# 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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## Contents

## Statement

It is possible to have a finite group , an abelian subgroup of , and an irreducible representation of over such that the degre of does *not* divide the index .

## Related facts

### Opposite divisibility facts

These are all for irreducible representations over , 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 be symmetric group:S3 and be a subgroup of order 2 in (i.e., S2 in S3). Then, , whereas has an irreducible linear representation of degree .