Homework1

SOLUTIONSFORHOMEWORKONE

1.Is(Z,◦)agroup?wheretheoperationisdefinedby

a◦b=a+b+1,∀a,b∈Z.

Solution.(Z,◦)isagroupsince◦isclearlyalawofcompositionand(Z,◦)satisfiesthefollowingconditions:

(1)◦isassociative:(a◦b)◦c=(a+b+1)◦c=a+b+c+2=a+(b+c+1)+1=a◦(b+c+1)=a◦(b◦c),foranya,b,c∈Z;

(2)(Z,◦)hasanidentity:(−1)◦a=(−1)+a+1=a,anda◦(−1)=a+(−1)+1=aholdforanya∈Z;

(3)eachelementa∈Zhasaninverse:(−a−2)◦a=(−a−2)+a+1=−1,anda◦(−a−2)=a+(−a−2)+1=−1holdforanya∈Z.

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2.Assumethattheequationxyz=1holdsinagroupG.Doesitfollowthatyzx=1?Thatyxz=1?

Solution.Theequationyzx=1holdssinceyzistheinverseelementofx.However,yxz=1isn’talwaysright.Forexample,considerthegroupGL2(R)whoseidentityischoose󰀁󰀁I2,and󰀂󰀂󰀁󰀂

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x=,y=,z=.

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Thenwehavexyz=I2,whileyxz=.

01/2

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3.LetGbeagroup,withmultiplicativenotation.WedefineanoppositegroupGowithlawofcompositiona◦basfollows:theunderlyingsetisthesameasG,butthelawofcompositionistheopposite,thatis,wedefinea◦b=ba.Provethatthisdefinesagroup.

Proof.Bydefinition,◦isclearlyalawofcompositionandisassociative:(a◦b)◦c=(ba)◦c=c(ba)=(cb)a=a◦(cb)=a◦(b◦c),foranya,b,c∈Go.Secondly,sinceGhasanidentity1,and1◦a=a1=a,a◦1=1a=a,so1isalsotheidentityofGo.Thirdly,sinceeveryelementahasaninversea−1inthegroupG,awehavea−1◦a=aa−1=1,a◦a−1=a−1a=1,soa−1isalsotheinverseofainGo.Thus,Goisagroup.

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4.Leta,bbeelementsofagroupG.Assurethatahasorder5,andthata3b=ba3.

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Provethatab=ba.

Proof.Sincetheorderofais5,i.e.,a5=1,sotheequation

ab=a5(ab)=a3(a3b)=a3(ba3)=(a3b)a3=(ba3)a3=(ba)a5=baholds.

5.Whichofthefollowingaresubgroups?

(a)GLn(R)⊂GLn(C).(b){−1,1}⊂R×.

(c)ThesetofpositiveintegersinZ+.

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(d)ThesetofpositiverealsinR.󰀁󰀂

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(e)Thesetofallmatrices,witha=0inGL2(R).

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Solution.(a)Notethattheproductofanytworealinvertiblematricesisstilla

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realinvertiblematrix,theidentityofGLn(C)istherealinvertiblematrixIn,andtheinverseelementofeveryrealinvertiblematrixisitsinversematrix,whichisalsoarealinvertiblematrix.SoGLn(R)isasubgroupofGLn(C).

(b)LetSbetheset{−1,1}.ClearlySisclosedundermultiplication,andcontainstheidentity1ofR×.AlsotheinverseofeachelementsofSissitself.SoSisasubgroupofR×.

(c)Sincethesetofpositiveintegersdoesn’tcontaintheidentity0ofZ+,soitisnotasubgroupofZ+.

(d)sincetheproductofanytwopositiverealsisstillpositive,theidentityofR×istheinteger1,andtheinverseelementofeverypositiverealisalsopositive,sothesetofpositiverealsisasubgroupofR×.

(e)Notethatthesetisn’tasubsetofGL2(R).Ofcourse,itisnotasubgroupofGL2(R).

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