Lemma 2

Assume that $k$ is of characteristic 0. Let $\mathfrak{g}$ be a semi-simple Lie algebra over $k$, B the Killing form of $\mathfrak{g},\mathfrak{m}$ a subalgebra of $\mathfrak{g}$. Assume that the following conditions are satisfied:

1) the restriction of B to $\mathfrak{m}$ is non-degenerate;

2) if $x\in \mathfrak{m}$, the semi-simple and nilpotent components$^1$ of $x$ in $\mathfrak{g}$ belong to $\mathfrak{m}$.

Then $\mathfrak{m}$ is reductive in $\mathfrak{g}$ (Chap. I, § 6, no. 6).

By Chap. I, §6, no. 4, Prop. $5d),\mathfrak{m}$ is reductive. Let $\mathfrak{c}$ be the centre of $\mathfrak{m}$. If $x\in \mathfrak{c}$ is nilpotent, then $x= 0$; indeed, for all $y\in \mathfrak{m}$, ad $x$ and ad $y$ commute, their composition ad $x\circ$ ad $y$ is nilpotent, and $B(x, y) = 0$, so $x= 0$. Now let $x$ be an arbitrary element of $\mathfrak{c}$; let $s$ and $n$ be its semi-simple and nilpotent components. We have $n\in \mathfrak{m}$. Since ad $n$ is of the form P(ad $x$), where P is a polynomial with no constant term, (ad $n$)$.\mathfrak{m}= 0$ and so $n\in \mathfrak{c}$, and then $n= 0$ by the above. Thus ad $x$ is semi-simple. Consequently, the restriction to $\mathfrak{m}$ of the adjoint representation of $\mathfrak{g}$ is semi-simple (Chap. I, §6, no. 5, Th. $4b$)$)$.