Irreducible character

An irreducible character is the character of an irreducible representation. Over the complex numbers, every finite-group character decomposes uniquely as a nonnegative integer combination of irreducible characters.

For an irreducible complex character of a representation, the indicator is $1$ for real type, $-1$ for quaternionic type and $0$ for complex type, the latter meaning $\chi \ne \overline{\chi }$. It detects the invariant bilinear form on an irreducible representation. For an odd-order group, squaring permutes its elements, so every nontrivial irreducible character has indicator zero by character orthogonality.

For a complex irreducible representation $D$ of a finite group $G$, the sum $A={\sum }_{g}D(g^2)$ commutes with every $D(h)$: conjugation sends its summands to $D((hg{h}^{-1}{)}^2)$, merely permuting the summation index. Schur's lemma gives $A=cI$. Taking the trace yields $c\dim D={\sum }_g\chi (g^2)=|G|\nu (D)$, proving the identity. Here $\nu (D)$ is the Frobenius-Schur indicator. For a reducible representation the sum is still central in its image, but need not be one scalar on all irreducible constituents.

An odd-order finite group has only odd irreducible degrees, by irreducible character degree divides the group order. Its nontrivial characters occur in distinct conjugate pairs, by the Frobenius-Schur indicator. Therefore $|G|-k(G)={\sum }_{\chi }(\chi (1{)}^2-1)$ is a sum of pairs $2(d^2-1)$ with $d$ odd. Since $d^2\equiv 1(\mathrm{mod}8)$, each pair is divisible by sixteen. Here $k(G)$ is the number of conjugacy classes.

For a real-valued irreducible character of a finite odd-order group, pairing $g$ with $g^{-1}$ gives $⟨\chi ,1_G⟩=(\chi (1)+2A)/|G|$, with $A$ an algebraic integer. If this inner product were zero, $\chi (1)$ would be even, since a rational algebraic integer is an integer. But an irreducible character degree divides the group order, so it is odd. Therefore the character is trivial.