 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. HenceBoth irreducibility over the algebraically closed field and faithfulness are used: scalar action alone would not embed the centre.