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 $H$?



Subgroups of finite simple groups $L(q^f)$ of Lie type normalized by $L(q)$
The answer is no. By a theorem of Burgoyne, Griess, and me ([Maximal subgroups and automorphisms of Chevalley groups](http://projecteuclid.org/euclid.pjm/1102811433), 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 $f$.


