I'm approaching to the study of representation theory and I'm following several texts, especially J. Serre's Linear Representations of Finite Groups. At page 20 of this book there is the following result:
Let $s\in G$, and let $c(s)$ be the number of elements in the conjugacy class of $s$.
- We have $\sum_{i=1}^h\overline{\chi_i(s)}\chi_i(s)=\frac{g}{c(s)}$ ($h$ is the number of irreducible representations, equal to the number of conjugacy classes of $G$).
- For $t\in G$ not conjugate to $s$, we have $\sum_{i=1}^h\overline{\chi_i(s)}\chi_i(s)$.
After the proof he explains how to determine the character table of $S_3$.
Let $G=S_3$. We have $|G|=6$, and there are three conjugacy classes: the element 1, the three transpositions, and the two cyclic permutations. Let $t$ be a transposition and $c$ a cyclic permutation. We have $t^2=1,c^3=1,tc=c^2t$; whence there are just two characters of degree 1: the unit character $\chi_1$ and the character $\chi_2$ giving the signature of a permutation. Since the number of irreducible characters is equal to the number of conjugacy classes, there exists one other irreducible character $\theta$; if $n$ is its degree we must have $1+1+n^2=6$, hence $n=2$. The values of $\theta$ can be deduced from the fact that $\chi_1+\chi_2+2\theta$ is the character of the regular representation of $G$.
I don't think this is hard, but there is a reasoning that I don't get: when he notices the order of $t$ and $c$ and computes their product, and then he says that there are just two one-dimensional characters. When I construct this character table, the sign permutation is easy to find, and we can consider it for all symmetric groups. But I don't understand this kind of reasoning. I have reported the result above because maybe he uses it, but I'm not sure.