MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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

I am interested in classical groups (in particular $SL_n$, $Sp_{2n}$, $SO_n^{+}$) over finite rings of the form $$R_k=\mathbb{F}_q[t]/(t^k)$$ for some prime power $q$ (where $q$ is odd in the orthogonal case) and $k \in \mathbb{N}$. Over a finite field, the maximal subgroups of the classical groups are known, and so it is for example known that any two semisimple elements of orders $q^n+1$ and $q^n-1$ generate the group $Sp_{2n}(\mathbb{F}_q)$ (and similar results for the other classical groups). I am wondering whether it is true that any two semisimple elements of orders $q^{n(k-1)}(q^n+1)$ and $q^{n(k-1)}(q^n-1)$ generate $Sp_{2n}(R_k)$. Is anyting like this known? Are there lists of maximal subgroups for classical groups over finite rings? I would also be interested in similar results over $\mathbb{Z}_p/(p^k)$.

share|cite|improve this question
I fixed a problem with your LaTeX caused by jSMath (there was nothing wrong with your LaTeX specifically though, just a bug. – Harry Gindi Mar 22 '10 at 10:58
There are definitely some people who have studied classical groups over commutative rings including local rings, for example N.A. Vavilov (St. Petersburg) and his collaborators. My impression is that results have been somewhat scattered and fragmentary. There are some connections with algebraic K-theory as well. A creative literature search using MathSciNet might be useful. Generation of the groups has been a standard theme, but I doubt that you will find definitive results on maximal subgroups for your rings. – Jim Humphreys Mar 22 '10 at 20:42

The answer is YES (at least if $n,k \ge 2$), because elements of the orders you prescribe do not exist! The problem is that the orders of elements in the kernel of $\operatorname{Sp}_{2n}(R_k) \to \operatorname{Sp}_{2n}(R_1) = \operatorname{Sp}_{2n}(\mathbf{F}_q)$ are quite small.

Let $A$ and $B$ be your two elements. I am assuming that by semisimple, you mean in particular that the images of $A$ and $B$ in $\operatorname{Sp}_{2n}(R_1)$ should be semisimple. You are requiring $A^{q^n+1}$ and $B^{q^n-1}$ to have order $q^{n(k-1)}$. By the semisimplicity, these powers would have to be of the form $I+tM$. But then they are killed by raising to the $q^r$ power already for $r:=n(k-1)-1$, since $q^r \ge 2^{2(k-1)-1} \ge k$ and hence $$(I+tM)^{q^r} = I + t^{q^r} M^{q^r} = I$$ in $\operatorname{Sp}_{2n}(R_k)$. This contradicts your specifications.

share|cite|improve this answer

ooh, yes. I misphrased my question. What I meant to ask ist whether any two maximal tori of order $q^{n(k-1)}(q^n+1)$ and $q^{n(k-1)}(q^n-1)$ generate $Sp_{2n}(R_k)$. Of course these tori are far from cyclic. I guess my head was still in the cyclic case $Sp_{2n}(\mathbb{F}_q)$. Sorry!

share|cite|improve this answer
Click "edit" and fix it. – Harry Gindi Mar 23 '10 at 10:58

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.