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Let $G$ be a connected solvable algebraic group. If $T$ is a maximal torus of $G$, then $C_G(T)$ is nilpotent.

A connected solvable algebraic group is nilpotent if and only if the set of semisimple elements form a subgroup. So, in order to prove $C_G(T)$ is nilpotent, I shall first determine whether it is connected. If it is, the next step is to prove that $(C_G(T))_s$, i.e., the set of semisimple elements of $C_G(T)$ is its subgroup.

As $G$ is connected solvable, $G$ can be imbeded in some $T(n,k)$, consisting of upper triangular matrices. Then $T=G \cap D(n,k)$, consisting of diagonal matrices. In order to prove $(C_G(T))_s$ is a subgroup, let $x_1,x_2$ be two elements in $G_s$, such that for any $y \in T$, $x_1y=yx_1$, $x_2y=yx_2$. Then how to prove $x_1x_2$ is still semisimple?

I have no idea. Thanks very much for any answer, hint or reference.

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up vote 2 down vote accepted

I would do it this way: $C=C_G(T)$

Let $a \in C_s$ an element which is semisimple. Then $at=ta$ for all $t \in T$. Let $H$ be ths closed subgroup spanned by $T$ and $a$. Then $H$ is commutative all its elements are semisimple therefore it is a torus und $T \subset H$ thus $T=H$ and $a \in T$. This proves, that $C_s=T$ is a subgroup thus C is nilpotent.

I hope this helps.

Greetings

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-_- I just tried to prove $(C_G(T))_s$ was a subgroup and didn't found out it in fact $=T$. @Daniel: Thank you very much. Sincere Greetings~ –  ShinyaSakai Aug 1 '11 at 17:38
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