Show that the maximal toral sublgebra is self noramalizing

Question

Let $L$ be a semisimple Let $H$ be the maximal toral subalgebra. I want to show that $H=N_L(H)$.

The elements in $H$ are all semisimple. The elements in $N_L(H)$ are $x$ such that $[x,H]=H$. I do not see how to relate these two things.

Answer 1

Since $H$ is a Lie-subalgebra, we get the inclusion $H\subseteq N_L(H)$. For the other inclusion consider the root space decomposition $L=H \oplus (\bigoplus_{\alpha \in \Phi}L_\alpha)$ and use the fact that $[L_\alpha, H]=L_\alpha$ (which holds since $H=L_0$ and $\alpha+0\neq 0$).