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# Honors Algebra I MTH 418H

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Date Created: 09/19/15

Honors Algebra Supplemental Lecture Notes for MTH 418419 20012001 Ulrich Meierfrankenfeld January 137 2002 1 Introduction These notes are about the material I covered in class but can not be found in the text book Herstein 2 AN EXAMPLE FOR THE FIRST ISOMORPHISM THEOREM 2 2 An example for the First Isomorphism Theorem Let S c E C l 0 1 Since labl aHbl for all 011 6 C S is a multiplciative subgroup of Ctt7 De ne gt 3 R7 H 57 7 T H 62 Then 1 clearly is an onto isomorphsim Moreover Mr 1 if an only if r is an integer So ker gt Z and the First lsomorphism Theorem tells us Rng 3 Group Action De nition 31 Let G be a group and S a set We say that G acts on S via j provided that a j G x SHS7 95 H9955 239s afunctz39on b ghs ghsfor allg t ES andsES c essforalls S For example SymQ acts on 9 via 9 w 9M for all g E SymQw E 9 Let G be any group Then G acts on itself by left multiplication g gtk s gs for all 95 6 G C also acts on itself by conjugation 9 s 989 1 for all 97 s E G Lets us verify that this is indeed an action Let 97175 6 G Then 9h 8 9h89h 1 ghsh lg 1 9h8h 1g 1 9 h 8 and esese 1esee Conjugation plays an important role in Group Theory For conVience we de ne 9h ghgil Note that we established the important formula 3 GROUP ACTION 3 ls right multiplication 98Sg an action ghssghandghsgshshg So we see that right multiplication is an action if and only if G is aloelian7 that is if and only if right multiplcation is the same as left multiplication But here is a little trick how to turn right multiplcation into an action which I will call twisted right multiplication De ne gssgil Then 9h 8 8gh 1 8h 19 18h 19 1 h 89 1 9 h 8 and so twisted right multiplication is an action For a further example let G be a group and H any subgroup of G The action of G by left multiplication on GH is de ned as g 8H 98W for all 95 6 G Let T 5H Then 9T98H98hlhEHg8h l hEHgtlt TgT In particular 7 7 is well de ned Also gh T ghT ghT 9 hT and e T T so this is indeed an action Let G be acting on the set S De ne I G a SymS by WW8 g 8 ls this well de ned That is7 is ltIgtg a bijection It is since ltIgtg 1 is its inverse ltIgtg 1o lt1gt98 9 1 9 s 9 1 98 6 8 8 So ltIgtg 1o ltIgtg ids Similarly ltIgtg o ltIgtg 1 ids and so ltIgtg E SymS Note that I is a homomorphism 3 GROUP ACTION 4 190 WWW lt1gt9 1gth8 9 h 8 9h 8 ltIgt9h8 Hence ltIgtg o ltIgth gh Conversely suppose that G is a group7 S is a set and I G a SymS is a homomor phism De ne g s ltIgtgs for all g E G and s E S Let us verify that this is an action of G on S g h 8 lt1gt9 1gth8lt1gt90 WWW lt1gt9h89h 8 and e gtk s es id5s 5 So we see that there is a one to one correpondence between the action of a group G on a set S and homomorphisms from G to Sym Let G be acting on the set S and let I G a SymS be the corresponding homo morphism Let G5 ltIgtG Let 73S be the set of all subsets of S7 the power set of G For a subset T of S so T E 73S and g E G de ne ggtkTggtktlt T It is readily veri ed that this de nes an action of G on 73S We de ne 00T gEGlgttVt T and N0T gEGlgTT Note that N0T C39GT7 where on the right hand side G is Viewd as acting on 73S Proposition 32 Centralizers and Normalizers Let G be acting on a set S and let T Q S 3 GROUP ACTION 5 Proof a Clearly e t t and so e E 00T Let gh E 00T and t E T Then ghtght gttand sotE 00T Alsog 1tg 1gt g lgt e t t Hence 9 1 E 00T and 00T is indeed a subgroup of G b Follows from a since NaT 00T c and d Since 9 t E for all g E N0T we obatin a well de ned map NCTXTEgtT7 gtgtgt Clearly this is an action for N0T on T Let I N0T a SymT be the corresponding homomorphism Then 19 idT if and only ifgt t for all t E T and so CG T ker 11 So c holds The second statement in d now follows from the First lsomorphism Theorem e follows form c and f from d applied with T S Note here