Corollary

Assume that $k$ is of characteristic zero and that $\mathfrak{g}$ is semi-simple. Then the subalgebra $\mathfrak{g}^1(a)$ satisfies conditions (1) and (2) of Lemma 2 ; it is reductive in $\mathfrak{g}$.

Condition (1) follows from part (ii) of Prop. 12; condition (2) follows from Prop. 4 of no. 1.