Midtermsolutions.Problem1. (1) Denewhatitmeansfortwogroupstobeisomorphic.(2) Denetheorderofanelementofagroup. Giveanexampletoshowthattheordercanbeinnite. Noproofisnecessary.Proof.(1) Two groups G and Hare isomorphic if there exists a bijective map f: G Hs.t. fisahomomorphism. Thatis, fisonetoone, ontoandsatisesf(xy) = f(x)f(y)foranytwoelementsx, y G.(2) LetGbeagroupandx G. Theorderofxisn Zifnisthesmallestpositivenumberforwhichxn=e. ThisisequivalenttosayingthatnistheorderofthesubgroupofGgeneratedbyx.Anexampleofanelementofinniteorderistheelement1inthegroupZofintegersunderaddition.

Problem2. LetG = {1, 2, 3, 4}withgrouplawmultiplicationmodulo5.(1) DescribeallthesubgroupsofG. Noproofisnecessary.(2) DescribeanisomorphismfromGtoitself,besides(x) = x. Noproofisnecessary.Proof.(1) Since any element of G other than 1 generates G,there are two subgroups:{e}andG.(2) Anisomorphism: G Gcanbedenedby(x)=x1so(1)=1,(2) = 3,(3) = 3and(4) = 4.

Problem3. LetGbeagroupandletA, BbesubgroupsofG. SetC= {a b|a A, b B}(1) Provethat,ifGisabelian,thenCisasubgroup.(2) GiveanexampletoshowthatCneednot beasubgroup.Proof.(1) SinceGisagroup,CisasubsetofG. Weneedtoshow:Closedundergrouplaw: Letc=a bandc

=a

b

beinCwitha, a

Aandb, b

B. Then(a b) (a

b

) = (a a

) (b b

)sinceG is abelian. Since A, Bare subgroups of G,a a

A and b b

B.Thusc c

CsoCisclosedunderthegrouplaw.Identity: SinceA, BaresubgroupsofG,e A, B. Thuse e = e C.Inverses: Let c = ab Cwith a A, b B. Since A, Bare subgroupsof G, we have that a1A, b1Bsoa1b1C. Then(a b) (a1 b1) = (a a1) (b b1) = esinceGis abelian. ThuseveryelementofChasaninverse.Associativity: SinceCisasubsetofG,multiplicationinCisassocia-tive.ThereforeCisagroupifGisabelian.12(2) LetG=S3, letA= {e, (12)}andletB= {e, (13)}. ThenC= {e, (12),(13), (12)(13)=(132)}. ButCisnotasubgroupofS3because(132)2=(123)innotinC.

Problem4. (1) LetGbeagroupoforder27andx G. Supposealsothatx9isnottheidentity. ProvethatGiscyclic.(2) ProvethatS4isnotisomorphictothedihedral groupoforder24. [Hint:howmanyelementsoforder3inbothgroups?]Proof.(1) Firstnotethatxisaxedelement(soxdoesntstandforanyarbitraryelementof G.) Theorderof xdividestheorderof G. SincetheorderofGis27,theorderofxis1,3,9or27. Iftheorderofxwere1,3,or9wecouldtake(x1)9,(x3)3orx9andseethattheresultwouldbetheidentity.But any of those three expressions are equal to x9. We are given that x9isnottheidentity. Sotheorderofxcannotbe1,3,or9. Thereforetheorderofxis27.Takethegroupgeneratedbyx. Itis= {e, x, x2, . . . , x26}. All oftheelementsinthissetaredistinct. Thereare27elementsinand27elementsinGsowemusthave=G. ThusGisgeneratedbyx.ThereforeGiscyclic.(2) Thereare8elements of order 3inS4. Theyarethethree-cycles (123),(132),(134),(143),(124),(142),(234),and(243).The dihedral group of order 24 is D12since Dnhas 2n elements. D12hastwo elements of order 2. They are r4and r8. (These are the only rotationsoforder3. Allreectionshaveorder2.)Iff: D4 S4wereanisomorphism,itwouldsendelementsoforder3toelementsof order3. Andsincef isabijection, itwouldhavetosenddistinct elements of order 3inS4todistinct elements of order 3inD4.Butthenumberofelementsoforder3inS4andD4aredierent,sotherecannotbeabijectionbetweenthem. SothereisnoisomorphismbetweenS4andD4. ThereforeS4andD4arenotisomorphic.

Problem5. Letf: Sn HbeahomomorphismwhereHisanabeliangroup.(1) Showthat if ,

are transpositions, thenf() =f(

). [Hint: Brieyexplain why every transposition is of the form (12)1for some Sn.(2) Provethatthereareexactlytwohomomorphismsf: Sn {1}.Proof.(1) Firstweshowthehint. Let= (ab)beatransposition. Let = (1a)(2b)so1=(2b)(1a). Thennote that the element (1a)(2b)(12)(2b)(1a) isactually(ab). (Youcangurethisoutbyseeingthatif sendsktok

forsomek,then1sendsk

tok. Ifkisneither1nor2,then(12)doesnothingtoit. Andthenwouldputthekbacktok

. So(12)1onlymoves numbers that 1sends to 1 or 2. Then you say, I want 1to sendato1andbto2, andyoubuildsuchan.) Inanycase, thisshowsthatforanytranspositionthereisans.t. = (12)1.3Letf: Sn Hbeahomomorphism. Wewill nowshowthatf()=f(12) for any transposition . From this we can conclude that f() = f(

)foranytwotranspositionsand

. Sowrite= (12)1. Thenf() =f((12)1). Sincefisahomomorphism,thisisjustf()f(12)f(1) =f()f(12)f()1. SinceHisabelian,wenallygetthatf() = f(12).Sincef() = f(12)foranytransposition,wehaveshownthatifHisabelian,f() = f(

).(2) NotethatH= {1}isanabeliangroup. Soiff: Sn Hisahomomor-phism,thenfsendsallthetranspositionstothesameplace. Soeitherallthetranspositionsgetmappedto1, orall thetranspositionsgetmappedto-1.Snisgeneratedbytranspositions. Soforany Sn,canbewrittenasaproductoftranspositions. Thatis, = 12 nwhereiisatrans-positionforalli. Sincefisahomomorphism,f() = f(1)f(2) f(n).If f sendsall transpositionsto1, thenf()=1forall . If f sendsalltranspositionsto-1,thenf()=(1)nwhichis1ifnisevenand-1ifnisodd.Sothereareexactlytwopossiblehomomorphisms f. Therst is theonethatsendseveryelementto1. Thesecondistheonethatsendsevenpermutationsto1andoddpermutationsto-1.