# Are the $\Gamma(N)$ the only normal congruence subgroups of $\mathrm{SL}_2(\mathbb{Z})$?

The answer to the original question is no, see JSE!

Are the $\Gamma(N)$ the only normal congruence subgroups of $\mathrm{PSL}_2(\mathbb{Z})$ with no finite subgroups (elliptic elements)? What about the normal subgroups of $\Gamma(4)$, which does not contain torsion elements. Here, $\gamma \in \Gamma(N)$, if $\gamma \cong 1 \mod N$.

I state the question in the local picture for general G: Let $o$ be a local ring and $G \subset GL(n)$ a group. Then it makes sense to speak about $G( p^r)$, which consists of matrices which are congruent to $1$ modulo $p^r$, where $p$ is the maximal ideal in $o$. From $o \rightarrow o/\p^r$, we get an exact sequence $$G(p^r) \rightarrow G(o) \rightarrow G( o / p^r).$$ Is it surjective? I guess the $G(p^r)$ form a basis of neighborhoods for $1$ and are normal open compact subgroups. Under which conditions are these all normal subgroups in $G(o)$ with no finite subgroups?

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You need your local ring to be compact in the $p$-adic topology to say that the congruence subgroups are compact. This is not true for local rings in general, but it is true for complete dvrs. –  S. Carnahan Apr 1 '11 at 5:27

Almost but not quite. A congruence subgroup has to contain some Gamma(N). So if it is normal its image in SL_2(Z) / Gamma(N) is a normal subgroup of SL_2(Z/NZ). That group is almost simple but not quite. A proper normal subgroup need not be trivial; it could be +-1 for instance. Or if N is very small you have a few more choices; for instance, the commutator subgroup of SL_2(Z) is a finite-index congruence subgroup of (if I recall correctly) index 12, containing Gamma(6).

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Sorry, I meant of course $PSL$ instead of $SL$. What for subgroup of $\Gamma(4)$ instead of $\Gamma(1)$? Do the elipptic elements/finite subgroups give the trouble? –  Marc Palm Mar 31 '11 at 15:47
Please forgive me that I changed the questions accordingly.... –  Marc Palm Mar 31 '11 at 16:08
Gamma(4) will have new normal subgroups; Gamma(4)/Gamma(8) is (Z/2Z)^3 so you have a whole bunch of normal subgroups in between those two. –  JSE Mar 31 '11 at 17:07

(i) Normal congruence subgroups of PSL(2, Z) has been classified by D. L. McQuillan, Amer. J. Math. 87 (1965), 285-296. (ii) The only torsion normal subgroups of PSL(2, Z) are the powers subgroups P_n= < x^n : x\in PSL(2,Z)>, where n = 2 or 3.

One may now list all normal congruence subgroups of level N.

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