Orthogonality of characters

Remarks

Proof idea for 1: if $\chi$ is nontrivial, pick $y$ with $\chi(y) \ne 1$. Since $x \mapsto yx$ is a bijection of $G$, the sum $S = \sum_x \chi(x)$ satisfies $S = \sum_x \chi(yx) = \chi(y) S$, so $S = 0$. Statement 3 is 1 applied to $\chi \overline{\psi} = \chi\psi^{-1}$, since $\overline{\psi(x)} = \psi(x)^{-1}$ on the unit circle. Statement 2 is proved in the same way as 1, once one knows that for $x \ne e$ some character has $\chi(x) \ne 1$; this follows from the classification of finite abelian groups.

Example: for $G = \Z/4\Z$ and the character $\chi(a) = i^a$, the sum is $1 + i - 1 - i = 0$.

Statement 2 needs $G$ abelian. For $G = S_3$, the only homomorphisms to $\C^\times$ are the trivial character and the sign. At a $3$-cycle $x \ne e$ both take the value $1$, so the sum is $2$ instead of $0$.

Going deeper

For a student

These relations are the finite abelian case of the Schur orthogonality relations. They are the engine of discrete Fourier analysis: the inversion formula for the Fourier transform on $G$ is a direct consequence. In number theory, the version for Dirichlet characters modulo $q$ is used to pick out an arithmetic progression, as in the proof of Dirichlet's theorem on primes in progressions.