Character does not determine representation in any prime characteristic

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Statement

Suppose G is a finite group and K is a field of characteristic p a prime number (it does not matter here whether p divides the order of G). It is possible to construct two representations \varphi_1,\varphi_2 of G over K that are inequivalent as representations but have the same character.

Related facts

Proof

Outrageous example

Take \varphi_1 as a sum of p copies of the trivial representation and \varphi_2 as a sum of 2p copies of the trivial representation.

Both \varphi_1 and \varphi_2 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.