or a generalized Hall plane of type 2, provided the nets defined by the two Baer axes are the same (see Section 2).

When the Baer-homology axes do not produce equal nets, the problem appears to be quite difficult in the general, arbitrary kernel case. Actually, there are geometric reasons for considering the general problem of classifying the planes of order q2 with spread in PG(3, q) that admit at least two Baer-homology groups of order q - 1 .

In [27], one of the authors shows that corresponding to a translation plane of order q2 and kernel GF(q) that admits a Baer-homology group of order q - 1, there is a partial flock of deficiency 1 of a hyperbolic quadric in PG(3, q), that is, a set of q mutually disjoint conics which lie on a hyperbolic quadric.

If the Baer homology group of order q - 1 is normal within the full collineation group of the translation plane then there is a homomorphism from the collineation group of the translation plane onto the collineation group of the partial flock and

MAXIMAL BAER GROUPS IN TRANSLATION PLANES 67

effectively the study of partial flocks of deficiency 1 is equivalent to the study of translation planes of order q2 with spread in PG(3, q) which admit a B aer-homology group of order q - 1.

In Sections 3, 4 and 5 we consider the following problem: Determine the translation planes of order q2 and kernel containing GF(q) which

admit at least two Baer groups of order q - 1. When the plane has kemel GF(q) then if there is one Baer group B of order q - 1

(or dividing q - 1) there is always another in BK* where K* denotes the kernel homology group. Let N denote the net of degree q + 1 whose components are the components fixed by B. Then there is exactly one other Baer subplane of N which is fixed by B. We shall call this subplane coFix B and the fixed point subplane Fix B. So, given B there is also a Baer group/3 such that coFix/~ = Fix B and both subplanes are subplanes of N. So, when we say two Baer groups B, B, we shall require that either coFix B is not Fix/~, or coFix/~ is not Fix B.

Actually, we are able to give a complete classification of the translation planes of order q2 with spread in PG(3, q) which admit at least two Baer groups of order q - 1. If q is not 3, 4 or 5, it turns out that the plane is always Hall. However, there are counterexamples when q -- 3, 4 or 5 and these are explicated in the article. Actually, in Section 5, we construct some new examples of maximal partial hyperbolic flocks in PG(3, 5) with four conics.

Foulser [11] has shown that in finite translation planes of odd prime characteristic, elations and Baer-elations cannot coexist. In Section 6, we consider whether there is a form of incompatibility between homologies and Baer-homologies. In particular, we consider the possible incompatibility between Baer groups of order q - 1 and homology groups of order q - 1 and show that such a situation always gives back the two Baer-homology group situation. Thus, we obtain some incompatibility results as corollaries to our analysis of the groups generated by two Baer groups of order q - 1 in translation planes of order q2.

In Section 7, we consider partial hyperbolic flocks and discuss the implications of Sections 3, 4 and 5. Finally, Section 8 deals with transitive partial hyperbolic flocks.

BACKGROUND RESULTS

Some of the results used in this paper shall be listed for convenience.

RESULT I (Foulser and Johnson [13]). Let 7r be a translation plane of order q2 that admits a collineation group isomorphic to SL(2, q). I f the Sylow p-groups for q = p~ are Baer then 7r is the Hallplane oforderq 2.

RESULT II (Jha [21]). Let 7r be a translation plane of order q2 that admits a Baer co lIineation group of order q(q - 1). Then 7r is a generalized Hall plane and thus derived from a semifield plane coordinatized by a semifield plane of order q2 with middle nucleus isomorphic to GF(q).


We recall that a generalized Hall plane of type i is a translation plane of order q2 that admits i Baer subplanes fixed pointwise by collineation groups which have the same infinite fixed points and such that the generated group contains a group of order q and acts transitively on the nonfixed infinite points.

RESULT Ill (Johnson [29]). Let 7r be a generalized Hall plane o f order q2 and type 1 or 2 and let N denote the associated net containing the Baer subplanes f x e d pointwise by the indicated groups. Then N is derivable and i f there is a Baer group of order q - 1 then 7r may be derived from a semifield plane o f order q2 whose coordinatizing semifield has either a middle or right nucleus isomorphic to GF(q) respectively as 7r is o f type 1 or 2.

RESULT IV (Thas [44], Bader [2], Johnson [30]). Aflock o f a hyperbolic quadric in PG(3, q) is equivalent to a translation plane of order q2 whose spread is a set o f q + 1 reguli in PG(3, q), mutually sharing two lines.

RESULT V (Thas [42, 43], Bader and Lunardon [3]). Let SF denote the set offlocks o f a hyperbolic quadric in PG(3, q) and 7r F the set o f corresponding translation planes (see IV). Then 7rF consists o f the Dickson nearfield planes with kernel containing GF(q) and the irregular nearfield planes o f order 112, 232, and 592 with central kernels.

RESULT VI (Thas [44], Gevaert et al. [14]). (1) A flock F of a quadratic cone in PG(3, q) is equivalent to a translation plane 7r F of order q2 whose spread is a set o f q reguli in PG(3~ q) mutually sharing one line.

(2) I f TrF of(1) is not Desarguesian then the only reguli in the spread of Tr F are the q reguli in the description o f Tr F.

RESULT VII (Payne and Thas [40]). A partial flock o f deficiency 1 o f a quadratic cone in PG(3, q) may be uniquely extended to a flock.

RESULT VI~ (Johnson [27]). (1) A partial flock o f deficiency I o f a quadratic cone in PG(3, q) is equivalent to a translation plane o f order q2 and kernel containing GF(q) that admits a Baer group of order q.

(2) A partial flock o f deficiency 1 of a hyperbolic quadric in PG(3, q) is equivalent to a translation plane o f order q2 and kernel containing GF(q) that admits a Baer group of order q - 1.

RESULT IX (Jha and Johnson [24]). Let 7r be a translation plane of order q2 that admits a Baer group B or order >_ 2q 1/2. Then one of the following alternatives must occur:

(i) B is normal in the full collineation group of Tr, (ii) 7r is Hall, or

(iii) 7r is a known plane o f order 16.

Further, i f the order is even and > 16 then any elation group has order <_ 2.

RESULT X (Foulser [11]). Let ~r be a translation plane o f odd order f f . Let B be a


Baer p-group o f Tr of order >_ 9. Then the net of degree pt/2 + 1 containing Fix B is invariant under the full collineation group.

RESULT XI (see, e.g., Ltineburg [35, (48.5), p. 248]). Let R be a regulus of the vector space V o f rank 4 over GF(q) and let G ~ GL(2, q) be the subgroup o f the stabilizer of R in GL(V) which fixes the opposite regulus of R linewise. Then G leaves invariant each Desarguesian spread which contains R.

RESULT XII (see, e.g., Ltineburg [35, (4.7), p. 20]). Let P be a projective plane and Q a proper subplane of P. I f a is a perspectivity of P then ~r fixes Q if and only if the center and the axis of or belong to Q and there is a point X of Q, which is distinct from the center and not on the axis of or, such that Xcr also belongs toQ.

RESULT XIII (Hering et al. [19]). Let G be a doubly transitive group whose one- point stabil&er admits a normal sharply transitive subgroup. I f the socle of G is a nonabelian simple group then the possibilities for N are:

(i) N - PSU(3, h), degree h 3 + 1, (ii) N ---- Sz(h), degree h 2 + 1 where h = 22a+1.

(iii) N --- PSL(2, h), degree h + 1,

(iv) N ~ R(h) , degree h 3 + 1.

ASSUMPTION. We shall assume in this paper that 7r is a translation plane of order q2 that admits at least two distinct Baer groups of order q - 1, where for at least one o f the groups the axis is not the coaxis of the other group.

2. When Baer Groups have the same Infinite Fixed Points

Recall that a generalized Hall plane of type 1 is a translation plane of order q2 that admits a Baer group of order q(q - 1) in the translation complement and a generalized Hall plane of type 2 is a translation plane of order q2 that admits a collineation group of order q(q - 1), all of whose elements are Baer, but not all fix the same Baer subplane.

Considering the problem of classifying the translation planes of order q2 that admit at least two Baer-homology groups of order q - 1 and such that the Baer groups have the same infinite fixed points, we may give a complete description of these planes.

(2.1) THEOREM. Let 7r be a translation plane of order q2 that admits at least two Baer groups of order q - 1 such that the axis of at least one group is not the coaxis of the other. Assume that the Baer groups have the same fixed points on the line at infinity. Then one of the following possibilities occurs:

(i) 7r is a Hall plane or the Desarguesian plane of order 9 and the group generated by the Baer groups is isomorphic to GL(2, q);


(ii) 7r is derivable from a semifield plane of order q2 with a coordinatizing semifield with middle nucleus isomorphic to GF(q); 7r is a generalized Hall plane (of type 1), and the group generated by the Baer groups is the semidirect product of a Baer-elation group of order q by a Baer-homology group of order q - 1, where all Baer groups have the same fixed point subspace;

(iii) 7r is derivable from a semifield plane of order q2 with a coordinatizing semifield with right nucleus isomorphic to GF(q); 7r is a generalized Hall plane of type 2, and the group generated by the Baer groups is the semidirect product of a Baer-elation group of order q by a Baer-homology group of order q - 1, where the Baer groups do not fix the same subspace pointwise;

(iv) if the kernel of the plane is isomorphic to GF(q) then 7r is Hall or Desarguesian o f order 9 in all of the cases.

