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

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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

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 \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.