generators of $SL_3(\mathbb{Z}/p\mathbb{Z})$

Question

Is there an explicit description (actually any nice characterization would do) of the generators of $SL_3(\mathbb{Z}/p\mathbb{Z})$? I found many references about the generators of $SL_2(\mathbb{Z}/p\mathbb{Z})$, but none about $SL_3$. I am particularly interested in the subset of $SL_3(\mathbb{Z}/p\mathbb{Z})$ consisting of matrices of the form $S = \{I + cE_{21}, I + cE_{32}, I + cE_{31}\} $ where $E_{ij}$ is the matrix with just the $(i,j)$-th entry having a value $c$ and the rest of the entries $0$. All the matrices are $3 \times 3$ matrices here. Do these matrices generate $SL_3(\mathbb{Z}/p\mathbb{Z})$? (If yes, can someone point me to a proof of this fact or sketch a proof here?)

Answer 1

Write $e_{ij}(r)=1+rE_{ij}$ ($i\neq j$) and $e_{ij}=e_{ij}(1)$.

Then standard reduction shows that for every field $K$ and $n\ge 0$, the group $\mathrm{SL}_n(K)$ is generated by the matrices $e_{ij}(r)$ when $i,j$ ranges over distinct pairs and $r$ ranges over $K$. In particular, for $K=\mathbf{Z}/p\mathbf{Z}$ ($p$ prime), the $n(n-1)$ matrices $e_{ij}$ generate $\mathrm{SL}_n(K)$.

Actually, the latter result holds for an arbitrary integer $p$. For $p$ prime power, this follows from a similar argument, and general $p\neq 0$ follows from the Chinese remainder theorem. The case $p=0$ is proved directly ($\mathrm{SL}_n(R)$ is generated by the matrices $e_{ij}(r)$ for every Euclidean domain $R$).