a mathematical toolkit for group theorists · of rank 2. hyperbolic groups, groups acting properly...

A Mathematical Toolkit for Group Theorists

Aditi Kar

March 14, 2019

2

These are Lecture Notes from a weekly class that I am offering for doc-toral students of Mathematics in the University of Southampton.

Chapter 1

Ping Pong Lemma and FreeProducts

1.1 Historical outline

Broadly speaking, the Ping Pong Lemma in group theory is a test for findingfree products inside a given group. The initial version of the Lemma isattributed to Felix Klein who used it to study subgroups of Kleinian groupsi.e. discrete groups of isometries of hyperbolic 3-space.

In 1972, Jacques Tits used the Ping pong Lemma to prove that, everysubgroup of a finitely generated linear group is either virtually solvable orcontains a non-abelian free group. This Theorem is now known as the TitsAlternative for linear groups and in contemporary group theory, has beengeneralised as follows: a group G is said to satisfy the Tits Alternative ifevery subgroup is either virtually solvable or contains F2, the free groupof rank 2. Hyperbolic groups, groups acting properly and co-compactly onCAT(0) cube complexes and subgroups of mapping class groups are a fewexamples of groups which satisfy the Tits Alternative.

Further applications of the Ping Pong Lemma include establishing non-amenability, exponential growth or Property Pnaive. Indeed, an amenablegroup does not contain F2 and to check for exponential growth, one simplyhas to find a free subgroup of rank 2 inside the group.

Property Pnaive is a rather strong version of the Tits Alternative intro-duced by Pierre de la Harpe in his study of group C∗-algebras. A group issaid to have property Pnaive if for any finite subset F ⊂ G\{1} there existsy ∈ G such that for each x ∈ F , the subgroup generated in G by x and y isthe free product 〈x〉 ∗ 〈y〉. Non-elementary hyperbolic groups and CAT(0)

3

4 CHAPTER 1. PING PONG LEMMA AND FREE PRODUCTS

cubical groups which do not have a normal amenable subgroup are knownto have property Pnaive. It is not known if SL(n,Z), n ≥ 3, has Pnaive ornot.

Lemma 1 (Ping Pong Lemma). Let G be a group acting on a set X andlet A and B be subgroups of G with max{|A|, |B|} ≥ 3. Set H to be thesubgroup generated by A and B in G. Suppose there exist non-empty subsetsXi, i = 1, 2 of X such that

• X2 is not contained in X1,

• aX2 ⊂ X1 for all a ∈ A\{1}, and

• bX1 ⊂ X2 for all b ∈ B\{1}.

Then, H is isomorphic to the free product A ∗B.

This is one version of the Ping pong Lemma. Different statements existin the literature.

1.2 Free products

A monoid is is a set endowed with a binary operation (we will call thismultiplication) which is associative and has an indentity element. Given aset A, let W (A) be the collection of all finite sequences of elements from A.Take concatenation to be the multiplication on W (A) so that the identityelement is the empty word. This makes W (A) into the free monoid on thealphabet A. A typical element in W (A) has the form a1a2 . . . ak, whereeach ai ∈ A; we say k is the length of the sequence and identify A with thesequences of unit length.

Consider now a family of groups {Gi}i∈I and choose A to be the disjointunion of the Gi’s. We obtain a group from the free monoid on A by definingan equivalence relation on W (A) as follows: for any w,w′ ∈W (A),

• wew′ ≈ ww′, if e is an identity element from some Gi

• wabw′ ≈ wcw′, if a, b, c ∈ Gi for some i and ab = c.

Let W := W (A)/ ≈. Then, W is a group in which the inverse of a1 . . . ak isa−1k . . . a−1

1 . W is the free product of the Gi, written ∗i∈IGi.

Example 2. If Gi is infinite cyclic for each i ∈ I, then ∗i∈IGi is simply thefree group with basis I.

1.3. PROOF OF PING PING LEMMA 5

Recall that the Gi’s are subgroups of the free monoid W (A). The ques-tion then arises whether these group survive as subgroups once we form thequotient W of W (A) by the equivalence relation ≈.

Definition 3. Let w = a1 . . . ak be an element of W (A) such that k ≥ 1 andai ∈ Gni. Then, w is said to be reduced if ni 6= ni+1 for all i = 1, . . . k − 1.

Proposition 4. Every element of ∗i∈IGi is equivalent to a unique reducedword in W (A).

Proof. Take w = a1 . . . ak to be a reduced word as in Definition 3. Leta ∈ Gj for some j. Define R(aw) as below.

R(aw) = w, if a = 1

(aa1)a2 . . . ak, if n1 = j and a 6= a−11

a2 . . . ak, if n1 = j and a = a−11

aa1a2 . . . ak, if n1 6= j.Evidently, this is a reduced word representing the product aw. One can nowargue by induction that every equivalence class is represented by a reducedword. To establish uniqueness, we use a classic argument due to van derWaerden.

Let X be the set of all reduced words in W (A). We now define a mapfrom ∗i∈IGi to SX , the symmetric group on X. For each a ∈ A, let σ(a) :X → X be the map that takes w to R(aw). Extend σ to all of ∗i∈IGi bysetting σ(a1 . . . ak to be the composition σ(a1) . . . σ(ak). To complete theargument, observe now that for any w ∈ X, σ(w) ∈ SX and σ(w)(1) = w.Therefore, for a pair w,w′ of reduced words, σ(w) = σ(w′) if and only ifw = w′.

Corollary 5. Each Gi is naturally a subgroup of ∗i∈IGi. Write γi for thecanonical injection.

Corollary 6. (Universal Property) Let H be a group and let fi : Gi → Hbe a family of homomorphisms. Then there exists a unique extension F :∗i∈IGi → H such that F ◦ γi = fi for all i ∈ I.

1.3 Proof of Ping Ping Lemma

Ping Pong Lemma. Let G be a group acting on a set X and let A and B besubgroups of G with max{|A|, |B|} ≥ 3. Set H to be the subgroup generatedby A and B in G. Suppose there exist non-empty subsets Xi, i = 1, 2 of Xsuch that


• X2 is not contained in X1,

• aX2 ⊂ X1 for all a ∈ A\{1}, and

• bX1 ⊂ X2 for all b ∈ B\{1}.

