Theorem 2

Let $\mathfrak{g}$ be a Lie subalgebra of $\mathfrak{g}\mathfrak{l}(V)$. The following conditions are equivalent:

(i) $\mathfrak{g}$ is decomposable;

(ii) every Cartan subalgebra of $\mathfrak{g}$ is decomposable;

(iii) $\mathfrak{g}$ has a decomposable Cartan subalgebra;

(iv) the radical of $\mathfrak{g}$ is decomposable.

(i) $=\Rightarrow$ (ii): This follows from Cor. 2 of Prop. 3.

(ii) $=\Rightarrow$ (i): This follows from Cor. 1 of Th. 1, since $\mathfrak{g}$ is generated by its Cartan subalgebras (§2, no. 3, Cor. 3 of Th. 1).

(ii) $=\Rightarrow$ (iii): This is clear.

(iii) $=\Rightarrow$ (ii): By Cor. 3 of Th. 1, we can assume that $k$ is algebraically closed. The Cartan subalgebras of $\mathfrak{g}$ are then conjugate under the elementary automorphisms of $\mathfrak{g}($§3, no. 2, Th. 1); in view of Remark 1 of §3, no. 1, it follows that, if one of these is decomposable, they all are.

(i) $=\Rightarrow$ (iv): This follows from Cor. 2 of Prop. 4.

(iv) $=\Rightarrow$ (i): Assume that the radical $\mathfrak{r}$ of $\mathfrak{g}$ is decomposable. Let $\mathfrak{s}$ be a Levi subalgebra of $\mathfrak{g}$; it is decomposable (Prop. 2). Hence $\mathfrak{g}=\mathfrak{s}+\mathfrak{r}$ is decomposable (Cor. 1 of Th. 1).