Theorem $3.1$ from Robert Griess' "Elementary abelian p-subgroups of algebraic groups".
Let p be a prime, $\mathbb{K}$ an algebraically closed field, G = SL($n+1$, $\mathbb{K}$), Z $\leqslant$ Z(G), E a p-subgroup of GL($n+1$, $\mathbb{K}$) such that EZ/Z is elementary abelian. If |Z| $\equiv 0$ (mod p) is not true, then E is toral. If it is true, nontoral groups exist and maximal ones are the images mod Z of a group of the form $A\circ B$, where A $\cong$ p$^{1+2r}_{+}$ for some r $\geqslant 1$ and where B is an abelian (and toral) subgroup of the commuting algebra of A, $n+1$ = p$^r$s. Given A, if B is as large as possible, rk(B) = s and B $\cong$ $\mathbb{Z}_{p^{c+1}} \times p^{s-1}$, where p$^{c}$ = |Z|$_p$.
Is there a problem with ${p^{c+1}}$ in B. Should it be ${p^{c}}$?
A counterexample: $n+1 = 12$ and Z = Z(G).
Then according to this theorem, for $12 = 2$ $\times$$6$, E = $2$$^{1+2}\circ \mathbb{Z}_{2^{2+1}}\times 2^5$ . Then EZ/Z is $2^2 \times 2 \times 2^5$, if I'm correct. But PGL$_6$($\mathbb{K}$) has no toral elementary abelian $2$-subgroup of rank 6.