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 "H"1, "H"2..., "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 pairwise disjoint nonempty subsets "X"1, "X"2...,"X""k" of "X" such that the following holds:
Then
Proof.
By the definition of free product, it suffices to check that a given (nonempty) reduced word represents a nontrivial element of formula_2. Let formula_3 be such a word of length formula_4, and let
where formula_6 for some formula_7. Since formula_8 is reduced, we have formula_9for any formula_10 and each formula_11 is distinct from the identity element of formula_12. We then let formula_3 act on an element of one of the sets formula_14. As we assume that at least one subgroup formula_15 has order at least 3, without loss of generality we may assume that formula_16 has order at least 3. We first make the assumption that formula_17and formula_18 are both 1 (which implies formula_19). From here we consider formula_3 acting on formula_21. We get the following chain of containments:
By the assumption that different formula_14's are disjoint, we conclude that formula_3 acts nontrivially on some element of formula_21, thus formula_3 represents a nontrivial element of formula_27. 
To finish the proof we must consider the three cases: 
In each case, formula_36after reduction becomes a reduced word with its first and last letter in formula_37. Finally, formula_36represents a nontrivial element of formula_2, and so does formula_40. This proves the claim.
The Ping-pong lemma for cyclic subgroups.
Let "G" be a group acting on a set "X". Let "a"1...,"a""k" be elements of "G" of infinite order, where "k" ≥ 2. Suppose there exist disjoint nonempty subsets
of "X" with the following properties:
Then the subgroup "H" = <"a"1, ..., "a""k"> ≤ "G" generated by "a"1, ..., "a""k" is free with free basis {"a"1, ..., "a""k"}.
Proof.
This statement follows as a corollary of the version for general subgroups if we let "X""i"= "X""i"+∪"X""i"− and let "H""i" = ⟨"a""i"⟩.
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
is free of rank two.
Proof.
Indeed, let "H"1 = <"A"> and "H"2 = <"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
and
Consider the standard action of SL(2,Z) on R2 by linear transformations. Put
and
It is not hard to check, using the above explicitly descriptions of "H"1 and "H"2 that for every nontrivial "g" ∈ "H"1 we have "g"("X"2) ⊆ "X"1 and that for every nontrivial "g" ∈ "H"2 we have "g"("X"1) ⊆ "X"2. Using the alternative form of the ping-pong lemma, for two subgroups, given above, we conclude that "H" = "H"1∗"H"2. Since the groups "H"1 and "H"2 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 = <"g""n", "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:
The ping-pong lemma now implies that "H" = <"g""n", "h""m"> ≤ "G" is free of rank two.