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 ⊂ Gnf1g there exists y 2 G such that for each x 2 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 maxfjAj; jBjg ≥ 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 2 Anf1g, and • bX1 ⊂ X2 for all b 2 Bnf1g. 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 2 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 fGigi2I 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 2 W (A), 0 0 • wew ≈ ww , if e is an identity element from some Gi 0 0 • wabw ≈ wcw , if a; b; c 2 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 ∗i2I Gi. Example 2. If Gi is infinite cyclic for each i 2 I, then ∗i2I 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 2 Gni . Then, w is said to be reduced if ni 6= ni+1 for all i = 1; : : : k − 1. Proposition 4. Every element of ∗i2I 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 2 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 ∗i2I Gi to SX , the symmetric group on X. For each a 2 A, let σ(a): X ! X be the map that takes w to R(aw). Extend σ to all of ∗i2I Gi by setting σ(a1 : : : ak to be the composition σ(a1) : : : σ(ak). To complete the argument, observe now that for any w 2 X, σ(w) 2 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 ∗i2I 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 : ∗i2I Gi ! H such that F ◦ γi = fi for all i 2 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 maxfjAj; jBjg ≥ 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 2 Anf1g, and • bX1 ⊂ X2 for all b 2 Bnf1g. 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 Anf1g and the bi's belong to Bnf1g. 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 2 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 jzj ≥ 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 C^ = C [ f1g by fractional linear transformations, i.e. a b az + b :z = : c d cz + d ^ ^ Choose X1 = fz 2 C : jzj < 1g and X2 = fz 2 C : jzj > 1g. Then, n M1 X1 ⊂ X2 and JX2 ⊂ X1. Invoke the Ping Pong Lemma to deduce the Proposition. 1.4 Exercises 1. Let W := W ([i2I 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 f[a; b] = a−1b−1ab : a 2 Anf1g; b 2 Bnf1gg. (b) If A = Z=2Z and B = Z=3Z, then A∗B is the modular group P SL(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 P SL(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 2 G, there exists a surjection f : G ! F to a finite group F such that f(g) 6= 1.