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Math Help - Image of a group's center under an irreducible representation

  1. #1
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    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.
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  2. #2
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    Quote Originally Posted by paulk View Post
    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.
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  3. #3
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    Yes its over the complex numbers. Thanks for your help.
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