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 dimV \geq 2 any 1-dimensional subspace of V constitutes a strict sub-representation of \rho. This violates irreducibility, so dimV =1.

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


One thought on “Progress on irreps of abelian groups

  1. Is this fact true if one replaces the complex numbers by, say, real numbers? That is, should we necessarily have that the underlying field is algebraically closed?

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