= Solution
In an irreducible complex <group representation> $\rho$, every $\rho(z)$ for $z$ in the <center of a group> commutes with the entire representation. <Schur lemma> therefore gives $\rho(z)=\lambda_z I$. Faithfulness makes $z\mapsto\lambda_z$ an injective homomorphism $Z(G)\to\mathbb C^\times$.
Its image is a finite subgroup. If its order is $m$, <Lagrange's theorem for finite groups> shows that every element is an $m$th root of unity. These roots form a <cyclic group>, and every subgroup of a <cyclic group> is cyclic. Hence
$$
\boxed{Z(G)\text{ is cyclic}}.
$$
Both irreducibility over the algebraically closed <field> and faithfulness are used: scalar action alone would not embed the centre.
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