Relation between congruence subgroups. $\Gamma(M)\Gamma(N) = \Gamma(\gcd(M,N))$ 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 $\Gamma(M)\Gamma(N)\subset \Gamma(G)$.
 A: As I don't get points for accepting my own answer, I suppose I'll add my solution to have something to accept :).
We have $\Gamma(M)\Gamma(N) = \{\gamma_m\gamma_n:\gamma_m\in\Gamma(M)$ and $\gamma_n\in\Gamma(N)\}$.
We know $\Gamma(\gcd(M,N))\supset\Gamma(M)$ and $\Gamma(\gcd(M,N))\supset\Gamma(N)$. This implies any element of $\Gamma(M)\Gamma(N)$ can be viewed as a product of elements of $\Gamma(\gcd(M,N))$, which must be in $\Gamma(\gcd(M,N))$, so we have 
\begin{align}
 \Gamma(\gcd(M,N))\supset\Gamma(M)\Gamma(N).\ \ \ \ \ \ \ \ \ \ \ \ (\ast)
\end{align}
Now we will show that in fact, $\Gamma(\gcd(M,N)) = \Gamma(M)\Gamma(N)$. We know the index of $\Gamma(N)$ in $\Gamma(1)$ is
$$ N^3 \prod_{p|N} \left(1 - \frac{1}{p^2}\right).$$
Define $i(N)$ to be given by this formula for any positive integer $N$. Then $i(N)$ is multiplicative as $i(1) = 1^3 = 1$ and if $N = qr$ for some positive, relatively prime integers $q$ and $r$, we have
\begin{align*}
 i(qr) = i(N) &= N^3\prod_{p|N} \left(1 - \frac{1}{p^2}\right)\newline
 &= (qr)^3\prod_{p|qr} \left(1 - \frac{1}{p^2}\right)\newline
 &= q^3r^3\prod_{p|q} \left(1 - \frac{1}{p^2}\right)\prod_{p|r} \left(1 - \frac{1}{p^2}\right)\newline
 &= \left(q^3\prod_{p|q} \left(1 - \frac{1}{p^2}\right)\right)\left(r^3\prod_{p|r} \left(1 - \frac{1}{p^2}\right)\right) = i(q)i(r).
\end{align*}
Now the index of $\Gamma(M)\Gamma(N)$ in $\Gamma(1)$ must be, from the second isomorphism theorem, $i(M)i(N)/i((\operatorname{lcm}(M,N)))$. But as $i$ is multiplicative, we have
$$ i(M)i(N) = i(MN) = i(\gcd(M,N))i(L),$$
so dividing by $i(L)$, we see $\Gamma(\gcd(M,N)) = \Gamma(M)\Gamma(N)$.
