 Solution

The derived series gives a direct test, with no structural theorem needed. By bilinearity and antisymmetry of the Lie bracket, every bracket in $\mathfrak{h}$ is a multiple of $z$, and $z=[x,y]$ actually occurs. ThereforeConsequently this Heisenberg Lie algebra is a nonzero solvable Lie algebra. It is an ideal of a Lie algebra in itself, since $[\mathfrak{h},\mathfrak{h}]\subseteq \mathfrak{h}$. Its solvable radical is therefore all of $\mathfrak{h}$, contradicting the first criterion in part (a). Hence