HOMOMORPHISMSANDISOMORPHISMSDenition:If(G, )and(H, )aregroups,thenafunctionf: G Hisahomomorphismiff(x y) = f(x) f(y)forallx, y G.Example:Let(G, )beanarbitrarygroupandH= {e},thenthefunctionf: G Hsuchthatf(x) = e forany x Gisahomomorphism. Infact,f(x y) = e = e e = f(x) f(y).Example:Let(G, )beanarbitrarygroup,thenthefunctionf: G Gsuchthatf(x) = x forany x Gisahomomorphism. Infact,f(x y) = x y= f(x) f(y).Example:Letf: Z+ Z+2beafunctionsuchthatf(x) =

[0]ifxiseven[1]ifxisodd.Thenfisahomomor-phism. Infact,ifx + yiseven,thenf(x + y) = [0] = f(x) + f(y).Similarly,ifx + yisodd,thenf(x + y) = [1] = f(x) + f(y).Example:Letf: GL(2, R) R=0beafunctionsuchthatf(M) = det M forany M GL(2, R).Thenfisahomomorphism. Infact,f(M1M2) = det(M1M2) = det(M1) det(M2) = f(M1)f(M2).1Denition:Letafunctionf:G Hbeahomomorphism. Iffisalsoaone-onecorrespondence, thenfiscalledanisomorphism. TwogroupsGandHarecalledisomorphic,denotedbyG = H,ifthereexistsanisomorphismbetweenthem.Example:Weshowthat R+ = R>0.Infact,letf(x) = ex.Toprovethatthisisanisomorphism,weshouldcheckthatf: R+ R>0isone-onecorrespondenceandthatf(x + y) = f(x)f(y)forallx, y R.Therstpartistrivial,sincef(x) = exisdenedforallx Randitsinverseg(x) = ln xisalsodenedforallx R>0.Thesecondpartisalsotrue,sincef(x + y) = ex+y= exey= f(x)f(y).Denition:LetG= {a1, a2, . . . , an}beanitegroup. AmultiplicationtableforGisann nmatrixwhoseijentryisaiaj:G a1a2. . . aj. . . ana1a1a2a1a2. . . a1aj. . . a1ana2a2a1a2a2. . . a2aj. . . a2an...............aiaia1aia2. . . aiaj. . . aian...............anana1ana2. . . anaj. . . ananWewillalsoagreethata1= 1.2Example:Multiplicativetablefor Z5isZ5[1] [2] [3] [4][1] [1] [2] [3] [4][2] [2] [4] [1] [3][3] [3] [1] [4] [2][4] [4] [3] [2] [1]Remark:It isclear that twogroups G= {a1, a2, . . . , an}andH= {b1, b2, . . . , bn}of thesameordernareisomorphicif andonlyif it ispossibletomatchelements a1, a2, . . . , anwithelementsb1, b2, . . . , bnsuchthatthisone-onecorrespondenceremainsalsoforcorrespondingentriesaiajandbibjoftheirmultiplicationtables.Example:1. ThemultiplicationtablesbelowshowthatP = Z+2= Z3 :P even oddeven even oddodd odd evenZ+2[0] [1][0] [0] [1][1] [1] [0]Z3[1] [2][1] [1] [2][2] [2] [1]2. Themultiplicationtablesbelowshowthat Z+4= Z5 :Z+4[0] [1] [2] [3][0] [0] [1] [2] [3][1] [1] [2] [3] [0][2] [2] [3] [0] [1][3] [3] [0] [1] [2]Z5[1] [2] [4] [3][1] [1] [2] [4] [3][2] [2] [4] [3] [1][4] [4] [3] [1] [2][3] [3] [1] [2] [4]3. Themultiplicationtablesbelowshowthat Z+6= Z7 :Z+6[0] [1] [2] [3] [4] [5][0] [0] [1] [2] [3] [4] [5][1] [1] [2] [3] [4] [5] [0][2] [2] [3] [4] [5] [0] [1][3] [3] [4] [5] [0] [1] [2][4] [4] [5] [0] [1] [2] [3][5] [5] [0] [1] [2] [3] [4]Z7[1] [3] [2] [6] [4] [5][1] [1] [3] [2] [6] [4] [5][3] [3] [2] [6] [4] [5] [1][2] [2] [6] [4] [5] [1] [3][6] [6] [4] [5] [1] [3] [2][4] [4] [5] [1] [3] [2] [6][5] [5] [1] [3] [2] [6] [4]3Theorem1:Letf: G Hisahomomorphismofgroups. Then(i)f(e) = e;(ii)f(x1) = f(x)1;(iii)f(xn) = [f(x)]nforalln Z.Proof:(i)Wehavee e = e = f(e e) = f(e) = f(e)f(e) = f(e).Multiplyingbothsidesby[f(e)]1,weget[f(e)]1f(e)f(e) = [f(e)]1f(e) = e f(e) = e = f(e) = e.(ii)Wehavex x1= e = f(x x1) = f(e) = f(x)f(x1) = f(e).Sincef(e) = eby(i),wegetf(x)f(x1) = e.Similarly,fromx1 x = eonecandeducethatf(x1)f(x) = e.So,f(x)f(x1) = f(x1)f(x) = e,whichmeansthatf(x1) = f(x)1.(iii)Ifn 1,onecanprovef(xn) = [f(x)]nbyinduction. Ifn < 0,thenf(xn) = f((x1)n) = [f(x1)]n,whichisequalto[f(x)]nby(ii). Theorem2:AnytwocyclicgroupsGandHofthesameorderareisomorphic.Proof(Sketch):SupposethatG = a = {1, a, a2, . . . , am1}andH= b = {1, b, b2, . . . , bm1}.Thenf: G Hwithf(ai) = bi, 0 i m1,isanisomorphismandG = H. .4Example:Weknowthatanygroupoftheprimeorderiscyclic. ThereforebyTheorem2anytwogroupsofthesameprimeorderareisomorphic.Example:Letpbeaprimenumber. Weknowthatthegroup Z+p1iscyclic,sinceZ+p1= [1].Itispossibletoprovethat Zpisalsocyclic. Also,|Z+p1| = |Zp| = p 1.ThereforefromTheorem2itfollowsthat Z+p1= Zp .Problem: ShowthatV = Z+4 .Solution:AssumetothecontrarythatV = Z+4 .Thenthereisaone-onecorrespondencef: V Z+4 .Fromthis,inparticularly,followsthatthereexistsx Vsuchthatf(x) = [1].Thisand(iii)ofTheorem1givef(x2) = [f(x)]2= [1]2= [1] + [1] = [2].Wenowrecallthatforanyelementx Vwehavex2= e.Bythisand(i)ofTheorem1wegetf(x2) = f(e) = e = [0].Thisisacontradiction. Remark:Onecanshowthatanygroupoforder4isisomorphictoeither Z+4orV.5Denition:Iff: G Hisahomomorphism,deneker f= {x G : f(x) = 1}.Theorem3:Letf: G Hbeahomomorphism. Thenker fisasubgroupofGProof:From(i)ofTheorem1itfollowsthat1 ker f,sincef(1) = 1.Next,ifx, y ker f,thenf(x) = 1 = f(y),hencef(xy) = f(x)f(y) = 1 1 = 1,soxy ker f.Finally,ifx ker f,thenf(x) = 1andsof(x1) = [f(x)]1= 11= 1,hencex1 ker f.Thereforeker fisasubgroupofG. 