= Solution
For a complex <irreducible representation>, every central element acts as a scalar by <Schur lemma>. If the representation is faithful, these scalars embed $Z$ in $\mathbb C^*$. Every finite subgroup of $\mathbb C^*$ is cyclic: if its order is $m$, all its elements satisfy $z^m=1$, and the $m$th <roots of unity> form a <cyclic group>. Thus \b[$Z$ is cyclic].
The trivial group satisfies the conclusion immediately. Otherwise let $G$ be a finite $p$-group with cyclic centre, and decompose a <faithful representation> as a direct sum of irreducibles using <Maschke's theorem>. The nontrivial cyclic $p$-group $Z$ has a unique subgroup $C$ of order $p$. Some component does not kill $C$, since otherwise the whole representation would kill it. If that component had a nontrivial kernel $N$, the kernel would be a <normal subgroup> and the allowed intersection theorem gives $N\cap Z\ne1$. Every nontrivial subgroup of cyclic $Z$ contains $C$, contradicting the choice of component. \b[That irreducible component is faithful.]
Finally the <regular representation> is faithful, since left multiplication by a nonidentity element does not fix the basis vector at the identity. A cyclic centre therefore supplies a faithful irreducible component of it. The first argument proves the converse, giving \b[a finite $p$-group has a faithful irreducible complex representation exactly when its centre is cyclic].
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