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

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,

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.

• So it means that $\phi$ can not be represented as a sum of two irreducible characters as $\chi+\chi$? Sep 4 '19 at 19:49
• Corrct, @Galymbek. Many ways to see that. If $\phi$ is the character of the regular representation, then its inner product with the trivial character $\chi_0$ is $\langle\phi,\chi_0\rangle=1$. Thus $\langle \frac12\phi,\chi_0\rangle=\frac12$ meaning that $\frac12\phi$ is not a character. Alternatively, the irreducible characters are linearly independent (in the space of class functions), so the presentation of a class function as a linear combination of irreducible characters is unique. Sep 5 '19 at 3:54