Number Theory: An Introduction to Mathematics

7 Applications 427

Φn(x)=

∏

k

(x−ωk)

has integer coefficients, it follows thatα∈Q. Consequently, by Lemma 12,α∈Z.
But|α|<1, since|χ(s)|<nand|χ(sk)|≤nfor everyk. Henceα=0, and thus
χ(sk)=0forsomekwith(k,m)=1. Ifg(x)is the monic polynomial inQ[x]of
least positive degree such thatg(ωk)=0, then any polynomial inQ[x] withωkas a
root must be divisible byg(x). Since we showed in Chapter II,§5 that the cyclotomic
polynomialΦn(x)is irreducible over the fieldQ, it follows thatg(x)=Φn(x)and
thatΦn(x)dividesf(x). Hence alsoχ(s)=f(ω)=0.

Before stating the next result we recall from Chapter I,§7 that a group is said to
besimpleif it contains more than one element and has no nontrivial proper normal
subgroup.

Proposition 18If a finite group G has a conjugacy classCof cardinality pa,forsome
prime p and positive integer a, then G is not a simple group.

Proof Ifs∈Cthen, by (9),

∑

μ

nμχμ(s)= 0.

Assume the notation chosen so thatχ 1 is the character of the trivial representation.
Ifχμ(s)=0foreveryμ>1forwhichpdoes not dividenμ, then the displayed
equation has the form 1+pζ=0, whereζis an algebraic integer. Since− 1 /pis not
an integer, this contradicts Lemma 12. Consequently, by Lemma 17, for somev> 1
we must haveA(v)(s)=ωInv,whereω∈C.ThesetKvof all elements ofGwhich
are represented by the identity transformation in thev-th irreducible representation is a
normal subgroup ofG. MoreoverKv={e},sinceKvcontains all elementsu−^1 s−^1 us,
andKv=G,sincev>1. ThusGis not simple.

Corollary 19If G is a group of order paqb,where p,q are distinct primes and a,b
non-negative integers such that a+b> 1 , then G is not simple.

Proof LetC 1 ,...,Crbe the conjugacy classes ofG, withC 1 ={e},andlethkbe the
cardinality ofCk(k= 1 ,...,r).Thenhkdivides the ordergofGand

g=h 1 +···+hr.

Suppose first thathj =1forsomej>1. ThenCj={sj},wheresjcommutes
with every element ofG. Thus the cyclic groupHgenerated bysjis a normal sub-
group ofG.ThenGis not simple even ifH =G,sincea+b>1 and any proper
subgroup of a cyclic group is normal.
Suppose next thathk=1foreveryk>1. IfGis simple then, by Proposition 18,
qdivideshkfor everyk>1. Sinceqdividesg, it follows thatqdividesh 1 =1, which
is a contradiction.