Proposition 1

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 $\mathfrak{g}$. Let

Consider the following conditions:

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

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

(iii) $\mathfrak{t}$ and $\mathfrak{t}'$ are G-conjugate.

We have (i) $\Leftarrow \Rightarrow$ (ii) $=\Rightarrow$ (iii). If $k$ is algebraically closed, the three conditions are equivalent.

(i) $\Leftarrow \Rightarrow$ (ii): This follows from Lemma 4.

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

We assume that $k$ is algebraically closed and prove that (iii) $=\Rightarrow$ (i). We treat first the case in which $\mathfrak{t}=\mathfrak{t}'=\mathfrak{g}=\mathfrak{s}\mathfrak{l}(2, k)$. Since ad$_{\mathfrak{g}}x$ is nilpotent, the endomorphism $x$ of $k^2$ is nilpotent (Chap. I, §6, Th. 3), so there exists a matrix $A\in \mathbf{G}\mathbf{L}(2, k)$ such that $AxA^{-1}=X$, and consequently an automorphism $\alpha$ of $\mathfrak{s}\mathfrak{l}(2, k)$ such that $\alpha (x) =x'$; now $\alpha \in$ Aut$_e(\mathfrak{g}) ($§5, no. 3, Cor. 2 of Prop. 5). We now pass to the general case; we assume that $\mathfrak{t}$ and $\mathfrak{t}'$ are G-conjugate and prove that $x$ and $x'$ are G-conjugate. We can assume that $\mathfrak{t}=\mathfrak{t}'$. By the preceding, there exists $\beta \in$ Aut$_e(\mathfrak{t})$ such that $\beta x=x'$. Now, if $t\in \mathfrak{t}$ is such that ad$_{\mathfrak{t}}t$ is nilpotent, then ad$_{\mathfrak{g}}t$ is nilpotent; so $\beta$ extends to an element of Aut$_e(\mathfrak{g})$.