Proposition 5

Let $\mathfrak{g}$ be a Lie algebra, $\mathfrak{r}$ its radical, $\mathfrak{h}$ and $\mathfrak{h}'$ Cartan subalgebras of $\mathfrak{g}$, and $\varphi$ the canonical homomorphism from $\mathfrak{g}$ to $\mathfrak{g}/\mathfrak{r}$. The following conditions are equivalent:

(i) $\mathfrak{h}$ and $\mathfrak{h}'$ are conjugate by an elementary automorphism of $\mathfrak{g}$;

(ii) $\varphi (\mathfrak{h})$ and $\varphi (\mathfrak{h}')$ are conjugate by an elementary automorphism of $\mathfrak{g}/\mathfrak{r}$.

(i) $=\Rightarrow$ (ii): This is clear.

(ii) $=\Rightarrow$ (i): We assume that condition (ii) is satisfied and prove (i). By Lemma 3, we are reduced to the case where $\varphi (\mathfrak{h}) =\varphi (\mathfrak{h}')$. Put $\mathfrak{k}=$ $\mathfrak{h}+\mathfrak{r}=\mathfrak{h}'+\mathfrak{r}$, which is a solvable subalgebra of $\mathfrak{g}$. Then $\mathfrak{h}$ and $\mathfrak{h}'$ are Cartan subalgebras of $\mathfrak{k}$, so there exists $x\in \mathscr{C}^{\infty}(\mathfrak{k})$ such that $e^{ad_{\mathfrak{k}}x}\mathfrak{h}=\mathfrak{h}'$ (Th. 3). Since $\mathfrak{k}/\mathfrak{r}$ is nilpotent, $\mathscr{C}^{\infty}(\mathfrak{k})\subset \mathfrak{r}$; on the other hand, $\mathscr{C}^{\infty}(\mathfrak{k})\subset [\mathfrak{k},\mathfrak{k}]\subset [\mathfrak{g},\mathfrak{g}]$, so $x\in \mathfrak{r}\cap [\mathfrak{g},\mathfrak{g}]$; by Chap. I, §5, no. 3, Th. 1, ad$_{\mathfrak{g}}x$ is nilpotent, so $e^{ad_{\mathfrak{g}}x}$ is an elementary automorphism of $\mathfrak{g}$ transforming $\mathfrak{h}$ to $\mathfrak{h}'$.