Proof Let B1 and Bz be Baer-homology groups of order q - 1 in the translation complement and such that they fix the same infinite points. Then Fix B1 and Fix B2 are Baer subplanes of the same net N of degree q + 1. Within N there is a coaxis Baer subplane coFix Bi which is fixed by Bi, i = 1, 2 (Foulser ([10])) but on which Bi induces the kernel homology group of order q - 1 so that the various subplanes are Desarguesian subplanes. We are assuming that {Fix B1, coFix B1 } {Fix B2, coFix B2}. Hence, B1 either moves Fix B2 or coFix B2 and in either case, forces there to be another Baer subplane in N incident with the zero vector. By Foulser ([10]), the net N is derivable.

We have the following possible situations:

(i) Fix B1 = Fix B2, but coFix B1 ~ coFix B2,

(ii) Fix B1 = coFix B2, but coFix B1 ¢ Fix B2,

(iii) coFix B1 = coFix B2, but Fix B1 ~ Fix B2.

(iv) {Fix B1, coFix B1} fq {Fix B2, coFix B2} is empty.

In cases (i), (ii), or (iii), there is a fixed Baer subplane and the group generated by the Baer groups is transitive on the remaining q Baer subplanes incident with the zero vector in the net N. Hence, there is a Baer group of order q in the generated group. Since there is now a group of order q(q - 1) such that each element of the group is Baer, it follows that the group acts transitively on the nonfixed infinite points. The result follows from Result II. Note that, in case (i), the plane may be derived from a semifield plane with coordinatizing semifield of dimension 2 over a middle nucleus; in case (iii), the plane may be derived from a semifield plane with coordinatizing semifield of dimension 2 over a right nucleus; and in case (ii) the plane may be derived from a semifield plane of dimension 2 over both a right and middle nucleus.

Assume case (iv). Then it follows that the Baer groups generate a group which acts transitively on the q + 1 Baer subplanes of the net N.

If q = 3 then the plane could be Hall or Desarguesian and both planes admit Baer groups of the type indicated.


Since the net N is derivable, it is well known that there is a field K ~- GF(q) such that the net corresponds to a regulus in PG(3, K). Clearly, the generated group is a linear subgroup of GL(2, K) generated by cyclic homology groups of order q - 1 acting on a particular component considered as a Desarguesian affine plane of order q. Since the group acts transitively on the q + 1 Baer subplanes and the stabilizer of a Baer subplane contains a group of order q - 1, it follows that the group induced in PGL(2, K) has order divisible by q2 _ 1.

First assume that there are no p-elements. Then either q = 2 or 3 or the group must be isomorphic to A4, $4 or A5 and hence q2 _ 1 divides 12, 24, or 60. The only possibilities are q = 2, 3, 4, or 5 by the divisibility restriction.

If q = 4 the group must, in fact, be A5. However, A5 does not contain cyclic

groups of order 4. If q = 5, the group must be $4. Since the stabilizer of a Baer subplane is cyclic

of order 4, there must be six cyclic groups of order 4 which must be pairwise disjoint (as each fixes exactly a Baer subplane (subline) pointwise). However, in $4, the six cyclic groups of order 4 are not mutually disjoint.

Thus, there must be p-elements and since we have a group of order divisible by q2 _ 1 in PGL(2, K) , it follows that the generated group in PGL(2 , / ( ) must be PSL(2, pS) or PGL(2, pS) for some integer s _< r where q = p~. Thus, q2 _ 1 divides (p2S _ 1)/2 or pZs _ 1 which implies that pS = q and the group induced must be PGL(2, K). Now there is a Baer group B of order q - 1 such that the normalizer of this group in PGL(2 , / ( ) is dihedral of order 2(q - 1) and interchanges the two Baer subplanes fixed by B. Thus, there is a Baer group of order q - 1 with axis the coaxis of B and coaxis the axis of B (where axis of B is taken to mean the Baer subplane fixed by B). Thus, the full scalar group (acting on a line of the net) is contained within the generated group on the translation plane.

Hence, the translation plane admits a collineation group G - GL(2, q) such that the p-elements are Baer collineations. By Result I, the translation plane must be Hall. This completes case (iv) listed above.

It remains to show that when the kernel contains a field isomorphic to GF(q), then only the Hall plane or the Desarguesian plane of order 9 can occur in any of the four cases (i)-(iv).

Assume that the kernel contains L ~ GF(q). By the above argument, in each of the four cases, there is a Baer group of order q. By Results VII and VIII(l), the derived plane corresponds to a flock of a quadratic cone in PG(3, q). But, also, it follows that the net N is a regulus net (corresponding to a regulus in PG(3, L)) so that by Result VIII(2), the derived plane also corresponds to a flock of a hyperbolic quadric in PG(3, q). By Result V, this plane is a nearfield plane which admits elations. Hence, the plane is either the Desarguesian plane or has order 9. This completes the proof of (1.1).

72 MAURO BILIOTT1 AND NORMAN L. JOHNSON

3. Baer-Homology Groups Translation Planes with Spreads in PG(3, q)

In this section, we completely determine the translation planes of order q2 and kernel containing GF(q) that admit at least two Baer groups of order q - 1. In [27], it is noted that the semifield plane of order 16 and kernel GF(4) admits such groups as the semifield admits an automorphism of order 3 which fixes a subfield isomorphic to GF(4). Since a homology group of order q - 1 which fixes a regulus net in a Desarguesian spread produces a Baer group in the corresponding Hall plane obtained by derivation of the indicated regulus net, the Hall planes also admit at least two Baer groups of order q - 1. We shall see that the irregular nearfield plane of order 25, and the exceptional Walker plane of order 25, with infinite orbits of lengths 10, 16 under the full group, also admit two Baer groups of order q - 1 = 4. However, these are exactly the possibilities.

Our main result is

(3.1) THEOREM. Let 7r be a translation plane of order q2 and kernel containing K ~ GF(q). If~r admits at least two Baer-homology groups of order q - 1 in the translation complement then one of the following situations must occur:

(i) :r is Hall, (ii) :r is Desarguesian of order 9,

(iii) 7r is the semifield plane o f order 16 with kernel GF(4), (iv) 7r has order 25 and is the irregular nearfield plane, (v) 7r has order 25 and is the exceptional Walker plane with orbit structure

{10, 16} on the line at infinity.

In the following, we assume the hypothesis of (3.1). We shall give the proof by a series of lemmas. However, we shall prove parts (iii), (iv) and (v) for order 16 and 25 in Sections 8 and 7 respectively.

NOTATION. In particular, let Bi, i = 1, 2, be Baer-homology groups of order q - 1 such that B1 is not contained within BzK*, where K* denotes the group of kemel homologies of order q - 1. Let Ni denote the net of degree q + 1 containing the fixed point subplane Fix Bi of Bi.

LEMMA (1). Coordinates may be chosen so that N1 has the form

z = O,y = z 0 f ( u )

for all u ~ K and f a function on K. In this representation B1 has the form: 1 } "lL

1 l u ~ K * .

u


Proof Choose coordinates so that 7r = {(Xl, x2, Yl, Y2) [ xi, Yi E K, i = 1, 2}, Fix B1 = {(xl,O, yl,O) iXl, Yl C K ) and 7r and Fix B1 share the components labeled x = 0, y = 0, and y = x (see Foulser [10] or Johnson [27]).

LEMMA (2). If Fix B2 is contained in N1 then 7r is Hall or the order is 9 and the plane is Desarguesian or Hall.

Proof Use Theorem (2.1) (iv). Hence, if N1 and N2 share all of their parallel classes, we are finished. In the

following, we show that for orders > 25, the nets share either all (q + l) or 0 or 1 parallel classes.

LEMMA (3). I f the order q2 > 25 then we may assume that [N1 fq N2[ = 0, 1 or q + 1 for all nets Ni of Tr which contain Baer groups of order q - 1.

Proof Assume that Fix B2 is not contained in N1. Let F denote the full collineation group of 7r. Assume further that N1 and Nlg share at least two components (parallel classes) L1, L2. Then there are two cases:

CASE I. For g ¢ F and B1 and B e leave both {Tr0 = Fix B1, 71"1 = coFix B1} f) Li for i = 1,2 invariant, then N19 = NI.

CASE II. On at least one of the lines, say L1, the group does not leave {Tr0 fq L1,71" 1 0 L 1 ) invariant.

CASE I. We prove that this cannot occur unless q2 _< 25. Proof Consider (B1,B~). Assume that 7r0 = Fix B1 # coFix B1 = 7rl =

({(0, x2, 0, Y2) [ x2, Y2 6 K}). Let N1 and N19 share the lines L1 and L2. Since B e is a faithful subgroup of GL(2, q) and is cyclic (see Lemma (1)) of order > 2 then B e fixes both r0 N L1 and 7rl f3 L1.

Further we are assuming that B e leaves {Tr0, 7rl} N Lz invariant. It then follows that B e leaves both 7r0 and 7rl invariant as they are 2-dimensional K-subspaces (note that we may consider L1 as a Desarguesian plane upon which B e acts).

