Number Theory: An Introduction to Mathematics

426 X A Character Study

Lemma 16Ifρ:s→A(s)is a representation of degree n of a finite group G, then
the characterχofρsatisfies

|χ(s)|≤n for any s∈G.

Moreover, equality holds for some s if and only if A(s)=ωIn,whereω∈C.

Proof Ifs∈Ghas orderm, there exists an invertible matrixTsuch that

T−^1 A(s)T=diag[ω 1 ,...,ωn],

whereω 1 ,...,ωnarem-th roots of unity. Henceχ(s)=ω 1 +···+ωnand

|χ(s)|≤|ω 1 |+···+|ωn|=n.

Moreover|χ(s)|=nonly ifω 1 ,...,ωnall lie on the same ray through the origin
and hence only if they are all equal, since they lie on the unit circle. But then
A(s)=ωIn.

Thekernelof the representationρis the setKρof alls∈Gfor whichρ(s)=In.
EvidentlyKρis a normal subgroup ofG. By Lemma 16,Kρmay be characterized as
the set of alls∈Gsuch thatχ(s)=n.

Lemma 17Letρ:s→A(s)be an irreducible representation of degree n of a finite
group G, with characterχ, and letCbe a conjugacy class of G containing h elements.
If h and n are relatively prime then, for any s∈C, eitherχ(s)= 0 or A(s)=ωInfor
someω∈C.

Proof Sincehandnare relatively prime, there exist integersa,bsuch thatah+bn=

1. Then

χ(s)/n=ahχ(s)/n+bχ(s).

Sincehχ(s)/nandχ(s)are algebraic integers, it follows thatχ(s)/nis an algebraic
integer. We may assume that|χ(s)| <n, since otherwise the result follows from
Lemma 16.
Supposeshas orderm.If(k,m)=1, then the conjugacy class containingskalso
has cardinalityhand thusχ(sk)/nis an algebraic integer, by what we have already
proved. Hence

α=

∏

k

χ(sk)/n,

wherekruns through all positive integers less thanmand relatively prime tom,isalso
an algebraic integer. Butχ(sk)=f(ωk),whereωis a primitivem-th root of unity and

f(x)=xr^1 +···+xrn

for some non-negative integersr 1 ,...,rnless thanm. Thusαis a symmetric function
of the primitive rootsωk. Since thecyclotomic polynomial