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It's not possible for extensions of ${\rm PSL}(3,q)$ to have the same maximal abelian subgroups as ${\rm PSL}(3,q)$ since these don't include the abelian subgroups not wholly contained in ${\rm PSL(3,q)}$.

However we can still try and classify the maximal abelian subgroups of extensions of ${\rm PSL}(3,q).$

Here's a rough sketch of how I'd approach this. Suppose that $H$ is a maximal abelian subgroup of a cyclic extension of $K$ where $K = {\rm PSL(3,q)}$. There are two cases:

• $H\leq K$. These can be read off from the known subgroups of $PSL(3,q)$. (See Mitchell, Hartley or Bloom for the first classification of these. Or see Kleidman & LibeckLiebeck, or the survey article by King for a modern treatment.)
• $H\not \leq K$. Then $H$ contains an outer automorphism $g$ and so, in particular $H \cap K$ lies in $C_K(g)$. Subcases that you are interested in:

(a) If $g$ is a graph automorphism of order $2$, then $C_K(g)\cong {\rm PSL}_2(q)$ or ${\rm PGL}_2(q)$ (since these are isomorphic to 3-dimensional orthogonal groups). The maximal subgroups of these groups are easy (Dickson gave the first proof), and the abelian subgroups can be read off.

(b) If $g$ is a graph automorphism of order greater than $2$, then we can assume it has order a multiple of $4$ (otherwise one of its powers is a graph aut of order $2$) and so $H$ contains an involution $h$ of $PSL(3,q)$. Now $C_K(h)=\hat GL(2,q)$ and, again, abelian subgroups can be read off.

(c) If $g$ is a field automorphism of order $3$, then $C_K(g)$ is a subfield subgroup. And so the classification of subgroups is the same as for $K$.

(d) If $g$ is a field automorphism of order greater than $3$. Again we can assume it has power a power of $3$, so $H$ contains an element of order $3$. Its centralizer can be calculated (its structure will depend on whether $3$ divides $q$, $q-1$ or $q+1$) and maximal abelians read off.

p.s. Some further thought made me wonder whether your comment about abelian subgroups being the same was supposed to be this:

A maximal abelian subgroup of ${\rm PSL}_3(q).2$ (graph aut) restricts to a maximal abelian subgroup of ${\rm PSL}_3(q)$.

I checked the ATLAS and this is not true: Let $K={\rm PSL}_3(5)$. Then $K.2 \backslash K$ contains an element of order $8$ whose centralizer $C$ is of order $8$, thus is maximal abelian. But $K\cap C$ (which has order $4$) is not maximal abelian in $K$.

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It's not possible for extensions of ${\rm PSL}(3,q)$ to have the same maximal abelian subgroups as ${\rm PSL}(3,q)$ since these don't include the abelian subgroups not wholly contained in ${\rm PSL(3,q)}$.

However we can still try and classify the maximal abelian subgroups of extensions of ${\rm PSL}(3,q).$

Here's a rough sketch of how I'd approach this. Suppose that $H$ is a maximal abelian subgroup of a cyclic extension of $K$ where $K = {\rm PSL(3,q)}$. There are two cases:

• $H\leq K$. These can be read off from the known subgroups of $PSL(3,q)$. (See Mitchell, Hartley or Bloom for the first classification of these. Or see Kleidman & Libeck, or the survey article by King for a modern treatment.)
• $H\not \leq K$. Then $H$ contains an outer automorphism $g$ and so, in particular $H \cap K$ lies in $C_K(g)$. Subcases that you are interested in:

(a) If $g$ is a graph automorphism of order $2$, then $C_K(g)\cong {\rm PSL}_2(q)$ or ${\rm PGL}_2(q)$ (since these are isomorphic to 3-dimensional orthogonal groups). The maximal subgroups of these groups are easy (Dickson gave the first proof), and the abelian subgroups can be read off.

(b) If $g$ is a graph automorphism of order greater than $2$, then we can assume it has order a multiple of $4$ (otherwise one of its powers is a graph aut of order $2$) and so $H$ contains an involution $h$ of $PSL(3,q)$. Now $C_K(h)=\hat GL(2,q)$ and, again, abelian subgroups can be read off.

(c) If $g$ is a field automorphism of order $3$, then $C_K(g)$ is a subfield subgroup. And so the classification of subgroups is the same as for $K$.

(d) If $g$ is a field automorphism of order greater than $3$. Again we can assume it has power a power of $3$, so $H$ contains an element of order $3$. Its centralizer can be calculated and maximal abelians read off.