If Fix B e = 7to or 7rl then clearly N1 = Nlg. Otherwise, B e induces a cyclic collineation group on 7r0 (Tr 1 ) which fixes at least two lines of 7r0 (Tr 1 ). Further, we may assume, without loss of generality, that B~ fixes 7to fq L1 and 71 1 N L2 pointwise as B~ induces a homology group on Li as a Desarguesian affine plane.

We may choose L1 to be x = 0 and L2 to be y = 0 with an appropriate basis change such that the basic form for the net N1 does not change. For example, assume that L1 is not x = 0 or y = 0. Then L1 is

ul 0 ] y = x 0 f ( u l )

with B1 represented as in Lemma (1). Then we may map L1 onto ff = 0 by the basis change

74 M A U R O B I L I O T T I A N D N O R M A N L. J O H N S O N

O" ----

[ o] I - 0 f ( u ) l

0 I

Then

"-+ y ---- X .

y = x 0 f(u 0 f ( u ) - f ( u l )

We have that B e = B1. Note that B~ fixes 7r0 N L2 pointwise if and only if a corresponding group in B~K* fixes 7rl O L2 pointwise. While it may not be the case that this group is actually B~, we only use the existence of a Baer group of order q - 1 fixing 7to N L1 and 7rl n L2 pointwise. So, we may assume that B e fixes 7r0 N L1 and 7rl O L2 pointwise. Indeed, the group must fix either 7r0 n L1 or 7r0 n L2 pointwise as it fixes only two 1-spaces on Li for i = 1,2, and by assumption cannot fix both 7r0 n L1 and 7r0 N L2 pointwise.

Hence,

U 1

1 u j

[ u E K* for a fixed integer j # 0 } .

But, since B e is a Baer collineation group, a component

y = x f

is fixed if and only if

e fu J J f for a l l u # 0 .

Hence, c = 0 so that d # 0 and then j = 1 and f = 0. Thus, applying the groups to a component of the form

we obtain a set of (q - 1) a components

[ t-lg vf]forall t , v inK_{O} . Y : X L v-1 t

MAXIMAL BAER GROUPS IN TRANSLATION PLANES

It follows that the components of Fix B~ are

x = O , y = x 0

Applying B1 to these components, we obtain:

0 U -1 V 0 0 U u- - l v

75

uh(v) 1 0 for all u ¢ 0 and for all v E K.

It follows directly that h(u) = u - lho for all u ~ 0 where h0 is a nonzero constant in K.

Applying B e to the components

[; o] y = x f ( v )

for all v ~ K of Fix B1, we obtain

[ u - ' v 0 1 o vI( ) which shows that f ( v ) = v - l f o for all nonzero v in K, where f0 is a constant in K. Since we may assume that y = x is a component of Fix B1, f0 = 1.

We shall show that the only possible planes have order < 25. We also have a partial spread

y = x v - 1 t

where g, f are constants in K and for all nonzero t, v in K. Hence, the difference of any two distinct indicated matrices is nonsingular.

Since there is an orbit of length (q - 1) 2, we consider

which is nonsingular if t and v are not both equal to 1. Hence, we must have:

(t -1 - 1)(t - 1)g - (v - 1)(v -1 - 1 ) f ¢ 0 for all nonzero t, v not both 1.

CASE (i): q is even.

76 MAURO BILIOTrI AND NORMAN L. JOHNSON

Then the requirement for a partial spread is:

(t -1 + t)g + (v -1 + v)f ¢ O.

Since clearly we may suppose fg ~ O, we must obtain:

V2 q- V(t -1 -Jr t)g + 1 ¢ O. f

t - 1 + t = 0 if and only if t = 1. So, assuming t ¢ 1, divide the above equation by ((t -1 + t )g/ f) 2 and let x = v/((t -1 + t)g/f) to obtain:

t f )2 X 2 q- X ~- (t 2 q- 1)g ~ 0 for all t ~ 0 or 1.

Hence, we must have the trace of t f~ ((t 2 + 1)g) = 1 for all t ?~ 0 or 1, t ~ K.

Let f / g = c. Let u = t 2 + 1 so that u 2r-1 + 1 = t for q = U. Then

c(u 2~-1 + 1)u q-2 must have trace 1 for all u ~ 0 or 1. But, this latter expression is ¢(u q/2+q-2 + u q-a) : c ( u (q-2)/2 q- u (q-2)) (note that q/2 + q - 2 is (q -

1) + (q - 2)/2). Furthermore, (q - 2) is relatively prime to (q - 1) so that letting U (q-2)/2 = y, we obtain:

t r a c e c ( y + y 2 ) = 1 for a l l y ¢ O o r l .

This is equivalent to:

{c(y q- y2) _{_ c2(y2 q_ y4) q_.,..~_ C2r-l(y2r-I q_ y) q_ l)}(y2 .~ y)

is identically zero, as we now obtain a polynomial in y for all values y E K. Now isolating on the coefficient of the y-term, we see that if 2 ~- 1 + 2 < q, then

there are no overlapping y-terms and the only possibility is a 1 for the coefficient of the y-term, which is a contradiction.

Hence, 2 ~-1 + 2 >_ q so that q = 2 or 4. We may stop here, but continuing with the special cases, note if q = 2 there is essentially nothing to prove. Thus, assume that q = 4. Then t -1 = t 2 and we obtain a partial spread if and only if t29t + v2vf ~ 0 for t or v are not zero. Since this reduces to g + f ~ 0 for tv ~ 0, we clearly may choose g, f appropriately. Note that this forces the plane to be the unique semifield plane of order 16 and kernel GF(4).

CASE (ii): q is odd. We still must have (t - 1 - 1 ) ( t - 1 ) g - (v - 1 - 1 ) ( v - 1 ) f ~ 0 for t , v ~ 0

where at least one of t and v is also not 1. Equivalently, (2 - (t + t -1))g - (2 - (v -1 + v))f ~ O. Form the quadratic

in v:

v2f + ( (2- - (t + t - 1 ) ) g - 2f)v + f ¢ O.


Let (t + t -1 ) = ct and note that ct = c~, for t ¢ 0 or 1 if and only if u = l i t or t so that {ct I t C K - {0, 1 }} has cardinality (q - 3 ) / 2 + 1 = (q - 1 ) /2 (note

l i t = t i f and only i f t = 4-1).

Hence, we have:

( ( 2 ) _ c t ) g 2 ) 2 _ 4 must be a nonsquare for all et.

Let g / f = d -1, so we obtain (2 - c t ) d - l ( ( 2 - ct)d -1 - 4 ) or also ( 2 - c t ) ( ( 2 -

ct) - 4d)d -2 is a lways a nonsquare, so that (2 - c t ) ( (2 - ct) - 4d) is a lways a nonsquare for t ~ 0 or 1. Since 2 - et = - ( t 2 - 2t + 1) / t = - ( t - 1)2/ t and

(2 - ct) - 4d = - ( t 2 - (2 - 4d)t + 1)/ t , we must have that:

(t 2 - (2 - 4d)t + 1) is a nonsquare for all t ¢ 0 or 1.

Let t = (2 - 4d) to obtain either 1 is a nonsquare or (2 - 4d) = 0 or 1.

First consider (2 - 4d) = 0 so that t 2 + 1 is a nonsquare for all t ¢ 0 or 1.

For t = - 1, we obtain 2 is a nonsquare and for t = 2, we obtain 5 is a nonsquare, so that 2 . 5 is a square. But for t = 3, 32 + 1 = 10 is a nonsquare. Hence, we have

a contradict ion unless poss ibly 3 is 0. In this case, 2 - 4d = 0 implies that d = 2

or rather that g = 2 f . Since 2 is a nonsquare, the map z ~ z + 1 maps the nonzero squares onto the

nonsquares. Hence, it maps 2 to 0, 0 to 1 and the nonsquares different f rom 2 onto the nonzero squares different f rom 1. Let w E K - GF(3) such that w 2 is not in

GF(3). Then w 2 + 1 is a nonsquare, w 2 + 2 is a nonzero square and hence w 2 + 3 = w 2

is a nonsquare, a contradiction. Hence, K = GF(3) and the order q2 = 9. I f 2 - 4 d = l t h e n ( t - 1)2 2 + 3 is a nonsquare for all t ¢ 0 or 1, we have

equivalently: w 2 + 3 is a nonsquare for all w ¢ 1 or - 1 (note w = 2t - 1 so that t -= 0 or 1

if and only if w = - 1 or 1 respectively).

Hence 3 is not a square, so the characteristic is not 3. Thus, 2 7~ - 1 and 22 + 3 = 7 is a nonsquare. Hence, in this case, 21 is a square and i f21 ~ 1 or - 1 ,

that is, the characteristic ~ 5 or 11, respectively, then 4 . 2 . 3 = 24 is a nonsquare

implies 2 is a square. This in t um implies that 2 + 3 = 5 is a nonsquare so that 3 • 5 = 15 is a square and 32 • 2 = 18 is a square. Since, 15 + 3 is a nonsquare,

we have a contradiction. I f K has characteristic 11 then 3 is a square. So, K has characteristic 5.

We show that i f K has characteristic 5, then K --- GF(5) . Let ~2 denote the set consist ing of the squares different f rom 1 including zero.

The map + 3 maps f~ onto the nonsquares of K (we have exactly (q - 1 ) /2 images).

Suppose that u E K is a nonsquare and assume that u + 3 is also a nonsquare. Since - 3 maps the nonsquares onto f~, we have that u = (u + 3) - 3 = a square.


