Is there a reasonably elementary and short proof that a subgroup of consisting of unipotent matrices over a field centralizes a flag?
By elementary, I mean accessible to students who have had a semester of linear algebra and a semester of group theory, but no commutative algebra or representation theory. By short I mean less than a page.
I think this is a result of Levi for nilpotent algebras.
An $n \times n$ matrix is called unipotent if it satisfies any of the following equivalent conditions (1) its minimum polynomial is a power of $x-1$ (2) its characteristic polynomial is $(x-1)^n$ (3) it centralizes a flag.
A set $H$ of $n \times n$ matrices over the field $K$ is said to centralize a flag if there are subspaces $0 = V_0 \leq V_1 \leq \ldots \leq V_k = K^n$ such that for every $h \in H$, $i=1,2,\ldots,k$, and $v \in V_i$, one has that $(h-1)v \in V_{i-1}$. A single matrix $h$ centralizes a flag iff $\{h\}$ centralizes a flag.
The key point is that if a subgroup consists only of elements that centralize flags (each element having its own personal flag) then there is a single flag that they all centralize.