Characters of irreducible representations of odd-order groups

Question

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?

Answer 1

Let $\chi \in Irr(G)$, with $\chi \neq 1_G$ (the trivial character). Apply the orthogonality relation to $\chi$ and $1_G$, then $$\sum_{x \in G}\chi(x)=0.$$ Now assume that $\chi$ is real-valued. Then $\chi(x)=\overline{\chi(x)}$ (complex conjugate) for all $x \in G$. From the orthogonality formula above one has $$\chi(1)=-\sum_{x \in G-\{1\}}\chi(x).$$ But $\chi(1) \mid |G|$, so the degree of $\chi$ is odd. The oddness of $|G|$ also implies that for all $x \in G-\{1\}$, $x\neq x^{-1}$, so $$\sum_{x \in G-\{1\}}\chi(x)=\sum \chi(x)+\chi(x^{-1}),$$ where the last sum is taken over a set of disjoint pairs $\{x,x^{-1}\}$. Now $\chi(x)$ is the sum of the eigenvalues of a matrix representing $x$, while $\chi(x^{-1})$ is the sum of the eigenvalues of the inverse of that same matrix. The eigenvalues of the matrix are roots of unity and for a root of unity $\zeta$, one has $\zeta^{-1}=\overline{\zeta}$. Since $\chi$ is real-valued one must have $\chi(x)=\chi(x^{-1})$. It follows that $$\frac{1}{2}\chi(1)=-\sum_{x}\chi(x).$$ We now have a contradiction: the left hand is half an odd positive integer, while the right hand side is an algebraic integer.

Note (added April $20^{th} 2019$) The above appears also as Problem (3.16) (a result of William Burnside) in Isaacs' book Character Theory of Finite Groups. It has the following very nice consequence, which is Problem (3.17) of the same book.

Theorem Let $|G|$ be odd and suppose the number of conjugacy classes of $G$ is $k$. Then $|G| \equiv k$ mod $16$.

Proof By the previous the non-principal irreducible characters come in $\frac{k-1}{2}$ pairs, $\chi$ and $\overline{\chi}$ and for all characters $\chi(1)=\overline{\chi(1)}$. Now we know that $$|G|=\sum_{\chi \in Irr(G)}\chi(1)^2= 1+\sum_{\chi \neq 1_G}\chi(1)^2=1+2\sum_{i=1}^{ \frac{k-1}{2}}\chi_i(1)^2$$ Since $\chi_i(1)$ divides $|G|$ it must be odd, and odd squares are $\equiv 1$ mod $8$. So for each $i$, we have $\chi_i(1)^2=1+8a_i$, with $a_i \in \mathbb{Z}$. Hence $|G|=1+2\sum_{i=1}^{ \frac{k-1}{2}}(1+8a_i)=k+\sum_{i=1}^{ \frac{k-1}{2}}16a_i$ and the result follows.

There are various generalizations of the result, some of them coming from Björn Poonen, American Mathematical Monthly 102 $(1995)$, no. $5$, pp. $440-442$ and Michael Reid, American Mathematical Monthly 105 $(1998)$, no. $4$, pp. $359-361$. See also the paper by Robert van der Waall, Elem. Math. 25 $(1970)$, $136–137$.

Answer 2

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.