Sanov subgroup in SL(2,Z) is free of rank two

Statement
Consider $$G = SL(2,\mathbb{Z})$$, the fact about::special linear group:SL(2,Z) (this is the subgroup of the general linear group:GL(2,Z) comprising the matrices of determinant one). Then, the subgroup of $$G$$ generated by the elements:

$$\begin{pmatrix}1 & 2 \\ 0 & 1\end{pmatrix}, \begin{pmatrix}1 & 0 \\ 2 & 1\end{pmatrix}$$

is a free group with these two elements forming a freely generating set.

Related facts

 * Dickson's theorem

Facts used

 * 1) uses::Ping-pong lemma (the form involving two elements): If $$G$$ acts on a set $$X$$, and $$a,b \in G$$ and $$A,B \subseteq X$$ are such that neither $$A$$ nor $$B$$ is contained in the other. Further, suppose that $$b^n(A) \subseteq B$$ and $$a^n(B) \subseteq A$$ for all nonzero integers $$n$$. Then, the subgroup of $$G$$ generated by $$a$$ and $$b$$ is a free group with $$\{ a,b \}$$ as a freely generating set.

Proof
We construct an action that satisfies the hypothesis for fact (1) (the ping-pong lemma).

Let $$X = \mathbb{Z}^2$$ with $$G$$ acting on it the usual way (by matrix multiplication with column vectors). Consider the subsets:

$$A = \{ (x,y) \mid |y| > |x| \}, B = \{ (x,y) \mid |x| > |y| \}$$.

We now prove the conditions for the ping-pong lemma:


 * Neither of $$A$$ and $$B$$ is contained in the other: In fact, it is cler from the definition that they are disjoint non-empty sets.
 * $$a^n(B) \subseteq A$$ for $$n \ne 0$$: Suppose $$(x,y) \in \mathbb{Z}^2$$ such that $$|x| > |y|$$. Then $$a^n((x,y)) = (x,2nx + y)$$. Since $$|x| > |y|$$, $$|2nx + y| \ge |2n||x| - |y| \ge 2|x| - |y| = |x| + (|x| - |y|) > |x|$$ by assumption.
 * $$b^n(A) \subseteq B$$ for $$n \ne 0$$: This is similar to the previous part.