# 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.

-

Let $A\equiv 1\mod ND$, then $$\alpha (A-1) \alpha^{-1}=\frac{1}{D}\alpha (A-1) \mathrm{adj}(\alpha)\equiv 0 \mod N.$$ Hence $\alpha A\alpha^{-1}\equiv 1 \mod N$. Here $\mathrm{adj}(\alpha)$ is the adjoint matrix (the matrix of co-factors); we use the fact that all entries of $\mathrm{adj}(\alpha)$ are integers, and all entries of $A-1$ are integers divisible by $ND$. Therefore $\alpha^{-1}$ conjugates the $ND$-congruence subgroup into the $N$-congruence subgroup, so the conjugation by $\alpha$ does what you want.