612 3 41 12 3 42 24 6 83 36 9 124 481216 I II III IVI I II III IVII II IV V I V IIIIII III V I IX XIIIV IV V III XII XV I12 3 41 12 3 42 24 6 83 36 9 124 481216 III II I IVII V I IV II V IIIIV XII V III IV XV IIII IX V I III XIII III II I IVZ+2[0] [1][0] [0] [1][1] [1] [0]Z3[1] [2][1] [1] [2][2] [2] [1]Z+2[0] [1][0] [0] [1][1] [1] [0]Z3[1] [2][1] [1] [2][2] [2] [1][0] [1][1] [2]Z+4[0] [1] [2] [3][0] [0] [1] [2] [3][1] [1] [2] [3] [0][2] [2] [3] [0] [1][3] [3] [0] [1] [2]Z5[1] [2] [3] [4][1] [1] [2] [3] [4][2] [2] [4] [1] [3][3] [3] [1] [4] [2][4] [4] [3] [2] [1][0] [1][1] [2][2] [3][3] [4]Z+4[0] [1] [2] [3][0] [0] [1] [2] [3][1] [1] [2] [3] [0][2] [2] [3] [0] [1][3] [3] [0] [1] [2]Z5[1] [2] [4] [3][1] [1] [2] [4] [3][2] [2] [4] [3] [1][4] [4] [3] [1] [2][3] [3] [1] [2] [4][0] [1][1] [2][2] [4][3] [3]Z+6[0] [1] [2] [3] [4] [5][0] [0] [1] [2] [3] [4] [5][1] [1] [2] [3] [4] [5] [0][2] [2] [3] [4] [5] [0] [1][3] [3] [4] [5] [0] [1] [2][4] [4] [5] [0] [1] [2] [3][5] [5] [0] [1] [2] [3] [4]Z7[1] [2] [3] [4] [5] [6][1] [1] [2] [3] [4] [5] [6][2] [2] [4] [6] [1] [3] [5][3] [3] [6] [2] [5] [1] [4][4] [4] [1] [5] [2] [6] [3][5] [5] [3] [1] [6] [4] [2][6] [6] [5] [4] [3] [2] [1]Z+6[0] [1] [2] [3] [4] [5][0] [0] [1] [2] [3] [4] [5][1] [1] [2] [3] [4] [5] [0][2] [2] [3] [4] [5] [0] [1][3] [3] [4] [5] [0] [1] [2][4] [4] [5] [0] [1] [2] [3][5] [5] [0] [1] [2] [3] [4]Z7[1] [3] [2] [6] [4] [5][1] [1] [3] [2] [6] [4] [5][3] [3] [2] [6] [4] [5] [1][2] [2] [6] [4] [5] [1] [3][6] [6] [4] [5] [1] [3] [2][4] [4] [5] [1] [3] [2] [6][5] [5] [1] [3] [2] [6] [4]Z+4[0] [1] [2] [3][0] [0] [1] [2] [3][1] [1] [2] [3] [0][2] [2] [3] [0] [1][3] [3] [0] [1] [2]V (1) (12)(34) (13)(24) (14)(23)(1) (1) (12)(34) (13)(24) (14)(23)(12)(34) (12)(34) (1) (14)(23) (13)(24)(13)(24) (13)(24) (14)(23) (1) (12)(34)(14)(23) (14)(23) (13)(24) (12)(34) (1)[0] (1)[1] (12)(34)[2] (13)(24)[3] (14)(23)Denition:If (G, )and(H, )aregroups, thenafunction f:G His a homomorphismiff(x y)=f(x) f(y)forallx, y G.Example:Let (G, ) be an arbitrary group andH= {e},thenthefunctionf:G Hsuchthatf(x)=e forany x Gisahomomorphism. Infact,f(x y)=e=e e=f(x) f(y).Example:Let (G, ) beanarbitrarygroup, thenthefunctionf:G Gsuchthatf(x)=x forany x Gisahomomorphism. Infact,f(x y)=x y=f(x) f(y).Example:Letf: Z+Z+2beafunctionsuchthatf(x)=

[0]ifxiseven[1]ifxisoddThenfisahomomorphism. Infact, ifx +yiseven,thenf(x +y)=[0]=f(x) +f(y).Similarly,ifx +yisodd,thenf(x +y)=[1]=f(x) +f(y).Example:Letf:GL(2, R) R=0beafunctionsuchthatf(M)=det MforanyM GL(2, R). Thenfisaho-momorphism. Infact,f(M1M2)=det(M1M2)=det(M1) det(M2)=f(M1)f(M2).Denition:Letafunctionf:G Hbeahomo-morphism. Iffisalsoaone-onecorre-spondence,then f