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 fG^

gGf(g)={nf=f0=id0otherwise

where f0 is the principal character.

Proof

If f is the principal character then clearly

gGf(g)=gG1=|G|=n.

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

f(a)gGf(g)=gGf(a)f(g)=gGf(ag).

Now, because a is an element of the group, and is thus invertible, the function xax is a bijection. Thus

gGf(ag)=gGf(g)

given they count the same terms.

This allows us to deduce that

f(a)gGf(g)=gGf(g)(f(a)1)gGf(g)=0.

This expression is in C and hence it is equal to zero if either f(a)=1 or gGf(g)=0, however f(a)1 by choice of a and therefore

gGf(g)=0.

Lemma

For any finite abelian group G, and element gid, there exists a character fi such that fi(g)1.

Proof

If fi(g)=1 then fi(gk)=fi(g)k=1 and the character fi 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 gG

fG^f(g)={ng=id0otherwise.
Proof

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

fG^f(g)=fG^f(id)=fG^1=|G^|=n

noting that |G^||G|.

Now suppose that gid. Therefore, by the previous lemma, there exists a character fi such that fi(g)1. Then we have that

fi(g)fG^f(g)=fG^fi(g)f(g).

Because fi is an element of the group and has an inverse fi×f also varies over G^ and hence

fG^fi(g)f(g)=fG^(fi×f)(g)=fG^f(g).

This lets us deduce that

fi(g)fG^f(g)=fG^f(g)(fi(g)1)fG^f(g)=0.

Since fi(g)1 by choice of fi, we can conclude that

fG^f(g)=0.

Corollary

Let G be a finite abelian group with |G|=n. For each g,hG

fG^f(h)f(g)={ng=h0otherwise.
Proof

This follows from the previous result by replacing g with h1g. That is we have

fG^f(h1g)={nh1g=id0otherwise.

and

fG^f(h1g)=fG^f(h1)f(g)=fG^f(h)1f(g)=fG^f(h)f(g)

where f(h)1=f(h) because they are roots of unity.

Then in the condition in the cases clearly h1g=idg=h, as required.