Then, H is isomorphic to the free product A ∗B.

Proof. Let w 6= 1 be a reduced word in A and B. Suppose that w =a1b1 . . . akbka, where a, a1, . . . , ak belong to A\{1} and the bi’s belong toB\{1}. Then, by hypothesis, wa(X2) ⊆ (X1) as X2 is not contained in X1,we have w 6= 1.

If w is of the form a1b1 . . . akbk, then choose a ∈ A which is different froma−1

1 . By the earlier argument, a−1wa is non-trivial and therefore, w 6= 1.Finally note that any word of the form b1a1 . . . bkak is also non-trivial as itsinverse is non-trivial.

Proposition 7. If |z| ≥ 2, then M1 =

(1 z0 1

)and M2 =

(1 0z 1

)generate a free subgroup of rank 2 in SL(2,C).

Proof. This proof is attributed to Sanov in Lyndon and Schupp.

Set J =

(0 11 0

)and observe that M2 = JM1J . This means any

element in the subgroup generated by M1 and M2 may be written in theform Mn1

1 JMn2J . . .. So it suffices to show that the subgroup generated byM1 and J is the free product 〈M1〉 ∗ 〈J〉. Consider the action of SL(2,C) onthe Riemann sphere C = C ∪ {∞} by fractional linear transformations, i.e.(

a bc d

).z =

az + b

cz + d.

Choose X1 = {z ∈ C : |z| < 1} and X2 = {z ∈ C : |z| > 1}. Then,Mn

1 X1 ⊂ X2 and JX2 ⊂ X1. Invoke the Ping Pong Lemma to deduce theProposition.

1.4 Exercises

1. Let W := W (∪i∈IGi)/ ≈ as in section 1.2. Verify that W is a group.

2. Let A,B be groups. Let φ denote the canonical map from A ∗ B tothe direct product A×B. For instance, φ takes the reduced word aba′b′ toaa′bb′. Let K := kerφ.

1.4. EXERCISES 7

(a) Prove that K is free with basis {[a, b] = a−1b−1ab : a ∈ A\{1}, b ∈B\{1}}.(b) IfA = Z/2Z andB = Z/3Z, thenA∗B is the modular group PSL(2,Z) =SL(2,Z)/ ± 1. Observe that in this case, K is the commutator subgroupof A ∗ B. What is the rank of K? This is another proof of the fact thatPSL(2,Z) and therefore SL(2,Z) has a free subgroup of finite index.

3. A group G is said to be residually finite if for every non-identity elementg ∈ G, there exists a surjection f : G → F to a finite group F such thatf(g) 6= 1. In this exercise, we will verify that if A and B are residually finitethen A ∗B is residually finite.(i) First, show that free groups are residually finite. A famous theorem ofMalcev says, every finitely generated linear group is residually finite. But,for free groups, we can prove this by elementary means. Recall that thefree group F2 of rank 2 contains free subgroups of all ranks and we know F2

embeds in SL(2,Z). It suffices therefore to show that SL(2,Z) is residuallyfinite. Verify this, using the standard surjections SL(2,Z)→ SL(2,Z/2nZ).

(ii) Using (2), show that the result holds if A and B are finite. Note here,that the kernel K of the map from A ∗ B to A × B is a free subgroup offinite index in A ∗B. Therefore, K is finitely generated. This helps us findfinite index characteristic subgroups in K as follows:

• there are only finitely many maps from a finitely generated group Gto a finite group;

• every subgroup H of index n in G gives rise to a homomorphism φHfrom G onto the symmetric group Sn on G/H;

• set L = ∩{H<G : [G:H]=n} ker(φH) and show that L is a characteristicsubgroup of finite index in G.

(iii) Show, that the general case follows from (ii).

4. (Harpe II.B.32) In the groupG = PSL(2,R), an element g =

(a bc d

)6=

1 is parabolic if it has a unique fixed point in R = R ∪ {∞} ⇔ the tracetr(g) of g is ±2⇔ g is conjugate to a translation in R.

The element g is said to be hyperbolic if it has two distinct fixed points inR⇔ |tr(g)| ≥ 2⇔ g is conjugate to a dilation x 7→ λx of R.

The element g is elliptic if it has a fixed point in the upper half plane ⇔|tr(g)| < 2 ⇔ g is conjugate to a rotation.


Let Γ be the subgroup of G generated by two elements γ1, γ2. Verifythat Γ contains a free group of rank 2 in the following cases.(i) Both γ1 and γ2 are parabolic with distinct fixed points in R.(ii) While γ1 is parabolic with fixed point x ∈ R, γ2(x) 6= x.(iii) Both γ1, γ2 are hyperbolic and FixR(γ1) ∩ FixR(γ1) = ∅.

5. Show that SL(2,Z) is generated by the elementary matrices i.e. the

matrices

(1 10 1

)and

(1 01 1

)generate SL(2,Z). This involves applying

the Euclidean algorithm to the entries of a matrix.

Further, show that Γ(2) = 〈(

1 20 1

),

(1 02 1

)〉 is a free subgroup of

finite index in SL(2,Z). This involves a variation on the argument showingthat Γ(1) ∼= SL(2,Z) and is a tedious undergraduate exercise!

Chapter 2

Groups acting on trees

Free products, Amalgams and HNN extensions all fall in the class of groupsacting on trees. The theory of groups acting on trees is popularly called BassSerre Theory in honour of its discoverers, Jean Pierre Serre and Hyman Bass.Serre gave a series of lectures about his ideas on the subject in College deFrance in Paris in 1968. Famously, Hyman Bass was in the audience andwrote up lecture notes. These appeared in an Asterique volume and werelater translated to English by Stillwell.

In what follows, we will first briefly review free groups, these being thefirst examples of groups acting on trees. Then, we will describe amalgamsand HNN extensions, leading upto the Fundamental Theorem of Bass Serretheory.

Trees. A tree is a connected graph without circuits. One can show thatany two points in a tree are joined by a unique geodesic (path withoutbacktracking) and moreover, a tree is contractible.

