Proposition 6

Assume that $k$ is algebraically closed and that $\mathfrak{g}$ is semi-simple. Let G be a group of automorphisms of $\mathfrak{g}$ containing Aut$_e(\mathfrak{g})$. Let $(x, h, y)$ and $(x', h', y')$ be $\mathfrak{s}\mathfrak{l}_2$-triplets in $\mathfrak{g}$. The following conditions are equivalent:

(i) $h$ and $h'$ are G-conjugate;

(ii) $(x, h, y)$ and $(x', h', y')$ are G-conjugate.

We only have to prove the implication (i) $=\Rightarrow$ (ii), and we are reduced immediately to the case in which $h=h'$. Introduce $\mathfrak{g}^2$ and $G_h$ as in Lemma 8. We have $x\in \mathfrak{g}^2$, and $[x,\mathfrak{g}^0] =\mathfrak{g}^2$ by §1, no. 2, Cor. of Prop. 2. Hence $G_hx$ contains a subset of $\mathfrak{g}^2$ that is dense and open in the Zariski topology, and so does $G_hx'$. So there exists $a\in G_h$ such that $a(x) =x'$. We have $a(h) =h$, and consequently $a(y) =y'$ (no. 1, Lemma 1).