Corollary

An element x of g is a semi-simple (resp. nilpotent) endomorphism of M if and only if $ad_g x$ is a semi-simple (resp. nilpotent) endomorphism of g.

Let s (resp. n) be the semi-simple (resp. nilpotent) component of $x \in g$. Then $s \in g$ and $n \in g$ (Proposition 3). Then $ad_g s$ (resp. $ad_g n$) is the semi-simple (resp. nilpotent) component of $ad_g x$, by Lemma 2 of $§ 5$, no. 4. If x is semi-simple (resp. nilpotent) so then is $ad_g x$. If now $ad_g x$ is semi-simple (resp. nilpotent), it is equal to $ad_g s$ (resp. $ad_g n$) and hence $x = s$ (resp. $x = n$) since the adjoint representation of g is faithful.