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Im having difficulties understanding a ceratain argument in the following proof, taken from T.A springer's linear algebraic groups.

In the following proof C is the intersection of the unipotent parts of all borel groups containing the maximal torus T, and C$\alpha$ is the intersection of the unipotent parts all such borel groups for wich $\alpha$ is a positive root. enter image description here

I dont understand why should the borel groups be contained in TC or TC$\alpha$

roy yanai
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1 Answers1

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The heuristic is as follows: The toral (and abelian) subgroup allows us to simultaneously diagonalize them, such that $C$ are all upper triangular matrices with ones on the diagonal (unipotent). Then $C_\alpha T=TC_\alpha$ is a Borel subgroup (maximal solvable). The fact that $\alpha$ is a positive root, fixes the root system and so the possible conjugates. Adding any other generator would mean to add a negative root and destroy solvability.

Marius S.L.
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