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?

  • $\begingroup$ What are the $n_i$ here? Are you looking at real or complex representations? $\endgroup$ – Qiaochu Yuan Dec 2 '14 at 22:41

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 Exercise (3.16) (a result of William Burnside) in Isaacs' book Character Theory of Finite Groups. It has the following very nice consequence, which is Exercise (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 $(1970)$, $136–137$.

  • $\begingroup$ Thank you Nicky. I actually ended up following a similar procedure, but I just computed the scalar product $(\chi, \epsilon)$ and showed it is different from 0. $\endgroup$ – user195887 Dec 4 '14 at 23:45

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.


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