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I was wondering if I could get some help with the following problem. I know how to prove it with Schur's Lemma but I'm having problems without it.

Let G be an abelian group. Let V be an irreducible faithful CG-module. Prove that dimV = 1 and G is cyclic without using Schur's Lemma.

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  • $\begingroup$ But $G=({\mathbb C} \setminus \{0\},\times)$ has such a representation and it is not cyclic. $\endgroup$ – Derek Holt Aug 14 '15 at 12:39
  • $\begingroup$ Sorry G is assumed to be finite. $\endgroup$ – jackwo Aug 14 '15 at 12:47
  • $\begingroup$ In that case you can find a common eigenvector of a set of generators of $G$ giving ${\rm dim }V = 1$, and then $G$ is a finite subgroup of ${\mathbb C}^*$ and hence cyclic. $\endgroup$ – Derek Holt Aug 14 '15 at 13:11
  • $\begingroup$ Ok, so I think that I first have to show that for every element of the group there is a subspace of dimension 1. Then I need to show that vg spans V, but how would I go about doing this? $\endgroup$ – jackwo Aug 17 '15 at 12:51
  • $\begingroup$ JackWo, why did you delete the question leading up to this (link viewable only to 10k+ users). The same here - the question discussing the non-triviality of the intersection of the zero sum submodule and the span of the all ones vector? If I were a cynic I might think that you want to cover your tracks or something. More likely it is a new user's partial shame reaction (which is nothing to worry about). $\endgroup$ – Jyrki Lahtonen Aug 17 '15 at 16:17

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