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I've a question regarding the congruence subgroup.

We have the following congruence subgroup

$\Gamma^* := \left\{ M \in \Gamma | XM \equiv X(mod \; \mathbb{Z}^2),\; m \cdot det|\begin {smallmatrix} X \\ XM \end{smallmatrix}| \in \mathbb{Z} \right\}$

of $\Gamma$.


$\Gamma$ is a subgroup of $SL_2(\mathbb{Z})$,

$M = \begin{pmatrix} a & b \\ c & d \\ \end{pmatrix} \in \Gamma$, $X = \begin{pmatrix} \lambda & \mu \end{pmatrix}\in \mathbb{Q}^2$, $m \in \mathbb{Z}$

This subgroup can also be written as

$\Gamma^* = \left\{ \begin{pmatrix} a & b \\ c & d \\ \end{pmatrix} \in \Gamma | (a-1)\lambda + c \mu ,\; b \lambda + (d-1)\mu,\; m(c \mu^2 + (d-a)\lambda\mu-b\lambda^2) \in \mathbb{Z} \right\}$

and hence contains $\Gamma \cap \Gamma \left(\frac{N^2}{(N,m)}\right)$ if $NX \in \mathbb{Z}^2$.

My question is, why is $\Gamma^* \supset \Gamma \cap \Gamma \left(\frac{N^2}{(N,m)}\right)$ true? What does $\Gamma \left(\frac{N^2}{(N,m)}\right)$ mean exactly? Actually I only have found references to the main congruence subgroup $\Gamma(N)$ (but never anything to a more complex form).

Thanks, Jan

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At a first glance, I'd guess $\Gamma(N^2/(N,m))$ means precisely that. The principal congruence subgroup of level $N^2/gcd(N,m)$. (I am supposing $m$ is a fixed integer, yes?) – Kimball Apr 10 '11 at 21:16
Hi Komball, thanks for your answer. Actually my question is not asked as precise as I would like to ask it. I was aware of the notion of the gcd. However, my question is how you could come to the conclusion that the subgroup $\Gamma^*$ contains $\Gamma \cap \Gamma \left(\frac{N^2}{(N,m)}\right)$. I am missing some steps why you can write this. – Jan Apr 10 '11 at 21:50
Actually the question is more like, what is the connection between $\Gamma^*$ and $\Gamma(\frac{N^2}{(N,m)})$. However, the missing part is, how you come from $\Gamma^*$ to $\Gamma \cap \Gamma \left(\frac{N^2}{(N,m)}\right)$ and how are the intermediate steps or what rules/definition do you apply in order to make the statement $\Gamma^* \supset \Gamma \cap \Gamma \left(\frac{N^2}{(N,m)}\right)$. Thanks a lot – Jan Apr 10 '11 at 21:56

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