This is absurd and hence either u + 3 = 1 or u + 3 E fL Let w E K such that/13 2

is not in GF(5). T h e n w 2 + 3 i s a n o n s q u a r e , w 2 + 3 + 3 = w 2 + 1 i sasquare , w 2 + 4 i s

a nonsquare, w 2 + 7 = w 2 + 2 is a square, w 2 + 5 = w 2 is a nonsquare - a contradiction.

Hence, we must have K ~ GF(5). This completes the proof of Case I.

CASE II. We prove that the plane is Hall. Proof. The assumptions imply that the orbit lengths of the 1-spaces on L1 are

either 1 and q or q + 1. In the latter case, since the kernel ~ GF(q), we may use the argument of (2.1)(iv)

analyzing the group acting on a component of the net N to conclude that there is a collineation group inducing GL(2, q) on L1. Since there are at least two fixed components, it follows that there must be a Baer group of order q but unless there are at least three fixed components, now we do not know if GL(2, q) acts as a collineation group of the translation plane as there could be affine homologies with axis L1.

In the case where there is an orbit structure of lengths q and 1, there must also be a Baer group of order q. Thus, we may argue both of these cases simultaneously.

By Result VII and Result VIII, the plane must be a derived conical flock plane. Moreover, by Result IX, B1 and B2 must normalize the Baer group E of order q or the plane is forced to be Hall, Thus, we see that the q + 1 orbit situation leads to the Hall plane or the order of the plane is 16.

Continuing with the {q, 1} orbit situation, in the corresponding derived plane ~, the group (B1, B2) must act and if the plane is not Desarguesian then the group must permute the q reguli (see Result VI(2)).

Since Bi must fix the subplane Fix E in ~r, it follows that both B1 and B2 must fix Fix E V/L1 pointwise (without loss of generality, as otherwise there is such a group within BiK*) but both cannot fix Fix E pointwise as otherwise the original two Baer subplanes Fix B1, Fix B2 (or coFix B1, coFix B2) would be equal. Actually, if either B1 or B2 fixes Fix E pointwise, then the plane is a generalized Hall plane (see Result II) and since the kernel is GF(q), the plane must be Hall (see (2.1)).

Hence, B1 and B2 act on Fix E as homology groups and thus in the derived plane ~, the groups Bi must still act as Baer groups since Fix E is a line of ~.

Note that E becomes x = 0 in the derived plane ~ and the spread for ~ may be represented in the form

x = O, Y = X [ u + g(t)t f(t)]u

for all u, t E K and f , g functions on K.


Let 7- E B1 of prime power order. Then since B1 must leave the derived net

[ 0 0 ] for all u C K)invariant, and 7- fixes x = 0 (that is, E), N ( x = O,y = x u T

I. A

must fix another component of/Y. But the components which 7- fixes are also fixed

[ o ° ° ] = x M . C h o o s e a n e w b a s i s b y by B1. Let B1 fix the component y = x u0

[ I - M ] and note that g r = 0

E =

o] ( ou

I

in the representation. Note that cr commutes with E, so that we may assume that B1 fixes y = 0 and x = 0.

Now we want to choose a basis for (y = 0) = {(xl, x2, 0, 0) I xi C K for i = 1, 2} so that B1 fixes exactly x~ = 0 and fixes Xl = 0 pointwise. Choose a change in basis T so that x = 0 is represented as required. Furthermore, consider the basis change in the plane # given by Diag{T, T} and note that E must commute with this matrix. This implies that we may represent B1 in the form

< u 0

0 1

al a2

a3 a4

l u C K - {0}>

where the ai depend on u. Note that y = x is mapped under B1 onto

y = x [ul ][ai a2]

] a 3 a4

SO that a2 = a3 = 0. B1 is Baer implies (Xl, x2, Yl, Y2) is fixed by B1 if and only if xl = 0, and

{a4u = 1 for a l l u o rx3 = 0} or {a4 = 1 for a l l u o rx4 = 0}. Note that (0, x2, 0, 0) E Fix B1 for all x2 E K implies that if there exists a nonzero element y~" such that (0, 0, y~', 0) C Fix B1 then Fix B1 = {(0, x2, ffl, 0 ) ] X2, ffl E K } = P3. Otherwise, it is only possible that Fix B1 = (0, x2, 0, Y2) I x2, Y2 E K} = P4 (since not both a4u and a4 can be equal to 1 for any particular value u).

80 MAURO BILIO'ffI'I AND NORMAN L. JOHNSON

First assume that Fix B1 = P3 so that the B1 is generated by an element

[ °° I ] 1 u - I 1 for some integer j . y = x maps to y = x us implies that

us

j = - 1 as the net .N is left invariant by B1. But, B1 maps

[ob] i la 2bl y = x onto y = x . c d c u - l d J

So, it fixes the component if and only if a = d = 0 and u -z = 1 since b ~ 0, and hence q - 1 = 2 or q = 3.

Thus, it must be that B1 fixes P4. However, this means that B1 is Baer in ~ with Baer subplane in one of the q

regulus nets of ~. Furthermore, there must be exactly q Baer groups of order q - 1 generated by B1 and B2 which normalize E.

So, we have actually shown the following: Each Baer group B of order q - 1 must normalize E and Fix E must lie within

one of the q regulus nets of ~ defined by E (i.e. each regulus net is defined by L E (3 (x = 0), where L is a component of the plane in the net). Since any Baer group must fix any such net which contains a fixed component, we may choose a

[ u O ] for al l representation so that this fixed net by B has the form x = O, y = x o u

u e If . However, we cannot claim that Fix B lies in the net used in the derivation from 7r.

Suppose that two of these Baer subgroups have Baer subplanes in the same regulus net in # which implies that the plane ~ is Hall. But, the Hall planes admit Baer groups of order q and E is an elation group in ~. By Result IX, the order is 2, 3 or4.

Thus, there is a Baer subplane in each of the q regulus nets of ~. If this is the case, then the original plane may be obtained by the derivation of one of these regulus nets which forces one of the Baer subgroups to be a homology group in 7r - a contradiction. This completes the proof of Case II and gives the proof to Lemma (3).

So, except for the orders 16 and 25, we may assume that the nets either share 0 or exactly one parallel class. Our next two lemmas (Lemmas 4 and 5) show that there are doubly transitive groups acting under these assumptions. We consider first that the nets share no parallel classes.

LEMMA (4). I f all o f the nets containing Baer subplanes fixed by Baer groups of order q - 1 are disjoint then there must be a group (B1, B2) containing exactly (q + 1) /2 groups Bi i f* and acting doubly transitive on a set of(q + 1)/2 nets N~ containing Fix Bi.

MAXIMAL BAER GROUPS IN TRANSLATION PLANES 8 1

Proof By Result VIII, if 7to, 711 are the Baer axis and coaxis of B1 and Fix B1 is contained in the net N1 of degree q + 1 and if LI is a component of rr - N1, then {fro, rq} U BIL1 is a regulus in PG(3, q).

Let Ri, i = 1 , 2 , . . . , q denote the q regulus nets defined above in rr - N1. If B2 is contained in N2 and N1 ¢ N2 then we must have IN2 f) Ri] <_

2 (since otherwise, B2 would have to intersect 7to and rrl - a contradiction by assumption).

Hence, as B1 acts regularly on each Ri - {fro, rq}, we have N2g ~ N2 for g B1 such that Igl # 2 or 1 and hence by our assumptions N2g cannot share any component with N2 so [N2g fq N2] = 0. Similarly, N2g and N2g* are disjoint on parallel classes unless possibly 9-1g * has order 2 or 1.

For q even, there are (q - 1) nets N2g for all g C B1 which are mutually disjoint on parallel classes, which implies that there must be at least (q+ 1 )(q - 1 ) = (q2 _ 1 ) components distinct from the q + 1 components of B1 - a contradiction.

For q odd, either we have a contradiction as above or there are exactly (q - 1)/2 mutually disjoint disinct images of N2.

In this case, there are (q2_ 1)/2 components which are also disjoint from N1. By our assumptions, it is not possible for N1 to be partially mapped into this

set of nets without being mapped onto one of these nets. But, then we would have Baer groups B~ and B1 h in the same net of degree q + 1 for g E B1 and h E B2. We are finished unless these Baer groups have the same Baer axis modulo the kernel.

gh - 1 , Hence, we may assume that B E < B~K* so that B1 is contained in B 2 K .

There are (q - 1 ) /2 distinct Bl-images of N2 and (q - 1)/2 distinct B2-images of N1. If any two of these images intersect, they are equal by the above argument and assumption.

Hence, in (B1, B2), there is a set o f t distinct groups MiK* and a set of mutually distinct nets Ni of degree q + 1 containing Fix Mi, where each Mi is a Baer group of order q - 1. Moreover, t _> 1 + (q - 1)/2. Hence, we obtain a net of degree (q + 1)t which leaves a net of degree q2 + 1 - (q + 1 )t which must be permuted semiregularly by Mi for any i = 1, 2 , . . . , t. Hence, (q - 1) must divide q2 _ 1 + 2 - (q + 1)~. Let t = 1 + (q - 1)/2 + s so that (q - 1) must divide

2 - ( q + l ) 1 + + s = 1 - q 2 ( q + l ) s

so that (q - 1) divides (q + 1)s and hence, (q - 1) divides 2s or rather (q - 1)/2 divides s. I f s 7~ 0 then s _> (q - 1)/2 so that t > 1 + (q - 1), but then the indicated net occupies _> q(q + 1) components which cannot be the case. Hence, s = 0 and t = 1 + ( q - 1)/2 = (q + 1)/2.

