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.


  • $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.