Generating Classical Groups over Finite Local Rings

Question

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

EDIT: The answer to the question as asked is usually vacuously yes. Instead, asm meant actually to ask about the group generated by two tori (not just two semisimple elements) of the specified orders, specifically they said (in a now deleted answer):

"I misphrased my question. What I meant to ask is 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 $Sp_{2n}(\mathbb{F}_q)$."

Answer 1

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.