Progress on irreps of abelian groups

I can’t sleep and I’ve had an idea, so here’s another post. We’ve discussed a few times the possibility of proving that every irreducible representation of an abelian group is 1-dimensional, by elementary means.

Alex proved this statement in his talk today (as a corollary of some wider considerations), using the character machinery that he and Pat outlined, however I think I’ve got a proof that circumvents the need for this. I claim very little originality – the key idea is stolen right from the proof of ‘Schur’s lemma’ in the form that it is presented in MATH3103. Here goes:

Suppose $G$ is an abelian group, $V$ is a finite dimensional vector space over $\mathbb{C}$ and $\rho: G \to GL(V)$ is an irreducible representation. Fix $g$ in $G$ and let $\lambda$ be an eigenvalue of $\rho(g)$ whose corresponding eigenspace is $E_\lambda \subset V$. Then if $v$ is an element of $E_\lambda$, for any $h$ in $G$ we have: $\rho(g)\rho(h)v = \rho(gh)v = \rho(hg)v = \rho(h)\rho(g)v = \lambda \rho(h)v$, so $\rho(h)v$ is also an element of $E_\lambda$. Therefore $E_\lambda$ is closed under the action of $G$ and forms a sub-representation of $\rho$. By irreducibility it must hold that $E_\lambda = V$, so $\rho(g) = \lambda I$, where $I$ is the identity on $V$. Since $g$ was arbitrary we may conclude that $\rho$ sends every group element to some scalar multiple of the identity and it’s clear then that if dim$V \geq 2$ any 1-dimensional subspace of $V$ constitutes a strict sub-representation of $\rho$. This violates irreducibility, so dim$V =1$.

The other nice thing about this argument is that it seems to work for infinite $G$. Thoughts?