 Solution

For $z\in Z$, the eigenspaces of $\rho (z)$ are invariant under $G$, since every representing matrix commutes with $\rho (z)$. Irreducibility forces one such eigenspace to be all of $W$, so $\rho (z)=\chi (z)I$. The multiplication law gives $\chi (z{z}^{\prime })=\chi (z)\chi (z^{\prime })$, making $\chi$ a one-dimensional, hence irreducible, group representation of $Z$. Therefore