Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 ?

share|cite|improve this question
Just a comment before I think about the question: $G$ is usually called a "section" of $H$. – user641 May 22 '11 at 3:52
up vote 6 down vote accepted

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).

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.