# Conjugate of congruence subgroup of level N contains congruence subgroup of level ND

I was wondering if anyone could give me tips on the following question:

Suppose $\alpha \in\text{GL}_2^{+}(\mathbb{Q})$ has integral entries and is such that det$(\alpha) = D > 0$.

If $\Gamma$ is a congruence subgroup of level $N$ then $\alpha^{-1}\Gamma\alpha$ contains a congruence subgroup of level $ND$.

I am really just starting out with modular forms so don't know a great deal of stuff. I hope I haven't missed something simple.

EDIT - Since yesterday the problem has been solved (on overflow). Basically it is easier to show that $\alpha\Gamma(ND)\alpha^{-1} \subseteq \Gamma$. The result follows.

In order to show this use the fact that:

$\alpha(\gamma - I)\alpha^{-1} = \frac{1}{D}\alpha(\gamma-I)\text{adj}(\alpha)^{T} \equiv 0$ mod $N$

for any $\gamma\in\Gamma(ND)$ (since $\gamma - I \equiv 0$ mod $ND$ and so all matrix entries in the "numerator" are integers divisible by $ND$...cancelling by $D$ still gives divisibility by $N$).

From this it follows that $\alpha\gamma\alpha^{-1} \equiv I$ mod $N$, so that:

$\alpha\gamma\alpha^{-1} \in \Gamma(N) \subseteq \Gamma$

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