Proposition 3

The radical $r$ of a Lie algebra $g$ is the smallest ideal of $g$ such that $g/r$ has radical $\{0\}$.

Let $a$ be an ideal of $g$ and $\phi$ the canonical mapping of $g$ onto $g/a$. If the radical of $g/a$ is zero, then $\phi(r)$, which is a solvable ideal of $g/a$, is zero; hence $r \subset a$.

On the other hand, the inverse image $\phi^{-1}(r')$ of the radical $r'$ of $g/r$ is an ideal of $g$ which is solvable by Proposition 1 and hence is equal to $r$; therefore $r' = \{0\}$.