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?



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


