Schur corollary: central elements act by scalars

If $z$ lies in the center $Z(G)$, then $\rho(z)$ is a scalar operator:

More generally, if $a$ lies in the center $Z(k[G])$ of the group algebra, then the $k$-linear operator “action by $a$” on $V$ commutes with $\rho(G)$ and hence is also scalar on $V$.

Consequence: abelian groups have only $1$-dimensional irreps

If $G$ is abelian, then every $g\in G$ is central, so every $\rho(g)$ is scalar. If $\dim(V)>1$, every line in $V$ would be $G$-stable, contradicting irreducibility. Hence any irreducible representation of an abelian group over $k$ has $\dim(V)=1$.

1. $C_n$. Since $C_n$ is abelian, every irreducible complex representation is $1$-dimensional. Concretely, if $C_n=\langle g\rangle$ and $\zeta=e^{2\pi i/n}$, the irreducibles are the characters
   $$\chi_j(g)=\zeta^j \qquad (j=0,1,\dots,n-1).$$

1. $D_8$ (order $8$). For the dihedral group of the square, the center is $Z(D_8)=\{1,r^2\}$ where $r$ is rotation by $90^\circ$. In the $2$-dimensional “geometric” irreducible representation on $\mathbb C^2$, $r^2$ acts as rotation by $180^\circ$, i.e.
   $$\rho(r^2) = -I,$$
   a scalar operator. In every $1$-dimensional representation, $\rho(r^2)=+1$ (also a scalar, necessarily).

1. $S_3$. $Z(S_3)=\{e\}$ is trivial, so the corollary imposes no nontrivial scalar constraints—consistent with the existence of a $2$-dimensional irreducible representation.