Proposition 7

Let $\mathfrak{g}$ be a decomposable Lie subalgebra of $\mathfrak{g}\mathfrak{l}(V)$.

(i) There exists a Lie subalgebra $\mathfrak{m}$ of $\mathfrak{g}$, reductive in $\mathfrak{g}\mathfrak{l}(V)$, such that $\mathfrak{g}$ is the semi-direct product of $\mathfrak{m}$ and $\mathfrak{n}_V(\mathfrak{g})$.

(ii) Any two Lie subalgebras of $\mathfrak{g}$ with the properties in (i) are conjugate under Aut$_e(\mathfrak{g})$.

The radical $\mathfrak{r}$ of $\mathfrak{g}$ is decomposable (no. 2, Cor. 2 of Prop. 4). By Cor. 2 of Prop. 6, there exists a commutative subalgebra $\mathfrak{t}$ of $\mathfrak{r}$, consisting of semi-simple elements, such that $\mathfrak{r}=\mathfrak{t}\oplus \mathfrak{n}_V(\mathfrak{r})$. Since ad$_{\mathfrak{g}}\mathfrak{t}$ consists of semi-simple elements, $\mathfrak{g}$ is the direct sum of $[\mathfrak{t},\mathfrak{g}]$ and the centralizer $\mathfrak{z}$ of $\mathfrak{t}$ (Chap. I, §3, no. 5, Prop. 6). Since $[\mathfrak{t},\mathfrak{g}]\subset \mathfrak{r},\mathfrak{g}=\mathfrak{z}+\mathfrak{r}$. Consequently, if $\mathfrak{s}$ is a Levi subalgebra of $\mathfrak{z}$ (Chap. I, §6, no. $8$)$,\mathfrak{g}=\mathfrak{s}+\mathfrak{r}$, so $\mathfrak{s}$ is a Levi subalgebra of $\mathfrak{g}$. Put $\mathfrak{m}=\mathfrak{s}\oplus \mathfrak{t}$. Since $[\mathfrak{s},\mathfrak{t}] =\{0\},\mathfrak{m}$ is a Lie subalgebra of $\mathfrak{g}$, reductive in $\mathfrak{g}\mathfrak{l}(V)$ by Chap. I, §6, no. 5, Th. 4. Moreover,

since $\mathfrak{n}_V(\mathfrak{g}) =\mathfrak{n}_V(\mathfrak{r})$. Hence (i).

Now let $\mathfrak{m}'$ be a Lie subalgebra of $\mathfrak{g}$ complementary to $\mathfrak{n}_V(\mathfrak{g})$ and reductive in $\mathfrak{g}\mathfrak{l}(V)$. We show that $\mathfrak{m}'$ is conjugate to $\mathfrak{m}$ under Aut$_e(\mathfrak{g})$. We have $\mathfrak{m}'=\mathfrak{s}'\oplus \mathfrak{t}'$, where $\mathfrak{s}'= [\mathfrak{m}',\mathfrak{m}']$ is semi-simple and the centre $\mathfrak{t}'$ of $\mathfrak{m}'$ consists of semi-simple elements. Then $\mathfrak{r}=\mathfrak{t}\oplus \mathfrak{n}_V(\mathfrak{g}) =\mathfrak{t}'\oplus \mathfrak{n}_V(\mathfrak{g})$. In view of Cor. 2 of Prop. 6, we are reduced to the case $\mathfrak{t}=\mathfrak{t}'$. Then $\mathfrak{s}'\subset \mathfrak{z}$; since dim$\mathfrak{s}'=$ dim$\mathfrak{s}$, $\mathfrak{s}'$ is a Levi subalgebra of $\mathfrak{z}$. By Chap. I, §6, no. 8, Th. 5, there exists $x\in \mathfrak{n}_V(\mathfrak{z})$ such that $e^{adx}(\mathfrak{s}) =\mathfrak{s}'$; since $x$ commutes with $\mathfrak{t}$, we also have $e^{adx}(\mathfrak{t}) =\mathfrak{t}$.