Number Theory: An Introduction to Mathematics

402 X A Character Study

Proposition 2Let G be a finite abelian group of order g andG its dual group. Thenˆ

(i)

∑

a∈G

χ(a)=

{

gifχ=χ 1 ,

0 ifχ=χ 1.

(ii)

∑

χ∈Gˆ

χ(a)=

{

gifa=e,

0 if a=e.

Proof Put

S=

∑

a∈G

χ(a).

Since it is obvious thatS=gifχ=χ 1 , we assumeχ=χ 1 .Thenχ(b)=1forsome
b∈G.Sinceabruns through all elements ofGat the same time asa,

χ(b)S=

∑

a∈G

χ(a)χ(b)=

∑

a∈G

χ(ab)=S.

Sinceχ(b)=1, it follows thatS=0.
Now put

T=

∑

χ∈Gˆ

χ(a).

EvidentlyT=gifa=esince, by Proposition 1,Gˆalso has orderg. Thus we now
assumea=e. By Proposition 1 also, for someψ∈Gˆwe haveψ(a)=1. Sinceχψ
runs through all elements ofGˆat the same time asχ,

ψ(a)T=

∑

χ∈Gˆ

χ(a)ψ(a)=

∑

χ∈Gˆ

χψ(a)=T.

Sinceψ(a)=1, it follows thatT=0.

Since the product of two characters is again a character, and sinceψis the inverse
of the characterψ, Proposition 2(i) can be stated in the apparently more general form

(i)′ ∑

a∈G

χ(a)ψ(a)=

{

gifχ=ψ,

0 ifχ=ψ.

Similarly, sinceχ( ̄ b)=χ(b−^1 ), Proposition 2(ii) can be stated in the form

(ii)′ ∑

χ∈Gˆ

χ(a)χ( ̄b)=

{

gifa=b,

0 if a=b.

The relations (i)′and (ii)′are known as theorthogonality relations, for the characters
and elements respectively, of a finite abelian group.