The following is taken from section 3.1 of $\textit{Représentations linéaires des groupes finis}$ by J.-P. Serre:
Exercise 2) Let $\rho$ be an irreducible representation of $G$ of degree $n$ and character $\chi$; let $C$ be the centre of $G$ ($\textit{i.e.}$ the set of $s \in G$ such that $st=ts$ for all $t \in G$.), and let $c$ be the order of $C$.
c) Show that if $\rho$ is faithful ($\textit{i.e.}$ $\rho_s \neq 1$ for $s \neq 1$), then $C$ is cyclic.
Now if we let $G=C_2\times C_3 = \{\sigma,\tau \mid \sigma^2=\tau^3=e, \sigma\tau=\tau\sigma \}$, we can define the following irreducible representation $\rho:G \to \mathbb{C}^{\times}$:
\begin{equation}
\rho(\sigma)=-1, \qquad \rho(\tau) = \omega
\end{equation}
where $\omega \in \mathbb{C}^{\times}$ is a primitive third root of unity.
$\rho$ is a faithful irreducible representation and $G$, being Abelian, coincides with its own centre. Hence its centre is not cyclic.
Why is this not a counterexample?
Best Answer
$G$ is a cyclic group. By the Chinese Remainder Theorem, if $m$ and $n$ are relatively prime, $C_m \times C_n \cong C_{mn}$.