I want to prove that if $G$ is a group and the order of $G$ is odd, and $\chi$ is a real-valued irreducible character of $G$, then $\chi$ must be the trivial representation, $\chi = \epsilon$.
So far, I know that $|G| = \sum_{i = 1}^{n} n_i\chi_i^2$, and that $|G|$ odd $\implies \exists j \in \mathbb{N}: |G| = 2j+1$. Therefore:
$2j+1 = n_1 \chi^2 + \sum_{i=2}^n n_i \chi_i^2$, and therefore $n_1\chi^2$ must be even and $\sum_{i=2}^n n_i \chi_i^2$ even, or both of them must be odd. Since I have no further information about the characters of the other irreducible representations of this group, I am not sure how to proceed. Any tips?
Best Answer
Another standard proof consists in working with the character table $T$, as a matrix whose $(\chi, g)$ entry is $\chi(g)$, where $\chi$ ranges in the irreducible characters of $G$, and $g$ ranges in a set of representatives of the conjugacy classes.
Consider the permutation of the rows that takes row $\chi$ to row $\overline{\chi}$. So this takes $T$ to $A T$, where $A$ is a suitable permutation matrix.
Consider the permutation of the columns that takes the column $g$ to the column of (the class of) $g^{-1}$. Once more, this takes $T$ to $T B$, for a suitable permutation matrix $B$.
Since $\overline{\chi(g)} = \chi(g^{-1})$, we have $A T = T B$. Since $T$ is invertible, $A$ and $B$ are conjugate, so in particular their traces, that is, the numbers of fixed points of the corresponding permutations, coincide.
Now if an element $g \in G$ is conjugate to its inverse, $g^{x} = g^{-1}$, for some $x \in G$, then $g^{x^{2}} = g$, so $x^{2} \in C_{G}(g)$, which yields $x \in C_{G}(g)$, as the order of $G$ is odd, so that $g = g^{-1}$, and $g = 1$, once more because the order of $G$ is odd.
Therefore the trace of $B$ is $1$. But then the trace of $A$ is also $1$, that is, there is only one character $\chi$ such that $\chi = \overline{\chi}$, and then it has to be the trivial character.