Proposition 3

Let $\mathfrak{a}$ be a decomposable Lie subalgebra of $\mathfrak{g}\mathfrak{l}(V)$ and let $\mathfrak{b}$ and $\mathfrak{c}$ be vector subspaces of $\mathfrak{g}\mathfrak{l}(V)$ such that $\mathfrak{b}\subset \mathfrak{c}$. Let $\mathfrak{a}'$ be the set of $x\in \mathfrak{a}$ such that $[x,\mathfrak{c}]\subset \mathfrak{b}$. Then $\mathfrak{a}'$ is decomposable.

Put $\mathfrak{g}=\mathfrak{g}\mathfrak{l}(V)$; the subalgebra $\mathfrak{h}'$ of $\mathfrak{g}\mathfrak{l}(\mathfrak{g})$ consisting of the $z\in \mathfrak{g}\mathfrak{l}(\mathfrak{g})$ such that $z(\mathfrak{c})\subset \mathfrak{b}$ is decomposable (Example 1). Let $\pi :\mathfrak{g}\rightarrow \mathfrak{g}\mathfrak{l}(\mathfrak{g})$ be the adjoint representation of $\mathfrak{g}$. Prop. 2 (iv), applied to $\pi$, shows that $\pi^{-1}(\mathfrak{h}')$ is decomposable. Hence so is $\mathfrak{a}'=\mathfrak{a}\cap \pi^{-1}(\mathfrak{h}')$.