Proposition 2

Assume that $k$ is algebraically closed. Let $\mathfrak{g}$ be a Lie algebra, $\mathfrak{h}$ and $\mathfrak{h}'$ Cartan subalgebras of $\mathfrak{g}$. There exist $u\in E(\mathfrak{h})$ and $u'\in E(\mathfrak{h}')$ such that $u(\mathfrak{h}) =u'(\mathfrak{h}')$.

We retain the notation of Lemma 2. From the fact that $\mathfrak{h}$ and $\mathfrak{h}'$ are Cartan subalgebras, it follows that $\mathfrak{g}^0(\mathfrak{h}) =\mathfrak{h}$ and $\mathfrak{g}^0(\mathfrak{h}') =\mathfrak{h}'$. By Lemma 2 and Prop. 3 of App. $I, E(\mathfrak{h})\mathfrak{h}_r$ and $E(\mathfrak{h}')\mathfrak{h}'_r$ contain open dense subsets of $\mathfrak{g}$ in the Zariski topology. Thus $E(\mathfrak{h})\mathfrak{h}_r\cap E(\mathfrak{h}')\mathfrak{h}'_r\not=\emptyset$. In other words, there exist $u\in E(\mathfrak{h}), u'\in E(\mathfrak{h}'), h\in \mathfrak{h}_r, h'\in \mathfrak{h}'_r$ such that $u(h) =u'(h')$; then