Theorem 1

Let $\mathfrak{g}$ be a Lie algebra.

(i) If $x$ is a regular element of $\mathfrak{g},\mathfrak{g}^0(x)$ is a Cartan subalgebra of $\mathfrak{g}$.

(ii) If $\mathfrak{h}$ is a maximal nilpotent subalgebra of $\mathfrak{g}$, and if $x\in \mathfrak{h}$ is regular in $\mathfrak{g}$, then $\mathfrak{h}=\mathfrak{g}^0(x)$.

(iii) If $\mathfrak{h}$ is a Cartan subalgebra of $\mathfrak{g}$, then dim($\mathfrak{h}$)$\geq$ rk($\mathfrak{g}$).

(iv) The Cartan subalgebras of $\mathfrak{g}$ of dimension rk($\mathfrak{g}$) are the $\mathfrak{g}^0(x)$ where $x$ is a regular element.

Let $x$ be a regular element of $\mathfrak{g}$ and let $\mathfrak{h}=\mathfrak{g}^0(x)$. Clearly $\mathfrak{h}^0(x) =\mathfrak{h}$. Since $x$ is regular in $\mathfrak{h}$ (Prop. 9), rk($\mathfrak{h}$) $=$ dim($\mathfrak{h}$), so $\mathfrak{h}$ is nilpotent. On the other hand, $\mathfrak{h}=\mathfrak{g}^0(x)\supset \mathfrak{g}^0(\mathfrak{h})\supset \mathfrak{h}$, so $\mathfrak{h}=\mathfrak{g}^0(\mathfrak{h})$ is a Cartan subalgebra of $\mathfrak{g}$ (Prop. 4). This proves (i).

If $\mathfrak{h}$ is a maximal nilpotent subalgebra of $\mathfrak{g}$, and if $x\in \mathfrak{h}$ is regular in $\mathfrak{g}$, then $\mathfrak{h}\subset \mathfrak{g}^0(x)$ and $\mathfrak{g}^0(x)$ is nilpotent by (i), so $\mathfrak{h}=\mathfrak{g}^0(x)$, which proves (ii).

If $\mathfrak{h}$ is a Cartan subalgebra of $\mathfrak{g}$, there exists $x\in \mathfrak{h}$ such that $\mathfrak{h}=\mathfrak{g}^0(x)$ (Cor. 1 of Prop. 4), so dim($\mathfrak{h}$)$\geq$ rk($\mathfrak{g}$), which proves (iii). If in addition dim($\mathfrak{h}$) $=$ rk($\mathfrak{g}$)$,x$ is regular. Finally, if $x'$ is regular in $\mathfrak{g},\mathfrak{g}^0(x')$ is a Cartan subalgebra by (i), and is obviously of dimension rk($\mathfrak{g}$). This proves (iv).

We shall see in §3, Th. 2 that, when $k$ is of characteristic zero, all the Cartan subalgebras of $\mathfrak{g}$ have dimension rk($\mathfrak{g}$).