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

Statement

Consider $G=SL(2,\mathbb {Z} )$, the 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. 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 $n\neq 0$: Suppose $(x,y)\in \mathbb {Z} ^{2}$ such that $|x|>|y|$. Then $a^{n}((x,y))=(x,2nx+y)$. Since $|x|>|y|$, $|2nx+y|\geq |2n||x|-|y|\geq 2|x|-|y|=|x|+(|x|-|y|)>|x|$ by assumption.
• $b^{n}(A)\subseteq B$ for $n\neq 0$: This is similar to the previous part.