Is the commutator subgroup $[B,B]$ contained in the unipotent radical $R_u(B)$?

Question

Suppose $B$ is a connected, solvable algebraic group, and $T$ a maximal torus of $B$. I read that if $S$ is a subtorus, then $N_B(S)=C_B(S)$, the normalizer is equal to the centralizer.

I know that $B$ is the semidirect product of $T$ and the unipotent radical $R_u(B)$. The proof is just:

If $n\in N_B(S)$, $s\in S$, then $[n,s]\in [B,B]\cap S\subset R_u(B)\cap S=1$.

I don't follow $[B,B]\cap S\subset R_u(B)\cap S$. I assume it follows from $[B,B]\subset R_u(B)$. Is this true, if so, why? Thanks.

Answer 1

Here $B$ is connected solvable, so $B/R_u(B)$ is a torus, in particular abelian. Thus $[B, B] \leq R_u(B)$.