Lecture Notes from a Weekly Class That I Am Oﬀering for Doc- Toral Students of Mathematics in the University of Southampton

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 Free Products

1.1 Historical outline

Broadly speaking, the Ping Pong Lemma in group theory is a test for finding free products inside a given group. The initial version of the Lemma is attributed to Felix Klein who used it to study subgroups of Kleinian groups i.e. discrete groups of isometries of hyperbolic 3-space. In 1972, Jacques Tits used the Ping pong Lemma to prove that, every subgroup of a finitely generated linear group is either virtually solvable or contains a non-abelian free group. This Theorem is now known as the Tits Alternative for linear groups and in contemporary group theory, has been generalised as follows: a group G is said to satisfy the Tits Alternative if every subgroup is either virtually solvable or contains F2, the free group of rank 2. Hyperbolic groups, groups acting properly and co-compactly on CAT(0) cube complexes and subgroups of mapping class groups are a few examples 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 amenable group does not contain F2 and to check for exponential growth, one simply has 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 is said to have property Pnaive if for any finite subset F ⊂ G\{1} there exists y ∈ G such that for each x ∈ F , the subgroup generated in G by x and y is the free product hxi ∗ hyi. 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 known to have property Pnaive. It is not known if SL(n, Z), n ≥ 3, has Pnaive or not.

Lemma 1 (Ping Pong Lemma). Let G be a group acting on a set X and let A and B be subgroups of G with max{|A|, |B|} ≥ 3. Set H to be the subgroup generated by A and B in G. Suppose there exist non-empty subsets Xi, 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 exist in the literature.

1.2 Free products

A monoid is is a set endowed with a binary operation (we will call this multiplication) which is associative and has an indentity element. Given a set 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 identity element is the empty word. This makes W (A) into the free monoid on the alphabet A. A typical element in W (A) has the form a1a2 . . . ak, where each ai ∈ A; we say k is the length of the sequence and identify A with the sequences of unit length. Consider now a family of groups {Gi}i∈I and choose A to be the disjoint union of the Gi’s. We obtain a group from the free monoid on A by deﬁning an equivalence relation on W (A) as follows: for any w, w0 ∈ W (A),

0 0 • wew ≈ ww , if e is an identity element from some Gi

0 0 • 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 is −1 −1 ak . . . a1 . W is the free product of the Gi, written ∗i∈I Gi.

Example 2. If Gi is inﬁnite cyclic for each i ∈ I, then ∗i∈I Gi is simply the free 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 the quotient W of W (A) by the equivalence relation ≈.

Deﬁnition 3. Let w = a1 . . . ak be an element of W (A) such that k ≥ 1 and ai ∈ 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∈I Gi is equivalent to a unique reduced word in W (A).

Proof. Take w = a1 . . . ak to be a reduced word as in Definition 3. Let a ∈ Gj for some j. Define R(aw) as below. R(aw) = w, if a = 1 −1 (aa1)a2 . . . ak, if n1 = j and a 6= a1 −1 a2 . . . ak, if n1 = j and a = a1 aa1a2 . . . ak, if n1 6= j. Evidently, this is a reduced word representing the product aw. One can now argue by induction that every equivalence class is represented by a reduced word. To establish uniqueness, we use a classic argument due to van der Waerden. Let X be the set of all reduced words in W (A). We now define a map from ∗i∈I Gi 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∈I Gi by setting σ(a1 . . . ak to be the composition σ(a1) . . . σ(ak). To complete the argument, observe now that for any w ∈ X, σ(w) ∈ SX and σ(w)(1) = w. Therefore, for a pair w, w0 of reduced words, σ(w) = σ(w0) if and only if w = w0.

Corollary 5. Each Gi is naturally a subgroup of ∗i∈I Gi. Write γi for the canonical injection.

Corollary 6. (Universal Property) Let H be a group and let fi : Gi → H be a family of homomorphisms. Then there exists a unique extension F : ∗i∈I Gi → 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 be subgroups of G with max{|A|, |B|} ≥ 3. Set H to be the subgroup generated by A and B in G. Suppose there exist non-empty subsets Xi, i = 1, 2 of X such that 6 CHAPTER 1. PING PONG LEMMA AND FREE PRODUCTS

• 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 to B\{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 from −1 −1 a1 . By the earlier argument, a wa is non-trivial and therefore, w 6= 1. Finally note that any word of the form b1a1 . . . bkak is also non-trivial as its inverse is non-trivial. 1 z 1 0 Proposition 7. If |z| ≥ 2, then M = and M = 1 0 1 2 z 1 generate a free subgroup of rank 2 in SL(2, C). Proof. This proof is attributed to Sanov in Lyndon and Schupp. 0 1 Set J = and observe that M = JM J. This means any 1 0 2 1 element in the subgroup generated by M1 and M2 may be written in the n1 n2 form M1 JM J.... So it suﬃces to show that the subgroup generated by M1 and J is the free product hM1i ∗ hJi. Consider the action of SL(2, C) on the Riemann sphere Cˆ = C ∪ {∞} by fractional linear transformations, i.e. a b az + b .z = . c d cz + d

ˆ ˆ Choose X1 = {z ∈ C : |z| < 1} and X2 = {z ∈ C : |z| > 1}. Then, n M1 X1 ⊂ X2 and JX2 ⊂ X1. Invoke the Ping Pong Lemma to deduce the Proposition.

1.4 Exercises

1. Let W := W (∪i∈I Gi)/ ≈ 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 to the direct product A × B. For instance, φ takes the reduced word aba0b0 to aa0bb0. Let K := ker φ. 1.4. EXERCISES 7

(a) Prove that K is free with basis {[a, b] = a−1b−1ab : a ∈ A\{1}, b ∈ B\{1}}. (b) If A = Z/2Z and B = Z/3Z, then A∗B is the modular group PSL(2, Z) = SL(2, Z)/ ± 1. Observe that in this case, K is the commutator subgroup of A ∗ B. What is the rank of K? This is another proof of the fact that PSL(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 element g ∈ G, there exists a surjection f : G → F to a finite group F such that f(g) 6= 1. In this exercise, we will verify that if A and B are residually finite then A ∗ B is residually finite. (i) First, show that free groups are residually finite. A famous theorem of Malcev says, every finitely generated linear group is residually finite. But, for free groups, we can prove this by elementary means. Recall that the free 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 residually n finite. Verify this, using the standard surjections SL(2, Z) → SL(2, Z/2 Z).

(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 of finite index in A ∗ B. Therefore, K is finitely generated. This helps us find finite index characteristic subgroups in K as follows: • there are only finitely many maps from a finitely generated group G to a finite group;

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

• set L = ∩{H

Let Γ be the subgroup of G generated by two elements γ1, γ2. Verify that Γ contains a free group of rank 2 in the following cases. ˆ (i) Both γ1 and γ2 are parabolic with distinct fixed points in R. ˆ (ii) While γ1 is parabolic with fixed point x ∈ R, γ2(x) 6= x. (iii) Both γ1, γ2 are hyperbolic and F ix (γ1) ∩ F ix (γ1) = ∅. Rˆ Rˆ 5. Show that SL(2, Z) is generated by the elementary matrices i.e. the 1 1 1 0 matrices and generate SL(2, ). This involves applying 0 1 1 1 Z the Euclidean algorithm to the entries of a matrix. 1 2 1 0 Further, show that Γ(2) = h , i is a free subgroup of 0 1 2 1 finite index in SL(2, Z). This involves a variation on the argument showing ∼ that 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 groups acting on trees. The theory of groups acting on trees is popularly called Bass Serre 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 de France in Paris in 1968. Famously, Hyman Bass was in the audience and wrote up lecture notes. These appeared in an Asterique volume and were later translated to English by Stillwell. In what follows, we will first briefly review free groups, these being the first examples of groups acting on trees. Then, we will describe amalgams and HNN extensions, leading upto the Fundamental Theorem of Bass Serre theory. Trees. A tree is a connected graph without circuits. One can show that any two points in a tree are joined by a unique geodesic (path without backtracking) and moreover, a tree is contractible. N.B. The action of the group on the tree is always assumed to be simplicial i.e. no group element takes an edge to its inverse so that the quotient of the tree 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 when the action is free: the existence of a free action on a tree characterises free groups. 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 Cayley graph 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 a free 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 free group with basis S and let H < G. The Cayley graph Γ of G with respect to 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 index in 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 group with basis S. Let H < G. Then, one one can choose a set T of representatives of G/H satisfying the following conditions:

1 2 n • If t ∈ T has the reduced form t = s1 s2 . . . sn , where si ∈ S, i = ±1 1 and i = i+1 whenever si = si+1, then all the partial products 1, s1 , 1 2 1 2 n s1 s2 , . . . , s1 s2 . . . sn belong to T .

