Proposition 6

Let $\mathfrak{g}$ be a decomposable Lie subalgebra of $\mathfrak{g}\mathfrak{l}(V)$. Let $\mathscr{T}$ be the set of commutative subalgebras of $\mathfrak{g}$ consisting of semi-simple elements, and $\mathscr{T}_1$ the set of maximal elements of $\mathscr{T}$. Let $\mathscr{H}$ be the set of Cartan subalgebras of $\mathfrak{g}$.

(i) For $\mathfrak{h}\in \mathscr{H}$, let $\varphi (\mathfrak{h})$ be the set of semi-simple elements of $\mathfrak{h}$. Then $\varphi (\mathfrak{h})\in \mathscr{T}_1$.

(ii) For $\mathfrak{t}\in \mathscr{T}_1$, let $\psi (\mathfrak{t})$ be the commutant of $\mathfrak{t}$ in $\mathfrak{g}$. Then $\psi (\mathfrak{t})\in \mathscr{H}$.

(iii) The maps $\varphi$ and $\psi$ are inverse bijections from $\mathscr{H}$ to $\mathscr{T}_1$ and from $\mathscr{T}_1$ to $\mathscr{H}$.

(iv) If $k$ is algebraically closed, Aut$_e(\mathfrak{g})$ operates transitively on $\mathscr{T}_1$.

Let $\mathfrak{h}\in \mathscr{H}$, and put $\mathfrak{t}=\varphi (\mathfrak{h})$. By Prop. 5 and Cor. 2 of Prop. $3,\mathfrak{t}\in \mathscr{T}$ and $\mathfrak{h}=\mathfrak{t}\times \mathfrak{n}_V(\mathfrak{h})$. For any subalgebra $\mathfrak{u}$ of $\mathfrak{g}$, we denote by $\psi (\mathfrak{u})$ the commutant of $\mathfrak{u}$ in $\mathfrak{g}$. Then $\mathfrak{h}\subset \psi (\mathfrak{t})$, and $\psi (\mathfrak{t})\subset \mathfrak{g}^0(\mathfrak{h})$ since the elements of $\mathfrak{n}_V(\mathfrak{h})$ are nilpotent, so $\mathfrak{h}=\psi (\mathfrak{t})$. If $\mathfrak{t}'\in \mathscr{T}$ and $\mathfrak{t}\subset \mathfrak{t}'$, we have $\mathfrak{t}'\subset \psi (\mathfrak{t}) =\mathfrak{h}$ so $\mathfrak{t}'=\mathfrak{t}$, and hence $\mathfrak{t}\in \mathscr{T}_1$.

Let $\mathfrak{t}\in \mathscr{T}_1$, and put $\mathfrak{c}=\psi (\mathfrak{t})$. Let $\mathfrak{h}$ be a Cartan subalgebra of $\mathfrak{c}$. By §2, no. 3, Prop. $10,\mathfrak{h}\in \mathscr{H}$ and $\mathfrak{t}\subset \mathfrak{h}$. Put $\mathfrak{t}_1=\varphi (\mathfrak{h})\in \mathscr{T}$. Then $\mathfrak{t}\subset \mathfrak{t}_1$ so $\mathfrak{t}=\mathfrak{t}_1$, and $\mathfrak{h}=\psi (\mathfrak{t}_1) =\psi (\mathfrak{t}) =\mathfrak{c}$ by the preceding. Thus, $\psi (\mathfrak{t})\in \mathscr{H}$, and $\varphi (\psi (\mathfrak{t})) =\mathfrak{t}$.

We have thus proved (i), (ii) and (iii). Assume that $k$ is algebraically closed. Since Aut$_e(\mathfrak{g})$ operates transitively on $\mathscr{H}($§3, no. 2, Th. 1), Aut$_e(\mathfrak{g})$ operates transitively on $\mathscr{T}_1$.