Let $\Gamma$ be a congruence subgroup of $SL_2(\mathbb Z)$ and $\chi:\Gamma \to \mathbb{C}^*$ a character of $\Gamma$ with finite image. Is then $\ker(\chi)$ also a congruence group? If not, can someone give me a counterexample?

The answer is NO, in general. For a specific counterexample, let $\Gamma$ be the principal congruence subgroup of level two in $SL(2,{\mathbb Z})$. Then, $\Gamma$ modulo $\pm 1$, is the free group on two generators, and hence there is a homomorphism from $\Gamma$ onto ${\mathbb Z}/5{\mathbb Z}$ (the latter realised as the quotient of the free abelian group on two generators modulo a suitable subgroup). The congruence closure of $\Gamma$ may easily be shown to be the product $$SL_2(2{\mathbb Z}_2)\times SL(2,{\mathbb Z}_3) \times \prod _{p\neq 2,3} SL(2,{\mathbb Z}_p)$$where ${\mathbb Z}_p$ denotes the ring of $p$-adic integers. Using the fact that for $p\geq 5$, the group $SL_2({\mathbb Z}_p)$ is its own commutator, it is easy to see that any Abelian quotient of this congruence closure consists only of two and three torsion. Hence the kernel of $\Gamma$ to ${\mathbb Z}/5{\mathbb Z}$ cannot be a congruence subgroup.

[Edit] I see that the link provided by Matthias Wendt answers this question completely.

Le probleme des groupes de congruence pour SL$_2$, Ann. of Math. (2) 92 (1970) 489-527. $\endgroup$ – Jim Humphreys Sep 13 '14 at 13:05