the sum of the values of a character of a finite group is 0


The following is an argument that occurs in many proofs involving characters of groups. Here we use additive notation for the group G, however this group is not assumed to be abelianMathworldPlanetmath.

Lemma 1.

Let G be a finite groupMathworldPlanetmath, and let K be a field. Let χ:GK× be a character, where K× denotes the multiplicative groupMathworldPlanetmath of K. Then:

gGχ(g)={G, if χ is trivial,0K, otherwise

where 0K is the zero element in K, and G is the order of the group G.

Proof.

First assume that χ is trivial, i.e. for all gG we have χ(g)=1K. Then the result is clear.

Thus, let us assume that there exists g1 in G such that χ(g1)=h1K. Notice that for any element g1G the map:

GG,gg1+g

is clearly a bijection. Define 𝒮=gGχ(g)K. Then:

h𝒮 = χ(g1)𝒮
= χ(g1)gGχ(g)
= gGχ(g1)χ(g)
= gGχ(g1+g),(1)
= jGχ(j),(2)
= 𝒮

By the remark above, sums (1) and (2) are equal, since both run over all possible values of χ over elements of G. Thus, we have proved that:

h𝒮=𝒮

and h1K. Since K is a field, it follows that 𝒮=0K, as desired.

Title the sum of the values of a character of a finite group is 0
Canonical name TheSumOfTheValuesOfACharacterOfAFiniteGroupIs0
Date of creation 2013-03-22 14:10:30
Last modified on 2013-03-22 14:10:30
Owner alozano (2414)
Last modified by alozano (2414)
Numerical id 6
Author alozano (2414)
Entry type Theorem
Classification msc 11A25