Proposition 3

Assume that $\mathfrak{g}$ is semi-simple. The group Aut$_e(\mathfrak{g})$ is equal to its derived group. If $\mathfrak{g}$ is splittable, Aut$_e(\mathfrak{g})$ is the derived group of Aut$_0(\mathfrak{g})$.

Let $x$ be a non-zero nilpotent element of $\mathfrak{g}$. Choose $h, y\in \mathfrak{g}$ be such that $(x, h, y)$ is an $\mathfrak{s}\mathfrak{l}_2$-triplet (Prop. 2). The subalgebra $\mathfrak{s}$ of $\mathfrak{g}$ generated by $(x, h, y)$ can be identified with $\mathfrak{s}\mathfrak{l}(2, k)$. Let $\rho$ be the representation $z \rightarrow$ ad$_{\mathfrak{g}}z$ of $\mathfrak{s}=\mathfrak{s}\mathfrak{l}(2, k)$ on $\mathfrak{g}$, and let $\pi$ be the representation of $\mathbf{S}\mathbf{L}(2, k)$ compatible with $\rho ($§1, no. 4). The image of $\pi$ is generated by the exp($t$ ad$_{\mathfrak{g}}x$) and the exp($t$ ad$_{\mathfrak{g}}y$) with $t\in k($Algebra, Chap. III, §8, no. 9, Prop. 17), hence is contained in Aut$_e(\mathfrak{g})$. Since $\mathbf{S}\mathbf{L}(2, k)$ is equal to its derived group (Lemma 7 and loc. cit.), exp(ad$_{\mathfrak{g}}x$) belongs to the derived group G of Aut$_e(\mathfrak{g})$. Hence Aut$_e(\mathfrak{g})$ is equal to G. Assume now that $\mathfrak{g}$ is splittable. Since Aut$_0(\mathfrak{g})/$Aut$_e(\mathfrak{g})$ is commutative (§5, no. 3, Remark 3), the preceding proves that the derived group of Aut$_0(\mathfrak{g})$ is Aut$_e(\mathfrak{g})$.