# Character values up to permutation of characters need not determine automorphism class

## Contents

## Statement

It is possible to have a finite group and a splitting field for in characteristic zero such that the following holds: There are elements in different conjugacy classes and a permutation on the set of characters of irreducible representations (up to equivalence) of over such that the following holds:

- For every irreducible character , we have .
- and are not in the same automorphism class, i.e., they are not in the same orbit under the action of the automorphism group on .

## Related facts

### Similar facts

### Opposite facts

- Character orthogonality theorem
- Column orthogonality theorem
- Character determines representation in characteristic zero
- Splitting implies characters separate conjugacy classes

## Proof

### Example of the dihedral group

`Further information: dihedral group:D8, linear representation theory of dihedral group:D8`

This character table works over characteristic zero:

Representation/Conj class | (size 1) | (size 1) | (size 2) | (size 2) | (size 2) |
---|---|---|---|---|---|

Trivial representation | 1 | 1 | 1 | 1 | 1 |

-kernel | 1 | 1 | 1 | -1 | -1 |

-kernel | 1 | 1 | -1 | 1 | -1 |

-kernel | 1 | 1 | -1 | -1 | 1 |

2-dimensional | 2 | -2 | 0 | 0 | 0 |

The same character table works over any characteristic not equal to 2 where the elements 1,-1,0,2,-2 are interpreted over the field.

Note that the element of order four and the element of order two have the same character values up to permutation of characters, but they cannot be in the same orbit under the action of the automorphism group because the elements have different orders.