Orthogonality Relations for Characters

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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:

\(\ds \sum_{x \mathop \in G} \map \psi x\) \(=\) \(\ds \sum_{x \mathop \in G} \map {\chi_0} x\) by hypothesis
\(\ds \) \(=\) \(\ds \sum_{x \mathop \in G} 1\) Definition of Trivial Character
\(\ds \) \(=\) \(\ds \order G\)


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:

\(\ds \sum_{x \mathop \in G} \map \psi x\) \(=\) \(\ds \sum_{x \mathop \in G} \map \psi {y x}\) by the above claim

\(\ds \) \(=\) \(\ds \sum_{x \mathop \in G} \map \psi y \map \psi x\) $\psi$ as a character is by definition a homomorphism
\(\ds \) \(=\) \(\ds \map \psi y \sum_{x \mathop \in G} \map \psi x\)

Since by assumption $\map \psi y \ne 1$, it must be true that:

$\ds \sum_{x \mathop \in G} \map \psi x = 0$

$\blacksquare$