= Solution
Suppose $\chi$ is an <irreducible character> of a group of odd order and $\chi=\overline\chi$. Write $d=\chi(1)$. Every nonidentity element pairs with its distinct inverse, since there is no element of order $2$. Also $\chi(g^{-1})=\overline{\chi(g)}=\chi(g)$. If $\chi$ were nontrivial, <character orthogonality> with the trivial <character of a representation> would give
$$
\langle\chi,1_G\rangle=\frac{d+2A}{|G|},\qquad 0=d+2A,
$$
where $A$ is the sum of one <character of a representation> value from each inverse pair. This $A$ is an <algebraic integer>, so $A=-d/2\in\mathbb Q$ is an <integer>. Hence $d$ is even.
But an <irreducible character degree divides the group order>, so $d$ must be odd. One quick justification of that divisibility is that the central class sum of a <conjugacy class> $C$ acts by the <algebraic integer> $\lambda_C=|C|\chi(g_C)/d$. Its integrality follows from its <integer> <matrix> on the regular <group representation>. <Character orthogonality> then gives
$$
\frac{|G|}{d}=\sum_C\lambda_C\overline{\chi(g_C)},
$$
a rational <algebraic integer>, hence an <integer>. The parity contradiction proves $\boxed{\chi=1_G}$: \b[the trivial character is the only self-conjugate <irreducible character> of an odd-order group].
Back to article page
