TITLE: Characters of irreducible representations of odd-order groups
QUESTION [6 upvotes]: 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?

REPLY [2 votes]: If you know the Frobenius–Schur indicator, you can argue as follows:
Suppose not, i.e., $\chi$ is not the principal character, then by the orthogonality relationship, we have $\frac{1}{|G|}\sum_{g\in G}\chi(g)=0$.
Note that the Frobenius-Schur indicator can be computed as:
\begin{align}
\nu_2(\chi)&=\frac{1}{|G|}\sum_{g\in G}\chi(g^2).
\end{align}
But $|G|\equiv 1\pmod 2$ implies $g\mapsto g^2$ is a bijection of $G$. Hence
$$\nu_2(\chi)=\frac{1}{|G|}\sum_{g\in G}\chi(g^2)=\frac{1}{|G|}\sum_{g\in G}\chi(g)=0.$$
The Frobenius-Schur indicator equals $0$ can only happen when $\chi\ne\overline\chi$ which is a contradiction.