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.