This is probably well know, and maybe even trivial, but not to me. Consider for concreteness the subgroup
$$
\pm\Gamma\_0(3)=\left\{\begin{pmatrix}a & b \\ c & d\end{pmatrix}:\;a,b,c,d\in\mathbb{Z},ad-bc=\pm1, c\equiv 0\pmod 3\right\}
$$
of $GL\_2(\mathbb{Z})$. This has of course index 4 in $GL\_2(\mathbb{Z})$. The first (possibly completely ridiculous) question is



> 
> Does $\pm\Gamma\_0(3)$ contain a subgroup isomorphic to $GL\_2(\mathbb{Z})$?
> 
> 
> 


It's not even obvious to me that the two are not isomorphic as abstract groups. The second question is



> 
> Does $GL\_2(\mathbb{Z})$ contain subgroups that are isomorphic to $\pm\Gamma\_0(3)$ with finite index other than 4? If the answer is yes, then what is the least common multiple of all such indices? E.g. is there a subgroup of index 3 (or 5, or 7, or...) in $GL\_2(\mathbb{Z})$ isomorphic to $\pm\Gamma\_0(3)$? Or will all such indices be multiples of 4?
> 
> 
> 


An answer or technique that is applicable to other congruence subgroups and to other values of 2 would be a great bonus, but for now I would happily settle for an answer to this concrete question.



Congruence subgroups as abstract groups
Junkie's comment answers both parts of the first question, since $\pm \Gamma\_{0}(3)$ contains no element of order $4$ (its image after reduction (mod 3) would still have order $4$). They also provide a suggestion to deal with other primes. For other $p> 3$, I think you can do something like this. The matrices congruent to the identity (entrywise) (mod p) form a torsion free normal subgroup $H$ of ${\rm GL}(2,\mathbb{Z})$. The image of the congruence subgroup (mod $H$) is solvable, 
and has a normal Sylow $p$-subgroup with Abelian factor group. However, 
if $X$ is a subgroup of the congruence subgroup isomorphic to ${\rm GL}(2,\mathbb{Z})$,
then $X/X \cap H$ contains a dihedral subgroup of order $8$.


