If $\phi$ is a character of $G$ such that $\langle \phi,\phi \rangle=4$, then there exists a character $\chi$ of $G$ such that $\phi=2\chi$

Question

I am stuck on the following problem that says:

If $\phi$ is a character of $G$ such that $\langle \phi,\phi \rangle=4$, then there exists a character $\chi$ of $G$ such that $\phi=2\chi$

My Attempt:

I know that if $\phi=2\chi$ and $\langle \phi,\phi \rangle=4$ it implies $\langle 2\chi,2\chi \rangle=4$ and $\langle \chi,\chi \rangle=1$, so we conclude that $\chi$ is an irreducible character.

But, here I couldn't find any idea further to proceed,

Can someone help me out? Thanks in advance for your time!

Answer 1

The claim is false. It is possible that $\phi=\chi_1+\chi_2+\chi_3+\chi_4$ for some four distinct irreducible characters of $G$.

The smallest groups with four distinct characters are the abelian groups of order four, that is $C_2\times C_2$ and $C_4$. For both these groups the regular representation $\phi$ is the sum of four distinct irreducible characters, giving a counterexample.