For $H=\left( \begin{smallmatrix} 0 & I_n \\ -I_n & 0 \end{smallmatrix} \right)$ and a commutative ring $F$, the symplectic group $Sp(2n,F)$ is the set of all matrices $M\in F^{2n\times 2n}$ such that $MHM^T = H$. (Here, $M^T$ means the transposed matrix.)
My question is: for $\mathbb{F}_p$ a finite field of prime order, is $Sp(2n,\mathbb{F}_p) = Sp(2n,\mathbb{Z})\otimes\mathbb F_p$, the reduction mod $p$ of $Sp(2n,\mathbb{Z})$?
Best Answer
Yes: $\operatorname{Sp}(2n, \mathbb F_p)$ is generated by its root subgroups. Each root subgroup is cyclic, and generated by an element that lifts in an obvious way to $\operatorname{Sp}(2n, \mathbb Z)$.
More concretely, if I haven't messed up the calculations in your form (I'm used to a different one), then, writing $E_{i j}$ for the matrix that has a 1 in its $(i, j)$th entry and 0 elsewhere, we have that $\operatorname{Sp}(2n, \mathbb F_p)$ is generated by:
The $e_i - e_j$ generators correspond to an elementary row operation in the upper left block, combined with its inverse transpose in the lower right block. The $e_i + e_j$ and $2e_i$ generators generate symmetric matrices in the upper right block.