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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 $\mathbb{Z}_p/(p^k)$.

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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

2 Answers 2

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.

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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!

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Click "edit" and fix it. –  Harry Gindi Mar 23 '10 at 10:58

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