TITLE: free subgroups of $SL(2,\mathbb{R})$
QUESTION [5 upvotes]: 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.

REPLY [4 votes]: 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 $N^3$.