By the modular group I mean either $SL(2,\mathbb{Z})$ or $PSL(2,\mathbb{Z})$.


Where can I find examples of these?


Another question: is there a good (ideally analytical, but possibly computer-aided) way to determine if a subgroup of the modular group generated by some given matrices has finite index, and possibly allowing to compute the index?



Are there noncongruence subgroups (of finite index) of the modular group generated only by 2 or 3 elements?
The answer to your Question 1. is that there are not rank 2 subgroups of $PSL\_2(\mathbb{Z})$, but there are rank 3 subgroups. This follows from a [result of Wohlfahrt](http://projecteuclid.org/euclid.ijm/1256059574). He shows in Theorem 5
that any non-congruence subgroup $\Gamma < PSL\_2(\mathbb{Z})$ has index $\geq 7$. 
Misha points out in a comment to his answer that a finite coarea fuchsian group of rank $N$ has $Area(\mathbb{H}^2/\Gamma)\leq 2\pi(N-1)$. Therefore in the case $\Gamma$ is non-congruence, $Area(\mathbb{H}^2/\Gamma)=[PSL\_2(\mathbb{Z}):\Gamma] Area(\mathbb{H}^2/PSL\_2(\mathbb{Z})) \geq 7\pi/3$. Thus, $N=rank(\Gamma) \geq 3$. Misha's claim (in the non-uniform case) follows from an application of the free product decomposition of non-uniform Fuchsian group into cyclic groups, [Grushko's theorem](http://en.wikipedia.org/wiki/Grushko_theorem), and Gauss-Bonnet. 


Wohlfahrt also gives an example of a non-congruence subgroup of index $7$ in $PSL\_2(\mathbb{Z})$. On p. 531, he points out that there exists an index $7$
subgroup with two cusps. One determines that the group has genus $0$,
and two cone points of orders $2$ and $3$ respectively. This group is therefore
rank $3$, isomorphic to $\mathbb{Z}\ast \mathbb{Z}/2\ast\mathbb{Z}/3$. I 
didn't check, but I think this also lifts to a rank 3 subgroup of $SL\_2(\mathbb{Z})$.


