 Solution

Let $\pi :L\to L/I$ be the quotient map. A full flag of ideals in $L/I$ lifts by inverse images to a flag from $I$ to $L$, with dimensions increasing by one. Prepending $0<I$ supplies the required full flag in $L$. Thus supersolubility lifts across a one-dimensional ideal.

Over $\mathbb{C}$, a nonzero soluble $L$ has a one-dimensional ideal: apply part (a) to its adjoint image acting on $L$. A common eigenvector spans an adjoint-invariant line, which is an ideal. The quotient is soluble and has smaller dimension, so induction and the lifting result give a full flag of ideals. Therefore a soluble complex Lie algebra is supersoluble.

Conversely this implication needs no field restriction. If $L$ has a full flag, its first nonzero ideal $I$ is one-dimensional and abelian, and $L/I$ has a full flag. By induction $L/I$ is soluble, so some term of the derived series of $L$ lies in $I$; its next derived term is zero. Hence