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 $Sp(2n,\mathbb{F}_p)$ given by reduction mod $p$ of $Sp(2n,\mathbb{Z})$?

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

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 $Sp(2n,\mathbb{F}_p) = Sp(2n,\mathbb{Z}) \otimes \mathbb{F}_p$?

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 $Sp(2n,\mathbb{F}_p)$ given by reduction mod $p$ of $Sp(2n,\mathbb{Z})$?