Irreducible rep, group centre: $\pi$(z) $=\lambda$(z)v

Question

Note: not sure if title is displaying well; formula is directly below

lambda is a scalar that I need to show exists

$\pi$(z) $=\lambda$(z)v lambda is a scalar that I need to show exists

I want to show this equality holds for all $z \in Z$ where :

• G is a finite group
• Z is its centre
• ($\pi, V$) is an irreducible representation of G

---

I believe that the action of $\pi$(z) on V is a G intertwiner, and I know that V has no G-invariant subsoaces but am not sure how one arrives at the equality

---

Any help would be much appreciated

Answer 1

Hint: this a Schur like result. I assume the ground field of $V$ is algebraically closed. Let $z\in Z$, consider the eigenspace $V_c$ associated to the eigenvalue $c$ of $\pi(z)$ show that $V_c$ is a submodule of $V$ since $\pi(z)$ commutes with $\pi(G)$ so it is $V$ since the representation is irreducible.