(Question 16.2 of Humphreys's book Introduction to Lie algebras) Let $L$ be semisimple, $H$ a Cartan Subalgebra, $\Delta$ a base for the root system $\Phi$. Prove that any subalgebra of $L$ consisting of nilpotent elements, and maximal with respect to this property, is cojugate under $\mathscr E(L)$ to $N(\Delta)$, the derived algebra of $B(\Delta)$.
Notation. Call $x\in L$ strongly ad-nilpotent if $x\in \{z\in L: (\mbox{ad }y-a.I)^k(z) = 0 \mbox{ for some
integer $k\in \mathbb Z_{>0}$}\}, a\neq 0.$ In particular, $x$ will be ad-nilpotent, so it makes sense to consider $\mbox{exp } \mbox{ad }x = \sum_n (\mbox{ad }x)^n/n!$ . Denote by $\mathscr E(L)$ the subgroup generated by all $\mbox{exp} \mbox{ ad } x$ for $x$ strongly ad-nilpotent.
Given a base $\Delta$ of $\Phi$, $B(\Delta)$ denotes the Borel subalgebra $B(\Delta) = H\oplus(\oplus_{\alpha \in \Phi^+} L_\alpha)$ and its derived algebra is $[B(\Delta),B(\Delta)] = N(\Delta) = \oplus_{\alpha \in \Phi^+} L_\alpha$.
Ideas. Let $K\subseteq L$ be a subalgebra consisting of nilpotent elements. Then, its elements will be ad-nilpotent and $K$ is nilpotent by Engel's theorem. In particular, $K$ is solvable, so its contained in some Borel subalgebra $B$ (maximal solvable). Now, Borel subalgebras are conjugate under $\mathscr E(L)$, so $\sigma(B(\Delta)) = B$ for some $\sigma \in \mathscr E(L)$.
Somehow I have to use the hyphotesis that $K$ is maximal wrt to the fact that its elements are nilpotent. I can't see how to argue that $K\subseteq \sigma(N(\Delta))$, because if that was true, then equality of sets must hold.
Any insight or help is very much appreciated.
Best Answer
Put $K' = \sigma^{-1}(K) \leq B(\Delta)$. Then it consists of nilpotent elements(i.e. for any $x\in K'$, $\operatorname{ad}_L x \in \operatorname{End}(L)$ is nilpotent), and is maximal with respect to this property. If $0 \neq h$ is an element of $H$ (which is a toral subalgebra of $L$), then it is not a nilpotent element of $L$ because $\operatorname{ad}_L h$ is not nilpotent. (Note that $\operatorname{ad}_H h $ is nilpotent, however.) Thus $K' \cap H = 0$, which implies $K' \subseteq B(\Delta) - \left( H \setminus\{ 0\} \right) = N(\Delta) $. It follows that $K=\sigma(K') \subseteq \sigma \left( N(\Delta) \right)$.