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 generate 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 statement

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

X₁⁺,...,X_k⁺ and X₁^–,...,X_k^–

of X with the following properties:

• a_i(X − X_i^–) ⊆ X_i⁺ for i = 1, ..., k;

• a_i⁻¹(X − X_i⁺) ⊆ X_i^– for i = 1, ..., k.

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

Proof

To simplify the argument, we will prove the statement under the following mild additional assumption:

• $X\ne \bigcup_{i=1}^k (X_i^+\cup X_i^-).$

The argument for the general case is similar to the one given below but requires more careful analysis.

Choose a point x in X such that

$x\not\in \bigcup_{i=1}^k (X_i^+\cup X_i^-).$

To show that H is free with free basis a₁,...,a_k it suffices to prove that every nontrivial freely reduced word in the alphabet

A = {a₁, ..., a_k, a₁⁻¹, ..., a_k⁻¹}

represents a nontrivial element of G.

Let w be such a freely reduced word, that is, w = b_nb_n−1...b₁, where n ≥ 1, where each b_j belongs to A and where w does not contain subwords of the form a_ia_i⁻¹ or a_i⁻¹a_i.

Induction on j shows that for every j = 1, ..., n we have

$b_j\dots b_2b_1\ x\in \bigcup_{i=1}^k (X_i^+\cup X_i^-).$

Thus

$wx\in \bigcup_{i=1}^k (X_i^+\cup X_i^-).$

Therefore wx ≠ x and hence w ≠ 1 in G, as required.

The name "ping-pong lemma" is motivated by the fact that, in the above argument, the point b_jb_j−1...b₁x bounces like a ping-pong between the sets X₁⁺, ..., X_k⁺, X₁^–,...,X_k^– as j varies over j = 1, ..., n.

Ping-pong lemma for several subgroups

There is also a version of the ping-pong lemma which ensures that several subgroups of a group acting on a set generate a free product.

A version for two subgroups

Let G be a group acting on a set X and let H₁, H₂ be two subgroups of G such that |H₁| ≥ 3 and |H₂| ≥ 2. Suppose there exist two non-empty subsets X₁ and X₂ of X such that the following hold:

• X₂ is not contained in X₁;
• for every h₁ ∈ H₁, h₁ ≠ 1 we have h₁(X₂) ⊆ X₁;
• for every h₂ ∈ H₂, h₂ ≠ 1 we have h₂(X₁) ⊆ X₂.

Then the subgroup H=<H₁, H₂>≤G of G generated by H₁ and H₂ is equal to the free product of H₁ and H₂:

H = H₁∗H₂.

A version for an arbitrary finite number of subgroups

The following version of the ping-pong lemma for several subgroups appears in .

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

• For any i≠j and for any h∈H_i, h≠1 we have h(X_j)⊆X_i.

Then

$\langle H_1,\dots, H_k\rangle=H_1\ast\dots \ast H_k.$

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

$\scriptstyle A=\begin{pmatrix}1 & 2\\ 0 &1 \end{pmatrix}$ and $\scriptstyle B=\begin{pmatrix}1 & 0\\ 2 &1 \end{pmatrix}$

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

$H_1=\{A^n|n\in \mathbb Z\}=\{\begin{pmatrix}1 & 2n\\ 0 & 1 \end{pmatrix} : n\in\mathbb Z\}$

and

$H_2=\{B^n|n\in \mathbb Z\}=\{\begin{pmatrix}1 & 0\\ 2n & 1 \end{pmatrix} : n\in\mathbb Z\}.$

Consider the standard action of SL(2,Z) on R² by linear transformations. Put

$X_1=\{\begin{pmatrix}x \\ y \end{pmatrix}\in \mathbb R^2 : |x|>|y|\}$

and

$X_2=\{\begin{pmatrix}x \\ y \end{pmatrix}\in \mathbb R^2 : |x|<|y|\}.$

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 = 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 nontrivial 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, h^m> ≤ 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 ∈ 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(∂G – U_–) ⊆ U₊
• g⁻ⁿ(∂G – U₊) ⊆ U_–
• h^m(∂G – V_–) ⊆ V₊
• h^−m(∂G – V₊) ⊆ V_–

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

Applications of the ping-pong lemma

• The ping-pong lemma is used in Kleinian groups to study their so-called Schottky subgroups. In the Kleinian groups context the ping-pong lemma can be used to show that a particular group of isometries of the hyperbolic 3-space is not just free but also properly discontinuous and geometrically finite.
• Similar Schottki-type arguments are widely used in geometric group theory, particularly for subgroups of word-hyperbolic groups and for automorphism groups of trees.
• Ping-pong lemma is also used for studying Schottki-type subgroups of mapping class groups of Riemann surfaces, where the set on which the mapping class group acts is the Thurston boundary of the Teichmüller space. A similar argument is also utilized in the study of subgroups of the outer automorphism group of a free group.
• One of the most famous applications of the ping-pong lemma is in the proof of Jacques Tits of the so-called Tits alternative for linear groups. (see also for an overview of Tits' proof and an explanation of the ideas involved, including the use of the ping-pong lemma).
• There are generalizations of the ping-pong lemma that produce not just free products but also amalgamated free products and HNN extensions. These generalizations are used, in particular, in the proof of Maskit's Combination Theorem for Kleinian groups.
• There are also versions of the ping-pong lemma which guarantee that several elements in a group generate a free semigroup. Such versions are available both in the general context of a group action on a set, and for specific types of actions, e.g. in the context of linear groups, groups acting on trees and others.

See also

• Free group
• Free product
• Kleinian group
• Tits alternative
• Word-hyperbolic group
• Schottky group