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Let $G$ be a finite group of Lie type. Let $H$ be a subgroup of $G$ which contains unipotent elements. I want to find a 'nice' subgroup of $G$ that contains $H$, for example a Levi subgroup of $G$ which is minimal with this property. Do you have an idea how we can do this?

My motivation comes from the following theorem:

Let $P$ be a $p$-subgroup of $G$. Then there exists a parabolic subgroup $Q$ of $G$ such that $P$ lies in the unipotent radical of $Q$.

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To attempt an answer, I'll replace your notation with my own. One method is to work inside a corresponding semisimple (or reductive) algebraic group $G$ over an algebraically closed field, relative to which your finite group of Lie type is constructed. There is a 1971 paper by Borel and Tits here. In their paper you can start with an arbitrary closed (say finite) unipotent subgroup $V$ of a Borel subgroup of $G$. Then associate to $V$ a parabolic subgroup $P \subset G$ whose unipotent radical contains $V$. At the same time, $P$ contains the entire normalizer in $G$ of $V$. This is done efficiently but a bit indirectly. (For an exposition, see 30.3 in my 1975 book Linear Algebraic Groups.)

I'm not sure this kind of construction is what you can use, but it's well to be aware of the Borel-Tits paper in any case.

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Thanks for your answer. Can we conclude that ıf H is any subgroup of G that contains V, then H must be contained also in P that you mentioned as above? –  albert cohen Nov 26 '12 at 23:34
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@albert cohen: The answer to your question in the comment is negative: consider $G = {\rm{PGL}}_n(k)$ and $H$ the image of ${\rm{SL}}_n(k)$ (and various unipotent $V$ inside $H$). Is your $G$ simply connected? (The answer is almost surely negative even in such cases.) Also, the MO question "homomorphism into reductive groups" and its answer address the relevant result from the Borel--Tits paper mentioned in the above answer. –  user28172 Nov 27 '12 at 3:34
    
@albert cohen: As nosr comments, the question in your comment has a negative answer. And there is some possibly relevant but quite lengthy discussion in the earlier MO question 104201. –  Jim Humphreys Nov 28 '12 at 16:02
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