Theorem 2

Let $\mathfrak{g}$ be a semi-simple Lie algebra, $\mathfrak{n}$ a subalgebra of $\mathfrak{g}$ consisting of nilpotent elements, $\mathfrak{q}$ the normalizer of $\mathfrak{n}$ in $\mathfrak{g}$. Assume that $\mathfrak{n}$ is the set of nilpotent elements of the radical of $\mathfrak{q}$. Then $\mathfrak{q}$ is parabolic.

Note first of all that $\mathfrak{q}$ is decomposable (Chap. VII, §5, no. 1, Cor. 1 of Prop. 3). By Th. 1, it suffices to prove that $\mathfrak{q}$ is the orthogonal complement $\mathfrak{n}^0$ of $\mathfrak{n}$ with respect to the Killing form $\Phi$ of $\mathfrak{g}$. We know that $\mathfrak{q}\subset \mathfrak{n}^0$ (Chap. I, §4, no. 3, Prop. $4d$)$)$. By Chap. VII, §5, no. 3, Prop. 7, there exists a subalgebra $\mathfrak{m}$ of $\mathfrak{q}$, reductive in $\mathfrak{g}$, such that $\mathfrak{q}$ is the semi-direct product of $\mathfrak{m}$ and $\mathfrak{n}$. We show that the restriction of $\Phi$ to $\mathfrak{m}$ is non-degenerate. Let $\mathfrak{c}$ be the centre of $\mathfrak{m}$. We have $\Phi ([\mathfrak{m},\mathfrak{m}],\mathfrak{c}) = 0$ by Chap. I, §5, no. 5, Prop. 5, and the restriction of $\Phi$ to $[\mathfrak{m},\mathfrak{m}]$ is non-degenerate by Chap. I, §6, no. 1, Prop. 1. It remains to see that the restriction of $\Phi$ to $\mathfrak{c}$ is non-degenerate. Let $\mathfrak{k}$ be a Cartan subalgebra of $[\mathfrak{m},\mathfrak{m}]$; then $\mathfrak{k}\oplus \mathfrak{c}$ is commutative and reductive in $\mathfrak{g}$. Let $\mathfrak{h}$ be a Cartan subalgebra of $\mathfrak{g}$ containing $\mathfrak{k}\oplus \mathfrak{c}$ (Chap. VII, §2, no. 3, Prop. 10). Then $\mathfrak{h}\cap \mathfrak{q}$ is a commutative subalgebra of $\mathfrak{q}$ containing $\mathfrak{k}\oplus \mathfrak{c}$, and ad$_{\mathfrak{q}}x$ is semi-simple for all $x\in \mathfrak{h}\cap \mathfrak{q}$; hence $\mathfrak{h}\cap \mathfrak{q}$ is contained in a Cartan subalgebra $\mathfrak{h}'$ of $\mathfrak{q}$ (Chap. VII, §2, no. 3, Prop. 10); let $f$ be the projection of $\mathfrak{q}$ onto $\mathfrak{m}$ with kernel $\mathfrak{n}$; then $f(\mathfrak{h}')$ is a Cartan subalgebra of $\mathfrak{m}$ (Chap. VII, §2, no. 1, Cor. 2 of Prop. 4) containing $\mathfrak{k}\oplus \mathfrak{c}$, and consequently equal to $\mathfrak{k}\oplus \mathfrak{c}$; this proves that $f(\mathfrak{h}\cap \mathfrak{q}) =\mathfrak{k}\oplus \mathfrak{c}$, and since every element of $\mathfrak{h}$ is semi-simple in $\mathfrak{g}$, we have $\mathfrak{h}\cap \mathfrak{q}=\mathfrak{k}\oplus \mathfrak{c}$. Thus,

$\mathfrak{c}=\{x\in \mathfrak{h}|[x,\mathfrak{n}]\subset \mathfrak{n}$ and $[x,[\mathfrak{m},\mathfrak{m}]] = 0\}$.

By Lemma 1, the restriction of $\Phi$ to $\mathfrak{c}$ is non-degenerate.

Let $\mathfrak{q}^0$ be the orthogonal complement of $\mathfrak{q}$ in $\mathfrak{g}$ relative to $\Phi$. The preceding proves that $\mathfrak{q}\cap \mathfrak{q}^0=\mathfrak{n}$. Assume that $\mathfrak{q}\not=\mathfrak{q}^0$, so $\mathfrak{q}^0\not=\mathfrak{n}$ (and $\mathfrak{q}^0\supset \mathfrak{n}$). Since ad$_{\mathfrak{g}}\mathfrak{n}$ leaves $\mathfrak{q}$ stable, ad$_{\mathfrak{g}}\mathfrak{n}$ leaves $\mathfrak{q}^0$ stable; Engel’s theorem proves that there exists $x\in \mathfrak{q}^0$ such that $x \notin \mathfrak{n}$ and $[x,\mathfrak{n}]\subset \mathfrak{n}$. But then $x\in \mathfrak{q}^0\cap \mathfrak{q}=\mathfrak{n}$, a contradiction. Hence $\mathfrak{q}=\mathfrak{n}^0$.