N.B. The action of the group on the tree is always assumed to be simpliciali.e. no group element takes an edge to its inverse so that the quotient of thetree by the action is always a simplicial graph.

2.1 Free groups: a review

The starting point of the theory of groups acting on trees is the case whenthe action is free: the existence of a free action on a tree characterises freegroups. To prove this, we need the notion of a Cayley graph.

Definition 8. Let G be a group and let S be a subset of G. Define the Cayleygraph Γ := Γ(G,S) of G with respect to S to be the oriented graph having G

9

10 CHAPTER 2. GROUPS ACTING ON TREES

as its vertex set and G× S = edge(Γ)+, the set of positively oriented edges.Here the origin o(g, s) of (g, s) is g and the terminal vertex t(g, s) is gs.

Clearly, the action of the group by left multiplication on itself induces afree action of the group on the vertex and edge sets of Γ. The graph Γ(G,S)is connected if and only if S generates G.

Proposition 9. Let Γ be the Cayley graph of G with respect to the set S.Then, Γ is connected ⇔ the set S generates G.

Proof of Proposition 9.

Proposition 10. A group is free if and only if it acts freely on a tree.

Proof of Proposition 10.

2.1.1 Applications

We can now use this theory to retrieve some basic results about free groups.

Theorem 11. Every subgroup of a free group is free.

This is the famous Nielsen-Schreier Theorem. Let G = F (S) be the freegroup with basis S and let H < G. The Cayley graph Γ of G with respectto S is free and H acts freely on Γ. Hence, by Proposition 10, H is also free.

Theorem 12 (Schreier Index Formula). Let H be a subgroup of finite indexin a (finitely generated) free group G = F (S). Then,

rank(H)− 1 = [G : H](rank(G)− 1).

Explicit form of Schreier’s Theorem: Let G = F (S) be a free groupwith basis S. Let H < G. Then, one one can choose a set T of representa-tives of G/H satisfying the following conditions:

• If t ∈ T has the reduced form t = sε11 sε22 . . . sεnn , where si ∈ S, εi = ±1

and εi = εi+1 whenever si = si+1, then all the partial products 1, sε11 ,sε11 s

ε22 , . . . , s

ε11 s

ε22 . . . sεnn belong to T .

• Let T be as above and define W = {(t, s) ∈ T × S : ts /∈ T}; forevery (t, s) ∈ W set hs,t = tsu−1, where u ∈ T and Hts = Hu. ThenR = {ht,s : (t, s) ∈W} generates H.

2.2. FREE PRODUCTS WITH AMALGAMATION 11

2.1.2 Exercises

1.) Let G = F (S) be the free group with basis S. Let x, y be a pair ofnon-identity elements in G; set H = 〈x, y〉. Verify that either H ∼= F2 orelse, x and y commute and are powers of the same element.

2.) Consider the standard projection from F2 = F (x, y) onto Z/2Z ∗ Z/2Z.Find a basis for the kernel.3.) Show that a finite index subgroup of a finitely presented group is alsofinitely presented. (Use Explicit form of the Schreier Theorem)

2.2 Free Products with amalgamation

Free products with amalgamation, or amalgams for short, arise naturally inthe context of the Seifert-van Kampen Theorem. The Seifert-van KampenTheorem says, if X = A∪B and the topological spaces A, B and A∩B arepath-connected, then the fundamental group of X is a push-out as in thecommutative diagram below.

π1(A ∩B) → π1(A)↓ ↓

π1(B) → π(X)

If the inclusions of A and B into X induce π1-injective maps then π1(X)is precisely the amalgam of π1(A) and π1(B) along π1(A ∩ B). This is ob-served for instance, when a genus 2 surface Σ is cut along a separating curveγ. Then, γ separates Σ into two surfaces each of genus 1 and with one punc-ture. And, we recover the well-known decomposition of the surface group asan amalgam of two free subgroups amalgamating the infinite cyclic group.Cutting the surface along a non-separating curves yields a decomposition ofits fundamental group as an HNN extension, we will leave this for later.

Definition 13. The amalgam A∗CB of A and B along a common subgroup

C is a push out of the directed system given by:C → A↓B

This means we can complete the square to a commutative diagram

CiA−→ A

↓iB ↓B −→ A ∗C B

Moreover A ∗C B satisfies the universal property that given any group Hand maps φA, φB : A,B → H that agree on the respective images of C,


there is a unique homomorphism from A ∗C B to H extending φA, φB. Thegroup A ∗C B is explicitly constructed using the presentation

〈A,B | iA(c) = iB(c), ∀ c ∈ C〉

and as is the nature of these things, is unique.

Reduced words

As in free products, we have reduced words in amalgams: any w ∈ G whichnot an element of C can be written as g1g2 . . . gk, where k ≥ 1 and each gicomes alternatively from A\C and B\C (in particular, is non-trivial).

Moreover, if we can also write w as h1h2 . . . hl with k ≥ 1 and eachgi coming alternatively from A\C and B\C, then k = l and g1C = h1C,Cgk = Chk and CgiC = ChiC for all i 6= 1, k.

These statements are proved using van der Waerden’s trick from the pre-vious chapter and is left to the reader. We note here that, one consequenceof this statement is that A and B are bona fide subgroups of the amalgam.

We now explore the Bass Serre tree associated to an amalgam and provethe existence of a amalgam structure on a group is charcaterised by a certainaction on a tree.

2.2.1 Trees and Amalgams

Recall that a tree is a connected graph with no non-trivial circuits. Whena group acts in a tree T so that the quotient space TmodG of the action isa tree, we can always find a fundamental domain in T .

Definition 14. Let G be a group acting on a graph X. A fundamental do-main of X/G is a subgraph Y of X such that Y → X/G is an isomorphism.

Proposition 15. Let G be a group acting on a tree X. A fundamentaldomain exists if and only if X/G is tree.

