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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.

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is there an element of order $4$ in $\pm\Gamma_0(3)$? –  Junkie Aug 3 '11 at 6:59

3 Answers 3

up vote 13 down vote accepted

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 $8$.

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Brilliant, thank you! I knew this should be doable by just looking at torsion, but the fact I was missing was that $\Gamma_1(N)$, the group of matrices congruent to the identity modulo $N$ is torsion free. Is this easy to see? –  Alex B. Aug 3 '11 at 8:13
    
@Alex Bartel: Maybe consider the trace and determinant (minimal polynomial) of an element that has finite order. –  Junkie Aug 3 '11 at 8:16
    
@Junkie: of course, that works. Thanks! –  Alex B. Aug 3 '11 at 8:24
    
@Alex: as long as $N \neq 2$, of course. –  Geoff Robinson Aug 3 '11 at 13:57
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Just a stupid remark, you use that the group is torsion free every time you say the modular curve (just defined as the quotient of the upper half plane modulo your group) is a complex curve, since you don't have elliptic points in these cases (and you do in level 1). –  A. Pacetti Aug 3 '11 at 20:18

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.

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Very nice, thank you! –  Alex B. Aug 25 '13 at 13:01

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).

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