The answer to the original question is no, see JSE's answer!
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 $\gamma \cong 1 \mod N$.
$\newcommand\o{\mathfrak o}$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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