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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 $H$.2011-05-22

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I think the question is not correctly worded. As it is currently worded, it is impossible.

Take m = 77, then as you say PSL(2,Z/77Z) ≅ PSL(2,7) × PSL(2,11). However, the non-abelian simple groups involved in PSL(2,Z/prZ) are at most PSL(2,p) and PSL(2,5). In particular, PSL(2,7) × PSL(2,11) is never involved in PSL(2,Z/prZ).

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    Thanks ! The question is taken from an exercise in a book in group theory. I guess I shouldn't always trust books...2011-05-20