# Degree of irreducible representation need not divide index of abelian subgroup

This fact is related to: linear representation theory
View other facts related to linear representation theoryView terms related to linear representation theory |

## 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).

## 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 $2$.