Proof. Let G be a group acting on a tree X. If a fundamental domain exists,then there is a subgraph Y of X such that Y is isomorphic to XmodG. As Yhas no circuits, X/G has no circuits either. Moreover X/G is connected andso must be a tree. Conversely, if X/G is a tree, then consider the directedset {T ⊂ X : T embeds in X/G}. This is non-empty and has a maximalelement T ′. We claim that T ′ is isomorphic to X/G. If not we can findan edge e in X such that e embeds in X/G and it originates in T ′ but itsterminal vertex is not contained in T ′. Clearly T ′ ∪ e is a tree that embedsin X/G and thus contradicts the maximality of T ′.

2.2. FREE PRODUCTS WITH AMALGAMATION 13

Notation. For any vertex or edge v in X, we write Gv for the stabiliserof v under the action of G on X.

Theorem 16. A group decomposes as an amalgam A ∗C B if and only ifG acts (simplicially) on a tree with fundamental domain an edge e = (v, w)such that Ge = C, Gv = A and Gw = B.

The proof of the theorem proceeds via the following two Lemmas.

Lemma 17. Let G be a group acting on a graph X. Let T : P Qbe an edge e of X. Suppose that T is a fundamental domain of X/G. LetGP , GQ and Ge be the stabilisers of the vertices and the edge of T . Then,the following are equivalent.

1. X is a tree

2. the map GP ∗Ge GQ → G induced by the inclusions GP ↪→ G andGQ ↪→ G is an isomorphism.

Proof. Claim 1. X is connected if and only if G is generated by GP ∪GQ.

Claim 2. X is no circuits if and only if GP ∗Ge GQ → G is injective.

Lemma 18. Let G ∼= A ∗C B. Then, there is a tree X and only oneupto isomorphism on which G acts with fundamental domain, a segmentP Q, with GP = A,GQ = B and G(P,Q) = C.

Proof. The (Bass Serre) tree for G = A ∗C B is a diagrammatic represen-tation of the cosets of A,B and C in G. Indeed, define a graph as follows:take its vertex set to be G/A ∪ G/B and its edge set to be G/C. Theinitial vertex of an edge gC is given by its image gA under the canonicalmap G/C → G/A. Similarly, the terminal vertex of the edge gC is gB.The graph so obtained clearly supports a G-action coming from the regularaction of G on the coset representatives. Moreover, under this action, thereare two orbits of vertices and one orbit of edges. Lemma 17 implies that Xis a tree.

2.2.2 Drawing the Bass Serre tree for an amalgam

In this section we describe the Bass-Serre tree for an amalgam G = A ∗C Bin further detail. Our strategy for drawing the tree is motivated by thefollowing lemma which equates pointed trees with inverse systems.


Note that an important consequence of the absence of circuits in trees isthat trees are uniquely geodesic. Assigning unit length to each edge inducesa global metric on the tree, thus making it into a geodesic metric space.Now, any two vertices are joined by a unique distance-minimising path.Indeed, if a path is made by concatenating distinct edges, then it is distanceminimising. If a path visits a vertex more than once, then as there are nocircuits, the path must involve backtracking along certain edges and thispart of the path can be removed to get a path of shorter length joining theoriginal pair of vertices. Therefore, in a tree, a path is a distance-minimisinggeodesic if and only if it does not involve any backtracking.

As a tree is connected, any two vertices can be joined by a path andtherefore by a distance-minimising geodesic. Such a path must be unique.If not, suppose c1 = (e1, . . . , ek) and c2 = (f1, . . . , fk) are two geodesicsjoining v and w. So, i(e1) = v = i(f1) and t(ek) = w = t(fk). If, ei 6= fi forall i, then, the path c1c

−12 contains a non-trivial circuit.

Let T be a tree, fix a vertex v in T . Define sets Xi setting Xi = {w ∈V (T ) : d(v, w) = i}. Observe that given any vertex w in Xn, there is aunique vertex fn(w) that belongs to Xn−1 and lies on the geodesic joining wand v. This defines a function fn : Xn → Xn−1 and hence an inverse system

. . . Xnfn−→ Xn−1 −→ . . . X1

f1−→ X0.

Conversely, given an inverse system {Xi, fi}, with X0 = {P}, we canbuild a tree by designating ∪iXi to be the set of vertices and {(Q, fn(Q))}nto be the set of edges. Note that if one of the fn’s fails to be surjective thenthe tree has some terminal vertices i.e. vertices of valency 1.

This establishes the following Lemma:

Lemma 19. There is an equivalence between pointed trees and inverse sys-tems indexed by the integers.

In Lemma 18, we saw that the tree is described in terms of the cosetrepresentatives for A,B and C in G.

A representative for a left coset of A in G is given by a reduced wordg1g2 . . . gk such that

• k ≥ 1

• each gi comes alternatively from A\C and B\C, and

• gk /∈ A

2.3. HNN EXTENSIONS 15

Similarly, a coset representative in G/B is given by a reduced word asabove, which does not end in a syllable from B.

If we fix coset representatives for A/C and B/C, and form the list L ofall reduced words in terms of these cosets, then the set of edges of the BassSerre tree are in one-one correspondence with L. Moreover, if Ln denotesthe set of all reduced words in A/C and B/C of syllable length n, thenL = ∪nLn. The fundamental domain for the action of the amalgam onits Bass Sere tree is a segment σ := P Q and ∪nLn.σ gives theentire tree. Indeed, Ln.σ contain precisely the vertices at distance n fromthe fundamental domain.

Recall that there is one orbit of edges; if e and f share a common vertexv, then clearly there is some g belonging to Gv but not to Ge such thatg.e = f . Therefore, each vertex has valency [A : C] or [B : C].

From the above discussion, we can build a strategy for drawing the tree,by starting with fundamental domain σ and then using the reduced words oflength 1, to get the vertices at unit length from σ, and repeating the processto generate the whole tree.

2.3 HNN Extensions

HNN Extensions were first studied by three mathematicians, Graham Hig-man, Bernhard Neumann and Hanna Neumann. They used the HNN con-struction to build a group Γ which contains a given group G in such a fashionthat a pre-specified subgroup A of G is conjugate in Γ to an isomorphic copyof itself. More precisely, given groups A < G, and isomorphisms φ : A→ A,the HNN extension associated to the data (G,A, φ) is

G∗A = 〈G, t | t−1at = φ(a) ∀a ∈ A〉.

