Theorem 3

Let $\mathfrak{g}$ be a solvable Lie algebra, and let $\mathfrak{h},\mathfrak{h}'$ be Cartan subalgebras of $\mathfrak{g}$. There exists $x\in \mathscr{C}^{\infty}(\mathfrak{g})$ such that $e^{adx}\mathfrak{h}=\mathfrak{h}'$.

We argue by induction on dim$\mathfrak{g}$, the case where $\mathfrak{g}= 0$ being trivial. Let $\mathfrak{n}$ be a minimal non-zero commutative ideal of $\mathfrak{g}$. Let $\varphi :\mathfrak{g}\rightarrow \mathfrak{g}/\mathfrak{n}$ be the canonical morphism. Then $\varphi (\mathscr{C}^{\infty}\mathfrak{g}) =\mathscr{C}^{\infty}(\mathfrak{g}/\mathfrak{n})$ (Chap. I, §1, no. 5, Prop. 4). Since $\varphi (\mathfrak{h})$ and $\varphi (\mathfrak{h}')$ are Cartan subalgebras of $\mathfrak{g}/\mathfrak{n}($§2, no. 1, Cor. 2 of Prop. 4), there exists, by the induction hypothesis, an $x\in \mathscr{C}^{\infty}(\mathfrak{g})$ such that $e^{ad\varphi(x)}\varphi (\mathfrak{h}) =\varphi (\mathfrak{h}')$. Replacing $\mathfrak{h}$ by $e^{adx}\mathfrak{h}$, we can assume that $\varphi (\mathfrak{h}) =\varphi (\mathfrak{h}')$, in other words that

Then $\mathfrak{h}$ and $\mathfrak{h}'$ are Cartan subalgebras of $\mathfrak{h}+\mathfrak{n}$. If $\mathfrak{h}+\mathfrak{n}\not=\mathfrak{g}$, the assertion to be proved follows from the induction hypothesis. Assume from now on that $\mathfrak{h}+\mathfrak{n}=\mathfrak{h}'+\mathfrak{n}=\mathfrak{g}$.

By the minimality of $\mathfrak{n}, [\mathfrak{g},\mathfrak{n}] =\{0\}$ or $[\mathfrak{g},\mathfrak{n}] =\mathfrak{n}$. If $[\mathfrak{g},\mathfrak{n}] =\{0\}$, then $\mathfrak{n}\subset \mathfrak{h}$ and $\mathfrak{n}\subset \mathfrak{h}'($§2, no. 1, Prop. 5), so $\mathfrak{h}=\mathfrak{h}+\mathfrak{n}=\mathfrak{h}'+\mathfrak{n}=\mathfrak{h}'$. It remains to consider the case where $[\mathfrak{g},\mathfrak{n}] =\mathfrak{n}$, so $\mathfrak{n}\subset \mathscr{C}^{\infty}(\mathfrak{g})$. The ideal $\mathfrak{n}$ is a simple $\mathfrak{g}$-module; since $\mathfrak{g}=\mathfrak{h}+\mathfrak{n}$, and since $[\mathfrak{n},\mathfrak{n}] =\{0\}$, it follows that $\mathfrak{n}$ is a simple $\mathfrak{h}$-module. If $\mathfrak{h}\cap \mathfrak{n}\not=\{0\}$, then $\mathfrak{n}\subset \mathfrak{h}$, so $\mathfrak{g}=\mathfrak{h}$ and $\mathfrak{h}'=\mathfrak{h}$. Assume now that $\mathfrak{h}\cap \mathfrak{n}=\{0\}$. Then $\mathfrak{g}=\mathfrak{h}\oplus \mathfrak{n}$ and hence $\mathfrak{g}=\mathfrak{h}'\oplus \mathfrak{n}$, since $\mathfrak{h}$ and $\mathfrak{h}'$ have the same dimension.

For all $x\in \mathfrak{h}$, let $f(x)$ be the unique element of $\mathfrak{n}$ such that $x-f(x)\in \mathfrak{h}'$; if $x, y\in \mathfrak{h}$,

so $f([x, y]) = [x, f(y)] + [f(x), y]$. By §1, no. 3, Cor. of Prop. 9, there exists $a\in \mathfrak{n}$ such that $f(x) = [x, a]$ for all $x\in \mathfrak{h}$. We have (ad $a$)$^2(\mathfrak{g})\subset$ (ad $a$)$(\mathfrak{n}) = 0$, so, for all $x\in \mathfrak{h}$,

Thus $e^{ada}(\mathfrak{h}) =\mathfrak{h}'$. Since $a\in \mathscr{C}^{\infty}(\mathfrak{g})$, this completes the proof.