• Let T be as above and deﬁne W = {(t, s) ∈ T × S : ts∈ / T }; for −1 every (t, s) ∈ W set hs,t = tsu , where u ∈ T and Hts = Hu. Then R = {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 of ∼ non-identity elements in G; set H = hx, yi. Verify that either H = F2 or else, 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 also finitely presented. (Use Explicit form of the Schreier Theorem)

2.2 Free Products with amalgamation

Free products with amalgamation, or amalgams for short, arise naturally in the context of the Seifert-van Kampen Theorem. The Seifert-van Kampen Theorem says, if X = A ∪ B and the topological spaces A, B and A ∩ B are path-connected, then the fundamental group of X is a push-out as in the commutative 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 curve γ. 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 as an amalgam of two free subgroups amalgamating the inﬁnite cyclic group. Cutting the surface along a non-separating curves yields a decomposition of its fundamental group as an HNN extension, we will leave this for later.

Deﬁnition 13. The amalgam A∗C B of A and B along a common subgroup C → A C is a push out of the directed system given by: ↓ B This means we can complete the square to a commutative diagram i C −→A A ↓iB ↓ B −→ A ∗C B

Moreover A ∗C B satisﬁes the universal property that given any group H and maps φA, φB : A, B → H that agree on the respective images of C, 12 CHAPTER 2. GROUPS ACTING ON TREES there is a unique homomorphism from A ∗C B to H extending φA, φB. The group A ∗C B is explicitly constructed using the presentation

hA, B | iA(c) = iB(c), ∀ c ∈ Ci 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 which not an element of C can be written as g1g2 . . . gk, where k ≥ 1 and each gi comes 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 each gi 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 consequence of 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 prove the existence of a amalgam structure on a group is charcaterised by a certain action on a tree.

2.2.1 Trees and Amalgams Recall that a tree is a connected graph with no non-trivial circuits. When a group acts in a tree T so that the quotient space T modG of the action is a tree, we can always find a fundamental domain in T . Definition 14. Let G be a group acting on a graph X. A fundamental domain 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 fundamental domain 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 Y has no circuits, X/G has no circuits either. Moreover X/G is connected and so must be a tree. Conversely, if X/G is a tree, then consider the directed set {T ⊂ X : T embeds in X/G}. This is non-empty and has a maximal element T 0. We claim that T 0 is isomorphic to X/G. If not we can find an edge e in X such that e embeds in X/G and it originates in T 0 but its terminal vertex is not contained in T 0. Clearly T 0 ∪ e is a tree that embeds in X/G and thus contradicts the maximality of T 0. 2.2. FREE PRODUCTS WITH AMALGAMATION 13

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

Theorem 16. A group decomposes as an amalgam A ∗C B if and only if G 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 Q be an edge e of X. Suppose that T is a fundamental domain of X/G. Let GP , 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 and GQ ,→ 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 one upto isomorphism on which G acts with fundamental domain, a segment P Q, with GP = A, GQ = B and G(P,Q) = C.

Proof. The (Bass Serre) tree for G = A ∗C B is a diagrammatic representation 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. The initial vertex of an edge gC is given by its image gA under the canonical map 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 regular action of G on the coset representatives. Moreover, under this action, there are two orbits of vertices and one orbit of edges. Lemma 17 implies that X is 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 B in further detail. Our strategy for drawing the tree is motivated by the following lemma which equates pointed trees with inverse systems. 14 CHAPTER 2. GROUPS ACTING ON TREES

Note that an important consequence of the absence of circuits in trees is that trees are uniquely geodesic. Assigning unit length to each edge induces a 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 distance minimising. If a path visits a vertex more than once, then as there are no circuits, the path must involve backtracking along certain edges and this part of the path can be removed to get a path of shorter length joining the original pair of vertices. Therefore, in a tree, a path is a distance-minimising geodesic if and only if it does not involve any backtracking. As a tree is connected, any two vertices can be joined by a path and therefore by a distance-minimising geodesic. Such a path must be unique. If not, suppose c1 = (e1, . . . , ek) and c2 = (f1, . . . , fk) are two geodesics joining v and w. So, i(e1) = v = i(f1) and t(ek) = w = t(fk). If, ei 6= fi for −1 all i, then, the path c1c2 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 a unique vertex fn(w) that belongs to Xn−1 and lies on the geodesic joining w and v. This defines a function fn : Xn → Xn−1 and hence an inverse system

fn f1 ...Xn −→ Xn−1 −→ ...X1 −→ X0.