that N0S SD Some warning about the notation Na T and 00T It assumes that we do have a xed action of G on S in mind To avoid confusion we will sometimes write NgT7 9 for NC T to indicate underlying action Recall that for G acting on G we de ned three different actions Left Multiplication Twisted right multiplication and conjugation If T Q G we will reserve the symbols N0T and 00T for the action of G on G by conjugation If the action is by left multiplication we use NCT7 LM and C39GT7 LM Let T be a non empty subset of G We claim that C39GT7 LM e lndeed let t E G and g E C39GT7 LM Then gt t and so 9 e Computing H NHT7 LM is a little bit more complicated Note that H g G Also ht E T for all t E T So Ht Q T and we see that H is the largest subgroup of G so that T is the union of right cosets of H In particular if T is a subgroup of G7 NCT7 LM T Let G be acting on S and st E G De ne 5 Et mod G ift gs for someg E G We claim that this is an equivalence relation lndeed s e s and so E mod G is re exive lf tgs then s es g lgs g 1gs g lsst and E mod G is symmetric Finally ifs hr and t 95 then t 9 hr ghr and so E mod G is transitive You might have noticed the similarity in this argument with equivalence realation we used to de ne cosets lndeed7 the cosets are just the specail case of a subgroup H acting by left multiplication on G The equivalence classes are E mod G are called the orbits of G on S The orbit of G on S containg sis G s 9 s l g E G Note that if T is on orbit for G on S then N0T G and G acts on T The set of orbits for G on S is denotes by orbGS We say that G acts transitively on S if G has exactly one orbit on G Lemma 33 transitive Let G be acting on the set S The the following are equivalent a G acts tranisitiuely on S b For all st E T there exists 9 E G with t g gtk s c There exists 5 E T such that for allt E S there exists 9 E G with t g gtk s 3 GROUP ACTION 6 Proof a gt b Let 572 E S Since G has only one orbit on S7 5 E 25 Clearly b implies Finally c implies that all elements of S are in the same orbits as 5 So c implies a D Example Let H g G Then the action of G on GH is transitive lndeed gH g H and so all cosets are in the same orbit as H If G acts on the set G7 we will call S an G set Let S and T be G sets A G morphism gtSHTisafunctionzjSHTsothat gtgs g gts forallge GandsES A G isomorphism is a G morphism which is a bijection S and T are called G isomorphic if there exists a G isomorphism j S a G Lemma 34 orbits and isomorphisms Let 1 S a T be an isomomhism of G sets LetA Q S and B Proof a Let g E G and t E T and put 5 gt 1t Then gtg s g gts g t and so 9 1405 gt71g 25 So a holds b Let g E NaA and b E B Let a 141 Then 9 a E A and applying 1 to both sides 9 gtk b E B So N0A N0B The other inequality follows from a and symmetry c Let g E 00A and b E B Let a 141 Then 9 a a and applying 1 to both sides 9 b I So CO A CaB The other inequality follows from a and symmetry d If A is an orbit for G then A Ga for some a E G Applying 1 we get B G gta and so B is an orbit for G on S Symmetry completes the proof of e By b B is an N0A set and e is now obvious f If A is an orbit G N0A So f follows from e D As an example let S G with G acting by left multiplcation and let T G with G acting by conjunjugation Are these to action isomomorphic Note that G acts transitively on S But geg 1 g for all g in G and so 5 is an orbit for G on S So S and T are not ismomorphic unless G e The next lemma determines all transitive action of G up to isomorphism Proposition 35 classifying transitive actions 3 GROUP ACTION 7 a Let G be acting transitiuely on the set S Lets E S and put H Cas Then the map ltjgtGHgtS7 gHgtggtks is a well de ned isomorphsim of G sets b Let H and K be subgroups of G Then GH and GK are isomorphic as G sets if and only H and K are conjugate in G that is H 9K for some 9 E G Proof a Let h EH Then hs s and so gh s g hs gs So I is well de ned Next we verify that j is a G