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May 10, 2013 at 5:06 comment added Emerton this gives many examples. (JSE's answer below is a particular case, applied to a (torsion free modification of) $\Gamma(2)$.) Regards, Matthew
May 10, 2013 at 5:05 comment added Emerton $\mathrm{SL}(2,\mathbb Z/q^r)$ for some finite list of primes $q$ (those that divide the level of $\Gamma$), and these images again admit only finitely many possible abelian quotients. So if $\Gamma$ is any torsion free congruence subgroup for which $\mathbb H/\Gamma$ has non-trivial $H^1$, then (identifying $\Gamma$ with $\pi_1$ of the quotient) we get a surjection $\Gamma \to \mathbb Z$, and if we choose $N$ huge, then the kernel of $\Gamma \to \mathbb Z/N$ will be a non-congruence subgroup. Now actually $\mathbb H/\Gamma$ always has non-trivial $H^1$ if it has more than one cusp, so ...
May 10, 2013 at 5:01 comment added Emerton Dear Marc, If $\Gamma$ is a congruence subgroup of $\mathrm{SL}(2,\mathbb Z)$, not containing $-I$ for simplicity, then there is an absolute bound on the size of an abelian quotient of $\Gamma$ whose kernel is again a congruence subgroup. The reason is that for all but finitely many primes $p$, the image of $\Gamma$ in $\mathrm{SL}(2,\mathbb Z/p^r)/{\pm I}$ is the whole group (here $p^r$ is any power of $p$), and for $p \geq 5$ (if I've not blundering) this group has no abelian quotients. So any abelian image of $\Gamma$ with congruence kernel is a quotient of the image of $\Gamma$ in ...
May 9, 2013 at 16:03 answer added BS. timeline score: 5
May 9, 2013 at 2:00 answer added JSE timeline score: 18
May 8, 2013 at 9:42 answer added Neil Hoffman timeline score: 5
May 8, 2013 at 8:24 vote accept Marc Palm
May 8, 2013 at 8:20 history edited David Loeffler CC BY-SA 3.0
Improved formatting, corrected spelling, added subject-class tag
May 8, 2013 at 8:18 answer added David Loeffler timeline score: 17
May 8, 2013 at 7:37 comment added Marc Palm This is my motivation: mathoverflow.net/questions/129637/…
May 8, 2013 at 7:34 history asked Marc Palm CC BY-SA 3.0