A question about congruence subgroups [closed]

Question

For which $N_1$ and $N_2$ and $N$ be the greatest common divisor of $N_1$ and $N_2$, it is true that a congruence subgroup in $\mathrm{SL}_2(\mathbb{Z})$ generated by $\Gamma(N_1)\cup\Gamma(N_2)$ contains $\Gamma(N)$, where $\Gamma(N)$ is the principal congruence subgroup of level $N$.
If $(N_1,N_2)=1$, the $\Gamma(N_1)$ and $\Gamma(N_2)$ can generate $\mathrm{SL}_2(\mathbb{Z})=\Gamma(1)$ because it's easy to show that they can generate the generators of $\mathrm{SL}_2(\mathbb{Z})$, which is to say the proposition is true in this case. But in general, I am not sure if this proposition is right.

Answer 1

Here is an alternate proof which generalizes to the case of the groups $\Gamma_1(N)$ and $\Gamma_0(N)$.

Let $N_1,N_2 \geq 1$ be integers. Put $N=\mathrm{gcd}(N_1,N_2)$ and $M=\mathrm{lcm}(N_1,N_2)$. To ease notation, denote also $G_n = \Gamma(n) \backslash \mathrm{SL}_2(\mathbb{Z}) \cong \mathrm{SL}_2(\mathbb{Z}/n\mathbb{Z})$ for any integer $n$.

We will show that $\Gamma = \langle \Gamma(N_1), \Gamma(N_2) \rangle$ is equal to $\Gamma(N)$. One inclusion is clear. Let $G=\Gamma \backslash \mathrm{SL}_2(\mathbb{Z})$. We have injective maps \begin{equation*} G_M \hookrightarrow G_{N_1} \times_G G_{N_2} \hookrightarrow G_{N_1} \times_{G_N} G_{N_2} \end{equation*} where the fiber product over the group $G$ means the set of pairs having the same image in $G$. The first map is the diagonal map, and the second map comes from the canonical projection $G \to G_N$.

It suffices to prove that the second map is bijective, and for this it is enough to show that $G_M$ and $G_{N_1} \times_{G_N} G_{N_2}$ have the same cardinality. Note that this does not involve $\Gamma$ anymore. By the Chinese remainder theorem, it is enough to consider the case where $N_1$ and $N_2$ are powers of a fixed prime. But this case is obvious: if say $N_1$ divides $N_2$, then $M=N_2$ and the fiber product reduces to $G_{N_2}$.

The same reasoning works for the groups $\Gamma_1(N)$, respectively $\Gamma_0(N)$, replacing the group $G_N$ by the set $E_N = \Gamma_1(N) \backslash \mathrm{SL}_2(\mathbb{Z})$, respectively $\Gamma_0(N) \backslash \mathrm{SL}_2(\mathbb{Z})$. The only property that we need is that for any two coprime integers $N_1,N_2$, the canonical map $E_{N_1 N_2} \to E_{N_1} \times E_{N_2}$ is bijective, which is not hard to show for the above groups (the map is injective and it suffices to compute the cardinalities).

Answer 2

This is true in general, by an argument similar to the one you used for the coprime case. The subgroup $\Gamma$ generated by $\Gamma(N_1) \cup \Gamma(N_2)$ contains the matrices $$ a = \begin{pmatrix} 1 & N_1 \\ 0 & 1 \end{pmatrix}, b = \begin{pmatrix} 1 & N_2 \\ 0 & 1 \end{pmatrix} $$ so it also contains $\begin{pmatrix} 1 & N \\ 0 & 1 \end{pmatrix}$ as by Bézout's theorem $N = uN_1 + vN_2$ for some $u, v \in \mathbb Z$ so $c = a^ub^v$. But it is also a normal subgroup, as both $\Gamma(N_i)$ are, so it contains all conjugates of $c$ in $\Gamma(1)$. Assume that $N=p^m$ (the general case can be deduced from this by Chinese remainders); then the images of $\Gamma$ and $\Gamma(N)$ in $\mathrm{SL}_2(\mathbb Z /p^{m+1}\mathbb Z)$ are the same (they are both gerenated by elements of the form $1+p^m U$ where $U$ is a nilpotent matrix), and for all primes $q \not= p$ the image of $\Gamma$ in $\mathrm{SL}_2(\mathbb F_q)$ is everything (because it contains a nontrivial unipotent and it is normal). As $\Gamma$ is assumed to be congruence it must be equal to $\Gamma(N)$.