Conversely, given an inverse system {Xi, fi}, with X0 = {P }, we can build a tree by designating ∪iXi to be the set of vertices and {(Q, fn(Q))}n to be the set of edges. Note that if one of the fn’s fails to be surjective then the 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 systems indexed by the integers.

In Lemma 18, we saw that the tree is described in terms of the coset representatives for A, B and C in G. A representative for a left coset of A in G is given by a reduced word g1g2 . . . 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 as above, 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 of all reduced words in terms of these cosets, then the set of edges of the Bass Serre tree are in one-one correspondence with L. Moreover, if Ln denotes the set of all reduced words in A/C and B/C of syllable length n, then L = ∪nLn. The fundamental domain for the action of the amalgam on its Bass Sere tree is a segment σ := P Q and ∪nLn.σ gives the entire tree. Indeed, Ln.σ contain precisely the vertices at distance n from the fundamental domain. Recall that there is one orbit of edges; if e and f share a common vertex v, then clearly there is some g belonging to Gv but not to Ge such that g.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 of length 1, to get the vertices at unit length from σ, and repeating the process to 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 fashion that a pre-speciﬁed subgroup A of G is conjugate in Γ to an isomorphic copy of itself. More precisely, given groups A < G, and isomorphisms φ : A → A, the HNN extension associated to the data (G, A, φ) is

−1 G∗A = hG, t | t at = φ(a) ∀a ∈ Ai.

Examples. HNN extensions arise naturally in our universe. A extension G o Z is an HNN extension in which G and A are equal. More interest- ing examples include manifold groups, Baumslag Solitar groups and Right angled Artin groups. Hyperbolic 3-manifolds Following recent work of Agol, Wise, et al, we know that hyperbolic 3-manifolds are essentially circle bundles over embed- ded surfaces i.e. their fundamental groups are virtually π1(Σg) o Z, where Σg 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! 16 CHAPTER 2. GROUPS ACTING ON TREES

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

−1 π1(Σ2) = ha, b, c, d | aba = [d, c]bi

• 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

hx, t | t−1xpt = xqi.

∼ 2 Observe that BS(1, 1) = Z and BS(1, −1) is the fundamental group of 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 := hV | [v, w] = 1 ∀ (v, w) ∈ Ei

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

−1 Ai = hAi−1, vi | vi avi = a ∀ a ∈ ALink(vi)∩{v0,...,vi−1}i Theorem 20. A group Γ is an HNN extension of a group G along A if and only if Γ acts without inversions on a (simplicial) tree with one orbit of vertices and one orbits of edges and such that vertex stabilisers are all conjugate 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 extensions. Let Γ denote the HNN extension of G along A:

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

e e en A sequence g0, t 1 , g1, t 2 , . . . , t , gn, n ≥ 0, ei = ±1 is reduced if there is −1 −1 no consecutive subsequence t , gi, t with gi ∈ A or t, gi, t 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. His result is usually called Britton’s Lemma and stated as follows. 2.3. HNN EXTENSIONS 17

e e en Theorem 21 (Britton’s Lemma). If a sequence g0, t 1 , g1, t 2 , . . . , t , gn, e e en n ≥ 1, ei = ±1 is reduced, then the element g0t 1 g1t 2 . . . t gn 6= 1 in Γ.

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

e e en Normal form. A sequence g0, t 1 , g1, t 2 , . . . , t , gn, n ≥ 0 in Γ is in normal form 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,→ Γ e e en e e en and if g0t 1 g1t 2 . . . t gn = 1, n ≥ 1 then the sequence g0, t 1 , g1, t 2 , . . . , t , gn is not reduced. e e en (ii) Every element w of Γ has a unique representation g0t 1 g1t 2 . . . t gn, e e en where g0, t 1 , g1, t 2 , . . . , t , gn is in normal form.

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

hx, t | t−1xpt = xqi.