What this means is that there are (q - 1)/2 distinct images of N2 under B1 where the remaining net under consideration must be N1. This forces Nlg = N2h for some g C B2 and h C B1.

This proves Lemma (4).


We now turn to the situation where two nets N1 and N2 share exactly one component.

LEMMA (5). I f two nets N1, N2 corresponding to Baer groups B1, B2 of order q - 1 share exactly one component, but no two such nets share two components then the group G = ( B1, B2) acts doubly transitive on a set of q nets Ni that mutually share a common component and there exists a set of q subgroups BiK* where Bi is a Baer group of order q - 1 such that Fix Bi is a subplane of the net Ni.

Proof. We consider the Bl-images of Nz. Let the Bj-reguli be denoted by Ri, i = 1 , 2 , . . . , q. We assert that 1N29 f) Ril _< 2 for all i and 9 e B1 since if not then Fix B2 and coFix B2 must intersect both Fix B1 and coFix B1.

We are assuming that we have exactly one common component, so it is impossible for Fix B2 to intersect both Fix B1 and coFix B1 on the common component L.

Now consider the cases:

CASE (i): q is even. Then there must be exactly q - 1 Bl-images of N2. Furthermore, either we are

reduced to a previous situation where there are two distinct nets admitting two Baer groups of order q - 1 and sharing two components, or the q - 1 images of N2 share exactly the common component L of N1 and N2. In this case, the q - 1 images of N1 under B2 must intersect the set of images of Nz under B1 (or be equal to N1) so that these images must be equal if they intersect or, again, we are reduced to a previous case.

Hence, we have exactly (q - 1) + 1 nets Ni, i = 1 , 2 , . . . , q which mutually share exactly one component and we must have a doubly transitive group acting on these q nets.

CASE (ii): q is odd. There are q components of N2 besides the common component, so that the

number of components in the intersection with any Ri is 1 or 2. Since q is odd, there must be some regulus, say R1, which shares exactly one component with N2. It follows, as in the argument for Case (i) above, that there are exactly (q - 1) + 1 nets that are permuted by (B1, B2).

This proves Lemma (5).

LEMMA (6). I f two nets N1, N2 share exactly one component then either the plane is Hall or the order o f the plane is 4, 9, 16 or 25.

Proof. If the plane is not Hall then Lemma (5) applies. We shall use the notation of Lemma (5). In this case, there must be a p-subgroup of order at least q where q = p r .

Consider the group action H of G = (B1 ~ B2) on the q nets. Note that since H is doubly transitive and the stabilizer of N1 contains a normal regular group ------- B1 (i.e. is B1K*) on the remaining nets, then we can apply Result XIII.


Let S denote the socle o f / / . If S is not elementary abelian of order q then the degree is h t + 1 = q for t = 1,2 or 3.

So, q - 1 does not have a p-primitive divisor. Hence, q = 26 or q is odd and is a prime p or p2 where p - 1 or p + 1 = 2 s for some integer s. If q = 2 6 then 2 6 - - 1 ~ h ~. If q = p2 then p - 1 or p + 1 = 2 as p2 _ 1 = h t and hence p - 1 = 2 and p = 3. Thus, q is an odd prime p and q - 1 = h t = 2 rt for some integer r and t = 1,2 or 3. If t = 3 then 2 r + 1 divides 2 ~3 + 1 = q but q is prime and hence, this is a contradiction. If t = 2 then the socle is isomorphic to Sz(2 a) where a is an odd integer > 1. But, 22 + 1 divides 22a + 1 which is a contradiction as 22~ + 1 is prime.

Thus, we can only have the situation that the socle is isomorphic to PSL(2, h) where h + 1 = q = p is a prime. Recall that the Baer groups fix a component and there is a p-subgroup of order at least q induced on this component. Furthermore, PSL(2, h) is a quotient group of a subgroup of the group G generated by the Baer groups. Note that if R fixes the common component pointwise then G/I~ is isomorphic to a subgroup of GL(2, q). Note also that IRI divides q2(q2 _ 1) since any p-subgroup of R is normal of order dividing q2, any homology subgroup (with fixed axis) has order dividing q2 _ 1 and any two homology groups with distinct axes are conjugate by the normal elation p-subgroup. Hence, IGI and thus, for h even, IPSL(2, h)l = h ( h 2 - 1 ) = ( q - 1 ) q ( q - 2 ) d i v i d e s q ( q 2 - 1 ) ( q - 1 ) q Z ( q 2 - 1 ) . Thus, q-2mus td iv ideq2(q2-1) 2. Since (q -2 , q - l ) = 1 a n d ( q + l - 3 , q + l ) = 1 or 3, it follows that q - 2 = 3 or 32. In the former case, q = 5. In the latter case, q = 32 + 2 = 2 z + 1 for some integer z, which is impossible.

Hence, if the order q2 > 25, we may assume that the socle S is elementary abelian of order q.

CASE A. We f irs t assume that the collineation group G admits an elation with axis z = 0 which moves N1.

Represent

Let

( [1 B1 = p ~ = D i a g 0

i 0 bl b2 0 1 b3 b4

e = 0 1 0

0 0 1

which moves Indeed

~z

N1. Since f is 1-1 (see Lemma(1)), we may assume that b2b3 ¢ O.

[u01 Iu+ l y = z 0 f ( u ) maps to y = x b3 f ( u ) + b 4

84 MAURO BILIOTH AND NORMAN L. JOHNSON

So, if b3 = 0 choose u so that f (u) + b4 = 0 or ifb2 = 0, choose u = - b l in order to have a contradiction.

For p~, C B1, then

cP u ~_

bl b2 ] 1 b3 u - 1 b4

0 1 = eu.

0 0

Note that

1 0

0 1 CuC_ v --- 0 0

0 0

0 b2(u- v) b3(u - t -- v - 1 ) 0

1 0

0 1

= a(~_,) .

Then y = x o f(t) maps under a(~_v) onto

[ y = x b3(_1_v_i) f(t)

and the set T of all such components accounts for the q(q - 1) components not in N1 (for example, let u - v = w).

Now, in general, there is a set of components in the plane of the form

[g(t) m(t)] f o r a l l t c K . y = x 1 t

The Bl - images are

y = x [ u _ l

From the set T , we see that re(t) is independent of t or rather re(t) is a constant m.

Since, in one form, we have the component

[ y = x (u - v) -1 t

and, in the other, we have the component

:-l(t) b:(~- ~,)] y = x b3 (u -1 - v - 1 ) t '


we see that if we let b2(u - v) = w then (u - v )m = we (where c = m/b2) and hence we must have:

b3(u -1 - v - 1 ) = ( ( u - v ) ¢ - l ) -1 .

For u ¢ v, this translates into (u - v) 2 = - r u v for some constant r and all u, v nonzero and u ¢ v. Dividing by v 2 and letting x = u/v , we obtain x 2 + (r - 2)x + 1 = 0 for all x C K and x ~ 0 or 1. Hence, IKI - 2 _< 2 so that q_<4.

Hence, we may assume that there are no elations which move N1. CASE B. Let the Sylow p-subgroup of G be denoted by C + and recall that there

is a quotient subgroup of order q which acts transitively on the q nets and fixes the common component (x = 0). We assert that C+[(x = 0) must have order q.

Proof For each of the q nets Ni, there exists an element 7-i c C + which maps N1 onto Ni, vi 7 ~ vj for i 7~ j and we know that 7-i is not an elation. IfT-irf I is an elation with axis (x = 0) then, by the above, this element must fix N1. However, then N17-irf 1 = N1 if and only if Nl'ri = N17-j, contrary to the properties of the elements 7-k.

Since C + l ( x = 0) is an elation group of the Desarguesian plane of order q defined by the 1-spaces of (x = 0), it follows from Hering ([17]) and Ostrom ([36]) that SL(2, q) must be generated on (x = 0) if Fix C+I(x = 0) is not left invariant by (B1, Bz). Since B1 is a cyclic group of order q - 1, B1 normalizes a p-group of order q acting on (x = 0).

Thus, in any case, we may assume that BII(x = 0) normalizes a p-group C + t(x = 0) of order q. Let M denote the full pointwise stabilizer of (x = 0). Then t31 normalizes C + M since B 1 M / M normalizes C + M / M .

By noticing the form of B1 and the fact that C + leaves (x = 0) invariant, we may assume that the elements of C + have the general form

Av By

r ~ = 0 0 1 v

0 0 0 1

where Av and By are 2 x 2 matrices over K.

Taking N1 as y = x , x = 0, then if ~-v does not fix N1, the elements

of the image of N1 cannot have a 0 in the (2, 1)-entry of the matrix representing the rv-image set. Let

a3 a4 ' b3 b4 "

Then we obtain N1 rv:

al a2 b~ b2 + y = x

a3 an b3 b4 0 0

86 MAURO BILIOTTI AND NORMAN L, JOHNSON

Now the (2, 1)-entry is a3(bl + ra(t)) + a4b3. Now re(t) is 1-1 so irregardless of the coefficients, if a3 ~ 0, there exists t E K which makes the (2, 1)-entry 0, which says that 7-~ fixes N1 contrary to assumption. Hence, a3 = 0.

