# 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?

-
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

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