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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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    @MarianoSuárez-Alvarez Maybe you want to convert your comment to an answer? – 2013-06-11

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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).