Now 7-~ has order a power of p, so that this forces al = a4 = 1. Take an element [ abi

0 1 b3 b4 P = 0 1

0 0

which does not fix N1. Then the above argument shows that b3 # 0 and conjugate by elements of B1 to obtain elements

D "lz 1 au bl b2u ]

I

000 001 b3 u-lO1 b 1 I "

Form the commutator pz, plp~ l p~ 1 . Since

PiPl =

1 a(u+l ) [ 1 au]O 1

0 0 0 0

bl b2 ] b3 b3 J +

1

0

bl b2u ] b3u -1 b4 J

1

E;;]

p~l =

[i_ou I [bl 1 -au - 0 1 b3 u-1

0 0 1 -cu 0 0 0 1

b2u lb4 [1 - C U o 1

then

p~l/911 =

1 -a(u + 1)

0 1 D

0 0 1 -c(u + 1)

0 0 0 1

where

D = - [1o a u+l ] [b ,1 b3 b462]['o c]1

MAXIMAL BAER GROUPS 1N TRANSLATION PLANES 87

[1-ac][ bl 0 1 b3u -I b4

+ 1) ] ] 1

o

[, , ]w ere Then p~plp~lp-{ 1 is an elation o

T = - a(u + 1) / ) + 1 0 1 b3 b4

[ b l b 2 u 1 + 1 ) ] .

b3 u-1 b4 ] [; ;]) [0 -¢(1 A short calculation shows that

T = [ a b 3 ( u - u -1) -cab3(u + 1 ) ( u - u - I ) + ( b x c - a b 4 ) ( 1 - u ) ]

0 -b3c(u + 1)(1 - u -~) "

Rec 'wehavean twithc°mp°nents x I0 ora. v Since the indicated elation maps y = 0 onto a component with 0 in the (2, 1)-entry of the associated matrix, by choosing v = -ab3(u - u -1) for any given u, we must have that the (1, 2)-entry in the matrix T must be zero for all values of u ¢ 0. Hence, we must have that u2(-cab3(u + 1 )(u - u -z ) + (blc - ab4 )( 1 - u)) is identically zero for all values of u. This implies that the elements of GF(q) satisfy a polynomial of degree 4 or less, and hence 4 _> q.

Hence, if q > 4 we must have all associated coefficients equal to zero and since the coefficient on the polynomial term u 4 is -cab3 and c and b3 are both nonzero, then a must be zero. But, a = 0 implies that -b3c(u + 1)(1 - u -z) = 0 as the commutator represents an elation so that T is nonsingular. Since b3c ~ 0 we must have (u + 1)(1 - u -1) = 0 for all nonzero u in GF(q). Hence, u = 1 o r - 1 and q < 3 .

Thus, we have the proof of Lemma (6). Finally, we still must consider the situation when there are (q + 1) /2 mutu-

ally disjoint nets and a group acting doubly transitively on these nets, as in Lemma (4).

LEMMA (7). I f the nets N~ and N2 do not share a parallel class then q = 4. Proof. If any of the nets share a component we have that the plane has order

< 25 or the plane is Hall by previous lemmas. Hence, we may assume that we have the situation of Lemma (4) and we shall use the notation developed within this lemma.

In this case, note that since we may assume that the 2(q + 1) /2 = (q + 1) Baer subplanes in the nets are permuted, it follows that q - 1 of them are images under the group B1. Hence, the set of (q + 1) Baer subplanes form a regulus in

8 8 MAURO BILIOTTI AND NORMAN L. JOHNSON

PG(3, K) with the Baer groups acting as homology groups of the regulus (see Result VIII(2)).

It is immediate that there must be a 2-dimensional subspace upon which the group (B1, B2) acts faithfully as B1 and B2 must fix each Baer subplane of the regulus net defined by the Baer subplanes fixed pointwise by the various Baer groups of order q - 1. The subspaces left invariant by (B1 ~ B2) are not necessarily Baer subplanes of the translation plane.

Note also, the group acting on the set of q + 1 subplanes must be now isomorphic to GL(2, q) as the group is generated by homology groups (see the proof of Theorem (2.1)). That is, embed the regulus in a Desarguesian affine plane. Any group of the partial spread of the Baer subplanes which fixes each line of the opposite regulus acts as a collineation group of the Desarguesian affine plane E (see Result XI). Hence, since the group is generated by a set of homologies as acting on the net, the group must act faithfully on E and, hence, (B1, B2) -- GL(2, q). But, we know that, in this group, there can be no p-elements which fix Fix B1 because any such p-element must then fix coFix B1 and obviously, then, this is a contradiction.

Hence, we have the proof of the theorem except that we need to analyze the planes of orders q = 16 and 25. We consider these orders in Sections 4 and 5.

4. Some New Partial Hyperbolic Flocks of Deficiency I and Order 25

(4.1) THEOREM. (1) Let 7r denote the irregular nearfield plane of order 25. Then there exists a Baer group of order 4 and hence two Baer groups of order 4 which fix exactly two components in common.

(2) There is a partial flock of a hyperbolic quadric of PG(3, 5) of deficiency 1 which may be obtained from 7r by forming reguli that consist of nontrivial component orbits union Fix B and coFix B, where B is any Baer group of order 4.

This particular partial spread is maximal and unembeddable and the partial flock is nonextendable.

Proof. By Foulser ([12, (17.4)]), there is a nearfield automorphism of order 4. This automorphism translates into a Baer group of order 4 since the automorphism must leave GF(5) pointwise fixed. Now apply Result VIII(2) to obtain a partial flock of a hyperbolic quadric of deficiency 1. Now if the partial flock is not maximal then the net containing the Baer axis must be derivable and hence, since q --- 5, the net must be a regulus net. But this forces the irregular nearfield plane to have a central kernel. Hence, we obtain a maximal and unembeddable partial spread and hence a nonextendable partial hyperbolic flock of deficiency 1.


(4.2) THEOREM. (1) Let K ~ GF(5). The following represents a spread of PG(3, 5):

vg] [u 0] y = x i v _ 1 t ( typel) , y = x 0 u -1 (type 2),

0 - s ] y = x s - 1 0 (type3), x = 0 , y = 0

for all nonzero t, v, u, s E K where g is 2 or 3.

Denote the two possible translation planes by 7r2, 7r 3 . Furthermore, the spread admits the following Baer collineation groups of

order 4:

BI={ u 1 1 lu E K* B2 = ' 1

U

l u C K *

u i ° The plane also admits the collineations

7" = 1 1 0 and a = . 0 0

1 1

The translation plane also admits the homology groups of order 8: {[:o U - 1

H~ = 1

1

0

1 - 1

0

1 1 lu~K*} and H v = H~.

(2) 71" 2 is not isomorphic to 7r3 and ~r3 is the irregular nearfield plane. (3) There exist two partial hyperbolic flocks of deficiency 1 in PG(3, 5) corre-

sponding to the two translation planes :r2, 7r3, neither of which is extendable. Proof. We first verify that we have a spread. To show that we obtain a spread, we must check that the determinants and their

differences of each type are nonzero. Note that under the group (B1, B2) the indicated subspaces of type i are orbits

for i = 1,2,3.


The determinant of type i: i = 1, det = 2g, i = 2 or 3, det -- 1. The determinant of differences (i, j ) :

must have nonzero determinant for all nonzero t, v not both equal to 1. The determinant is (4 - (t + t -1) - (v + v-1))g . Multiply by t r i g .

The resultant vzt + v( t 2 -{- t + 1) + t • 0 if and only if ( . ) (t 2 + t + 1) 2 - 4t 2 is a nonsquare for each t (other than possibly t = 0 or t = 1):

A direct check:

t = 2 ,

t = 3 ,

t = 4 ,

(*) = ( 4 + 2 + 1) 2 - 1 6 = 3

( , ) = 3

( . ) = 2.

Now let t -- 1, then v 2 + 3v + 1 = (v - 1) 2 = 0 if and only if v = 1, which is not allowed by hypothesis.

{I °l } (2 ,2) det 0 u -1 - I z : ( u - 1 ) ( u - l - 1 ) # O f o r a l l u # 1.

{I [ 1]) (3 ,3) det - = ( s - 1 ) ( s - l - 1 ) ¢ 0 f o r a l l s ¢ l . s -1 1

{[u o] io 11) (2, 3) det 0 u -1 - 1 0 = 2 .

(1 ,2) 0 u -1 - 9 ] has determinant

( u - g ) ( u -1 - 1) + g = ( * * ) 2 g + 1 - (9 u-1 q- U).

A direct check:

u = 1, (**) = g

u = 2 , (**) = - ( g + 1) ¢ 0 s ince9 = 2 or 3

u = 3, ( * * ) = 3

u = 4 , ( * * ) = 3 9 + 2 ¢ 0 s i n c e g = 2 o r 3 .

M A X I M A L B A E R G R O U P S IN T R A N S L A T I O N PLANES 9 1

,13, [gm g]:l 1 g ]has eteoin t, 2g + 1 - (gs -1 + s) as in (1,2), so nonzero for g = 2 or 3.

