Corollary 4

Let g be a Lie algebra, r its radical, s a Levi subalgebra of g and m an ideal of g. Then m is the direct sum of m ∩ r which is its radical and m ∩ s which is a Levi subalgebra of m.

We know that m ∩ s is the radical of m (§ 5, no. 5, Corollary 3 to Proposition 5). Let h be a Levi subalgebra of m and s' a Levi subalgebra of g containing h (Corollary 1). The algebra m ∩ s' is an ideal of s', is therefore semi-simple, and contains h and is therefore equal to h. Hence m is the direct sum of $m \cap r$ and $m \cap s'$. There exists a special automorphism mapping $s'$ onto $s$; this automorphism leaves $r$ and $m$ invariant; hence $m$ is the direct sum of $m \cap r$ and $m \cap s$ and $m \cap s$ is a Levi subalgebra of $m$.