Proposition 2 (Jacobson-Morozov)

Assume that $\mathfrak{g}$ is semi-simple. Let $x$ be a non-zero nilpotent element of $\mathfrak{g}$. There exist $h, y\in \mathfrak{g}$ such that $(x, h, y)$ is an $\mathfrak{s}\mathfrak{l}_2$-triplet.

Let $\mathfrak{n}=$ Ker(ad $x$)$^2$. If $z\in \mathfrak{n}$, then [ad $x$,[ad $x$, ad $z]] =$ ad([$x,[x, z]]$) $= 0$. By Lemma 5, ad $x\circ$ ad $z$ is nilpotent, so Tr(ad $x\circ$ ad $z$) $= 0$. This shows that $x$ is orthogonal to $\mathfrak{n}$ with respect to the Killing form $\Phi$ of $\mathfrak{g}$. Since

$\Phi$((ad $x$)$^2y, y'$) $=\Phi (y$, (ad $x$)$^2y')$

for all $y, y'\in \mathfrak{g}$, and since $\Phi$ is non-degenerate, the orthogonal complement of $\mathfrak{n}$ is the image of (ad $x$)$^2$. Hence there exists $y'\in \mathfrak{g}$ such that $x=$ (ad $x$)$^2y'$. Put

we have $[h, x] = 2x$ and $h\in$ (ad $x$)$(\mathfrak{g})$. It now suffices to apply Lemma 6.