The maximal Abelian subgroups of PSL(3,q).2 the extension of PSL(3,q) by the graph automorphismAutomorphism - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T04:54:33Z http://mathoverflow.net/feeds/question/119639 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/119639/the-maximal-abelian-subgroups-of-psl3-q-2-the-extension-of-psl3-q-by-the-grap The maximal Abelian subgroups of PSL(3,q).2 the extension of PSL(3,q) by the graph automorphismAutomorphism darya 2013-01-23T10:10:30Z 2013-01-23T11:21:35Z <p>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)?</p> <p>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)?</p> <p>Thank you very much With best regards</p> http://mathoverflow.net/questions/119639/the-maximal-abelian-subgroups-of-psl3-q-2-the-extension-of-psl3-q-by-the-grap/119644#119644 Answer by Nick Gill for The maximal Abelian subgroups of PSL(3,q).2 the extension of PSL(3,q) by the graph automorphismAutomorphism Nick Gill 2013-01-23T11:01:48Z 2013-01-23T11:21:35Z <p>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 ${\rm PSL(3,q)}$. </p> <p>However we can still try and classify the maximal abelian subgroups of extensions of ${\rm PSL}(3,q).$</p> <p>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:</p> <ul> <li>$H\leq K$. 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 &amp; Liebeck, or the survey article by King for a modern treatment.)</li> <li><p>$H\not \leq K$. 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:</p> <p>(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.</p> <p>(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.</p> <p>(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 $K$.</p> <p>(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></li> </ul> <p>p.s. Some further thought made me wonder whether your comment about abelian subgroups being the same was supposed to be this:</p> <blockquote> <p>A maximal abelian subgroup of ${\rm PSL}_3(q).2$ (graph aut) restricts to a maximal abelian subgroup of ${\rm PSL}_3(q)$.</p> </blockquote> <p>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 $K$.</p>