Orthogonality of Group Characters

The following relations are often called the orthogonality relations of characters.

Theorem

Let $G$ be a finite abelian group with $| G | = n$ . For each $f \in \hat{G}$

\sum_{g \in G} f (g) = {\begin{cases} n & f = f_{0} = id \\ 0 & otherwise \end{cases}

where $f_{0}$ is the principal character.

Proof

If $f$ is the principal character then clearly

\sum_{g \in G} f (g) = \sum_{g \in G} 1 = | G | = n .

If $f$ is not the principal character, then there is some $a \in G$ for which $f (a) \neq 1$ . Fix such an $a$ and consider that

\begin{aligned} f (a) \sum_{g \in G} f (g) & = \sum_{g \in G} f (a) f (g) \\ = \sum_{g \in G} f (a g) . \end{aligned}

Now, because $a$ is an element of the group, and is thus invertible, the function $x \mapsto a x$ is a bijection. Thus

\sum_{g \in G} f (a g) = \sum_{g \in G} f (g)

given they count the same terms.

This allows us to deduce that

f (a) \sum_{g \in G} f (g) = \sum_{g \in G} f (g) ⟹ (f (a) - 1) \sum_{g \in G} f (g) = 0.

This expression is in $C$ and hence it is equal to zero if either $f (a) = 1$ or $\sum_{g \in G} f (g) = 0$ , however $f (a) \neq 1$ by choice of $a$ and therefore

\sum_{g \in G} f (g) = 0.

Lemma

For any finite abelian group $G$ , and element $g \neq id$ , there exists a character $f_{i}$ such that $f_{i} (g) \neq 1$ .

Proof

If $f_{i} (g) = 1$ then $f_{i} (g^{k}) = f_{i} (g)^{k} = 1$ and the character $f_{i}$ evaluates to $1$ on the cyclic subgroup generated by $g$ . If this group is non-trivial, then it has a non-principal character, which can be extended to a character of $G$ , however this would contradict the fact that all characters evaluate to $1$ on this cyclic subgroup.

%% TODO change this to use splitting into product of cyclic groups and fix all the characters on each group to be the principal character except 1

Theorem

Let $G$ be a finite abelian group with $| G | = n$ . For each $g \in G$

\sum_{f \in \hat{G}} f (g) = {\begin{cases} n & g = id \\ 0 & otherwise . \end{cases}

Proof

Suppose that $g = id$ . Then from the properties of a group homomorphism

\sum_{f \in \hat{G}} f (g) = \sum_{f \in \hat{G}} f (id) = \sum_{f \in \hat{G}} 1 = | \hat{G} | = n

noting that $| \hat{G} | ≅ | G |$ .

Now suppose that $g \neq id$ . Therefore, by the previous lemma, there exists a character $f_{i}$ such that $f_{i} (g) \neq 1$ . Then we have that

f_{i} (g) \sum_{f \in \hat{G}} f (g) = \sum_{f \in \hat{G}} f_{i} (g) f (g) .

Because $f_{i}$ is an element of the group and has an inverse $f_{i} \times f$ also varies over $\hat{G}$ and hence

\begin{aligned} \sum_{f \in \hat{G}} f_{i} (g) f (g) & = \sum_{f \in \hat{G}} (f_{i} \times f) (g) \\ = \sum_{f \in \hat{G}} f (g) . \end{aligned}

This lets us deduce that

f_{i} (g) \sum_{f \in \hat{G}} f (g) = \sum_{f \in \hat{G}} f (g) ⟹ (f_{i} (g) - 1) \sum_{f \in \hat{G}} f (g) = 0.

Since $f_{i} (g) \neq 1$ by choice of $f_{i}$ , we can conclude that

\sum_{f \in \hat{G}} f (g) = 0.

Corollary

Let $G$ be a finite abelian group with $| G | = n$ . For each $g, h \in G$

\sum_{f \in \hat{G}} \overset{―}{f (h)} f (g) = {\begin{cases} n & g = h \\ 0 & otherwise . \end{cases}

Proof

This follows from the previous result by replacing $g$ with $h^{- 1} g$ . That is we have

\sum_{f \in \hat{G}} f (h^{- 1} g) = {\begin{cases} n & h^{- 1} g = id \\ 0 & otherwise . \end{cases}

and

\begin{aligned} \sum_{f \in \hat{G}} f (h^{- 1} g) & = \sum_{f \in \hat{G}} f (h^{- 1}) f (g) \\ = \sum_{f \in \hat{G}} f (h)^{- 1} f (g) \\ = \sum_{f \in \hat{G}} \overset{―}{f (h)} f (g) \end{aligned}

where $f (h)^{- 1} = \overset{―}{f (h)}$ because they are roots of unity.

Then in the condition in the cases clearly $h^{- 1} g = id ⟺ g = h$ , as required.