Normalizer of the unipotent radical is the normalizer of the Borel?

Question

Let $G$ be a connected, reductive group, and let $B$ be a Borel subgroup of $G$. Let $U = \mathscr R_u(B)$ be the group of unipotent elements of $B$. Is it true that $N_G(U) = B$?

Since Borel subgroups are self normalizing, it is enough to show that everything which normalizes $U$ also normalizes $B$. It's easy to check this is true for $\textrm{GL}_2$.

If $x$ is a regular semisimple element which normalizes $U$, then the question of whether $x$ lies in $B$ comes down to the following: does there exist a maximal torus $T'$ of $G$ which normalizes $U$, such that $T'U$ is a Borel subgroup of $G$ which is different from $B$?

Answer 1

This is true. Let $T$ be a maximal torus of $B$, and let $\Delta$ be the simple roots of $T$ in $U$. Then $$(G,B,N_G(T),W)$$ is a BN-pair, where $W = N_G(T)/T$ is the Weyl group. The only subgroups of $G$ containing $B$ are the standard parabolic subgroups. Since $U$ is a normal subgroup of $B$, $N_G(U)$ contains $B$.

If we suppose the containment $B \subseteq N_G(U)$ is strict, then $N_G(U)$ contains a parabolic subgroup of the form $P_{\alpha}$ for some $\alpha \in \Delta$. Then $P_{\alpha}$ has a Levi decomposition as $M_{\alpha}N_{\alpha}$, where $M_{\alpha} = Z_G((\textrm{Ker}(\alpha)^0)$.

$T$ has index two in $N_G(T) \cap M_{\alpha}$. Let $n$ be an element of $N_G(T) \cap M_{\alpha}$ which is not in $T$. Then $n \in N_G(U)$, but $nU_{\alpha}n^{-1} = U_{-\alpha} \not\subseteq U$, contradiction.

By similar reasoning, if $P$ is any parabolic subgroup of $G$, and $N = \mathscr R_u(P)$, then the normalizer of $N$ is $P$.