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 the 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 Olijnyk and Suchchansky (2004), and the proof is from de la Harpe (2000). Let G be a group acting on a set X and let H1, H2, ..., Hk be 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: i ≠ s and for any h in Hi , h ≠ 1 we have h(Xs) ⊆ Xi . Then

Proof

By the definition of free product, it suffices to check that a given (nonempty) reduced word represents a nontrivial element of G. Let w be such a word of length m\geq 2, and let where for some. Since w is reduced, we have for any and each w_i is distinct from the identity element of. We then let w act on an element of one of the sets X_i. As we assume that at least one subgroup H_i has order at least 3, without loss of generality we may assume that H_1 has order at least 3. We first make the assumption that \alpha_1and \alpha_m are both 1 (which implies m \geq 3). From here we consider w acting on X_2. We get the following chain of containments: By the assumption that different X_i's are disjoint, we conclude that w acts nontrivially on some element of X_2, thus w represents a nontrivial element of G. To finish the proof we must consider the three cases: In each case, hwh^{-1} after reduction becomes a reduced word with its first and last letter in H_1. Finally, hwh^{-1} represents a nontrivial element of G, and so does 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 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⟩ ≤ SL2(Z) , generated by the matrices and is free of rank two.

Proof

Indeed, let H1 = ⟨A⟩ and H2 = ⟨B⟩ be cyclic subgroups of SL2(Z) generated by A and B accordingly. It is not hard to check that A and B are elements of infinite order in SL2(Z) and that and Consider the standard action of SL2(Z) on R2 by linear transformations. Put and It is not hard to check, using the above explicit 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 = H1 ∗ H2 . 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 nonidentity elements of finite order. Let g, h ∈ 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 boundary ∂G by homeomorphisms. It is known that if a in G is a nonidentity 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, 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(∂G – U–) ⊆ U+ g−n(∂G – U+) ⊆ U– hm(∂G – V–) ⊆ V+ h−m(∂G – V+) ⊆ V– The ping-pong lemma now implies that H = ⟨gn, hm⟩ ≤ G is free of rank two.

Applications of the ping-pong lemma