Subgroups of $PSL_2(\mathbb Z/m\mathbb Z)$

Let $G$ and $H$ be two (finite) groups. Say that $G$ is "involved" in $H$ if $G$ is a quotient of a subgroup of $H$.

The question is the following : prove that, for every $m$, there is a prime $p$ and $r\geq 1$ such that any group involved in $PSL_2(\mathbb Z/m\mathbb Z)$ is involved in $PSL_2(\mathbb Z/p^r\mathbb Z)$.

Of course, it is enough to do it only for $PSL_2(\mathbb Z/m\mathbb Z)$ itself. It's tempting to use the Chinese Remainder Theorem : this group is a product of groups of the form $PSL_2(\mathbb Z/p_i^{r_i}\mathbb Z)$. And then I'm blocked... Am I missing something ?

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Just a comment before I think about the question: $G$ is usually called a "section" of $H$. – user641 May 22 '11 at 3:52

How exactly do you know that the only simple non-abelian groups involved in PSL(2,$\mathbf Z/p^r\mathbf Z$) are PSL(2,p) and PSL(2,5)? – Myself May 19 '11 at 16:01