I'm hoping there's a pleasant way to solve this one.
Prove that $\Gamma(M)\Gamma(N) = \Gamma(\gcd(M,N))$.
Showing that $\Gamma(M)\Gamma(N) \subset \Gamma(\gcd(M,N))$ is rather straight forward, but the other direction seems to involve solving a system of two systems of equations, which could be rather messy. I was wondering if any knows of a nicer way to attack this problem?
Thanks!
For completeness, I may as well show the direction I said was straight forward, though this is not necessarily important to my question,
Let $\gamma_M\in\Gamma(M)$ and $\gamma_N\in\Gamma(N)$ be given by $ \gamma_M = \begin{pmatrix} a & b\\ c & d \end{pmatrix},$ $ \gamma_N = \begin{pmatrix} e & f\\ g & h \end{pmatrix}.$ We can compute their product, $ \gamma_M\gamma_N = \begin{pmatrix} ae + bg & af + bh\\ ce + dg & cf + dh \end{pmatrix},$ and notice that by the restrictions of $a,b,c,d$ modulo $M$ and $e,f,g,h$ modulo $N$ we know, letting $G = \gcd(M,N)$, $ ae + bg \equiv 1(\bmod\ G)$ $ af + bh \equiv 0(\bmod\ G)$ $ ce + dg \equiv 0(\bmod\ G)$ $ cf + dh \equiv 1(\bmod\ G)$ and so $\gamma_M\gamma_N\in\Gamma(G)$. This shows $\Gamma(M)\Gamma(N)\subset \Gamma(G)$.