morphism Let g E G and kH E GH Then gt9 MU gt9kH 910 s 9 lM 8 g gtkH Next let t E S Since G is transitive t g s for some 9 E G Thus gtgH t and j is onto Now suppose that gtgH Then 9 s k s7 k lg s s and fly 6 H Thus gH kH and j is one to one b We will rst compute CG a 9K gK if and only if agK gK Multiplying the last equation by g 1 from the right7 we get a E CagK if and only if a9K 9K Thus 009K 9K Suppose now that H and K are conjugate in K Then H 9K for some 9 E G By the preceeding calculation H 00gK and so by a GH and GK are isomorphic as G sets Suppose now that j GHtoGK is a G isomorphism Then gtH gK for some 9 E G By 34 00H 00gK and thus H 9K So b is proved D De nition 36 Let G be acting on the set S Then a set of representaiues for the orbits of G on S is a subset calR ofS such thatR contains exactly one element from each orbit for G on S Proposition 37 Orbit Equation Let G be acting on the set S and let R be a set of representaiues for the orbits of G on S Then lSl Z lOlZlGCGT OeorbSG r673 Proof Since S is the disjoint union of its orbits7 the rst equation holds For each orbit O for G on S there exists a unique r E R with r E R By 35 O E GCGT and so 0 lGCGr Thus also the second equation For the special case that G is acting on G by conjugation7 the orbit equation is called the class equation 4 S YLOW S THEOREM 8 Proposition 38 Class Equation Let G be a group and R a set of representatives for the conjugacy classes of G Then W ZGCGW ZG Z GCaT r673 reRZG Proof The rst follows from 37 Let g E ZG Then Cg Since R contains an element from the orbit Cg we get 9 E R and so ZG Q R Also 00g G and hence GCGg 1 Thus 2 90060 WWW r6ZG and the second equation follows from the rst D Lemma 39 Products of subgroups Let A and B be subgroups of G a AB is a subgroup ofG if and only if AB BA b IfA N0B then AB is a subgroup of G a Suppose AB is a subgroup of G Then AB BA 1 B lA l BA Suppose next that AB BA Then e ee 6 AB7 ABAB ABAA AABB AABB AB and AB 1 B lA l BA AB So AB is a subgroup of G b If A N0B then aB Ba for all a E A Thus AB BA and AB is a subgroup by a D 4 Sylow s Theorem Let n be a positive integer and p a prime Then np is the p part of n7 that is the largerst power dividing n So np pk for non negative integer k pk diVides n but pk1 does not Lemma 41 n Choose pk Let n and k be positiue integer and p a prime Suppose that pk diuz39des n Then 12 Proof Since kil p 71 i 1 71 7 2 pk pk H pk 72 it suffices to show that the p parts of nit and pk it are identical for all 1 g i lt pk 71 Let 0 lt i lt pk Then pk diVides n and pk and so diVides neither nit nor pkii So the p parts of nit and pkii are at most pk l The lemma now follows since nii7 pkii m71pk D 4 S YLOW S THEOREM 9 De nition 42 An automorphism of a group G is an isomorphism from G to G AutG denotes the set of all automorphism of G Lemma 43 inner automorphism Let G be a group a AutG 239s subgroup of SymG b Forg E G de ne lg G gt G7 a gt 9a Then lg E AutG C Then map G gt AutG 9 Hi9 is a homomorphism with kernel ZG d my My for all 6 AutG and g e G e Inn f Inn G i9 l g E G is a normal subgroup of AutG G g GZG Proof a Clearly id 6 AutG Let mg 6 AutG and 071 6 G Then 11 O gtab ab gta b a gtb 11 O a O gtb Thus 1 o f E AutG Let a 1a and b gt 1b Then gta b gtlta gt ltbgt ab Applying jfl to this equation we obtain 1a gt 1b WWII Thus jfl is a homomorphism and so jfl E AutG and AutG is a subgroup of SymG 0 9W9 gabg l gag lgbg l gay 6 9ha ghagh 1 ghoth 1g 1 9ha and so igh 2 9 0 ih lg id if and only if 9a a for all a E G This is the case if and only if ya ag for all a E G Thus the kernel Ofg Hi9 is ZG 1 Let ag E G and j E AutG Then ma a gt 0 i9 0 VIM gt9 gt 1a9 1 gt9a gt9 1 Ma a i ga SO my i lt gt Since lnnG is the image of the homomorphism g a 2 9 lnnG is a subgroup of AutG By d7 lnnG is normal in AutG f This follows from d and the First lsomorphism Theorem A D 4 SYLOW S THEOREM 10 Theorem 44 Existence of Sylow Subgroups Let G be