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Apr 6, 2012 at 15:19 comment added Marc Palm Okay, I recognize them now as the other maximal compact subgroup, which are useful when classifying supercuspidals.
Jan 27, 2012 at 10:58 vote accept Marc Palm
Jan 27, 2012 at 10:50 comment added François Brunault 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.
Jan 27, 2012 at 10:26 history answered David Loeffler CC BY-SA 3.0