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Congruence subgroups of $SL_2(\mathbb{Z})$ usually seem to be defined as a subgroup that contains $\Gamma(N) = \left\{\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \mod N \right\}$ for some positive integer $N$. But, I have also seen it defined as a subgroup defined by congruence conditions on the entries. The problem is, this isn't very specific. For example, what if I look at the set $\left\{\begin{pmatrix} a & b \\ c & d \end{pmatrix} : a^b \equiv c^2 d^{15} \mod 17 \right\}$? I just made that up off the top of my head and odds are it's not closed under multiplication and thus not a subgroup. Maybe the solution set is empty. But, assuming it is a nonempty subgroup, would that be considered a congruence subgroup? Or, more generally, are there more specific conditions on what types of congruences are allowed, i.e., is there a rigorous definition somewhere?

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I don't think the notion of "congruence relation on the entries" is supposed to include exponential functions. I think it's meant to be linear congruence relation on the entries, or perhaps some larger subset. – Arturo Magidin Apr 5 '11 at 19:01
I think the only examples I have ever seen have been of the form $\left\{\begin{pmatrix} a & b \\ c & d \end{pmatrix} \equiv x : x \in X\right\}$, where $X$ is some set of matrices. – Graphth Apr 5 '11 at 19:16
up vote 3 down vote accepted

If the subgroup $\Gamma$ contains $\Gamma(N)$, then we can find a coset decomposition $\Gamma = \displaystyle \cup_{\alpha \in A} \alpha \Gamma(N)$ where $A$ is a set of coset representatives.

Then we have $\Gamma = \left\{ \begin{pmatrix} a & b \\ c & d \end{pmatrix} \equiv \alpha \pmod{N}: \alpha \in A \right\}$.

If you have any set of linear congruence relations on $a, b, c, d$ that define a subgroup of $SL_2(\mathbb{Z})$, you can just take the lcm of the moduli of congruence and construct a set $A$ of coset representatives.

I believe that this is the content of the comment that "congruence subgroups are those defined by congruence relations." I don't know of anything to say about other forms of congruences, such as the one you suggest.

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