 Solution

Let $\mathfrak{g}={\mathfrak{sl}}_2(\mathbb{C})$ and let $V={\mathbb{C}}^2$ be its defining irreducible representation. SetSince $[\mathfrak{g},\mathfrak{g}]=\mathfrak{g}$ and $\mathfrak{g}V=V$,If $(x,v)$ is central, commuting with every $(0,w)$ gives $xw=0$ for all $w$, so faithfulness of the defining representation gives $x=0$. Commuting with every $(y,0)$ then gives $yv=0$ for all $y$; irreducibility and nontriviality give $v=0$. Thus $Z(\mathfrak{h})=0$.

The nonzero abelian subspace $V$ is a proper ideal of a Lie algebra, so $\mathfrak{h}$ is not simple. It is not a direct product of simple Lie algebras either, because such a product is semisimple and has no nonzero solvable ideal, whereas $V$ is one.