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.

  • $\begingroup$ Well I can do it do it for finite fields! $\endgroup$ – Derek Holt Aug 2 '13 at 7:30
  • $\begingroup$ I think I know the proof you mean: a counting argument like for proving the center of a p-group is non-trivial. By induction we just need to show $H$ (simultaneously) fixes at least one nonzero vector (or even just stabilizes a proper subspace, with induction for large dimension and conclusion if the dimension is 1, since the eigenvalues are 1). Right now I'm looking at Alperin-Bell's textbook where they use Wedderburn's result that an algebra with a k-basis of nilpotent elements is nilpotent. Engel also has a similar theorem for nilpotent Lie algebras (might have Engel and Levi mixed). $\endgroup$ – Jack Schmidt Aug 2 '13 at 16:28

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