Hello
For some part of my research I want to know that PSL(3,q).2, the extension of PSL(3,q) by the graph automorphism, has the same maximal abelian subgroups as PSL(3,q)?


Also if f is a field automorphism of GF(p^3) where q=p^3, then the maximal abelian subgroups of PSL(3,q).f are the same as PSL(3,q)?


Thank you very much
With best regards



The maximal Abelian subgroups of PSL(3,q).2 the extension of PSL(3,q) by the graph automorphismAutomorphism
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 & Liebeck, 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 $K$.


