Degree of irreducible representation need not divide index of abelian subgroup

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This fact is related to: linear representation theory
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Statement

It is possible to have a finite group G, an abelian subgroup H of G, and an irreducible representation \rho of G over \mathbb{C} such that the degre of \rho does not divide the index [G:H].

Related facts

Opposite divisibility facts

These are all for irreducible representations over \mathbb{C}, or more generally, over splitting fields of characteristic zero (and more generally, characteristic coprime to the order of the group).

Opposite bounding facts

Similar non-divisibility/non-bounding facts

Proof

Further information: linear representation theory of symmetric group:S3, subgroup structure of symmetric group:S3

Let G be symmetric group:S3 and H be a subgroup of order 2 in G (i.e., S2 in S3). Then, [G:H] = 6/2 = 3, whereas G has an irreducible linear representation of degree 2.