This proves that we have a spread. It is direct to verify that the above mappings provide collineations of the trans-

lation plane. Under the group (B1, B2, 7-, or, Hz), the orbits lengths are 2, 8, 16. Now we verify that 7r2 and 7r3 are not isomorphic. Let p be an isomorphism from 7r2 onto 7r3. Let F denote the full collineation

group. Now note that

[2 3]2 [2 ,] [,_12 3v] 1 1 = 3 - 1 v -1 t

if and only if t --- plane.

Note that

- 1 and - 2 = 2, a contradiction. Thus, 7r2 is not a nearfield

[9,1 V - 1 t 1 = 3v -1 + t 2v -1 + t "

Now (2v -1 + t) -1 • 3 = 4t -1 - 3v if and only i f (2v -1 + t)(4t -1 - 3v) = 3 if and only if 3(vt) -1 + 4 - 6 - 3(vt) = 3 if and only if (vt) -1 = (vt), provided 3v -1 + t ~ 0 and 2v -1 + t ~ 0.

if t = 2v_l then the above product is [ -V 0 ] 0 - - v -1 "

if t = 3v_l, the above product is [ 0 - v ] v -1 0 . Hence, the matrices of type (1)

permit multiplication by [ ~ - ~ ] .

Note that the order of [ 31 -31 ] is 3. Thus, there is a homology group of order k A

k with axis y = 0 where 3 and 8 divide k. Hence, 24 divides the order of this homology group and thus the plane 7r3 is a nearfield plane.

Note that the plane 7r3 must be the irregular nearfield plane since it contains the matrices

[o_1][1::] r,-,3 2v] 1 0 ' - 1 = [ v - 1 t

for t = - 2 , v = - 1 (see [15, p. 391]).

(4.3) THEOREM. I f a translation plane of order 25 and kernel GF(5) admits two


Baer groups B1 and B2 of order 4 such that the fixed point subplanes share exactly two axes L, M and the group B2 is generated by subgroups which fix Fix B1 fq L and coFix B1 fq M then the plane is isomorphic to one of the two planes of (4.2).

Proof (Sketch). Use Section 3, and the proof to Lemma (3), to show that the spread may be

represented as follows:

[o o] x = O, y = O, y : 3~ U- 1 ,

y - x 8 - 1 ~ y : X V - 1

for constants g and f and for all u, v E K*. Then form various differences to show that we must have h = - 1, f = -g - Then show that a spread for g = 1 or - 1 is impossible.

(4.4) THEOREM. If 7r is a translation plane of order 25 which admits two Baer groups of order 4 then 7r is either Hall or one of the planes of(4.1).

Proof If the Baer nets share >_ 2 components then we may use the proof of Lemma (3) of Section 3 to show that only the Hall planes occur in the situation stated in Lemma (4). Now use the lemmas of Section 3 and (4.3) to see that only the planes of (4.2) can occur.

Thus, if two nets corresponding to Baer groups of order 4 share >_ 2 components, the planes are completely determined. Hence, assume some pair of nets shares exactly one component.

Considering the proof to Lemma (6) of Section 3, we see that either the socle S is isomorphic to PSL(2, 4) - A5 or S is elementary abelian of order 5. However, the proof to Lemma (6) of Section 3 when S is elementary abelian also applies in the case q - 5 to provide a contradiction.

So, we are left with the case that As is induced on the common component. But, since q = 5 is a prime and the group is transitive on the five nets, it follows that a Sylow 5-subgroup of the group generated by the Baer groups is also transitive on the nets. If the proof of Lemma (6) of Section 3 is reread, it may be seen that the above property is all that is used in Case (b) of Lemma (6). Hence, this situation does not occur.

Finally, we assume that each pair of nets corresponding to Baer groups do not share any parallel classes. However, for q = 5, Lemma (7) applies to give a contradiction.

Now to complete the classification of translation planes of order q2 and kernel containing GF(q) which contain at least two Baer groups of order q - 1, we need to show that the above translation planes of order 25 are the irregular nearfield plane and the Walker exceptional plane of order 25 with orbit structure 10, 16.


Each of the planes of (4.2) admits a group of order 64 when restricted to the line at infinity (the group generated by the homologies has order 32 and there is an element of order 2 which interchanges centers with cocenters of the two homology groups) and under this group there are orbits of lengths 2, 8, 16.

The translation planes of order 25 are completely determined via computer in Cerwinski and Oakden [4]. Furthermore, the orbit lengths of the collineation groups on the line at infinity are determined as well (see also Cerwinski [5] and Charnes [6]).

By Table 2 of Charnes [6, p. 54], there are but two planes which admits subgroups on the line at infinity of order 64 and orbits under these subgroups of lengths 2, 8, 16. These are the planes numbered 1 and 11 in Table 1 and have groups and orbits as follows:

Group order 1920 and orbits 10, 16. Group order 1152 and orbits 2, 24.

By remarks in Cerwinski [5] or Charnes [6], it is known that the plane with orbit 10, 16 is an exceptional Walker plane.

Note that the fact that the irregular nearfield and the exceptional Walker plane admit groups of this type was not previously observed.

5. Order 16

There are exactly three translation planes of order 16 which have kemel GF(4), the Desarguesian and Hall planes and a semi field plane. It has been shown in Johnson [30] that the semifield plane does admit Baer groups of order 3. Also, see Lemma (7) when q = 4. Thus, the translation planes of order 16 and kernel GF(4) which admit Baer groups of order 3 are exactly the Hall and semifield planes.

6. The Compatibility of Baer Groups of Order q - 1 and Affine Homologies

Suppose there exists a translation plane of order q2 and kernel containing t ( ~- GF(q) which admits a Baer group B of order q - 1 and a homology group/ / . Then either H fixes Fix B and thus centralizes B or there exists h e H such that h does not leave Fix B invariant. In this case, either we may apply the results of Section 1 or h 2 must leave Fix B invariant, ff h 2 ~ 1 then the center and axis o f / / a r e in the subplane Fix B and, again, Section 1 applies or a c t u a l l y / / m u s t leave invariant the net of degree q + 1 containing Fix B. But it now follows t h a t / / m u s t fix each Baer subplane of this net so tha t / / l eaves Fix B invariant.

If h 2 = 1 and h interchanges Fix B and coFix B then h maps the infinite points of Fix B onto the infinite points of coFix B. If the axis and center of h are within the net then h must fix Fix B. Assume that either the axis or center of h are not in the net defined by Fix B. In this case, it follows that the axis and coaxis are both disjoint from the net in question.


Hence, we obtain the following theorem:

(6.1) THEOREM. Let 7r be a translation plane of order qa and kernel containing K ~ GF(q). Assume that 7r admits a Baer group B of order q - 1.

I f q > 5 then either

(i) 7r is Hall (ii) any homology group must either leave Fix B invariant (and so have axis and

coaxis within the net defined by Fix B) or (iii) q is odd and the homology group must have order 2 and interchange Fix B

and coFix B. In this case, the axis and coaxis o f the homology group are dis joint from the net defined by the Baer group.

(6.2) COROLLARY. Let 7r be a translation plane of order q2 and kernel containing K ---- GF(q) where q > 5. Then 7r cannotadmitbothaBaergroup B oforderq - 1 and an affine homology group of order q - 1.

Proof. By (6.1), the homology group must leave the Baer axis and coaxis invariant. So the center and axis of the homology group are in the projective extension of Fix B. We may represent the Baer group B by

[loo ] o o r u = 0 0 1 u E K* .

0 0 0

It then follows that the homology group H acts faithfully on the Desarguesian affine plane Fix B and hence must be cyclic of order q - 1 as H I Fix B _< GL(2, q). It also follows that the net containing the Baer subplane must be of the form

x = O, y = x 0 f ( t ) for a l l t E K

and some function f on K. But, it also follows that since H is cyclic, we may represent H in the form

0oo 0 1 0 0

P~= 0 0 u 0 u E K * ,

O 0 0 u j

where j is some integer relative prime to q - 1 by appropriate choice of a basis.

[1 b] . Applying B, we obtain the compo- There exists a component y -- x 0

nents:

y = x for all nonzero u E K. u -1 0


Applying H, we obtain the components:

au bu j ] y = x

u 0

Hence, clearly a = 0 (so b ~ 0) and j = - 1. Multiplying the group elements T~, and p~, we obtain the Baer group

1

/ ° 0

0

0 0 0

u 0 0

0 u 0

0 0 1

u E GF(q)*>.

We may now apply Sections 3, 4 and 5 as we now have two Baer groups of order q - 1.

7. Partial Hyperbolic Flocks

From Result VIII(2), every translation plane of order q2 and kemel containing K ~ GF(q) that admits a Baer group of order q - 1 corresponds to a partial hyperbolic flock in PG(3, q) with deficiency 1. We define the automorphism group of a partial flock to be the subgroup of f~+ (4, q) which permutes the conics of the partial flock.

(7.1) THEOREM. If 7r is not Hall or Desarguesian (of order 9), the semifield plane of order 16 and kernel GF(4), the irregular nearfield plane or the Walker exceptional plane with orbit 10, 16 of order 25, then the full collineation group of a partial flock of deficiency 1 is isomorphic to the group induced by the coUineation group of the corresponding translation plane which admits a Baer group of order q - 1 .

Proof Any automorphism group of the partial flock permutes the conics. The points of the conics correspond to lines of the reguli that are opposite to those defined by the images of a component by the Baer group of order q - 1, together with the fixed point subplane and the cofixed point subplane.

