Degree of irreducible representation need not divide index of abelian subgroup

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).

• Degree of irreducible representation divides group order
• Degree of irreducible representation divides index of center
• Degree of irreducible representation divides index of abelian normal subgroup

Opposite bounding facts

• Degree of irreducible representation is bounded by index of abelian subgroup
• Sum of squares of degrees of irreducible representations equals group order
• Order of inner automorphism group bounds square of degree of irreducible representation

Similar non-divisibility/non-bounding facts

• Degree of irreducible representation need not divide exponent
• Square of degree of irreducible representation need not divide group order

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 $2$.