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By definition, an algebraic group $G$ is reductive if its unipotent radical is $\{e\}$. The radical of an algebraic group is the identity component of its maximal normal solvable subgroup. The unipotent radical of $G$ is the set of all unipotent elements in the radical of $G$. Is there a general method to compute radical and unipotent radical of $G$?

We know that $GL_n$ is reductive so the unipotent radical of $GL_n$ is $\{e\}$. In particular, how to compute the radical and unipotent radical of $GL_n$? Thank you very much.

LJR
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1 Answers1

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For $GL_n$ the typical way I know is to apply Lie-Kolchin's theorem.

A consequence of that theorem is that a connected solvable closed subgroup is conjugate to a subgroup of upper triangular matrices. Then since the subgroup is also normal, it must actually be upper triangular, but by symmetry also lower triangular and thus it must consist of diagonal matrices.

Then one easily checks that any normal subgroup consisting of diagonal matrices must actually consist of scalar matrices (by conjugating with matrices of the form $I + E_{i,j}$ where $E_{i,j}$ is the matrix with a $1$ at the $(i,j)$ position and $0$ elsewhere), so in this case the radical is just the center, and thus the unipotent radical is trivial.

Tobias Kildetoft
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  • Why can the Lie-Kolchin Theorem be applied here? Because $GL_n$ is not solvable.. – rae306 Oct 23 '21 at 11:41
  • @rae306 You apply the theorem to the proposed connected solvable closed subgroup (seeing the inclusion into $GL_n$ as a representation of it). – Tobias Kildetoft Oct 23 '21 at 18:06
  • @rae306 I am not sure what you mean. To show that a group is reductive, we want to show that the radical is a torus or equivalently that the unipotent radical is trivial. – Tobias Kildetoft Oct 24 '21 at 11:17