Character does not determine representation in any prime characteristic

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

• Character determines representation in characteristic zero

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.