# Degrees of irreducible representations need not cover all prime factors

## Contents

## Statement

It may happen for a finite group that the set of prime factors of the Degrees of irreducible representations (?) of the group is a *proper* subset of the set of all prime factors of the group. Here, *degrees of irreducible representations* stands for the degrees of all the irreducible linear representations of the finite group over .

## Related facts

### Partial opposite fact

- Ito-Michler theorem states that if a particular prime does not divide the degree of any irreducible representation, then the -Sylow subgroup must be an abelian normal subgroup.

- Degree of irreducible representation divides order of group
- Degree of irreducible representation divides index of center
- Degree of irreducible representation divides index of Abelian normal subgroup
- Degree of irreducible representation need not divide exponent

## Proof

### Example of an abelian group

For a finite Abelian group, the degrees of irreducible representations are all , which certainly does not cover all prime factors of the group.

### Example of the symmetric group

`Further information: symmetric group:S3`

Consider the symmetric group on three letters. This is a group of order six, and the degrees of its irreducible representations are . The prime factor of the order of the group does not occur as a prime factor of the degree of any irreducible representation.

### Example of dihedral group, or any group with a very big Abelian normal subgroup

`Further information: Degree of irreducible representation divides index of Abelian normal subgroup`

Any dihedral group has an Abelian normal subgroup of index two. Thus, all its degrees of irreducible representations are either or , and so if the group is not a -group, there are certainly prime factors of the order of the group that are not covered in the degrees of irreducible representations.