Let $\Gamma = PSL(2,\mathbb{Z}) = \langle S,T \ | \ S^2=(ST)^3=1 \rangle$. Let $G$ be some mystery normal subgroup of $\Gamma$ that we happen to think may be congruence. Recall that a subgroup of $\Gamma$ is a congruence subgroup of level $N$ if it contains the principal congruence subgroup of level $N$.


Are there methods for establishing upper bounds on the possible levels of $G$, say given information about $\Gamma/G$? Establishing *lower* bounds is trivial: the minimal $N$ such that $T^N \in G$ (if such an $N$ exists) is the *minimal* possible level of $G$. 


For example, suppose $\Gamma/G \cong (\mathbb{Z}\_3 \times \mathbb{Z}\_9) \rtimes S\_3$ or more generally of the form (finite abelian group)-semidirect product-(symmetric group). I'm generally thinking about the case where $G$ is the kernel of some unitary representation with finite image. 



Upper bound on level of a congruence subgroup of the modular group
If you only look at upper bounds, you can use the result by Lubotzky: For every algebraic group G, there is some $c$, such that the level of a congruence subgroup $U$ is bounded by the $c(G:U)$.


If you want to determine the level algorithmically, you can look at the image of the subgroup in $PSL(\mathbb{Z}/p^n\mathbb{Z})$. Determine the cusp split of the subgroup, if there are two cusp width $c\_1, c\_2$, and $p$ a prime with $p\nmid c\_1c\_2$, then $U$ contains different parabolic elements of order coprime to $p$, hence $U\Gamma(p)/\Gamma(p)$ contains two different parabolic subgroups, hence $U\Gamma(p)/\Gamma(p)=PSL(\mathbb{F}\_p)$, thus $p$ does not divide the level. 


