Symplectic group over $\mathbb{Z}/p\mathbb{Z}$ is generated by its root subgroups This is a question about the answer in this other post: Symplectic group over integers and finite fields.
In general, for any ring $R$, the symplectic group $\text{Sp}(2n,R)$ is generated by its root subgroups and a maximal torus $T$.
Why is $\text{Sp}(2n,\mathbb{Z}/p\mathbb{Z})$ generated only by its root subgroups, where $p$ is a prime number?
A reference to a book that discusses this would already make me very happy, but I haven't been able to find one...
 A: I'll spell out Andrei Smolensky's argument a bit (and remove this from the unanswered list).
(1) For any field $k$, the group $\operatorname{SL}_2(k)$ is generated by its elementary subgroups. In particular, you can see that the torus is generated by elementary subgroups because
$$
\begin{bmatrix} 1&0 \\ a-a^2&1 \\ \end{bmatrix}
\begin{bmatrix} 1&-1/a \\ 0&1 \\ \end{bmatrix}
\begin{bmatrix} 1&0\\ -1+a&1 \\ \end{bmatrix}
\begin{bmatrix} 1&1 \\ 0&1 \\ \end{bmatrix}=
\begin{bmatrix} 1/a&0 \\ 0&a \\ \end{bmatrix}.$$
(2) For each pair of roots $\pm \alpha$ of a reductive group $G$, we get a map from $\operatorname{SL}_2$ to $G$. In particular, the torus of $\operatorname{SL}_2$ maps to the one parameter subgroup in the torus of $G$ corresponding to $\pm \alpha^{\vee}$. So, if the co-roots span the co-weight lattice (which is equivalent to $G$ being simply connected), then the tori of these various $\operatorname{SL}_2$'s will generate the torus of $G$.
Concretely, the co-roots of the symplectic group are $\pm e_i \pm e_j$ and $\pm e_k$, inside the co-weight lattice $\mathbb{Z}^n = \bigoplus \mathbb Z e_i$, and it is clear that these span.
