Congruence subgroups as abstract groups

Question

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.

Answer 1

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 $8$.

Answer 2

Your question is a great example of the usefulness of the multiplicativity of Euler characteristic. The fractional Euler characteristic of $\textrm{GL}_2\mathbb{Z}$ is

$$\chi(\textrm{GL}_2\mathbb{Z})=\ \ {-\frac{1}{24}}$$ This means by definition that any torsion-free index-$k$ subgroup of $\textrm{GL}_2\mathbb{Z}$ has Euler characteristic $-\frac{k}{24}$. For a proof (indeed a number of different proofs), see this Math Overflow question.

The Euler characteristic is multiplicative for finite-index subgroups, and it's immediate from the definition that the same is true for fractional Euler characteristics. Therefore since $\pm\Gamma_0(3)$ has index 4 in $\textrm{GL}_2\mathbb{Z}$, we have

$$\chi\big(\!\pm\!\Gamma_0(3)\big)=\ \,-\frac{1}{6}$$

This demonstrates that $\pm\Gamma_0(3)$ is not isomorphic to $\textrm{GL}_2\mathbb{Z}$, and moreover any finite-index subgroup isomorphic to $\pm\Gamma_0(3)$ must be of index 4. You can certainly continue this analysis: for example, whenever $p$ is prime we have $\chi(\pm\Gamma_0(p))=-\frac{p+1}{24}$, so none of the groups $\pm\Gamma_0(p)$ are isomorphic to each other, or to $\Gamma_0(p)$, since $\chi(\Gamma_0(p))=-\frac{p+1}{12}$ (I leave it to you to work out what happens with $\Gamma_0(n)$ when $n$ is composite). However, you can only take it so far: there are certainly congruence subgroups that are non-isomorphic but can't be distinguished by their Euler characteristic.

Answer 3

For $n\geq 3$ (and in higher-rank in general) the question is basically settled by Super-rigidity. For example, if you have two lattices of $\mathrm{SL}_n(\mathbb{R})$ which are abstractly isomorphic then they must be obtained from each other by an automorphism of $\mathrm{SL}_n(\mathbb{R})$ (inverse transpose and/or inner conjugation).