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 $Sp(2n,\mathbb{Z})$?



Symplectic group over integers and finite fields
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.


