Levi decomposition theorem

If $\mathfrak{r}$ is abelian, including $\mathfrak{r}=0$, the exact sequence $0\to \mathfrak{r}\to \mathfrak{g}\to \mathfrak{q}\to 0$ is an abelian extension for the induced adjoint $\mathfrak{q}$-action, so it defines a class in $H^2(\mathfrak{q},\mathfrak{r})$ under the bijection of [L5]. That class is zero by [L1]–[L2], and the zero class is exactly the split case by [L5]; hence the extension has a Lie section $\sigma :\mathfrak{q}\to \mathfrak{g}$. Its image $\mathfrak{s}$ is semisimple, intersects $\mathfrak{r}$ trivially, and complements it. This is the derived-length induction base.

Suppose $\mathfrak{r}$ is nonabelian and put $\mathfrak{t}=[\mathfrak{r},\mathfrak{r}]$. This is a characteristic ideal of $\mathfrak{r}$ and hence an ideal of $\mathfrak{g}$. The radical of $\mathfrak{g}\mathrm{/}\mathfrak{t}$ is $\mathfrak{r}\mathrm{/}\mathfrak{t}$: it is solvable, and any larger solvable ideal would have a solvable inverse image by [L3], contradicting maximality of $\mathfrak{r}$. Since this radical is abelian, step 1.1 supplies a Levi factor $\bar{\mathfrak{h}}$ in $\mathfrak{g}\mathrm{/}\mathfrak{t}$.

Let $\mathfrak{h}$ be the inverse image of $\bar{\mathfrak{h}}$. Then $\mathfrak{h}\mathrm{/}\mathfrak{t}\cong \mathfrak{q}$ is semisimple and $\mathrm{rad}(\mathfrak{h})=\mathfrak{t}$: the inclusion $\mathfrak{t}\subseteq \mathrm{rad}(\mathfrak{h})$ is clear, while every solvable ideal of $\mathfrak{h}$ maps to a solvable ideal of the semisimple quotient and hence lies in $\mathfrak{t}$. Since $\mathfrak{t}={\mathfrak{r}}^{(1)}$ has smaller derived length, apply the induction hypothesis to $\mathfrak{h}$ to obtain a semisimple complement $\mathfrak{s}$ to $\mathfrak{t}$.

Since $\mathfrak{h}=\mathfrak{t}\oplus \mathfrak{s}$ and $\mathfrak{g}=\mathfrak{r}+\mathfrak{h}$, we have $\mathfrak{g}=\mathfrak{r}+\mathfrak{s}$. Their intersection lies in $\mathfrak{r}\cap \mathfrak{h}=\mathfrak{t}$ and is zero, so $\mathfrak{s}$ is the required Levi factor by [L4]. The zero algebra is included, and all choices are finite-dimensional basis or subspace choices.