For which $n$ is it true that all surjections $SL_2(\mathbb{Z})\rightarrow SL_2(\mathbb{Z}/n\mathbb{Z})$ have kernel $\Gamma(n)$?

Question

For which integers $n$ does every surjection $SL_2(\mathbb{Z})\twoheadrightarrow SL_2(\mathbb{Z}/n\mathbb{Z})$ have kernel $\Gamma(n)$?

(this is the usual kernel, ie, the subgroup of matrices congruent to 1 mod $n$).

A positive answer (for some $n$) would certainly imply that $\Gamma(n)$ is characteristic inside $SL_2(\mathbb{Z})$, which is true for even $n$ by David Speyer's beautiful answer (pointing to a result of Hua and Reiner) here:

are the congruence subgroups $\Gamma(n)$ characteristic inside $\mathrm{SL}_2(\mathbb{Z})$?

From his answer, it's clear that our statement is false for odd $n$. However, even for $n\equiv 0\mod 2$, what's strange is that after doing some computations, the statement is true for $n = 2,4,6,\ldots, 24$, is false for $n = 26$, is true for $n = 28,30,32$, and is false again for $n = 34$.

Has there been any work on this? (I always find questions about $SL_2(\mathbb{Z})$ hard to google, due to the difficulty of making queries involving symbols).

EDIT: So far, after computing examples for even $n$ ranging from 2 through 58, the even $n$ for which the statement is false are: $$26,\ldots, 34,\ldots, 38,\ldots, 46,\ldots, 52, 54,\ldots, 58$$ where "$\ldots$" indicate gaps.

Answer 1

The following mostly answers the question:

Claim 1: If every surjection $SL_{2}(\mathbb{Z}) \to SL_{2}(\mathbb{Z}/n\mathbb{Z})$ has kernel $\Gamma(n)$, then all prime factors $p | n$ have $p \leq 11$.

To prove Claim 1, I need another claim.

Claim 2: If $p > 11$ is prime, there are elements $x, y \in PSL_{2}(\mathbb{F}_{p})$ so that $\langle x, y \rangle = PSL_{2}(\mathbb{F}_{p})$, $x$ has order $2$, $y$ has order $3$, and $xy$ does not have order $p$.

Proof of Claim 1 (using Claim 2): Assume that $p^{k} \parallel n$ with $p > 11$. Choose elements $X, Y \in SL_{2}(\mathbb{Z}/n\mathbb{Z})$ so that

$\bullet$ the images of $X$ and $Y$ in $PSL_{2}(\mathbb{F}_{p})$ have orders $2$ and $3$, respectively, $XY$ does not have order $p$, and $X$ and $Y$ generate $PSL_{2}(\mathbb{F}_{p})$, and

$\bullet$ replacing $X$ and $Y$ by $p$th powers of themselves (if necessary), assume that $X \bmod p^{k}$ and $Y \bmod p^{k}$ have order $4$ and $6$, respectively, in $SL_{2}(\mathbb{Z}/p^{k} \mathbb{Z})$, and

$\bullet$ take $X \equiv \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \pmod{n/p^{k}}$ and $Y \equiv \begin{bmatrix} 0 & -1 \\ 1 & 1 \end{bmatrix} \pmod{n/p^{k}}$.

It then follows that $X^{2} \equiv Y^{3} \equiv \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} \pmod{n}$ and $X^{4} \equiv I \pmod{n}$. Note that $\langle X, Y \rangle = SL_{2}(\mathbb{Z}/n\mathbb{Z})$ because the map $\langle X, Y \rangle \to SL_{2}(\mathbb{Z}/(n/p^{k})\mathbb{Z})$ is surjective and the map $\langle X, Y \rangle \to SL_{2}(\mathbb{Z}/p^{k} \mathbb{Z})$ is also surjective (because $PSL_2(\mathbb{F}_{p})$ is a proper quotient of the image mod $p^{k}$, and every proper subgroup of $SL_{2}(\mathbb{Z}/p^{k} \mathbb{Z})$ has all composition factors abelian or isomorphic to $A_{5}$).

Since $\langle x, y | x^{2} = y^{3}, x^{4} = 1 \rangle$ is a presentation for $SL_{2}(\mathbb{Z})$, there is a homomorphism $\phi : SL_{2}(\mathbb{Z}) \to SL_{2}(\mathbb{Z}/n\mathbb{Z})$ so that $\phi\left(\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}\right) = X$ and $\phi\left(\begin{bmatrix} 0 & -1 \\ 1 & 1 \end{bmatrix}\right) = Y$.

Now, $$\left(\begin{bmatrix} 0 & -1\\ 1 & 0 \end{bmatrix}^{3} \begin{bmatrix} 0 & -1 \\ 1 & 1 \end{bmatrix}\right) = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}, $$ has order $n$ in $SL_{2}(\mathbb{Z}/n\mathbb{Z})$, but the power of $p$ dividing the order of $X^{3} Y$ is at most $p^{k-1}$, because $(X^{3} Y)^{r} \equiv 1 \pmod{p}$ for some $r$ with $p \nmid r$ and $A^{r} \equiv 1 \pmod{p}$ implies that $A^{rp^{k-1}} \equiv 1 \pmod{p^{k}}$. Thus $(x^{3} y)^{rp^{k-1}} \in \ker \phi$, but $(x^{3} y)^{rp^{k-1}} \not\in \Gamma(n)$. QED Claim 1

Proving Claim 2 should not be too difficult, but it is a bit tedious. This requires carefully counting pairs $x, y$ (of orders $2$ and $3$ respectively) so that $xy$ has order $p$, and also counting pairs $x,y$ that both lie in one of the proper (Borel, dihedral, $A_{4}$, $S_{4}$ or $A_{5}$) subgroups of $PSL_{2}(\mathbb{F}_{p})$.

Note that there are some unusual things that happen for $n$ all of whose prime factors are $\leq 11$. In particular, there are examples with $n = 54$, $n = 98$, and $n = 108$. It seems plausible to me that there are only finitely many $n$ for which the kernel must be $\Gamma(n)$.