Theorem 1

Let $\mathfrak{g}$ be a Lie subalgebra of $\mathfrak{g}\mathfrak{l}(V)$ and let X be a subset of $\mathfrak{g}$ generating $\mathfrak{g}$ as a Lie algebra over $k$. If every element of X is either semi-simple or nilpotent, $\mathfrak{g}$ is decomposable.

a$)\mathfrak{g}$ is commutative.

The semi-simple (resp. nilpotent) elements of $\mathfrak{g}$ form a vector subspace $\mathfrak{g}_s$ (resp. $\mathfrak{g}_n$). The assumption is equivalent to $\mathfrak{g}=\mathfrak{g}_s\oplus \mathfrak{g}_n$, hence the fact that $\mathfrak{g}$ is decomposable.

b$)\mathfrak{g}$ is reductive.

Then $\mathfrak{g}=\mathfrak{g}'\times \mathfrak{c}$ with $\mathfrak{g}'$ semi-simple and $\mathfrak{c}$ commutative. By Prop. $2,\mathfrak{g}'$ is decomposable. Let $x=a+b\in \mathfrak{g}$ with $a\in \mathfrak{g}',b\in \mathfrak{c}$. Let $a_s, a_n, b_s, b_n$ be the semi-simple and nilpotent components of $a, b$. Since $a_s, a_n, b_s, b_n$ mutually commute, the semi-simple and nilpotent components of $x$ are $a_s+b_s, a_n+b_n$. Now $a_s, a_n\in \mathfrak{g}'$. If $x$ is semi-simple, $x=a_s+b_s$; since $a_s\in \mathfrak{g}'$, we have $b_s\in \mathfrak{g}$, so $b_s\in \mathfrak{c}$ since $b_s$ commutes with $\mathfrak{g}$; consequently, $a=a_s$ and $b=b_s$. Similarly, if $x$ is nilpotent, $a=a_n$ and $b=b_n$. It follows that the projections on $\mathfrak{c}$ of the elements of X are either semi-simple or nilpotent; by a), this implies that $\mathfrak{c}$ is decomposable. Retaining the preceding notation, but without the assumption on $x$, we now have $b_s, b_n\in \mathfrak{c}$, so $a_s+b_s, a_n+b_n\in \mathfrak{g}$, which proves the theorem in this case.

c) General case.

We assume that the theorem is proved for Lie algebras of dimension $<$ dim $\mathfrak{g}$ and prove it for $\mathfrak{g}$.

Let $\mathfrak{n}$ be the largest ideal of nilpotency of the identity representation of $\mathfrak{g}$. If $\mathfrak{n}= 0,\mathfrak{g}$ has an injective semi-simple representation, and so is reductive. Assume that $\mathfrak{n}\not= 0$. Let $\mathfrak{p}$ be the normalizer of $\mathfrak{n}$ in $\mathfrak{g}\mathfrak{l}(V)$. There exist $E, \rho ,F$ satisfying the conditions of Prop. 8. Since $\mathfrak{g}\subset \mathfrak{p},\rho (\mathfrak{g})$ leaves F stable; let $\rho_0$ be the representation $u \rightarrow \rho (u)|F$ of $\mathfrak{g}$ on F; we have $\mathfrak{n}=$ Ker$\rho_0$. The image under $\rho$ of every semi-simple (resp. nilpotent) element of $\mathfrak{g}\mathfrak{l}(V)$ is semi-simple (resp. nilpotent) (Prop. 2). The algebra $\rho_0(\mathfrak{g})$ is thus generated by its semi-simple elements and its nilpotent elements. By the induction hypothesis, $\rho_0(\mathfrak{g})$ is decomposable.

Let $x\in \mathfrak{g}$, and let $x_s, x_n$ be its semi-simple and nilpotent components. By Prop. 2, the semi-simple and nilpotent components of $\rho (x)$ are $\rho (x_s), \rho (x_n)$. Since $\rho_0(\mathfrak{g})$ is decomposable, there exist $y, z\in \mathfrak{g}$ such that

Then $x_s\in y+\mathfrak{n}, x_n\in z+\mathfrak{n}$, so $x_s, x_n\in \mathfrak{g}$. Q.E.D.