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Let $V$ be an irreducible $\mathbb CG$ module. We define $Z(G)$ to be the centre of $G$. For $z\in Z(G)$ show that there exists $\lambda_z\in\mathbb C$ such that $z\cdot v=\lambda_z\cdot v$ for all $v \in V$.

I might have done this but I am not sure if my answer is correct.

Recall Schur's Lemma:

Let $V$ and $W$ be irreducible $\mathbb CG$-modules.

i) Suppose $\theta \colon V \to W$ is a $\mathbb CG$-homomorphism. Then either $\theta = 0$ or $\theta$ is an isomorphism.

ii) If $\theta\colon V \to V$ is an isomorphism then there exists $\lambda \in \mathbb C$ such that $\theta(v)=\lambda v$ for all $v\in V$.

So we take a mapping $p\colon V \to V$ where $v\mapsto zv$. This is a $\mathbb CG$ homomorphism. We use Schur's Lemma, using part (i); Since $p$ is a $\mathbb CG$ homomorphism we can say that $p=0$ or $p$ is an isomorphism. If we take $p,g\in G$, then we have $gp=pg$ for all $g\in G$, so $p$ is in $Z(G)$, the centre of $G$. Hence $Z(G)$ is not empty. So $p$ is an isomorphism.

Next we use (ii). There exists $\lambda_z$ in $\mathbb C$ such that $p(V)=\lambda_z$ and since $p(v)=zv$, then we have $zv=\lambda_z$ as required.

Thanks

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  • $\begingroup$ Your argument is lacking some things. How did you conclude that $Z(G)$ was non-empty? And how did that imply that the map was an isomorphism? The place you need that the element given is in $Z(G)$ is for the map to be a homomorphism in the first place. $\endgroup$ – Tobias Kildetoft Mar 5 '14 at 11:50
  • $\begingroup$ because p exists in Z(G)? $\endgroup$ – ZZS14 Mar 5 '14 at 11:58
  • $\begingroup$ I don't understand what you mean by; The place you need that the element given is in Z(G) is for the map to be a homomorphism in the first place. $\endgroup$ – ZZS14 Mar 5 '14 at 11:59
  • $\begingroup$ If you take some arbitrary element in $G$, then the map you define from $V$ to $V$ will not usually be a homomorphism of $G$-modules. Your $p$ is not an element of $G$, it is a homomorphism of $G$-modules. $\endgroup$ – Tobias Kildetoft Mar 5 '14 at 12:00
  • $\begingroup$ oh, how would i show that Z(G) is not empty? $\endgroup$ – ZZS14 Mar 5 '14 at 12:03
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If you want to show that for every element in the centre $z\in Z(G)$ there is a complex number $\lambda_z\in \mathbb C$ such that $zv = \lambda_zv$ for all $v\in V$, you should take the following steps. (I can see them from in your argument but I want to help you structure them a bit.)

  1. Show that $p(v) := zv$ is indeed a $\mathbb CG$ homomorphism. ($z$ being in the centre is playing a central role here)

  2. Assume $p$ is zero. This yields $\lambda_z =0$.

  3. Assume $p$ is not zero. Use (ii) from Schur's Lemma.

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