Lemma D.6 of Fulton & Harris claims
If $H$ is regular, then $\mathfrak g_0(H)$ is abelian,
where $\mathfrak g$ is a complex Lie algebra, $\mathfrak g_0(H)=\ker\operatorname{ad}(H)^m$ for large $m$, and "regular" has been temporarily defined to mean $\dim\mathfrak g_0(H)\leqslant\dim\mathfrak g_0(X)$ for all $X \in \mathfrak g$. Though it is a bit unclear what is being assumed, it seems that $\mathfrak g$ is semisimple (this fact is used explicitly in the proof), while $H\in\mathfrak g$ can be arbitrary (i.e., $H$ is not necessarily semisimple/diagonalizable).
The first two paragraphs show that the Killing form $B$, when restricted to $\mathfrak g_0(H)$, is nondegenerate. I understand this part. The third paragraph is as follows.
Consider the Jordan decomposition $X = X_s + X_n$ of an element $X\in\mathfrak g_0(H)$. Since $\operatorname{ad}(X_n)=\operatorname{ad}(X)_n$ is nilpotent, $X_n$ belongs to $\mathfrak g_0(H)$, so $X_s = X - X_n$ does also. Then $\operatorname{ad}(X_s) = \operatorname{ad}(X)_s$ is nilpotent and semisimple on $\mathfrak g_0(H)$, so it vanishes there.
I don't understand the implications
- ad$(X_n)$ is nilpotent $\implies X_n\in\mathfrak g_0(H)$
- $X_s\in\mathfrak g_0(H)\implies X_s$ is nilpotent.
If this follows immediately from the definitions, or from the surrounding propositions in the chapter, I cannot see it. Please help clarify these claims.