Any automorphism of the partial flock must leave invariant the set of common lines of the reguli (see Johnson [30]). Hence, the automorphism induces a collineation of the partial spread defined by the q2 _ 1 components not in the net containing the two Baer subplanes (common lines of the reguli). However, by Bruck [1], there is a unique extension of this partial spread to an affine plane since there is an extension. Hence, it follows that the induced collineation of the partial spread induces a collineation of the translation plane.

Now let g be a collineation of the translation plane corresponding to the partial flock of deficiency 1. By the results of Section 3, either the plane is Hall or Desarguesian of order 9 or semifield of order 16 or a plane of order 25, as indicated,


or the full collineation group of the plane fixes the set which contains the axis or coaxis subplanes of the Baer group.

The stabilizer of the axis and coaxis induces a collineation group of the partial spread. It is at least conceivable that there may be a collineation which interchanges the axis and coaxis subplanes. Such a collineation still would induce a collineation in the partial spread because the line in common of the conics obtained from the polar planes would be fixed and the common points interchanged.

8. Transitive Partial Hyperbolic Flocks

(8.1) DEFINITION. Let F be a partial hyperbolic flock in PG(3, q). The automorphism group of F shall be the subgroup of PrL(4, q) which leaves the hyperbolic quadric invariant and which permutes the conics of F. We shall say that F is transitive if and only if there is a collineation group in PGL(4, q) which fixes the hyperbolic quadric and acts transitively on the conics of F. (One could, of course, study the situation where the transitive group is not necessarily assumed to be within PGL(4, q)).

Previously known are exactly two partial hyperbolic flocks of deficiency 1 which may not be extended to flocks. These have orders 4 and 9 and are described in Johnson [30] and Johnson and Pomareda [32] and both are transitive partial flocks. In this section, we describe the set of transitive partial flocks of deficiency 1 in terms of the spread set.

(8.2) THEOREM. Let F be a partial flock of deficiency 1 in PG(3, q) and let :r F denote the corresponding translation plane of order q2 which admits a Baer group B of order q - 1. Let N denote the net of degree q + 1 which contains Fix B. Let G F denote the full collineation group of the partial flock and let G~ denote the full collineation group of the translation plane. Denote the kernel homology group of order q - 1 by K*.

Assume that ~r F is not Hall and that q ~ 2, 3, 4 or 5.

(i) Then G~ leaves {Fix B, coFix B} invariant and normalizes BK*. (ii) G ~ / B K * -~ GF.

Proof. By Sections 2 and 3, if ~rF admits two Baer groups of order q - 1 where the second group B is not within BK* then either the plane is Hall or q - 2, 3, 4 or 5. Since K* is normal and B g must be within BK*, we have the proof of (i).

Using the Klein quadric construction of the partial flock from the translation plane (see [27]), it follows that a collineation of the translation plane which leaves the net N invariant must induce a collineation on the partial flock F. Since the full collineation group of the plane must leave N invariant, we see that the full group of F is induced from the collineation group of the plane. Let M denote the kernel of this representation. Since, looking at the Klein quadric construction, the points of the conics of the partial flock correspond to the set of q(q + 1) Baer subplanes


incident with the zero vector of the regulus nets defined by the component images under B (and the subspaces Fix B and coFix B), it follows that the kernel M is the collineation group of the plane which fixes each such Baer subplane. Clearly, BK* < M.

Choose

1 u

1

u

uE K*} (note the two uses of K*).

So, we choose

Fix B = {(Xl, O, y], O) I Xl, Yl E K}

and

coFix B = {(O, xz, O, y2) lx2, Y2 E K}.

Let cr E M. Since Fix B and coFix B are fixed, then

al 0 bl 0

0 a4 0 b4

el 0 dl 0

0 C4 0 d4

Note that there are q(q + 1) Baer subplanes of these nets, each of which intersects both Fix B and coFix B in 1-dimensional K-spaces. If we let the net N be denoted by

x ~ O ~ o]

Y = 0 for all u ~ K and rn some function on K,

then given any 1-space of Fix B, there are exactly q 1-spaces of coFix B such that the 2-space generated by two of these 1-spaces is not one of the 2-spaces of N. Since this allows for at most (q + 1 ) q 2-dimensional subspaces, it follows that these must be the q(q + 1) Baer subplanes which correspond to the partial hyperbolic flock.

For example, given the I-space ((0, 0, 1,0)) of Fix B and a 1-space of coFix B other than ((0, 0, 0, 1)), the generated 2-space must be left invariant by or.

Letting the 1-space be ((0, 1,0, 0)) forces Cl = 0 since the image of (0, 0, 0, 1)

[c, ° ] i s e i t h e r under cr is (ci, 0, all, 0) E {(0, a,/3, 0) l a,/3 E 1~ }. But, clearly 0 c4


zero ornonsingular, since the image of x = 0 is y o c4 , y o d4 "

C 4 = 0 .

Similarly, (( 1,0, 0, 0)) + ((0, 0, 0, 1)) must be fixed by cr so that bl = 0 and it follows that b4 = 0 (by taking the cr-image of y -- 0).

Taking (( 1,0, 0, 0)) + ((0, 1,0, 1)) forces a4 - - d 4 and since we have K* within M, we must take a4 = 1, without loss of generality.

Then crT-,~ where u = d~ -1 forces al = dl for otherwise crT-u fixes at least q3 distinct points. Hence, M <_ BK*.

Now assume that g is an automorphism of the partial flock F. Then, using the Klein quadric construction, there is an induced collineation of the partial spread of degree q2_ q+ 2 which corresponds to the components of the associated plane which are not in N union Fix B and coFix B. Moreover, the induced collineation must leave {Fix B, coFix B} invariant as these subplanes correspond to the common points of the set of conics obtained by taking the intersections with the Klein quadric of the polar planes of the planes containing the conics of the flock.

By Result VIII(2), there is a unique translation plane E which contains the components of partial spread as components with the exception of Fix B and coFix B and contains these 2-spaces as Baer subplanes. Clearly, if 9 + is the collineation of the partial spread above induced from g, then E9 + is a translation plane which contains the components of the partial spreads -{F ix B, coFix B} as components and contains Fix B and coFix B as Baer subplanes. Thus, E = E9 +. That is, every collineation of the partial flock induces a collineation of the associated translation plane. This proves (ii).

(8.3) THEOREM. Let F be a transitive hyperbolic partial flock of deficiency 1 in PG(3, q) which cannot be extended to a linear flock.

(i) I f q is odd then the corresponding translation plane 7r F of order q2 admits an elation group E of order q which fixes the net N containing the Baer subplane fixed pointwise by a group of order q - 1.

(ii) I f q is even then 7rF admits an elation group of order q/2 and either admits an elation group of order q or admits an element whose square is an elation and which fixes a component L and interchanges Fix B f) L and coFix B fq L.

Proof. From (8.1), there is an induced collineation group fixing N and acting transitively on the q reguli corresponding to the q orbits of the Baer group B. Let E be a Sylow p-subgroup of the induced group where q = pr. Hence, q IIE[ • Since E is a p-group acting on N and the degree of N is q + 1, E must fix a component L of N and a 1-dimensional subspace on L pointwise. Since FixB fq L must be fixed or interchanged with coFix B N L, it follows that E is an elation group with axis L or q is even and there is an element p which performs the interchange and there is an elation subgroup of order q/2 where p2 is an elation with axis L. This proves (i) and (ii).


(8.4) COROLLARY. Let F be a transitive partial hyperbolic flock o f deficiency 1 in PG(3, q). Let 7r V denote the corresponding translation plane.

I f q > 5 then either F may be extended to a linear flock or F is not extendable. Proof If F is extendable and q > 5 then either the translation plane 7rE is Hall

and hence F extends to a linear flock or there is an elation group of order q/(2, q) which becomes a Baer group in the derived plane 7r*. So 7r* is either Desarguesian (and is, if q is even) or a nearfield plane of odd order. In this case, there is a Baer group of order q and, by Foulser ([11 ]), the net of degree q + 1 containing the Baer axis must be left invariant or q = 3. Since this cannot occur in a nearfield plane, we have the proof to the corollary.

(8.5) THEOREM. Let F be a transitive partial hyperbolic flock in PG(3, q) for q odd where q > 5 (or for q even where there is an elation group o f order q) and assume that the partial flock may not be extended to a linear flock. Then the corresponding translation plane 7r F may be represented in the following form:

[ u + a bv ] x = O , y = x v - 1 m(v) '

y = x 0 re(t) for al lu , t, v E K andvnonzero,

where a and b are constants in K and m is an additive function on K. Proof. By (8.3), there is an elation group E of order q which must fix N. If we

represent the axis of the group by x = 0, it follows that N is

x = 0, y = x 0 re(u) for all u E K and m some function on K.

The elation group must be { ouo } 0 1 0 m(u) u C K 0 0 1 0

0 0 0 1

is tive ow et o ima esofacom onent = • to

obtain the representation.

Acknowledgement

This work was initiated during the spring of 1990 when the second author was visiting the University of Lecce. The authors gratefully acknowledge the support of the University of Lecce and the C.N.R.

] 00 MAURO BILIOTTI AND NORMAN L. JOHNSON

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maximal baer groups in translation planes and compatibility with homology groups

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