Symplectic group over integers and finite fields 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 $Sp(2n,\mathbb{Z})$?
 A: 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:

*

*$\{1 + t(E_{i j} - E_{(j + n)(i + n)}) : t \in \mathbb F_p\}$ (corresponding to the root $e_i - e_j$) for $1 \le i < j \le n$,

*$\{1 + t(E_{i(j + n)} + E_{j(i + n)}) : t \in \mathbb F_p\}$ (corresponding to the root $e_i + e_j$) for $1 \le i \ne j \le n$,

*$\{1 + t E_{i(i + n)} : t \in \mathbb F_p\}$ (corresponding to the root $2e_i$) for $1 \le i \le n$,
and

*the transposes of the various groups above (corresponding to the negative roots).

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.
