Proposition 3

Let $\mathfrak{g}$ be a Lie algebra, $\mathfrak{g}'$ a subalgebra of $\mathfrak{g}$. The following conditions are equivalent:

(i) $\mathfrak{g}'$ contains a regular element of $\mathfrak{g}$, and rk($\mathfrak{g}$) $=$ rk($\mathfrak{g}'$);

(ii) $\mathfrak{g}'$ contains a Cartan subalgebra of $\mathfrak{g}$;

(iii) every Cartan subalgebra of $\mathfrak{g}'$ is a Cartan subalgebra of $\mathfrak{g}$.

(i) $=\Rightarrow$ (ii): Assume that rk($\mathfrak{g}$) $=$ rk($\mathfrak{g}'$), and that there exists $x\in \mathfrak{g}'$ regular in $\mathfrak{g}$. Put $\mathfrak{h}=\mathfrak{g}^0(x),\mathfrak{h}'={\mathfrak{g}'}^0(x) =\mathfrak{h}\cap \mathfrak{g}'$. Then

rk($\mathfrak{g}'$)$\leq$ dim$\mathfrak{h}'\leq$ dim$\mathfrak{h}=$ rk($\mathfrak{g}$) $=$ rk($\mathfrak{g}'$)

so $\mathfrak{h}=\mathfrak{h}'\subset \mathfrak{g}'$. This proves (ii).

(ii) $=\Rightarrow$ (iii): Assume that $\mathfrak{g}'$ contains a Cartan subalgebra $\mathfrak{h}$ of $\mathfrak{g}$, and let $\mathfrak{h}_1$ be a Cartan subalgebra of $\mathfrak{g}'$. To prove that $\mathfrak{h}_1$ is a Cartan subalgebra of $\mathfrak{g}$, we can assume that $k$ is algebraically closed. Let $E(\mathfrak{h})$ and $E'(\mathfrak{h})$ be the groups of automorphisms of $\mathfrak{g}$ and $\mathfrak{g}'$ associated to $\mathfrak{h}$ (no. 2). By Th. 1, there exists $f\in E'(\mathfrak{h})$ such that $f(\mathfrak{h}) =\mathfrak{h}_1$. Now every element of $E'(\mathfrak{h})$ is induced by an element of $E(\mathfrak{h})$; indeed, it suffices to verify this for $e^{adx}$, with $x\in {\mathfrak{g}'}^{\lambda}(\mathfrak{h}),\lambda \not= 0$, in which case it follows from the inclusion ${\mathfrak{g}'}^{\lambda}(\mathfrak{h})\subset \mathfrak{g}^{\lambda}(\mathfrak{h})$. Thus $\mathfrak{h}_1$ is a Cartan subalgebra of $\mathfrak{g}$.

(iii) $=\Rightarrow$ (i): Assume that condition (iii) is satisfied. Let $\mathfrak{h}$ be a Cartan subalgebra of $\mathfrak{g}'$. Since this is a Cartan subalgebra of $\mathfrak{g}$, it contains a regular element of $\mathfrak{g}$ (Th. 2 (ii)), and on the other hand rk($\mathfrak{g}$) $=$ dim($\mathfrak{h}$) $=$ rk($\mathfrak{g}'$).