# Centre of a group with a faithful irreducible representation is cyclic

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 $$\rho:G \to \mathbb{C}^{\times}$$: $$$$\rho(\sigma)=-1, \qquad \rho(\tau) = \omega$$$$ where $$\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?

$$G$$ is a cyclic group. By the Chinese Remainder Theorem, if $$m$$ and $$n$$ are relatively prime, $$C_m \times C_n \cong C_{mn}$$.