# How we characterize a subgroup of finite group of Lie type with unipotent elements.

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.)
@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