4
$\begingroup$

The following is a question asked to me these days by Gülin Ercan.

Let $G = L(q^f)$ be a finite simple group of Lie type, and let $L(q) \cong H \le G$ be the group of fixed points of the automorphisms of $G$ induced by field automorphisms. In case $H$ is nonsolvable, can there exist a nontrivial subgroup $N$ of $G$ which is normalized by $H$ and which has trivial intersection with $H$?

Improve this question
$\endgroup$
1
  • $\begingroup$ If $L = \operatorname{GL}_2$ and $\epsilon$ is a non-square in $\mathbb F_q^\times$, then $L(q)$ is normalised by $E = \begin{pmatrix} 0 & \sqrt\epsilon^{-1} \\ \sqrt\epsilon & 0 \end{pmatrix}$. This is the opposite of the normalisation you asked for, but I wonder if $E$ is contained in a group normalised by $L(q)$. $\endgroup$
    – LSpice
    Commented Feb 27, 2021 at 17:16

1 Answer 1

Reset to default
5
$\begingroup$

The answer is no. By a theorem of Burgoyne, Griess, and me (Maximal subgroups and automorphisms of Chevalley groups, Pacific J. Math. 71 (1977), 365-403, Theorem 1), there would exist an integer $f_0$ such that $G_0\mathrel{:=}L(q^{f_0})\le HN\le G_0^*$, where $G_0^*$ is isomorphic to a group of automorphisms of $G_0$ generated by $\operatorname{Inn}(G_0)$ and (some) diagonal automorphisms. The theorem asserts that all finite overgroups of $H$ in the algebraic group overlying $G$ lie in such a "sandwich" (but exceptions can occur when $H$ is solvable). In your situation $f_0$ would divide $f$, and $f_0<f$ by simplicity of $G$. If $f$ is prime, then $G_0=H$ and $[G_0,N]\le G_0\cap N=1$, so $N=1$ as $N$ acts faithfully on $G_0$. If $f$ is not prime, then $N=1$ by induction on $f$.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .