Lemma 8

Let $\alpha \in B\cup (-B)$. Then ad $X_{\alpha}$ is locally nilpotent.$^1$

Assume that $\alpha \in B$. Let $\mathfrak{g}'$ be the set of $z\in \mathfrak{g}$ such that (ad $X_{\alpha}$)$^pz= 0$ for sufficiently large $p$. Since ad $X_{\alpha}$ is a derivation of $\mathfrak{g},\mathfrak{g}'$ is a subalgebra of $\mathfrak{g}$. By (21), $X_{\beta}\in \mathfrak{g}'$ for all $\beta \in B$. By (17), (19), (20), $H_{\beta}\in \mathfrak{g}'$ and $X_{-\beta}\in \mathfrak{g}'$ for all $\beta \in B$. Hence $\mathfrak{g}'=\mathfrak{g}$ and ad $X_{\alpha}$ is locally nilpotent. Since ad $X_{-\alpha}=\theta$(ad $X_{\alpha}$)$\theta^{-1}$, we see that ad $X_{-\alpha}$ is locally nilpotent.

We shall see that $\mathfrak{g}$ is finite dimensional, so that ad $X_{\alpha}$ is actually nilpotent.