 Normalizer-centralizer equality for a toral abelian subalgebra

If $\mathfrak{a}$ is a toral subalgebra and an abelian Lie algebra, its normalizer of a Lie subalgebra equals its centralizer of a Lie subalgebra. For $x$ in the former and $a\in \mathfrak{a}$, one has $[a,x]\in \mathfrak{a}$, hence $[a,[a,x]]=0$. A diagonalizable linear map has the same kernel as its square, by expansion in an eigenbasis. Applying this to $\mathrm{ad}a$ gives $[a,x]=0$ for every $a$. The reverse containment follows directly from the definitions. No assumption that $\mathfrak{g}$ is a semisimple Lie algebra is needed.