Examples. HNN extensions arise naturally in our universe. A extensionG o Z is an HNN extension in which G and A are equal. More interest-ing examples include manifold groups, Baumslag Solitar groups and Rightangled Artin groups.

Hyperbolic 3-manifolds Following recent work of Agol, Wise, et al, weknow that hyperbolic 3-manifolds are essentially circle bundles over embed-ded surfaces i.e. their fundamental groups are virtually π1(Σg) o Z, whereΣg denotes the orientable surface of genus g.

One relator groups. A theorem of Magnus shows how to write any one-relator group as an HNN extension. Explicit examples abound!


• Surface groups. Fundamental groups of surfaces π1(Σg) are evi-dently HNN extensions: for instance,

π1(Σ2) = 〈a, b, c, d | aba−1 = [d, c]b〉

• Baumslag Solitar Groups. The Baumslag Solitar groups BS(p, q),p, q 6= 0 are rather exotic. Given p, q 6= 0 the corresponding BS(p, q)is given by the presentation

〈x, t | t−1xpt = xq〉.

Observe that BS(1, 1) ∼= Z2 and BS(1,−1) is the fundamental groupof the Klein bottle. They are all HNN extensions of the form Z∗Z.

Right-angled Artin groups. Let Γ = (V,E) be a graph. The Right-angled Artin group associated to Γ is given via the following presentation.

AΓ := 〈V | [v, w] = 1 ∀ (v, w) ∈ E〉

AΓ can constructed via a sequence of HNN extensions: let v0, v1, . . . , vnbe some ordering of V . Let Ai be the Artin subgroup of AΓ generated byv0, v1, . . . , vi. Then, we have the hierarchy 1 < A0 < . . . < An = AΓ where

Ai = 〈Ai−1, vi | v−1i avi = a ∀ a ∈ ALink(vi)∩{v0,...,vi−1}〉

Theorem 20. A group Γ is an HNN extension of a group G along A ifand only if Γ acts without inversions on a (simplicial) tree with one orbitof vertices and one orbits of edges and such that vertex stabilisers are allconjugate to G and edge stabilisers are conjugate to A.

2.3.1 Reduced and normal forms in HNN extensions

We follow Lyndon and Schupp in describing normal forms in HNN exten-sions. Let Γ denote the HNN extension of G along A:

Γ := G∗A = 〈G, t | t−1at = φ(a) ∀a ∈ A〉.

A sequence g0, te1 , g1, t

e2 , . . . , ten , gn, n ≥ 0, ei = ±1 is reduced if there isno consecutive subsequence t−1, gi, t with gi ∈ A or t, gi, t

−1 with gi ∈ φ(A).Higman, Neumann and Neumann proved that G embeds in Γ. Later,

Britton established the non-triviality results for reduced forms in Γ. Hisresult is usually called Britton’s Lemma and stated as follows.

2.3. HNN EXTENSIONS 17

Theorem 21 (Britton’s Lemma). If a sequence g0, te1 , g1, t

e2 , . . . , ten , gn,n ≥ 1, ei = ±1 is reduced, then the element g0t

e1g1te2 . . . tengn 6= 1 in Γ.

To refine Britton’s Lemma and write normal forms for elements of Γ weneed to select coset representatives for A in G and for B := φ(A) in G.

Normal form. A sequence g0, te1 , g1, t

e2 , . . . , ten , gn, n ≥ 0 in Γ is in normalform if

1. g0 ∈ G\{1} is arbitrary

2. if ei = 1 then gi ∈ G/A

3. if ei = −1, then gi ∈ G/B

4. there is no subsequence of the form t, 1, t−1 or t−1, 1, t.

Theorem 22. (i) The homomorphism g 7→ g gives an embedding G ↪→ Γand if g0t

e1g1te2 . . . tengn = 1, n ≥ 1 then the sequence g0, t

e1 , g1, te2 , . . . , ten , gn

is not reduced.

(ii) Every element w of Γ has a unique representation g0te1g1t

e2 . . . tengn,where g0, t

e1 , g1, te2 , . . . , ten , gn is in normal form.

2.3.2 Exercises

Baumslag Solitar Groups. Recall that for p, q 6= 0 the correspondingBS(p, q) is given by the presentation

〈x, t | t−1xpt = xq〉.

1.) Draw a diameter 4 subtree of the Bass Serre tree for BS(p, q). What isthe valency of each vertex?

2.) Show that BS(1, q) is isomorphic to the semi-direct product Z[1q ] o Z.

To do this, consider the canonical map from BS(1, q) onto Z that kills thebase element x. Prove that the kernel of this map, which is apparently thenormal subgroup generated by x, is isomorphic to Z[1

q ]. (Therefore, BS(1, q)is metabelian.)

3.) A group G is Hopfian if every surjective homomorphism from G ontoitself is also injective. A famous theorem of Malcev says that every finitelygenerated residually finite group is Hopfian.

Show that for min{|p|, |q|} ≥ 2, BS(p, q) is not Hopfian and hence, notresidually finite.


2.4 General Case

A graph of groups (G,X) consists of a graph X = (V,E), a group GP foreach vertex P ∈ V and a group Ge for each e ∈ E along with monomor-phisms Ge ↪→ Gt(e). Note that we stipulate Ge = Ge.

Theorem 23 (Fundemental Theorem of Bass-Serre theory). A group G isthe fundamental group of a graph of groups (G,X) if and only if G acts without inversions on a tree T without a global fixed point and such that TmodGis isomorphic to X and ...

2.5 Applications

Theorem 24 (Kurosh Subgroup Theorem). Every subgroup of a free prod-uct A ∗B is either conjugate to a subgroup of A or B or else decomposes asa non-trivial free product.

Lemma 25. Let H be a finite group. Any action of H on a tree T has a(global) fixed point.