1.) Draw a diameter 4 subtree of the Bass Serre tree for BS(p, q). What is the valency of each vertex? 1 2.) Show that BS(1, q) is isomorphic to the semi-direct product Z[ q ] o Z. To do this, consider the canonical map from BS(1, q) onto Z that kills the base element x. Prove that the kernel of this map, which is apparently the 1 normal subgroup generated by x, is isomorphic to Z[ q ]. (Therefore, BS(1, q) is metabelian.) 3.) A group G is Hopfian if every surjective homomorphism from G onto itself is also injective. A famous theorem of Malcev says that every finitely generated residually finite group is Hopfian. Show that for min{|p|, |q|} ≥ 2, BS(p, q) is not Hopfian and hence, not residually finite. 18 CHAPTER 2. GROUPS ACTING ON TREES

2.4 General Case

A graph of groups (G, X) consists of a graph X = (V,E), a group GP for each 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-S´erretheory). A group G is the fundamental group of a graph of groups (G, X) if and only if G acts with out inversions on a tree T without a global ﬁxed point and such that T modG is isomorphic to X and ...

2.5 Applications

Theorem 24 (Kurosh Subgroup Theorem). Every subgroup of a free product A ∗ B is either conjugate to a subgroup of A or B or else decomposes as a 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. H fixes v then we are done. Otherwise, consider the subtree X ⊆ T generated by S. This is convex closure of S in T and comprises the union of geodesics joining any pair of points in S. Observe that X is a finite tree. As HS = S and any element h ∈ H carries a geodesic [s, t], s, t ∈ S to a geodesic joining the pair hs and ht belonging to S, H takes X to itself. Therefore it suffices to consider the case when T = X and T is finite. Set T 0 to be the set of terminal vertices i.e. the vertices of valency 1 in T . Observe that T 0 is preserved by H and therefore, the non-terminal vertices vert(T )\vert(T 0) also form an H-invariant set. We now argue by induction. The result clearly holds if T has one edge and assume that the result holds for any tree with fewer vertices than T . Applying the induction hypothesis to T 0, we deduce that H fixes a vertex in T 0 and hence in T .

Corollary 26. Suppose a group G decomposes as the fundamental group of a graph of groups (G, X). Then every finite subgroup of G is conjugate to a finite subgroup of Gv for some v ∈ V (X). Proof. Let H be a finite subgroup of G. By Lemma 25, H fixes a vertex ν of 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 of −1 X. Therefore, gHg < 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 is torsion-free.

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

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

For example the kernel of the standard projection of Z/2Z ∗ Z/3Z onto Z/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 group satisﬁes the Tits Alternative if every subgroup of G is either virtually solvable or contains a free subgroup of rank at least 2. Suppose a group G is acting without inversions on a tree such that all vertex stabilisers associated to the action satisfy the Tits Alternative. Show that G also satisﬁes the Tits Alternative. 20 CHAPTER 2. GROUPS ACTING ON TREES Chapter 3

Property (FA) and Geometric Aspects

Suppose that a group G is acting without inversions on a tree X. Consider the set of (global) fixed points XG = {P ∈ X | gP = P ∀g ∈ G}. Evidently XG has no circuits. In addition, XG is connected; given any pair of vertices v, w in XG, the geodesic joining v and w must be sent by any g ∈ G to the geodesic joining gv = v and gw = w! As the tree is uniquely geodesic and the group acts without inversions, the entire geodesic from v to w is also contained 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 following three 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 ﬁnitely generated.

Caveat. Amalgams do not necessarily map onto Z. An amalgam of torsion groups cannot map onto Z but restricting to non-torsion vertex groups doesn’t help. Burger and Moses have constructed examples of ﬁnitely presented 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 that finite groups have property (FA). Theorem 30 further shows that a finitely generated torsion group has property (FA). Indeed, let G be a finitely generated 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 only homomorphism 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’t hold and G decomposes as an amalgam, then G acts on the Bass Serre tree for this decomposition. If on the other hand, G maps onto Z, then G acts on the real line via this epimorphism. Therefore it suﬃces to prove the following lemma. Lemma 32. If G is a union of a strictly increasing chain of subgroups then G does not have property (FA).

