# If maximal tori are self centralizing in $R(G)$, why do Levi subgroups exist?

Let $G$ be a connected algebraic group, $L$ a connected subgroup. We say that $L$ is a Levi subgroup of $G$ if the product map $L \times R_u(G) \rightarrow G$ is an isomorphism of varieties.

Then $L$ is necessarily reductive, the restriction of $G \rightarrow G/R_u(G)$ to $L$ is a bijective morphism of algebraic groups, $R(L) = Z(L)^0$ is a maximal torus of $R(G)$, and $R(L)$ can be recovered as $L \cap R(G)$.

The following proposition is from Borel, Linear Algebraic Groups.

I understand the proof of (i), but (ii) I do not at all. Borel does not give a proof of (ii). He just says it follows from (i) and the following two facts:

Let $H$ be a connected solvable group.

(1) Maximal tori in $H$ are conjugate under $\bigcap\limits_{n=1}^{\infty} C^n$, where $C^1 \supseteq C^2 \supseteq \cdots$ denotes the descending central series.

(2) If $S \subseteq H$ is a subgroup of $H$ consisting of semisimple elements, then $S$ is contained in a torus, and $Z_G(S)$ is connected and equal to $N_G(S)$.

I don't understand how (ii) follows so easily. I let $T$ be a maximal torus of $R(G)$, and I'm trying to show that the product map $Z_G(T) \times R_u(G) \rightarrow G$ is an isomorphism of varieties.

So far, all I've got is that $Z_G(T) \cap R_u(G)$ is trivial. This is because of the hypothesis that $T = Z_{R(G)}(T) = Z_G(T) \cap R(G)$, whence $Z_G(T) \cap R_u(G)$ is a unipotent group contained in $Z_G(T) \cap R(G) = T$, which implies that $Z_G(T) \cap R_u(G)$ is both diagonalizable and unipotent, hence trivial.