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:



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> 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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How we characterize a subgroup of finite group of Lie type with unipotent elements.
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](http://gdz.sub.uni-goettingen.de/dms/load/img/?PPN=GDZPPN002088894). 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.


