I am trying to prove the following fact:

If $G$ is a finite group, $\chi$ is a complex irreducible character of $G$ and $\{g_1,\cdots,g_r\}$ is a complete set of representatives of conjugacy classes in $G$ then $\chi(g_1)+\cdots + \chi(g_r)$ is a non-negative integer.

It is easy to prove that the above quantity is integer. Sketch of proof: if $|G|=m$ then the components in the above expression lie in $\mathbb{Q}(\zeta_m)$ and automorphisms of this field correspond to bijections from $G$ to $G$ which permute conjugacy classes (maps $G\rightarrow G$, $g\mapsto g^l$ where $(l,|G|)=1$ and correspondingly there are Galois automorphisms); hence above quantity is invariant under all automorphisms of $\mathbb{Q}(\zeta_m)$ over $\mathbb{Q}$ and is an algebraic integer.

I didn't get why it is non-negative? Any hint?


Hint: let $G$ act on $G$ by conjugation and consider the permutation character $\pi$ of this action. Show that the Frobenius product $[\chi,\pi]$ is exactly the sum you want to calculate.


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