Proof. Let S := H.v be the orbit of a vertex v under H. If S = {v} i.e. Hfixes v then we are done. Otherwise, consider the subtree X ⊆ T generatedby S. This is convex closure of S in T and comprises the union of geodesicsjoining any pair of points in S. Observe that X is a finite tree. As HS = Sand any element h ∈ H carries a geodesic [s, t], s, t ∈ S to a geodesic joiningthe pair hs and ht belonging to S, H takes X to itself. Therefore it sufficesto consider the case when T = X and T is finite.

Set T ′ to be the set of terminal vertices i.e. the vertices of valency 1in T . Observe that T ′ is preserved by H and therefore, the non-terminalvertices vert(T )\vert(T ′) also form an H-invariant set. We now argue byinduction. The result clearly holds if T has one edge and assume that theresult holds for any tree with fewer vertices than T . Applying the inductionhypothesis to T ′, we deduce that H fixes a vertex in T ′ and hence in T .

Corollary 26. Suppose a group G decomposes as the fundamental group ofa graph of groups (G,X). Then every finite subgroup of G is conjugate to afinite subgroup of Gv for some v ∈ V (X).

Proof. Let H be a finite subgroup of G. By Lemma 25, H fixes a vertex νof the Bass Serre tree for (G,X) and is contained in the stabiliser of ν. But,ν is a translate (by an element g) of a vertex that projects to a vertex v ofX. Therefore, gHg−1 < Gv.

2.5. APPLICATIONS 19

Corollary 27. Suppose that G is the fundamental group of a graph of groups(G,X), such that for every vertex v of X, Gv is torsion-free, then G istorsion-free.

Proposition 28. Suppose a group G is acting without inversions in a treeT . If H is a subgroup of G that does not intersect any of the vertex stabilisersof the G-action, then H is free.

Proof. The subgroup H evidently acts freely on the tree T and therefore His free.

For example the kernel of the standard projection of Z/2Z ∗ Z/3Z ontoZ/2Z×Z/3Z is free. This is precisely the commutator subgroup of PSL(2,Z)∼= Z/2Z ∗ Z/3Z.

2.5.1 Exercises

Tits Alternative for groups acting on trees. We will say that a groupsatisfies the Tits Alternative if every subgroup of G is either virtually solv-able or contains a free subgroup of rank at least 2. Suppose a group G isacting without inversions on a tree such that all vertex stabilisers associatedto the action satisfy the Tits Alternative. Show that G also satisfies theTits Alternative.

Chapter 3

Property (FA) andGeometric Aspects

Suppose that a group G is acting without inversions on a tree X. Considerthe set of (global) fixed points XG = {P ∈ X | gP = P ∀g ∈ G}. EvidentlyXG has no circuits. In addition, XG is connected; given any pair of verticesv, w in XG, the geodesic joining v and w must be sent by any g ∈ G to thegeodesic joining gv = v and gw = w! As the tree is uniquely geodesic andthe group acts without inversions, the entire geodesic from v to w is alsocontained in XG and XG is a subtree of X.

Definition 29 (Serre). A group G is said to have property (FA) if XG 6= ∅for any tree X on which G acts.

The following theorem characterises groups with property (FA):

Theorem 30. A group has property (FA) if and only if all of the followingthree conditions are satisfied.

1. G is not an amalgam

2. G does have an epimorphism onto Z

3. G is not a union of a strictly increasing sequence of subgroups G1 <G2 < G3 < . . .; in particular, if G is countable, then G is finitelygenerated.

Caveat. Amalgams do not necessarily map onto Z. An amalgam of tor-sion groups cannot map onto Z but restricting to non-torsion vertex groupsdoesn’t help. Burger and Moses have constructed examples of finitely pre-sented simple groups which are amalgams of finitely generated free groups.

21

22 CHAPTER 3. PROPERTY (FA) AND GEOMETRIC ASPECTS

Finitely generated torsion groups. We have seen in Lemma 25 thatfinite groups have property (FA). Theorem 30 further shows that a finitelygenerated torsion group has property (FA). Indeed, let G be a finitely gen-erated torsion group, then G cannot decompose as an amalgam A ∗C B:the element ab, where a ∈ A\C, b ∈ B\C has infinite order. The onlyhomomorphism from a torsion group onto Z is the trivial one.

Corollary 31. A countable group with property (FA) is finitely generated.

Proof of Theorem 30. Assume first that G has property (FA). If (1) doesn’thold and G decomposes as an amalgam, then G acts on the Bass Serre treefor this decomposition. If on the other hand, G maps onto Z, then G actson the real line via this epimorphism. Therefore it suffices to prove thefollowing lemma.

Lemma 32. If G is a union of a strictly increasing chain of subgroups thenG does not have property (FA).

Proof of Lemma. LetG1 < G2 < G3 < . . . < G be a strictly increasing chainof subgroups such that ∪iGi = G. Then, we have the canonical surjectionsφi : G/Gi → G/Gi+1. Define a graph X with vertex set V =

∐iG/Gi

and edges given by (gGi, φi(gGi)). Evidently, there are no circuits in thegraph. Moreover, the graph is connected. Given a pair gGi, hGj of cosets,either g = h in G or else, g−1h 6= 1 but there exists some k ≥ i, j, suchthat g−1h ∈ Gk ⇔ gGk = hGk. Clearly the path (gGi, gGi+1, . . . , gGk)concatenated with the inverse of the path (hGj , hGj+1, . . . , hGk) connectsgGi and hGj . Thus, X is a tree and the action of G on the cosets G/Giextends to an action of G on the tree. If G fixes a vertex gGi, then for allx ∈ G, xgGi = gGi ⇔ G = Gi, which contradicts the hypothesis that G isa union of a strictly increasing chain of subgroups Gj .

This completes the proof of the forward implication in the statement ofTheorem 30. Suppose now that (1)-(3) hold and let G act without inversionson a tree X. Set Y to be the quotient graph XmodG. If π1(Y ) 6= 1, thenG maps onto Z which contradicts (2). Therefore Y is a tree and choosinga basepoint v ∈ Y , we can write Y as a directed union of its subtrees∪nYn, where Yn is the ball of radius n about v. Now, each tree Yn liftsto a subtree of X and identifying stabilisers we associate graphs of groups(Gn, Yn). Clearly, G is the directed union of the Gn, which contradicts (3)and therefore for some n large enough, X = Xn. Let Z be subtree of Xsuch that no proper subtree of Z provides a graph of group decompositionof the whole of G. Fix a terminal edge (leaf) e of Z. Clearly Gt(e) 6= 1 and

23

Z = Z\{e, t(e)} ∪ {e, t(e)}. If G′ denotes the fundamental group of the treeof groups associated to Z\{e, t(e)}, then G ∼= G′ ∗GeGt(e), which contradicts(1). Therefore, G has property (FA).

