free subgroups of $SL(2,\mathbb{R})$

Question

In the example section of the wikipedia article on the the Ping Pong lemma, you can see how to construct a free subgroup of $SL(2,\mathbb{R})$ with two generators $$ a_1 = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}, \ \ \ \ \ a_2 = \begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix}. $$ Is it possible to construct free subgroups of $SL(2,\mathbb{R})$ with an arbitrary number of generators in a similar way (using the Ping Pong lemma)? Do such subgroups even exist?

Sorry if this is a stupid question, I'm a noob at group theory.

Answer 1

Subgroups of free groups are free (this is well-known theorem by Nilsen-Schreier). Free group $H$ on two generators $F_2$ contains a subgroup of countable rank (for example, its commutator subgroup). This implies that $F_2$ contains any subgroup of finite rank. Indeed, if $H=\langle a_1,a_2,a_3,\dots\rangle$, then $\langle a_1,\dots, a_n\rangle$ gives you a free subgroup in $F_2$ of rank $n$.

So the free subgroup on $2$ generators that you've found in $SL(2,\mathbb{R})$ gives you automatically free subgroups of $SL(2,\mathbb{R})$ of arbitrary countable rank.

Answer 2

It seems worth noting that one can also explicitly construct free subgroups of $\mathrm{SL}(2,\mathbb R)$ of arbitrarily large (finite) rank.

Namely, if we let $\Gamma(N)$ denote the subgroup of $\mathrm{SL}(2,\mathbb Z)$ consisting of matrices that are $\equiv 1 \bmod N$, then for $N \geq 3$, the group $\Gamma(N)$ is free, on a number of generators that grows something like $N^3$.