iscalledanisomor-phism. TwogroupsGandHarecalledisomorphic,denotedbyG =H,if thereexistsanisomorphismbetweenthem.Example:WeshowthatR+ =R>0.Infact,letf(x)=ex.Toprove that this is anisomorphism,weshouldcheckthatf:R+R>0isone-onecorrespondenceandthatf(x +y)=f(x)f(y)forallx, y R.Therstpartistrivial,sincef(x)=exisdenedforall x Randits inverse g(x) =ln xis alsode-nedforall x R>0. Thesecondpartisalsotrue,sincef(x +y)=ex+y=exey=f(x)f(y).Denition:Let G= {a1, a2, . . . , an} be a nite group.AmultiplicationtableforGisannnmatrixwhoseijentryisaiaj:G a1a2. . . aj. . . ana1a1a2a1a2. . . a1aj. . . a1ana2a2a1a2a2. . . a2aj. . . a2an...............aiaia1aia2. . . aiaj. . . aian...............anana1ana2. . . anaj. . . ananWewillalsoagreethata1=1.Example:MultiplicativetableforZ5isZ5[1] [2] [3] [4][1] [1] [2] [3] [4][2] [2] [4] [1] [3][3] [3] [1] [4] [2][4] [4] [3] [2] [1]Remark:ItisclearthattwogroupsG= {a1, a2, . . . , an}andH= {b1, b2, . . . , bn}of the same order nare isomorphic ifandonlyif it is possibletomatchele-mentsa1, a2, . . . , anwithelementsb1, b2, . . . , bnsuchthat this one-one correspondenceremains also for corresponding entriesaiajandbibjoftheirmultiplicationta-bles.Example:1. The multiplication tables below showthatP =Z+2=Z3:P even oddeveneven oddodd odd evenZ+2[0] [1][0] [0] [1][1] [1] [0]Z3[1] [2][1] [1] [2][2] [2] [1]2. The multiplication tables below showthatZ+4=Z5:Z+4[0] [1] [2] [3][0] [0] [1] [2] [3][1] [1] [2] [3] [0][2] [2] [3] [0] [1][3] [3] [0] [1] [2]Z5[1] [2] [4] [3][1] [1] [2] [4] [3][2] [2] [4] [3] [1][4] [4] [3] [1] [2][3] [3] [1] [2] [4]3. The multiplication tables below showthatZ+6=Z7:Z+6[0] [1] [2] [3] [4] [5][0] [0] [1] [2] [3] [4] [5][1] [1] [2] [3] [4] [5] [0][2] [2] [3] [4] [5] [0] [1][3] [3] [4] [5] [0] [1] [2][4] [4] [5] [0] [1] [2] [3][5] [5] [0] [1] [2] [3] [4]Z7[1] [3] [2] [6] [4] [5][1] [1] [3] [2] [6] [4] [5][3] [3] [2] [6] [4] [5] [1][2] [2] [6] [4] [5] [1] [3][6] [6] [4] [5] [1] [3] [2][4] [4] [5] [1] [3] [2] [6][5] [5] [1] [3] [2] [6] [4]Theorem1:Letf:G Hisahomomorphismofgroups. Then(i)f(e)=e;(ii)f(x1)=f(x)1;(iii)f(xn)=[f(x)]nforalln Z.(i)Wehavee e=e = f(e e)=f(e),thereforef(e)f(e)=f(e).Multiplyingbothsidesby[f(e)]1, weget[f(e)]1f(e)f(e)=[f(e)]1f(e),whichgivese f(e)=e,sof(e)=e.(ii)Wehavex x1=e = f(x x1)=f(e),sof(x)f(x1)=f(e).Sincef(e)=eby(i),wegetf(x)f(x1)=e.Similarly, fromx1 x=eonecande-ducethatf(x1)f(x)=e.So,f(x)f(x1)=f(x1)f(x)=e,whichmeansthatf(x1)=f(x)1.(iii)Ifn 1,onecanprovef(xn)=[f(x)]nbyinduction. Ifn