For each character $\chi : Z\to \Bbb{C}$ let $$ T_\chi v = \frac1{|Z|}\sum_{z\in Z} \chi(z)^{-1} \rho(z)v\in \Bbb{C}^n, \qquad v\in \Bbb{C}^n$$ Because $\rho(g)T_\chi = T_\chi \rho(g)$ then the restriction to $T_\chi \Bbb{C}^n$ is a subrepresentation.
That $\rho$ is irreducible means only one of those subrepresentations is non-trivial and equal to the whole original representation.
If $\rho(z) \
e \chi(z)$ then $\rho(z)$ has an eigenvalue $e^{2i\pi a/b}, b= order(z)$ different from $\chi(z)$ so that $\ker(\rho(z)-e^{2i\pi a/b} I)\
i v$ and $T_\chi v=0$, a contradiction. Whence $\rho(z)v = \chi(z)v$.
Take $v\
e 0\in V$ and let $W$ be the subspace of $\Bbb{C}^n $ generated by the $\rho(g)v$. The restriction to $W$ is a non-trivial subrepresentation, because $\rho$ is irreducible then $W = \Bbb{C}^n$.
On the other hand $G = \bigcup_{j=1}^{|G|/|Z|} g_j Z$ and $W$ is generated by the $\rho(g_j) v $ thus $\dim(W) \le |G|/|Z|$.