Thread: Image of a group's center under an irreducible representation

1. Image of a group's center under an irreducible representation

For an irreducible representation $\rho$ of a finite group $G$ what can be said about the image of the center $Z$ of $G$?

I have run into several problems telling me to apply Schur's lemma to this but I have not had success with this approach. Any help is appreciated.

2. Originally Posted by paulk
For an irreducible representation $\rho$ of a finite group $G$ what can be said about the image of the center $Z$ of $G$?

I have run into several problems telling me to apply Schur's lemma to this but I have not had success with this approach. Any help is appreciated.
you didn't menstion but i guess your base field is $\mathbb{C}.$ well, this is what we can say: $\forall g \in Z(G), \ \exists \lambda \in \mathbb{C}: \ \rho (g)=\lambda I,$ where $I$ is the identity map. the reason is this:

let $\rho : G \longrightarrow \text{GL}(V)$ and $g \in Z(G).$ then for all $h \in G$ and $v \in V$ we'll have $\rho (g)(\rho(h)v)=\rho (gh)v=\rho (hg)v = \rho (h)(\rho (g)v).$ that means $\rho (g) : V \longrightarrow V$ is a $\mathbb{C}[G]$ module homomorphism.

also, since $\mathbb{C}$ is algebraically closed, $\rho(g)$ has an eigenvalue, say $\lambda.$ thus $\rho(g) - \lambda I : V \longrightarrow V$ is a $\mathbb{C}[G]$ module homomorphism which is not an isomorphism because it has a non-zero kernel.

thus, since $V$ is irreducible, we must have $\rho(g) - \lambda I = 0,$ by Schur's lemma.

3. Thanks

Yes its over the complex numbers. Thanks for your help.