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Let $\rho: G \to GL_n(\mathbb{C})$ be an irreducible representation, and $g\in Z(G)$. Show $\rho(g)$ is a scalar multiple of the identity matrix $I$.

I think I have it, here is my solution:

Since $\rho(g) \in Hom_G(\mathbb{C}, \mathbb{C})$ and $\rho$ is irreducible, consider a nonzero eigenvalue of $\rho(g)$, say $\lambda$, we have $\rho(g) -\lambda $ is a zero map by Schur's lemma, as the map contains a non-trivial kernel (i.e. the eigenvectors associated to $\lambda$ are in the kernel).

But I didn't use the condition that $g\in Z(G)$, so are there something wrong with my solution?

  • 0
    How did you prove that $\rho(g)-\lambda$ is the zero map?2012-05-07
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    Since $\rho(g) -\lambda$ is either a zero map or isomorphism by Schur's lemma, but it has a non-trivial kernel, so that it must be zero.2012-05-07
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    But how did you apply Schur's lemma to that map? How do you know the map is a homomorphisms of representations?2012-05-07
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    @MarianoSuárez-Alvarez I got it.2012-05-07
  • 0
    Great! :) Please be nice and write an answer to your own question, then!2012-05-07
  • 0
    @MarianoSuárez-Alvarez Maybe you want to convert your comment to an answer?2013-06-11

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