Proof of Lemma. Let G1 < G2 < G3 < . . . < G be a strictly increasing chain of subgroups such that ∪iGi = G. Then, we have the canonical surjections ` φi : G/Gi → G/Gi+1. Deﬁne a graph X with vertex set V = i G/Gi and edges given by (gGi, φi(gGi)). Evidently, there are no circuits in the graph. 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, such −1 that g h ∈ Gk ⇔ gGk = hGk. Clearly the path (gGi, gGi+1, . . . , gGk) concatenated with the inverse of the path (hGj, hGj+1, . . . , hGk) connects gGi and hGj. Thus, X is a tree and the action of G on the cosets G/Gi extends to an action of G on the tree. If G ﬁxes a vertex gGi, then for all x ∈ G, xgGi = gGi ⇔ G = Gi, which contradicts the hypothesis that G is a union of a strictly increasing chain of subgroups Gj. This completes the proof of the forward implication in the statement of Theorem 30. Suppose now that (1)-(3) hold and let G act without inversions on a tree X. Set Y to be the quotient graph XmodG. If π1(Y ) 6= 1, then G maps onto Z which contradicts (2). Therefore Y is a tree and choosing a 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 lifts to 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 X such that no proper subtree of Z provides a graph of group decomposition of 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 G0 denotes the fundamental group of the tree ∼ 0 of groups associated to Z\{e, t(e)}, then G = G ∗Ge Gt(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 B have property (FA) then G has property (FA).

3. If a ﬁnite index subgroup H of G has property (FA) then G has property (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 induces an 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) and the fact that finite groups have property (FA). Otherwise, take a set of e transversals E for G/H and set K = ∩g∈EH . Then K is a normal subgroup of 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 group 4(3, 3, 3) generated by reflections in the sides of an equilateral triangle in 2 the plane has property (FA) (by Corollary 36 below) but it is virtually Z 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 and these will provide us with criteria for establishing property (FA) and ﬁnding Ping pong partners.

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

1. There exists a unique pair (P,Q) of vertices, P ∈ T1 and Q ∈ T2 such that 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. 24 CHAPTER 3. PROPERTY (FA) AND GEOMETRIC ASPECTS

3. Every sub-tree of X intersecting both the Ti contains [P,Q]. Lemma 34 (Fixed points for tree automorphisms). Suppose that G is acting on 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 the geodesic joining P to gP is obtained by concatenating the geodesic joining P to Q followed by the geodesic joining Q to gP . The lemma is proved by induction on the distance of P from Xg. Let d = d(P,Xg). If d = 0 there is nothing to prove. If d = 1 then there is an edge e with o(e) = P and t(e) = Q ∈ Xg. Now, t(ge) = Q and o(ge) = gP 6= P else P would have to belong in Xg. The general case is now clear and g[P,Q] = [gP, Q]. Very Useful Observation. The mid-point of the geodesic from P to gP is fixed by g. Lemma 35 (Criterion for Property (FA)). Let G be a group acting on a tree X. Let A, B be subgroups of G such that G = hA, Bi, A = ha1, . . . , aki a b and B = hb1, . . . , bli. If ∀i = 1, . . . , k and j = 1, . . . , l, X i 6= ∅, X j 6= ∅ and Xaibj 6= ∅, then XG 6= ∅. Proof. Evidently, XA 6= ∅ and XB 6= ∅. As G = hA, Bi, we have XG = XA ∩ XB. Suppose XA ∩ XB = ∅, then using the Bridge Lemma, we know there is a unique pair of points P ∈ XA and Q ∈ XB such that A B A d(P,Q) = d(X ,X ) ≥ 1. As Q/∈ X , there is some ai such that aiQ 6= Q. But, ai fixes P and so, the geodesic joining Q to aiQ = aibjQ (for any j) is the concatenation of the geodesic joining P and Q and the geodesic joining P and aiQ. By hypothesis, the aibj are elliptic and therefore each aibj fixes the mid-point of [Q, aibjQ], namely, the point P . But, aibjP = P for all j implies that bjP = P for all j, which contradicts the assumption that P/∈ XB. Therefore, XA and XB must intersect and consequently, XG 6= ∅.

