Proposition 1

Let $\mathfrak{g}$ be a decomposable Lie subalgebra of $\mathfrak{g}\mathfrak{l}(V),x\in \mathfrak{g},s$ and $n$ the semi-simple and nilpotent components of $x$.

(i) The semi-simple and nilpotent components of ad$_{\mathfrak{g}}x$ are ad$_{\mathfrak{g}}s$ and ad$_{\mathfrak{g}}n$, respectively.

(ii) $x$ is regular in $\mathfrak{g}$ if and only if $s$ is.

(iii) If $\mathfrak{g}'$ is a subalgebra of $\mathfrak{g}\mathfrak{l}(V)$ containing $\mathfrak{g}$, every elementary automorphism of $\mathfrak{g}$ extends to an elementary automorphism of $\mathfrak{g}'$.

Put $\mathfrak{a}=\mathfrak{g}\mathfrak{l}(V)$. By Chap. I, §5, no. 4, Lemma 2, the semi-simple and nilpotent components of ad$_{\mathfrak{a}}x$ are ad$_{\mathfrak{a}}s$ and ad$_{\mathfrak{a}}n$; assertion (i) follows from this. We deduce that the characteristic polynomials of ad$_{\mathfrak{g}}x$ and ad$_{\mathfrak{g}}s$ are the same; hence (ii). If ad$_{\mathfrak{g}}x$ is nilpotent, ad$_{\mathfrak{g}}x=$ ad$_{\mathfrak{g}}n$, so ad$_{\mathfrak{g}'}n$ extends ad$_{\mathfrak{g}}x$, and $n$ is a nilpotent element of $\mathfrak{g}'$, hence (iii).

Let $\mathfrak{g}$ be a Lie subalgebra of $\mathfrak{g}\mathfrak{l}(V)$. We know (Chap. I, §6, no. 5, Th. 4) that the following conditions are equivalent:

(i) the identity representation of $\mathfrak{g}$ is semi-simple;

(ii) $\mathfrak{g}$ is reductive and every element of the centre of $\mathfrak{g}$ is a semi-simple endomorphism.

These conditions are actually equivalent to the following:

(iii) $\mathfrak{g}$ is a reductive subalgebra in $\mathfrak{g}\mathfrak{l}(V)$.

Indeed, (i) $=\Rightarrow$ (iii) by Chap. I, §6, no. 5, Cor. 3 of Th. 4, and (iii) $=\Rightarrow$ (i) by Chap. I, §6, no. 6, Cor. 1 of Prop. 7. We are going to show that if $\mathfrak{g}$ satisfies these conditions, $\mathfrak{g}$ is decomposable. More generally: