Proposition 11

Let $\mathfrak{n}$ be a nilpotent subalgebra of a Lie algebra $\mathfrak{g}$. There exists a Cartan subalgebra of $\mathfrak{g}$ contained in $\mathfrak{g}^0(\mathfrak{n})$.

Put $\mathfrak{a}=\mathfrak{g}^0(\mathfrak{n})$. Then $\mathfrak{n}\subset \mathfrak{a}$ since $\mathfrak{n}$ is nilpotent. If $x\in \mathfrak{a}$, let $P(x)$ be the determinant of the endomorphism of $\mathfrak{g}/\mathfrak{a}$ defined by ad $x$. Denote by $\mathfrak{a}'$ the set of $x\in \mathfrak{a}$ such that $P(x)\not= 0$, which is an open subset of $\mathfrak{a}$ in the Zariski topology; the relations $x\in \mathfrak{a}'$ and $\mathfrak{g}^0(x)\subset \mathfrak{a}$ are equivalent. By Prop. 7 (ii) of §1, no. 2, there exists $y\in \mathfrak{n}$ such that $\mathfrak{g}^0(y) =\mathfrak{a}$, and $y\in \mathfrak{a}'$ so $\mathfrak{a}'$ is non-empty. Since $\mathfrak{a}'$ is open, its intersection with the set of regular elements of $\mathfrak{a}$ is non-empty. Let $x$ be an element of this intersection. Then $\mathfrak{g}^0(x)\subset \mathfrak{a}$ and $\mathfrak{g}^0(x)$ is a Cartan subalgebra of $\mathfrak{a}$, hence is nilpotent. On the other hand, Prop. 10 (i) of §1, no. 3, shows that $\mathfrak{g}^0(x)$ is its own normalizer in $\mathfrak{g}$; it is therefore a Cartan subalgebra of $\mathfrak{g}$, which completes the proof.