# Maximal nilpotent and solvable Lie subalgebras

If $\mathfrak g$ is a finite dimensional complex semi-simple Lie algebra with maximal toral subalgebra $\frak h$.If $(E, ( , ),\Phi )$ is the corresponding root system. Fix a fundamental system $R$ of $\Phi$ with corresponding set of positive roots $\Phi^+$. how to prove that:

1. $N(R)$ is a maximal nilpotent subalgebra of $\mathfrak g$, where $N(R)=\bigoplus_{\alpha\in\Phi^+}\mathfrak g_{\alpha}$
2. $B(R)$ is a maximal solvable subalgebra of $\frak g$, where $B(R)=H\oplus N(R)$?

Let us first show the second claim. It is clear that $B=B(R)$ is a solvable subalgebra of $L$, since $[B,B]=N$ is nilpotent. Now let $B'$ be any solvable subalgebra of $L$ containing $H$. We will show that $B'\subseteq B$ for some base $\Delta$ of $\Phi$, so that $B$ is maximal solvable.
$B'$ has root space decomposition $$B'=H\oplus \bigoplus_{\beta \in S}L_{\beta}$$ for some subset $S\subseteq \Phi$. But $S$ cannot contain contain both $\beta$ and $-\beta$ since, otherwise, $B'$ would contain the semisimple Lie algebra $S_{\beta}:=L_{-\beta}\oplus \mathbb{C}h_{\beta}\oplus L_{\beta}$ which is not possible since all subalgebras of solvable algebras are solvable. But this implies that $S \subseteq \Phi_+$ with respect to some base $\Delta$ and thus $B' \subseteq B$.