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Does there exist Fuchsian groups, which is not conjugated in $SL(2, \mathbb{R})$ to a subgroup of $SL(2, \mathbb{Z})$, but still contains a congruence subgroup?

edited Jan 27 2012 at 11:04

asked Jan 26 2012 at 12:50

Marc Palm
6,946●7●20

David Loeffler · Accepted Answer · 2012-01-27 10:26:24Z UTC

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How about the normalizer of $\Gamma_0(N)$ in $PSL_2(\mathbb{R})$ for $N > 1$, which is the subgroup generated by $\Gamma_0(N)$ and $\begin{pmatrix} 0 & -1 \\ N & 0 \end{pmatrix}$ ?

(I wouldn't call these "non-congruence subgroups" as you do in your title: that technical term is usually parsed as "(non-congruence) subgroup". This isn't a subgroup of $PSL_2(\mathbb{Z})$ at all, but it certainly contains a principal congruence subgroup; perhaps we should call it a "congruence non-subgroup"?)

answered Jan 27 2012 at 10:26

David Loeffler
8,603●15●40

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This normalizer is indeed a discrete subgroup of $\mathrm{PSL}_2(\mathbf{R})$. For the property that no conjugate is contained in $\mathrm{PSL}_2(\mathbf{Z})$, I think one needs some assumption on $N$. This is true for example if $N$ is prime (this is Exercise 1.46 in Shimura's book Intro. to the arithm. theory of autom. functions). But in general I think there are counterexamples e.g. for $N=p^4$ with $p$ prime. – François Brunault Jan 27 2012 at 10:50

Okay, I recognize them now as the other maximal compact subgroup, which are useful when classifying supercuspidals. – Marc Palm Apr 6 2012 at 15:19

Non congruence subgroups containing congruence subgroups.

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