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!


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.

  • $\begingroup$ So it means that $\phi$ can not be represented as a sum of two irreducible characters as $\chi+\chi$? $\endgroup$
    – Galymbek
    Sep 4 '19 at 19:49
  • $\begingroup$ 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. $\endgroup$ Sep 5 '19 at 3:54

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