ping-pong lemma
Theorem (Ping Pong Lemma).
Let and let be a group acting on a space . Supposewe are given a class of pairwise disjoint subsets of and suppose are elements of such that
( is the complement of in ). Then, the subgroupof generated by is free.
Before turning to prove the lemma let’s state threesimple facts:
Fact 1.
For all we have and
Proof.
∎
Fact 2.
If then .
Proof.
and are disjoint therefore .Similarly, so .In the same way, so .∎
Fact 3.
If then
Proof.
Assume by contradiction that . Then, and therefore any element of intersects with either or. However, the elements of are pairwise disjoint and thereare at least 4 elements in so this is a contradiction.∎
Using the above 3 facts, we now turn to the proof of the Ping Pong Lemma:
Proof.
Suppose we are given such that and .and suppose further that is freely reduced, namely, if then . We want to showthat in . Assume by contradiction that . We get acontradiction by giving such that and therefore contradicting Fact 3above since .
The set is chosen as follows. Assume that then:
Define the following subsets of :
To complete the proof we show by induction
that for if then:
- 1.
if then .
- 2.
if then .
For the above follows from Fact 1 and thespecific choice of . Assume it is true for andassume that . We have two cases to check:
- 1.
: by the induction hypothesis is a subset of for some .Therefore, by Fact 2 we get that is a subset of .Consequently, we get the following:
Hence, if then:
And if then:
- 2.
: by the fact that is freely reduced weget an equality between and . Hence, if then andtherefore:
Similiarly, if then and therefore:
∎