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:

$\displaystyle \sum_{x \mathop \in G} \psi \left({x}\right) = \begin{cases} \left\lvert{G}\right\rvert & : \psi = \chi_0 \\ 0 & : \psi \ne \chi_0 \end{cases}$

and:

$\displaystyle \sum_{\chi \mathop \in G^*} \chi \left({y}\right) = \begin{cases} \left\lvert{G^*}\right\rvert & : y = e \\ 0 & : y \ne e \end{cases}$


Proof

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

\(\ds \sum_{x \mathop \in G} \psi \left({x}\right)\) \(=\) \(\ds \sum_{x \mathop \in G} \chi_0 \left({x}\right)\) Assumption
\(\ds \) \(=\) \(\ds \sum_{x \mathop \in G} 1\) Definition of Trivial Character
\(\ds \) \(=\) \(\ds \vert G \vert\)


If $\psi \neq \chi_0$, then $\exists y \in G$ such that $\psi \left({y}\right) \neq 1$.

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

So:

\(\ds \sum_{x \mathop \in G} \psi \left({x}\right)\) \(=\) \(\ds \sum_{x \mathop \in G} \psi \left({y x}\right)\) by the above claim
\(\ds \) \(=\) \(\ds \sum_{x \mathop \in G} \psi \left({y}\right) \psi \left({x}\right)\) $\psi$ as a character is by definition a homomorphism
\(\ds \) \(=\) \(\ds \psi \left({y}\right) \sum_{x \mathop \in G} \psi \left({x}\right)\)

Since by assumption $\psi \left({y}\right) \ne 1$, it must be true that:

$\displaystyle \sum_{x \mathop \in G} \psi \left({x}\right) = 0$

$\blacksquare$