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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?

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  • $\begingroup$ How did you prove that $\rho(g)-\lambda$ is the zero map? $\endgroup$ – Mariano Suárez-Álvarez May 7 '12 at 18:47
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    $\begingroup$ 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. $\endgroup$ – CC_Azusa May 7 '12 at 18:52
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    $\begingroup$ But how did you apply Schur's lemma to that map? How do you know the map is a homomorphisms of representations? $\endgroup$ – Mariano Suárez-Álvarez May 7 '12 at 19:13
  • $\begingroup$ @MarianoSuárez-Alvarez I got it. $\endgroup$ – CC_Azusa May 7 '12 at 23:18
  • $\begingroup$ Great! :) Please be nice and write an answer to your own question, then! $\endgroup$ – Mariano Suárez-Álvarez May 7 '12 at 23:28
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Let me extend the comments to an answer:

You need $g\in Z(G)$ in order to conclude that $\rho(g)\in \operatorname{Hom}_G(\mathbb{C}^n,\mathbb{C}^n)$: Let $v\in \mathbb{C}^n$ and $h\in G$, then $$h\cdot \rho(g)(v)=\rho(h)(\rho(g)(v))=\rho(hg)(v),$$ but $$\rho(g)(h\cdot v)=\rho(g)(\rho(h)(v))=\rho(gh)(v).$$ In order for these to coincide you need $gh=hg$ in general (for specific $\rho$ of course not).

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