Proposition 13

Let $\mathfrak{p}=\mathfrak{h}+\mathfrak{g}^P$ be a parabolic subalgebra of $(\mathfrak{g},\mathfrak{h})$, Q the set of $\alpha \in P$ such that $-\alpha \notin P$, and $\mathfrak{s}=\mathfrak{h}+\mathfrak{g}^{P\cap(-P)}$. Then $\mathfrak{p}=\mathfrak{s}\oplus \mathfrak{g}^Q,\mathfrak{s}$ is reductive in $\mathfrak{g}$, and $\mathfrak{g}^Q$ is the largest nilpotent ideal of $\mathfrak{p}$ and the nilpotent radical of $\mathfrak{p}$. The centre of $\mathfrak{p}$ is zero.

By Prop. $2,\mathfrak{s}$ is reductive in $\mathfrak{g}$ and $\mathfrak{g}^Q$ is a nilpotent ideal of $\mathfrak{p}$. If $\mathfrak{n}$ is the largest nilpotent ideal of $\mathfrak{p},\mathfrak{g}^Q\subset \mathfrak{n}\subset \mathfrak{h}+\mathfrak{g}^Q$ (Prop. 2 (i)); if $x\in \mathfrak{n}\cap \mathfrak{h}$, ad$_{\mathfrak{p}}x$ is nilpotent, so $\alpha (x) = 0$ for all $\alpha \in P$, and hence $x= 0$; this proves that $\mathfrak{n}=\mathfrak{g}^Q$. Since $[\mathfrak{h},\mathfrak{g}^Q] =\mathfrak{g}^Q$, the nilpotent radical of $\mathfrak{p}$ contains $\mathfrak{g}^Q$ and consequently is equal to $\mathfrak{g}^Q$. Let $z=h+\sum_{\alpha\in P}u_{\alpha}$ (where $h\in \mathfrak{h}, u_{\alpha}\in \mathfrak{g}^{\alpha}$) be an

element of the centre of $\mathfrak{p}$. For all $h'\in \mathfrak{h}, 0 = [h', z] =\sum\alpha (h')u_{\alpha}$, so $u_{\alpha}= 0$ for all $\alpha \in P$; it follows that $[h,\mathfrak{g}^{\beta}] = 0$ for all $\beta \in P$, so $h= 0$.