Proposition 4

Let $\mathfrak{g}$ be a Lie subalgebra of $\mathfrak{g}\mathfrak{l}(V)$ and $\mathfrak{n}$ an ideal of $\mathfrak{g}$. Then $\mathfrak{n}$ and $e(\mathfrak{n})$ are ideals of $e(\mathfrak{g})$, and $[e(\mathfrak{g}), e(\mathfrak{n})] = [\mathfrak{g},\mathfrak{n}]$.

Let $\mathfrak{g}_1$ be the set of $x\in \mathfrak{g}\mathfrak{l}(V)$ such that $[x,\mathfrak{n}]\subset [\mathfrak{g},\mathfrak{n}]$. This is a decomposable Lie subalgebra of $\mathfrak{g}\mathfrak{l}$(V), containing $\mathfrak{g}$ and hence $e(\mathfrak{g})$, cf. no. 1, Prop. 3; in other words, $[e(\mathfrak{g}),\mathfrak{n}]\subset [\mathfrak{g},\mathfrak{n}]$. Let $\mathfrak{n}_1$ be the set of $y\in \mathfrak{g}\mathfrak{l}(V)$ such that

This is a decomposable Lie subalgebra of $\mathfrak{g}\mathfrak{l}(V)$ containing $\mathfrak{n}$ by the preceding, and hence containing $e(\mathfrak{n})$; in other words $[e(\mathfrak{g}), e(\mathfrak{n})]\subset [\mathfrak{g},\mathfrak{n}]$, so

It follows that $[e(\mathfrak{g}),\mathfrak{n}]\subset [e(\mathfrak{g}), e(\mathfrak{n})]\subset \mathfrak{n}$, so $\mathfrak{n}$ and $e(\mathfrak{n})$ are ideals of $e(\mathfrak{g})$.