One way of interpreting the question might be: *Is the property of being congruence a topological property? Ie, is it detected at the level of Riemann surfaces $\mathcal{H}/\Gamma$?*

My motivation is that I'm looking for an efficient algorithm to test if a given finite index subgroup of $SL_2(\mathbb{Z})$ is congruence.

The best algorithm I'm aware of is Hsu's algorithm, but this only tests if a finite index subgroup of $\text{PSL}_2(\mathbb{Z})$ is congruence.

Given a finite index $\Gamma\le\text{SL}_2(\mathbb{Z})$, the image of $\Gamma$ in $\text{PSL}_2(\mathbb{Z})$ is congruence if and only if $\pm\Gamma$ is congruence. Here, $\pm\Gamma$ is the subgroup generated by $\Gamma$ and $-I$.

Let $\ell$ be the "Wolhfart level" of $\Gamma$ - that is, $\ell$ is the least common multiple of the cusp widths of $\mathcal{H}/\Gamma$. Then, a classical theorem of Wolhfart/Klein says that $\pm\Gamma$ is congruence if and only if it contains $\Gamma(\ell)$, equivalently if it contains $\pm\Gamma(\ell)$ (this theorem is actually quoted incorrectly in many sources, where the $\pm$ in $\pm\Gamma$ is ignored!!)

**The first question is:** Is testing for congruence in $\text{PSL}_2(\mathbb{Z})$ enough? Ie, if $\pm\Gamma$ is congruence, then must $\Gamma$ be congruence?

Of course this is only relevant if $-I\notin\Gamma$, in which case $\pm\Gamma = \Gamma\times\{\pm I\}$, and $\pm\Gamma(\ell) = \Gamma(\ell)\times\{\pm I\}$. Intersecting $\Gamma$ with $\pm\Gamma(\ell)$, there are two cases:

A) If $\ell = 2$, then if $\pm\Gamma$ is congruence, it must contain $\Gamma(2) = \pm\Gamma(2)$, so $\Gamma\cap\Gamma(2)$ has index 2 in $\Gamma(2)$, with quotient an 2-torsion abelian group, but the maximal 2-torsion abelian quotient of $\Gamma(2)$ is $\Gamma(2)/\Gamma(4) = C_2\times C_2\times C_2$, hence $\Gamma$ contains $\Gamma(4)$ and is congruence.

In particular, this shows that the Sanov subgroup - the index 12 subgroup of $\text{SL}_2(\mathbb{Z})$ generated by $$\begin{bmatrix} 1&2\\0&1 \end{bmatrix} \quad\text{and}\quad\begin{bmatrix} 1&0\\2&1 \end{bmatrix}$$ is congruence, even though the cusp widths are all 2 and it doesn't contain $\Gamma(2)$.

B) If $\ell\ge 3$, then again $\Gamma\cap\pm\Gamma(\ell)$ has index 2 inside $\pm\Gamma(\ell)$, and certainly any index 2 subgroup of $\pm\Gamma(\ell)$ can appear, and so the question reduces to:

**Reformulation of question:** For any $\ell\ge 3$, is every index 2 subgroup of $\pm\Gamma(\ell)$ a congruence subgroup?

**or equivalently:** For any $\ell\ge 3$, is the composite $\Gamma(\ell)'\Gamma(\ell)^2$ congruence? (the first is the commutator subgroup, the second is the subgroup of squares)

I don't even know of a reasonable way to computationally test this, since without Wolhfart/Klein's result, one might a priori have to test non-containment of infinitely many $\Gamma(n)$'s to prove that some $\Gamma$ is noncongruence.

Of course, if the answer is negative, then naturally one might ask:

**Is there an efficient algorithm to test whether a given finite index subgroup of $\text{SL}_2(\mathbb{Z})$ is congruence?**