# Why is $\textrm {ad}_{L}t_{\alpha}$ nilpotent?

This question is for users who own and are familiar with Humphreys' Introduction to Lie Algebras and Representation Theory.

I need help understanding a line in the proof of part (e) on page 38. It reads "$\textrm {ad}_{L}s$ is $nilpotent$ for all $s\in [SS]$".

Here, $S$ is the subalgebra of $L$ spanned by $x,y,t_{\alpha}$ with $[xt_{\alpha}]=0,[yt_{\alpha}]=0,[xy]=t_{\alpha}$. I realize that $[SS]=Ft_{\alpha}$, so I need to see why $\textrm {ad}_{L}t_{\alpha}$ is nilpotent. Humphreys says it's due to Lie's Theorem. I see that $S\simeq {ad}_{L}S\subseteq \mathfrak gl(L)$ is solvable, so there is a basis of $L$ relative to which $\textrm {ad}_{L}t_{\alpha}$ is upper triangular, but I don't see how this helps.

Does anyone have this figured out that could explain to me why $\textrm {ad}_{L}t_{\alpha}$ is nilpotent?

$ad_{L}S$ is a solvable subalgebra of $\mathfrak gl(L)$. By Lie's Theorem, the matrices of $ad_{L}S$ relative to a suitable basis of $L$ are upper triangular. In particular, $ad_{L}x$ and $ad_{L}y$ are upper triangular relative to this basis. So $ad_{L}t_{\alpha}=ad_{L}[xy]=[ad_{L}x,ad_{L}y]$ is strictly upper triangular, hence nilpotent.