Sanov subgroup in SL(2,Z) is free of rank two
is a free group with these two elements forming a freely generating set.
- Ping-pong lemma (the form involving two elements): If acts on a set , and and are such that neither nor is contained in the other. Further, suppose that and for all nonzero integers . Then, the subgroup of generated by and is a free group with as a freely generating set.
We construct an action that satisfies the hypothesis for fact (1) (the ping-pong lemma).
Let with acting on it the usual way (by matrix multiplication with column vectors). Consider the subsets:
We now prove the conditions for the ping-pong lemma:
- Neither of and is contained in the other: In fact, it is cler from the definition that they are disjoint non-empty sets.
- for : Suppose such that . Then . Since , by assumption.
- for : This is similar to the previous part.