3.0.1 Closure Properties of property (FA)

1. Every quotient of a group with property (FA) has property (FA).

2. Suppose that A,G and B fit into a s.e.s. A ↪→ G � B. If A and Bhave property (FA) then G has property (FA).

3. If a finite index subgroup H of G has property (FA) then G has prop-erty (FA).

The proof of (1) is clear. To prove (2), observe that if G acts on a tree X,then so does A. By hypothesis, XA 6= ∅ and the action of G on X inducesan action of B on XA. Finally XG = (XA)B and as B has property (FA),this is non-empty.

If H is a normal finite index subgroup, then (3) follows from (2) andthe fact that finite groups have property (FA). Otherwise, take a set oftransversals E for G/H and set K = ∩g∈EHe. Then K is a normal subgroupof finite index. If G acts on a tee X, then ∅ 6= XH ⊂ XK . Moreover,XG = (XK)G/K is non-empty and so G has property (FA).

We will see that the converse to (3) is not true. The Coxeter group4(3, 3, 3) generated by reflections in the sides of an equilateral triangle inthe plane has property (FA) (by Corollary 36 below) but it is virtually Z2

and hence possesses a finite index subgroup that violates (2) of Theorem 30.

3.0.2 Geometric Aspects

We will now concentrate on geometric aspects of groups acting on trees andthese will provide us with criteria for establishing property (FA) and findingPing pong partners.

Lemma 33 (Bridge Lemma). Let T1, T2 be non-empty disjoint sub-trees ofa tree X. Then,

1. There exists a unique pair (P,Q) of vertices, P ∈ T1 and Q ∈ T2 suchthat d(P,Q) = d(T1, T2) = min{d(v, w) | v ∈ T1, w ∈ T2}.

2. A path joining a vertex of T1 to a vertex of T2 contains the geodesic[P,Q] joining P and Q.


3. Every sub-tree of X intersecting both the Ti contains [P,Q].

Lemma 34 (Fixed points for tree automorphisms). Suppose that G is actingon a tree and 1 6= g ∈ G has a fixed point i.e. Xg 6= ∅. Let P ∈ X\Xg.SUppose Q ∈ Xg is the point such that d(P,Q) = d(P,Xg). Then thegeodesic joining P to gP is obtained by concatenating the geodesic joiningP to Q followed by the geodesic joining Q to gP .

The lemma is proved by induction on the distance of P from Xg. Letd = d(P,Xg). If d = 0 there is nothing to prove. If d = 1 then thereis an edge e with o(e) = P and t(e) = Q ∈ Xg. Now, t(ge) = Q ando(ge) = gP 6= P else P would have to belong in Xg. The general case isnow clear and g[P,Q] = [gP,Q].

Very Useful Observation. The mid-point of the geodesic from P to gPis fixed by g.

Lemma 35 (Criterion for Property (FA)). Let G be a group acting on atree X. Let A,B be subgroups of G such that G = 〈A,B〉, A = 〈a1, . . . , ak〉and B = 〈b1, . . . , bl〉. If ∀i = 1, . . . , k and j = 1, . . . , l, Xai 6= ∅, Xbj 6= ∅and Xaibj 6= ∅, then XG 6= ∅.

Proof. Evidently, XA 6= ∅ and XB 6= ∅. As G = 〈A,B〉, we have XG =XA ∩ XB. Suppose XA ∩ XB = ∅, then using the Bridge Lemma, weknow there is a unique pair of points P ∈ XA and Q ∈ XB such thatd(P,Q) = d(XA, XB) ≥ 1. As Q /∈ XA, there is some ai such that aiQ 6= Q.But, ai fixes P and so, the geodesic joining Q to aiQ = aibjQ (for anyj) is the concatenation of the geodesic joining P and Q and the geodesicjoining P and aiQ. By hypothesis, the aibj are elliptic and therefore eachaibj fixes the mid-point of [Q, aibjQ], namely, the point P . But, aibjP = Pfor all j implies that bjP = P for all j, which contradicts the assumptionthat P /∈ XB. Therefore, XA and XB must intersect and consequently,XG 6= ∅.

Corollary 36. Let W = (S, (mst)) denote the Coxeter group generated bythe reflections S and with associated Coxeter matrix (mst). If mst is finitefor all s, t, then W has property (FA).

Proof. The Coxeter group W is given by the presentation below.

〈 S | s2 = 1 ∀ s ∈ S; (st)mst = 1 ∀s 6= t ∈ S 〉

Recall that finite groups have property (FA). If W acts on a tree andthe mst are all finite, then for all s, t, Xs 6= ∅, Xst 6= ∅. Hence, by Lemma35, W has property (FA).

25

An automorphism of a tree X that fixes a point is said to be elliptic. Anautomorphism that is not elliptic is called hyperbolic. Given an automor-phism g without fixed points, define

|g| = minP∈vert(X)

d(P, gP )

T (g) = {P ∈ vert(X) | d(P, gP ) = |g|}

Theorem 37 (Translation axis). The set T (g) forms the vertex set for abi-infinite geodesic in X on which g acts by translations of length |g|. Everysub-tree of X preserved by g±1 contains T (g). If Q is a vertex outside T (g)then d(Q, gQ) = |g|+ 2d(Q,T (q)).

