= Solution
{c}
The <normalizer of a Lie subalgebra> and the <centralizer of a Lie subalgebra> are
$$
N_{\mathfrak g}(\mathfrak a)=\{x:[x,\mathfrak a]\subseteq\mathfrak a\},\qquad C_{\mathfrak g}(\mathfrak a)=\{x:[x,\mathfrak a]=0\}.
$$
The second is always contained in the first. Conversely, let $x\in N_{\mathfrak g}(\mathfrak a)$ and $a\in\mathfrak a$. Then $[a,x]\in\mathfrak a$, and, because $\mathfrak a$ is an <abelian Lie algebra>,
$$
(\operatorname{ad}a)^2x=[a,[a,x]]=0.
$$
By the definition of a <toral subalgebra>, $\operatorname{ad}a$ is <diagonalizable>. A <diagonalizable> <linear map> $T$ satisfies $\ker T^2=\ker T$: in an <eigenbasis>, both <kernels> consist of exactly the <eigenvectors> whose <eigenvalue> is zero. Thus $[a,x]=0$. This holds for every $a\in\mathfrak a$, so $x\in C_{\mathfrak g}(\mathfrak a)$ and
$$
\boxed{N_{\mathfrak g}(\mathfrak a)=C_{\mathfrak g}(\mathfrak a).}
$$
This is the <normalizer-centralizer equality for a toral abelian subalgebra>; it does not require $\mathfrak g$ to be a <semisimple Lie algebra>.
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