# Character does not determine representation in any prime characteristic

From Groupprops

## Statement

Suppose is a finite group and is a field of characteristic a prime number (it does not matter here whether divides the order of ). It is possible to construct two representations of over that are inequivalent as representations but have the same character.

## Related facts

## Proof

### Outrageous example

Take as a sum of copies of the trivial representation and as a sum of copies of the trivial representation.

Both and have character value zero everywhere, but they are not equivalent -- in fact, they do not even have the same degree.

It's possible to construct examples of inequivalent representations with the same degree by using two distinct irreducible representations.