# 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 $\Gamma(M)\Gamma(N)\subset \Gamma(G)$.

• I'm not sure, but it seems $\Gamma(G) \subset \Gamma(M)$ and $\Gamma(G) \subset \Gamma(N)$. Oct 10, 2011 at 18:52
• Unfortunately not, it goes in the other direction. For any positive integer $d$ dividing $M$ we have $\Gamma(d)\supset\Gamma(M)$.
– Alex
Oct 10, 2011 at 18:57
• One way to do it would be to simply compute the indices of these subgroups. This is certainly not the preferred way, however. But if you define the function $\theta(p^n)=(p^2-1)p^{3n-2}$, and extend it to be multiplicative over the natural numbers, then $\theta(n)$ is the index of $\Gamma(n)$ in $SL(2,\mathbb{Z})$. Then since you already have one containment, you want $\theta(m)\theta(n)=\theta(GCD)\theta(LCM)$, which is true by the multiplicative structure.
– user641
Oct 10, 2011 at 19:44
• @SteveD I'm afraid I don't see how this helps. If we let $c$ be the index of $\Gamma(M)\Gamma(N)$ in $SL(2,\mathbb{Z})$, then we know $\theta(GCD)$ must divide $c$, but I don't see the connection with $\theta(M)\theta(N)$ showing $c = \theta(GCD)$.
– Alex
Oct 10, 2011 at 22:41
• @Alex: The index of $\Gamma(M)\Gamma(N)$ is $\theta(M)\theta(N) / \theta(LCM)$.
– user641
Oct 10, 2011 at 22:48

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 $\Gamma(\gcd(M,N)) = \Gamma(M)\Gamma(N)$.