# Ping-pong lemma

In mathematics, the ping-pong lemma, or table-tennis lemma, is any of several mathematical statements that ensure that several elements in a group acting on a set freely generates a free subgroup of that group.

## History

The ping-pong argument goes back to late 19th century and is commonly attributed to Felix Klein who used it to study subgroups of Kleinian groups, that is, of discrete groups of isometries of the hyperbolic 3-space or, equivalently Möbius transformations of the Riemann sphere. The ping-pong lemma was a key tool used by Jacques Tits in his 1972 paper containing the proof of a famous result now known as the Tits alternative. The result states that a finitely generated linear group is either virtually solvable or contains a free subgroup of rank two. The ping-pong lemma and its variations are widely used in geometric topology and geometric group theory.

Modern versions of the ping-pong lemma can be found in many books such as Lyndon&Schupp, de la Harpe, Bridson&Haefliger and others.

## Formal statements

### Ping-pong lemma for several subgroups

This version of the ping-pong lemma ensures that several subgroups of a group acting on a set generate a free product. The following statement appears in, and the proof is from.

Let G be a group acting on a set X and let H1, H2,...., Hk be nontrivial subgroups of G where k≥2, such that at least one of these subgroups has order greater than 2. Suppose there exist pairwise disjoint nonempty subsets X1, X2,....,Xk of X such that the following holds:

• For any is and for any hHi, h≠1 we have h(Xs)⊆Xi.

Then

$\langle H_{1},\dots ,H_{k}\rangle =H_{1}\ast \dots \ast H_{k}.$ #### Proof

By the definition of free product, it suffices to check that a given reduced word is nontrivial. Let w be such a word, and let

$w=\prod _{i=1}^{m}w_{\alpha _{i},\beta _{i}}.$ Where ws,βjHs for all such βj, and since w is fully reduced αi≠ αi+1 for any i. We then let w act on an element of one of the sets Xi. As we assume for at least one subgroup Hi has order at least 3, without loss of generality we may assume that H1 has order at least 3. We first make the assumption that α1 and αm are both 1. From here we consider w acting on X2. We get the following chain of containments and note that since the Xi are disjoint that w acts nontrivially and is thus not the identity element.

$w(X_{2})\subseteq \prod _{i=1}^{m-1}w_{\alpha _{i},\beta _{i}}(X_{1})\subseteq \prod _{i=1}^{m-2}w_{\alpha _{i},\beta _{i}}(X_{\alpha _{m-1}})\subseteq \dots \subseteq w_{1,\beta _{1}}w_{\alpha _{2},\beta _{2}}(X_{\alpha _{3}})\subseteq$ $w_{1,\beta _{1}}(X_{\alpha _{2}})\subseteq X_{1}$ To finish the proof we must consider the three cases:

• If $\alpha _{1}=1;\alpha _{m}\neq 1$ , then let $h\in H_{1}\setminus \{w_{1,\beta _{1}}^{-1},1\}.$ (Such $h$ exists since by assumption H1 has order at least 3.)
• If $\alpha _{1}\neq 1;\alpha _{m}=1$ , then let $h\in H_{1}\setminus \{w_{1,\beta _{m}},1\}$ • And if $\alpha _{1}\neq 1;\alpha _{m}\neq 1$ , then let $h\in H_{1}\setminus \{1\}$ In each case, hwh−1 is a reduced word with α1' and αm '' both 1, and thus is nontrivial. Finally, hwh−1 is not 1, and so neither is w. This proves the claim.

### The Ping-pong lemma for cyclic subgroups

Let G be a group acting on a set X. Let a1,...,ak be elements of G of infinite order, where k ≥ 2. Suppose there exist disjoint nonempty subsets

X1+,...,Xk+ and X1,...,Xk

of X with the following properties:

• ai(X − Xi) ⊆ Xi+ for i = 1, ..., k;
• ai−1(X − Xi+) ⊆ Xi for i = 1, ..., k.

Then the subgroup H = <a1, ..., ak> ≤ G generated by a1, ..., ak is free with free basis {a1, ..., ak}.

#### Proof

This statement follows as a corollary of the version for general subgroups if we let Xi= Xi+Xi and let Hi = ⟨ai⟩.

## Examples

### Special linear group example

One can use the ping-pong lemma to prove that the subgroup H = <A,B>≤SL(2,Z), generated by the matrices

$A={\begin{pmatrix}1&2\\0&1\end{pmatrix}}$ and $B={\begin{pmatrix}1&0\\2&1\end{pmatrix}}$ is free of rank two.

#### Proof

Indeed, let H1 = <A> and H2 = <B> be cyclic subgroups of SL(2,Z) generated by A and B accordingly. It is not hard to check that A and B are elements of infinite order in SL(2,Z) and that

$H_{1}=\{A^{n}|n\in \mathbb {Z} \}=\left\{{\begin{pmatrix}1&2n\\0&1\end{pmatrix}}:n\in \mathbb {Z} \right\}$ and

$H_{2}=\{B^{n}|n\in \mathbb {Z} \}=\left\{{\begin{pmatrix}1&0\\2n&1\end{pmatrix}}:n\in \mathbb {Z} \right\}.$ Consider the standard action of SL(2,Z) on R2 by linear transformations. Put

$X_{1}=\left\{{\begin{pmatrix}x\\y\end{pmatrix}}\in \mathbb {R} ^{2}:|x|>|y|\right\}$ and

$X_{2}=\left\{{\begin{pmatrix}x\\y\end{pmatrix}}\in \mathbb {R} ^{2}:|x|<|y|\right\}.$ It is not hard to check, using the above explicitly descriptions of H1 and H2 that for every nontrivial g ∈ H1 we have g(X2) ⊆ X1 and that for every nontrivial g ∈ H2 we have g(X1) ⊆ X2. Using the alternative form of the ping-pong lemma, for two subgroups, given above, we conclude that H = H1H2. Since the groups H1 and H2 are infinite cyclic, it follows that H is a free group of rank two.

### Word-hyperbolic group example

Let G be a word-hyperbolic group which is torsion-free, that is, with no nontrivial elements of finite order. Let gh ∈ G be two non-commuting elements, that is such that gh ≠ hg. Then there exists M≥1 such that for any integers n ≥ M, m ≥ M the subgroup H = <gn, hm> ≤ G is free of rank two.

#### Sketch of the proof

The group G acts on its hyperbolic boundaryG by homeomorphisms. It is known that if a ∈ G is a nontrivial element then a has exactly two distinct fixed points, a and a−∞ in ∂G and that a is an attracting fixed point while a−∞ is a repelling fixed point.

Since g and h do not commute, the basic facts about word-hyperbolic groups imply that g, g−∞, h and h−∞ are four distinct points in ∂G. Take disjoint neighborhoods U+, U, V+ and V of g, g−∞, h and h−∞ in ∂G respectively. Then the attracting/repelling properties of the fixed points of g and h imply that there exists M ≥ 1 such that for any integers n ≥ M, m ≥ M we have:

• gn(∂GU) ⊆ U+
• gn(∂GU+) ⊆ U
• hm(∂GV) ⊆ V+
• hm(∂GV+) ⊆ V

The ping-pong lemma now implies that H = <gn, hm> ≤ G is free of rank two.