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 $q^n1$ 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 $q^{n(k1)}(q^n+1)$ and $q^{n(k1)}(q^n1)$ 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 $\mathbb{Z}_p/(p^k)$.

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 $B^{q^n1}$ to have order $q^{n(k1)}$. 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 $r:=n(k1)1$, since $q^r \ge 2^{2(k1)1} \ge k$ and hence $$(I+tM)^{q^r} = I + t^{q^r} M^{q^r} = I$$ in $\operatorname{Sp}_{2n}(R_k)$. This contradicts your specifications. 


ooh, yes. I misphrased my question. What I meant to ask ist whether any two maximal tori of order $q^{n(k1)}(q^n+1)$ and $q^{n(k1)}(q^n1)$ 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! 

