Corollary 2

Let $\mathfrak{a}$ be a complex Lie algebra. The following conditions are equivalent:

(i) $\mathfrak{a}$ is reductive.

(ii) There exists a compact real Lie algebra $\mathfrak{g}$ such that $\mathfrak{a}$ is isomorphic to $\mathfrak{g}_{(\mathbf{C})}$.

(iii) There exists a compact Lie group G such that $\mathfrak{a}$ is isomorphic to $L(G)_{(\mathbf{C})}$.

By Def. 1 of §1, no. 3, conditions (ii) and (iii) are equivalent and imply (i). If $\mathfrak{a}$ is reductive, it is the direct product of a commutative algebra, which clearly has a compact real form, and a semi-simple algebra which has one by Th. $1a)$, hence (i) implies (ii).