Corollary 1

Let s be a Levi subalgebra of g and h a semi-simple subalgebra of g. (a) There exists a special automorphism of g mapping h onto a subalgebra of s. (b) h is contained in a Levi subalgebra of g.

Let r be the radical of g and a = h + r, which is a subalgebra of g. Then a/r is semi-simple and r is solvable, hence r is the radical of a and h is a Levi subalgebra of a. On the other hand, a ∩ s = h' is a supplementary subalgebra to r in a and hence also a Levi subalgebra of a. Then there exists (Theorem 5) a ∈ [a, r] such that e^{ada} maps h onto h'. Now a ∈ [g, r]; e^{ada} maps h onto a subalgebra of s and e^{-ada}(s) is a Levi subalgebra of g containing h.