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Let $G$ be a finite group and $H$ be a nilpotent subgroup of $Aut(G)$. If $C_{G}(H)=1$, is $G$ solvable?

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The question is not well formulated. What do you know about the problem? Works of Shult, Gross, Khukhro and many others? What if $G$ is an Abelian group? – Mark Sapir Apr 30 '13 at 11:52
I've been musing on a this question under the (very strong) extra assumption that $G$ is simple, but even this seems a little tricky. I guess that $C_G(H)$ is never trivial here, but how to prove it? One could do an exhaustive analysis of the outer automorphism group, but that seems very unsatisfactory... I may well be missing an easy solution though... – Nick Gill Apr 30 '13 at 15:05
@Nick: The outer automorphism groups of all finite simple groups are well known (and very small). – Mark Sapir Apr 30 '13 at 16:59
Yes.The outer automorphism groups of all finite simple groups are well known and very small. But how to prove it rigorously, especially when $G$ is a simple group of Lie type? – sife Apr 30 '13 at 18:21
Yes. When $H$ is a p-group, we can use Theorem 6.2.3 in the book "finite groups (gorenstein)" to deduce that $(|G|,|H|)=1$. So by the Classification Theorem of Finite Simple Groups, $G$ is solvable. But I don't know how to check when $G$ is a simple group of Lie type. – sife Apr 30 '13 at 21:36

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