Orthogonality Relations for Characters

Theorem
Let $G$ be a finite abelian group with identity $e$.

Let $G^*$ be the dual group of characters $\chi : G \to \C_{\ne 0}$.

Let $\chi_0$ be the trivial character on $G$.

Let $\psi: G \to \C_{\ne 0}$ be any character.

Let $y \in G$ be arbitrary.

Then:


 * $\ds \sum_{x \mathop \in G} \map \psi x = \begin {cases}

\order G & : \psi = \chi_0 \\ 0 & : \psi \ne \chi_0 \end {cases}$

and:


 * $\ds \sum_{\chi \mathop \in G^*} \map \chi y = \begin {cases}

\order {G^*} & : y = e \\ 0 & : y \ne e \end {cases}$

Proof
If $\psi = \chi_0$, then it is straightforward that:

If $\psi \ne \chi_0$, then $\exists y \in G$ such that $\map \psi y \ne 1$.

As $x$ runs through $G$ in the summation, $y x$ also runs through $G$.

So:

Since by assumption $\map \psi y \ne 1$, it must be true that:
 * $\ds \sum_{x \mathop \in G} \map \psi x = 0$