Schaum's Outline of Discrete Mathematics, Third Edition (Schaum's Outlines)

456 ALGEBRAIC SYSTEMS [APP. B

(c) Sincee(a)=a, we havee∈Ha. Supposeg, g′∈Ha. Then(gg′)(a)=g(g′(a))=g(a)=a; hencegg′∈Ha.

Also,g−^1 (a)=asinceg(a)=a; henceg−^1 ∈Ha. ThusHais a subgroup ofG.

B.13.Prove Theorem B.6: LetHbe a subgroup of a groupG. Then the right cosets Ha form a partition ofG.

Sincee∈H, we havea=ea∈Ha; hence every element belongs to a coset. Now supposeHaandHbare not

disjoint. Sayc∈Ha∩Hb. The proof is complete if we show thatHa=Hb.

Sincecbelongs to bothHaandHb, we havec=h 1 aandc=h 2 b, whereh 1 ,h 2 ∈H.Thenh 1 a=h 2 b, and

soa=h− 11 h 2 b. Letx∈Ha. Then

x=h 3 a=h 3 h− 11 h 2 b

whereh 3 ∈H. SinceHis a subgroup,h 3 h− 11 h 2 ∈H; hencex∈Hb. Sincexwas any element ofHa, we have

Ha⊆Hb. Similarly,Hb⊆Ha. Both inclusions implyHa=Hb, and the theorem is proved.

B.14.LetHbe a finite subgroup ofG. Show thatHand any cosetHahave the same number of elements.

LetH={h 1 ,h 2 ,...,hk}, whereHhaskelements. ThenHa={h 1 a, h 2 a,...,hka}.

However,hia=hjaimplieshi=hj; hence thekelements listed inHaare distinct. ThusHandHahave the same

number of elements.

B.15.Prove Theorem B.7 (Lagrange): LetHbe a subgroup of a finite groupG. Then the order ofHdivides the
order ofG.
SupposeHhasrelements and there aresright cosets; say

Ha 1 ,Ha 2 ,...,Has

By Theorem B.6, the cosets partitionGand by Problem B.14, each coset hasrelements. ThereforeGhasrselements,

and so the order ofHdivides the order ofG.

B.16.Prove: Every subgroup of a cyclic groupGis cyclic.

SinceGis cyclic, there is an elementa∈Gsuch thatG=gp(a). LetHbe a subgroup ofG.IfH={e}, then

H=gp(e)andHis cyclic. Otherwise,Hcontains a nonzero power ofa. SinceHis a subgroup, it must be closed

under inverses and soHcontains positive powers ofa. Letmbe the smallest positive power of a such thatambelongs

toH. We claim thatb=amgeneratesH. Letxbe any other element ofH; sincexbelongs toGwe havex=anfor

some integern. Dividingnbymwe getaquotientqandaremainderr, that is,

n=mq+r

where 0≤r<m. Then

an=amq+r=amq·ar=bq·ar so ar=b−qan

Butan,b∈H. SinceHis a subgroup,b−qan∈H, which meansar∈H. However,mis the smallest positive power

of a belonging toH. Therefore,r=0. Hencex=an=bq. ThusbgeneratesH, andHis cyclic.

B.17.Prove Theorem B.8: LetHbe a normal subgroup of a groupG. Then the cosets ofHinGform a group
under coset multiplication defined by (aH)(bH )=abH.
Coset multiplication is well-defined, since

(aH )(bH )=a(Hb)H=a(bH)H=ab(HH)=abH

(Here we have used the fact thatHis normal, soHb=bH, and, from Problem B.57, thatHH=H.) Associativity

of coset multiplication follows from the fact that associativity holds inG.His the identity element ofG/H, since

(aH )H=a(HH)=aH and H(aH)=(H a)H=(aH )H=aH

Lastly,a−^1 His the inverse ofaHsince

(a−^1 H )(aH )=a−^1 aHH=eH=H and (aH )(a−^1 H)=aa−^1 HH=eH=H

ThusG/His a group under coset multiplication.