0
$\begingroup$

I will be so thankful for any comment or answer.

Suppose $S$ is a simple Lie type group of characteristic $p$ and $S\subseteq G \subseteq Aut(S)$ and $G_0$ is a subgroup of $G$ generated by all inner and diagonal automorphism of $S$ that lies in $G$. If $G$ has not any graph, field and graph-field automorphism of prime order $r$, Is it true that $\gcd(|G:G_0|,r)=1$?

Improve this question
$\endgroup$
6
  • $\begingroup$ You probably need to define graph, field and graph-field automorphisms explicitly as there is no universal terminology. (For instance Gorenstein, Lyons & Solomon have two definitions of field and graph auts in their volume 3 - see Warning 2.5.2 of that book.) $\endgroup$
    – Nick Gill
    Commented Nov 28, 2014 at 21:42
  • $\begingroup$ Possible counterexample: consider $S=PSL_2(p^2)$ for some odd prime $p$. Let $G$ be the group $\langle S, h\rangle$ where $h$ is the product of a diagonal automorphism $x$ and a field automorphism $y$ (both of order $2$). Depending on your definitions, this would be a counter-example so long as $y$ and $xy$ are not conjugate in $Aut(S)$. I guess they aren't but would need to check to be sure. $\endgroup$
    – Nick Gill
    Commented Nov 28, 2014 at 21:47
  • $\begingroup$ Checking the atlas for $p=5$ implies that $y$ and $xy$ are not conjugate in that case. So there's one counterexample (with Def 2.5.13 of GLS3). $\endgroup$
    – Nick Gill
    Commented Nov 28, 2014 at 21:48
  • $\begingroup$ One last comment: I presume you mean "generated by all inner and diagonal automorphisms of $S$ that lie in $G$", otherwise I'm not sure that the definition of $G_0$ makes sense. $\endgroup$
    – Nick Gill
    Commented Nov 29, 2014 at 1:33
  • $\begingroup$ @Nick, I edit base on your last comment. Exept $PSL_2(q)$ is there any counterexample? $\endgroup$
    – Hamid
    Commented Nov 29, 2014 at 4:32

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .