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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?

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  • $\begingroup$ Sorry, i forget to assume that $\chi$ is a finite character. $\endgroup$ – Abdullah.Y Aug 17 '14 at 13:22
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    $\begingroup$ I guess you can find some counterexamples in the answers to this question: mathoverflow.net/questions/130062 You should be more specific about $\Gamma$ or the group in which it is a congruence subgroup. If the congruence subgroup problem has a positive solution, then $\ker(\chi)$ will automatically be congruence again if the image of $\chi$ is finite. $\endgroup$ – Matthias Wendt Aug 17 '14 at 13:43
  • $\begingroup$ $\Gamma$ is a congruence subgroup of $SL_2(\mathbb{Z})$ $\endgroup$ – Abdullah.Y Aug 17 '14 at 21:02
  • $\begingroup$ Dear Abdullah, I have slightly edited your question in order to introduce the precisions you gave in comments. $\endgroup$ – Joël Aug 18 '14 at 3:57
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    $\begingroup$ This rank 1 case is quite complicated for the congruence subgroup problem, but the standard reference is Serre's paper: 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
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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.

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