Proof. Consider the geodesic [P, gP ] = (P = P0, P1, . . . , Pd), where d = |g|.Then, g[P, gP ] = [gP, g2P ] and we claim that the concatenation of [P, gP ]and [gP, g2P ] gives a path without backtracking and this is precisely thegeodesic joining P to g2P . Suppose not. If d = 1, this would make gP = P ,which is impossible as the action is without inversions. If d ≥ 2, thengP1 = Pd−1. But then d(P1, gP1) = d(P1, Pd−1) ≤ d − 2. This contradictsthe fact that |g| = d. Clearly now, ∪n∈Z gn[P, gP ] defines a bi-infinitegeodesic in X.

Suppose now that Q is a point at a distance of n ≥ 1 from T (g). Let Q′

be the vertex in T (g) closest to Q. Then, we argue that the geodesic joiningQ and gQ is produced by juxtaposing [Q,Q′], [Q′, gQ′] and [gQ′, gQ], thusgiving d(Q, gQ) = |g| + 2d(Q,T (q)). This means in turn, that, T (g) =∪n∈Zgn[P, gP ]! Moreover, suppose that a subtree T of X is preserved byg and its inverse. If Q is any vertex in T , then the geodesic [Q, gQ] iscontained in T . Unless Q is already a vertex of T (g), [Q, gQ] contains thevertex Q′ ∈ T (g) as above. This implies that T (g) ⊂ T .

Definition 38. The bi-infinite geodesic T (g) is called the translation axisfor g.

Commuting elements. One consequence of Theorem 37 is that commut-ing hyperbolic automorphisms have the same translation axes. Supposeg, h are hyperbolic automorphisms of a tree X such that [g, h] = 1. Then,for P ∈ T (g), d(h±1P, gh±1P ) = d(h±1P, h±1gP ) = d(P, gP ). Therefore,h±1T (g) = T (g) and by Theorem 37, T (h) ⊂ T (g). Similarly g±1T (h) =T (h) and we conclude that T (g) = T (h).


3.0.3 Finding Ping Pong Partners

Suppose a group is acting without inversions on a tree X. Let a, b be twoelements of G which are hyperbolic automorphisms of the tree. One of thefollowing may happen:

1. T (a) and T (b) coincide.

2. T (a) and T (b) share a common end.

3. T (a) ∩ T (b) is a finite path, possibly just a point or even empty.

When (3) holds, we can use a and b to find Ping Pong partners thatwill generate a copy of F2 inside G. Here, we will consider the case whenT (a) ∩ T (b) is non-empty and of finite length. The remaining cases are leftas exercises. Note that unless their intersection is empty, T (a) ∩ T (b) is ageodesic of finite length joining a pair of points, say P and Q. We mayassume by replacing a or b by their inverses if needed, that the directionof translation in both cases is from P to Q. Removing the geodesic [P,Q]from the tree X yields four connected components which we call A±1, B±1

so that aQ ∈ A+, a−1P ∈ A−, bQ ∈ B+ and b−1P ∈ B−. Assume that

d(P,Q) ≤ min{|a|, |b|} − 1 (∗)

Clearly, aA+ ⊂ A+. As (*) holds, aP belongs to A+ and thereforeaB− ⊂ A+. Also, aP ∈ A+ implies that aB+ ⊂ A+. Arguing as above wededuce the following.

• a(A+ ∩B+ ∩B−) ⊂ A+

• a−1(A− ∩B+ ∩B−) ⊂ A−

• b(B+ ∩A+ ∩A−) ⊂ B+

• b−1(B− ∩A+ ∩A−) ⊂ B−

We deduce that the subgroup 〈a, b〉 generated by a and b in G is F2 usingthe following version of the Ping Pong Lemma.

Lemma 39 (Ping Pong Lemma for cyclic subgroups). A group G is act-ing on a set Y ; let a, b ∈ G\{1}. Suppose there exists disjoint non-emptysubsets of Y , called A±1, B±1 such that a(Y \A−) ⊂ A+, a−1(Y \A+) ⊂ A−,b(Y \B−) ⊂ B+ and b−1(Y \B+) ⊂ B−. Then a and b generate a copy of F2

inside G.

27

We leave the reader to deduce the above version of the Ping Pong Lemmafrom Lemma 1. The discussion above proves

Lemma 40. Let G act without inversions on a tree; let a, b ∈ G be hyperbolicautomorphisms such that T (a) ∩ T (b) is non-empty and the length of theintersection is strictly smaller than both |a| and |b|. Then, 〈a, b〉 ∼= F2.

Corollary 41. Let G act without inversions on a tree; let a, b ∈ G behyperbolic automorphisms such that T (a) ∩ T (b) has finite length. Then,there exist n,m ∈ Z such that 〈an, bm〉 ∼= F2.

3.0.4 Exercises

Irreducible actions. We say, a group acts irreducibly on a tree if thereis no global fixed point for the action and the whole group does not preservea line or an end of the tree. Show that, if a group G acts irreducibly on atree, then G contains a free group of rank 2.

(Use the fact: if all the elements of the group are elliptic automor-phisms of the tree then the group fixes an end. This implies, if the actionis irreducible, then at least one element of the group acts as a hyperbolicautomorphism.)

This implies, if a group G is solvable or more generally amenable, then anaction of G on a tree must be reducible: if the action is non-trivial (withouta global fixed point), then the group will preserve a line or an end of thetree.

Translation Length functions. Suppose that a group G is acting with-out inversions on a tree X. Define a function | . | : G → R as follows. Forany g ∈ G,

| g | = min{d(v, gv) : v ∈ vert(X) }

For instance, if g is elliptic then | g | = 0 and | g | ≥ 1 for any hyperbolicg. The function | . | is called a translation length function of G and wepresent here some basic properties of | . |.

1. | 1 | = 0

2. | g | = | g−1 |

3. | h−1gh | = | g | for all g, h ∈ G

4. If g, h ∈ G are both hyperbolic and T (g) ∩ T (h) is either empty or asingle vertex, then | gh | = | gh−1 | = | g |+ | h |+ 2d(T (g), T (h)).


5. If g, h are hyperbolic and T (g) ∩ T (h) 6= ∅, then | gh | ≤ | g |+ | h |.

6. If g, h are elliptic and Xg ∩Xh = ∅, then gh, gh−1 are hyperbolic.

For a detailed treatment of translation length functions, see Culler andMorgan’s Group Actions on R-trees.

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