Let $\mathfrak{g}$ be a Lie algebra and consider $\operatorname{Rad}(\mathfrak{g})$, the radical of $\mathfrak{g}$, that is, the sum of all solvable ideals in $\mathfrak{g}$. Suppose that we have the decomposition $\mathfrak{g} = \mathfrak{h} \oplus \left(\bigoplus \mathfrak{g}_{\alpha}\right)$ where $\mathfrak{h}$ is an abelian subalgebra and $\mathfrak{g}_{\alpha}$ are root spaces. Why does $\operatorname{Rad}(\mathfrak{g}) \cap \mathfrak{g}_{\alpha} \neq 0$ imply $\mathfrak{g}_{\alpha} \oplus \mathfrak{g}_{-\alpha} \oplus [\mathfrak{g}_{\alpha}, \mathfrak{g}_{-\alpha}] \subset \operatorname{Rad}(\mathfrak{g})$?
you have to prove simply that $\mathfrak{g}_{\alpha}$ and $\mathfrak{g}_{-\alpha}$ is in $\operatorname{Rad}(\mathfrak{g})$, because $\operatorname{Rad}(\mathfrak{g})$ is also an ideal in the sens of lie algebra you will have $[\mathfrak{g}_{\alpha}, \mathfrak{g}_{-\alpha}] \subset \operatorname{Rad}(\mathfrak{g})$ and the you conclude by somme. –  Abdelmajid Khadari Apr 24 '12 at 23:32