Prove that if G has a faithful complex irreducible representation

I am struggling with the proof that if G has a faithful complex irreducible representation then $Z(G)$ is cyclic:

Let $\rho:G \rightarrow GL(V)$ be a faithful complex irreducible representation. Let $z \in Z(G)$.

Consider the map $\phi_z: v \mapsto zv$ for all $v \in V$. This is a G-endomorphism on $V$, hence is multiplication by a scalar $\mu_z$"

I keep coming across the term G-homomorphism. For instance in Schur's Lemma ..."Then any G-homomorphism $\theta:V \rightarrow W$ is 0 or an isomorphism".

What exactly is G-homomorphism?

Then the map $Z(G) \rightarrow \mathbb{C}^\times, z \mapsto \mu_z$, is a representation of $Z$ and is faithful (since $\rho$ is). Thus $Z(G)$ is isomorphic to a finite subgroup of $\mathbb{C}^\times$, hence is cyclic.

What is the justification for this last sentence?

• Such maps are also called intertwiners and $G$-equivariant maps. It means the map commutes with the action of $G$, i.e. if $f:X\to Y$ intertwines an action of $G$ on $X$ and $Y$ then $f(gx)=gf(x)$ for all $x\in X$ and $g\in G$. In this way the class of $G$-sets becomes a category, and the class of representations of $G$ becomes a category. – anon May 1 '15 at 12:50

If $\rho : G \to GL(V)$ and $\sigma : G \to GL(W)$ are representations then a $G$-homomorphism from $V$ to $W$ is a linear map $f:V \to W$ such that $f( \rho(g) v) = \sigma(g) f(v)$ for all $g \in G$ and $v \in V$. It is the same thing as a module homomorphism $V \to W$ if you think of $V$ and $W$ as $\mathbb{C}G$-modules.
The last sentence works like this. If $\rho : Z(G) \to \mathbb{C}^\times$ is injective then the first isomorphism theorem for groups says $Z(G) \cong \operatorname{im} \rho$ which is a finite (as $Z(G)$ is finite) subgroup of $\mathbb{C}^\times$. Any finite subgroup of $\mathbb{C}^\times$ is cyclic -- see for example:
so $Z(G)$ is cyclic.