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
|
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\(\ds \) | \(=\) | \(\ds \sum_{x \mathop \in G} \map \psi y \map \psi x\) | $\psi$ 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:
- $\ds \sum_{x \mathop \in G} \map \psi x = 0$
$\blacksquare$
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