# 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

## 1 Answer

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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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
PSL(2,Z/p^rZ) has a normal p-subgroup with quotient PSL(2,p), so any simple section of PSL(2,Z/p^rZ) is already a simple section of PSL(2,p). The subgroups of PSL(2,p) are dihedral, subgroups of AGL(1,p), or the "platonic" subgroups A4, A5, S4. The only one with a simple section is A5. – Jack Schmidt May 19 '11 at 16:21
Wow, thanks. That was slightly more nontrivial than I expected. – Myself May 19 '11 at 16:56
Thanks ! The question is taken from an exercise in a book in group theory. I guess I shouldn't always trust books... – Jean Lecureux May 20 '11 at 7:55