Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $p$ be a prime number. By a Cartan subgroup of $GL_n(\mathbb{F}_p)$ I mean an absolutely semisimple maximal abelian subgroup.

When $n=2$, it is well-known* that, for $G \subset GL_n(\mathbb{F}_p)$ of order prime to $p$, either $G$ is contained in a Cartan subgroup, or it is contained in the normalizer of a Cartan subgroup, or its image in $PGL_n(\mathbb{F}_p)$ is isomorphic to $A_4$, $A_5$, or $S_4$.

I would like to know if there is a similar result for larger even values of $n$;

where a subgroup of order prime to $p$ would either be contained in the normaliser of a Cartan, or its projective image be one in a finite list of groups.

*See e.g. section 2 of Serre's "Propriétés galoisiennes..." paper.

share|improve this question
1  
Presumably we should read "A_4, A_5, or S_4" as "a finite subgroup of PGL_2(\mathbb C)". –  Will Sawin Apr 26 '12 at 20:45
    
@Will: Actually no, I really do mean in $PGL_2(\mathbb{F}_p)$. Now you might ask for which $p$ this is possible. Restricting to odd $p$, $A_4$ and $S_4$ always embed in $PGL_2(\mathbb{F}_p)$, and $A_5$ embeds if and only if $p \equiv \pm 1$ mod 5. –  Barinder Banwait Apr 26 '12 at 21:23
    
@Will: Your comment is in the right direction, though it omits the cyclic groups. This comment does reinforce my sense that higher values of $n$ will get arbitrarily hard for list-making. –  Jim Humphreys Apr 26 '12 at 21:34
    
@Barinder: I don't understand your definition of "Cartan subgroup", since Serre is considering both the split and non-split maximal tori in a finite group of Lie type (not just the split tori). Aside from that, I doubt very much that anything as simple as the case $n=2$ will occur for larger $n$. The subgroup structure rapidly gets much more complicated. –  Jim Humphreys Apr 26 '12 at 21:36
    
@Jim: Thanks for your comment. But does my definition of Cartan subgroup really force me into the split case? I didn't think that it did. –  Barinder Banwait Apr 26 '12 at 22:13

Your Answer

 
discard

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

Browse other questions tagged or ask your own question.