Corollary 36. Let W = (S, (mst)) denote the Coxeter group generated by the reflections S and with associated Coxeter matrix (mst). If mst is finite for all s, t, then W has property (FA). Proof. The Coxeter group W is given by the presentation below. h S | s2 = 1 ∀ s ∈ S;(st)mst = 1 ∀s 6= t ∈ S i Recall that finite groups have property (FA). If W acts on a tree and s st the mst are all finite, then for all s, t, X 6= ∅, X 6= ∅. Hence, by Lemma 35, W has property (FA). 25

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

|g| = min d(P, gP ) P ∈vert(X)

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

Theorem 37 (Translation axis). The set T (g) forms the vertex set for a bi-inﬁnite geodesic in X on which g acts by translations of length |g|. Every sub-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 the geodesic 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, then gP1 = Pd−1. But then d(P1, gP1) = d(P1,Pd−1) ≤ d − 2. This contradicts n the fact that |g| = d. Clearly now, ∪n∈Z g [P, gP ] deﬁnes a bi-inﬁnite geodesic in X. Suppose now that Q is a point at a distance of n ≥ 1 from T (g). Let Q0 be the vertex in T (g) closest to Q. Then, we argue that the geodesic joining Q and gQ is produced by juxtaposing [Q, Q0], [Q0, gQ0] and [gQ0, gQ], thus giving d(Q, gQ) = |g| + 2d(Q, T (q)). This means in turn, that, T (g) = n ∪n∈Zg [P, gP ]! Moreover, suppose that a subtree T of X is preserved by g and its inverse. If Q is any vertex in T , then the geodesic [Q, gQ] is contained in T . Unless Q is already a vertex of T (g), [Q, gQ] contains the vertex Q0 ∈ T (g) as above. This implies that T (g) ⊂ T .

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

Commuting elements. One consequence of Theorem 37 is that commuting hyperbolic automorphisms have the same translation axes. Suppose g, 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). 26 CHAPTER 3. PROPERTY (FA) AND GEOMETRIC ASPECTS

3.0.3 Finding Ping Pong Partners Suppose a group is acting without inversions on a tree X. Let a, b be two elements of G which are hyperbolic automorphisms of the tree. One of the following 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 that will generate a copy of F2 inside G. Here, we will consider the case when T (a) ∩ T (b) is non-empty and of finite length. The remaining cases are left as exercises. Note that unless their intersection is empty, T (a) ∩ T (b) is a geodesic of finite length joining a pair of points, say P and Q. We may assume by replacing a or b by their inverses if needed, that the direction of 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 therefore aB− ⊂ A+. Also, aP ∈ A+ implies that aB+ ⊂ A+. Arguing as above we deduce 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 ha, bi generated by a and b in G is F2 using the following version of the Ping Pong Lemma. Lemma 39 (Ping Pong Lemma for cyclic subgroups). A group G is acting on a set Y ; let a, b ∈ G\{1}. Suppose there exists disjoint non-empty subsets of Y , called A±1,B±1 such that a(Y \A−) ⊂ A+, a−1(Y \A+) ⊂ A−, − + −1 + − b(Y \B ) ⊂ B and b (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 Lemma from Lemma 1. The discussion above proves

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

Corollary 41. Let G act without inversions on a tree; let a, b ∈ G be hyperbolic automorphisms such that T (a) ∩ T (b) has ﬁnite length. Then, n m ∼ there exist n, m ∈ Z such that ha , b i = F2.

3.0.4 Exercises Irreducible actions. We say, a group acts irreducibly on a tree if there is no global fixed point for the action and the whole group does not preserve a line or an end of the tree. Show that, if a group G acts irreducibly on a tree, then G contains a free group of rank 2. (Use the fact: if all the elements of the group are elliptic automorphisms of the tree then the group fixes an end. This implies, if the action is irreducible, then at least one element of the group acts as a hyperbolic automorphism.) This implies, if a group G is solvable or more generally amenable, then an action of G on a tree must be reducible: if the action is non-trivial (without a global fixed point), then the group will preserve a line or an end of the tree. Translation Length functions. Suppose that a group G is acting without inversions on a tree X. Define a function | . | : G → R as follows. For any g ∈ G, | g | = min{d(v, gv): v ∈ vert(X) } For instance, if g is elliptic then | g | = 0 and | g | ≥ 1 for any hyperbolic g. The function | . | is called a translation length function of G and we present 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 a single vertex, then | gh | = | gh−1 | = | g | + | h | + 2d(T (g),T (h)). 28 CHAPTER 3. PROPERTY (FA) AND GEOMETRIC ASPECTS

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 and Morgan’s Group Actions on R-trees.