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.