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In Serre's Linear Representations of Finite Groups, there's a simple yet useful statement

(Section 8.1 Corollary to Prop 24): Let $G$ be a finite group and $A$ an abelian normal subgroup. Then any finite dimensional complex irreducible representation $\rho$ has dimension dividing $(G:A)$.

To me, this result is very satisfying (because it reduces the complexity of the spectrum of $G$ a lot if $G$ is not far away from being abelian, in which case the spectrum is simplest!), so I would like to understand it more. Unfortunately, in the book it's proved by induction. I hope to see if there's a more natural or constructive proof. In particular, I wonder if it's possible to naturally associate $\rho$ to a subgroup of $G/A$, with matching dimension and order.

Student
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    A special case is when $A={e}$ which is still rather non-trivial. But I'm not aware of any proof of this that follows your idea of finding a subgroup of order $\dim\rho$. – Angina Seng Dec 21 '19 at 19:22
  • @LordSharktheUnknown yes, perhaps this is the point. I did not say what this corollary is following from: there is a subgroup $H$ that contains $A$ so that $\rho$ is induced from an irreducible module $\pi$ of $H$; also the restriction of $\pi$ to $A$ breaks down to several isomorphic irreducible representation! That means after passing down to $H$, the representation is “almost” a representation of $H/A$ because the irreps of $A$ are almost trivial. Then using your comment, the degree should be dividing $(H:A)$ thus also $(G:A)$! I will be more careful and carry it out completely. Thanks! – Student Dec 22 '19 at 01:40
  • related: https://math.stackexchange.com/questions/243221/proofs-that-the-degree-of-an-irrep-divides-the-order-of-a-group – Student Dec 22 '19 at 01:48

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