Given a finite group $G$, let $\rho:G \to GL(n,\mathbb{C})$ be an irreducible representation of $G$.
Now consider the group ring $\mathbb{C}G$. There is a very natural way of extending $\rho$ to $\rho^{*}$ a representation of $\mathbb{C}G$:
Let $\sum_{g\in G}\lambda_{g}g\in \mathbb{C}G$ with $\lambda \in \mathbb{C}$, then:
$$\rho^{*}(\sum_{g\in G}\lambda_{g}g)=\sum_{g\in G}\lambda_{g}\rho*(g)$$
Now, denote $Z(\mathbb{C}G)=\{z\in\mathbb{C}G : zr=rz \hspace{0.5cm}\forall r \in \mathbb{C}G\}$.
I am wondering whether there is a method to obtain an irreducible representation of $Z(\mathbb{C}G)$ given an irreducible representation of the group ring.
I have noticed that since $Z(\mathbb{C}G)$ is a commutative ring, it follows that all its irreducible representations must be 1-dimensional, but I do not know how they relate (if they do) to the representations of the $\mathbb{C}G$.
Any thoughts?
Since the representation $\rho$ is irreducible and the field $\mathbf{C}$ is algebraically closed, any $G$-module endomorphism of $\rho$ must be a multiplication by a scalar. This in particular applies to each central element $z \in Z(\mathbf{C}G)$ of the group algebra. It follows that there is a homomorphism $\alpha_\rho:Z(\mathbf{C}G) \to \mathbf{C}$ of $\mathbf{C}$-algebras sending $z$ to the scalar by which it acts on the vector space underlying the representation $\rho$. This homomorphism $\alpha_\rho$ is called the central character associated with $\rho$. In particular, as a representation of $Z(\mathbf{C}G)$, the space $\mathbf{C}^n$ is the direct sum of $n$ copies of the same $1$-dimensional irreducible representation given by $\alpha_\rho$.
You can compute the values of $\alpha_\rho$ in terms of the character $$\chi_\rho(g)=\mathrm{tr}(\rho(g))$$ of $\rho$ as follows: if $$z=\sum_{g \in G} c_g g$$ then the trace of $\rho(z)$ is the dimension $n$ of $\rho$ times the scalar value $\alpha_\rho(z)$ by which $z$ acts. Thus $$\alpha_\rho(z)=\frac{1}{n} \sum_{g\in G} c_g \chi_\rho(g).$$