a nite group p a prime and h a positive integer such that pk divides Then there exists a subgroup S of G with W pk Note that G acts via left multiplication on PpkG7 the set of all subsets of G of size pk Let R be a set of representatives for the orbits of G on 73 By the orbit equation7 lekGl Z lGCGR7LMl R672 Let lGlp kaT By 41 the p part of letGl by 19 Hence there exists R E R so that pT1 does not divide lGGGRLMl Let S 00RLM Then pT1 does not divide and so pk divides S Let r E R Then Sr Q R Since lRl pk we conclude that lSl pk Thus 5 pk a a The maximal p subgroups of a group G are called Sylow p subgroups of G SylpG denotes the set of all Sylow p subgroups of G Theorem 45 Sylow7s Theorem Let G be a nite group of oder n pkm where k and m are positive integers p is a prime andp does not divide m SylpG 75 0 a b Let S g G Then S is Sylow p subgroup of G if and only if lSl pk c Every p subgroup of G is contained in a Sylow p subgroup of G d Any two Sylow p subgroups of G are conjugate in G e The number of Sylow p subgroups of G divides m and is congruent to 1 mod p f Let S E SylpG Then lGNGSl lSylpGl Proof By Lagrange Theorem7 every p subgroup of G has order dividing pk So any subgroup of order pk is a maximal p subgroup This proves the rst part of Also by 447 a holds Since G is nite7 the set of p subgroups containing a given p subgroup of G has a maximal element Hence any p subgroup lies in a maximal p subgroup and c holds We will now show that G acts on SylpG by conjugation Let g E G and S E SylpG Then lgSl lSl and thus 9S is a p subgroup of G Let T be a p subgroup of G with 9S g T Then S g 971T and so by maximality of S7 S 971T and 9S T Thus 9S is indeed a maximal p subgoup of G Let S7 T E SylpG and consider the orbit ST of S on SylpG If S N5T7 then ST is a subgroup of order Emil So ST is a p group and the maximalities of S and T imply S ST S Thus S has a unique orbit of length 1 on SylpG7 namely lfS 7 N5T 4 SYLOW S THEOREM 11 then p divides SN5T So all other orbits have length divisible by p Thus e holds Let 9 be an orbit for G on SylpG Let S E 9 Then 9 is a union of S orbits and so 0 E 1 mod p Suppose that O 7 SylpG and pick S E SylpG 9 Then all orbits of S on 9 have length diVisble by p a contradiction to pf Hence SylpG that is d holds By 44 G has a Sylow p subgroup of order pk By d all Sylow p subgroups have order pk and so also the second direction in b holds Finally f follows from c and 35 D Example Let G be a nite group of order 21 Then the number of Sylow 7 subgroups has has to diVide 3 and is 1 mod 7 So G has exactly one Sylow 7 subgroup S7 and S7 31 G The number 53 of Sylow 7 subgroups diVides 7 and is 1 mod 3 Hence 53 1 or 53 7 Suppose rst that 53 1 Then G has a unique Sylow 3 subgroup S3 S7 S3 1 and so G S7S3 Also S7 S3 S7 S3 1 Hence G is abelian Suppose next that 53 7 Let S3 6 Syl3G Then GNGS3 7 and so N0S3 S3 Since A O B 1 for distinct Sylow 3 subgroup of order we get that 00Sy13G NGA n A 1 AeSy13G AeSy13G Hence G is isomorphic to the image of G in SymSy13G Sym7 So lets us assume that G SymZ7Z Since Sym7 has a unique class of elements of order 7 namely the seven cycles we may assume that S7 0123456 Also since S3 has a unique xed point on Sy137 we may assume that S3 xes 0 and S3 abedef Let s 0123456 and t abedef Then 15 6 S7 and so 15 5139 for some 1 g i g 6 Then 5 tas 5i3 and so i3 E1 mod 7 Thus i 124 lfi 1 s and t commute and s E Nat N0S3 a contradiction Thus i 2 or 4 Replacing t by t 1 we may assume that i 2 So ts sZt Let 1 g k g 7 Then tsk 521 and 5130 k Thus 750 75820 82kt0 823740 8230 2k Hence t 241365 To summarize If G is a group of order 21 with seven Sylow 3 subgroups then G E st SymZ7Z where skgtk1andtkgt2k Corollary 46 Cauchy Let G be a nite group dndp a prime dividing the order of G Then G has an element of order p 4 SYLOW S THEOREM 12 Proof By 44 ther exists a subgroup S of order p in G Let 1 7 s E S Then 1 7 151 and lsl divides lSl p So 5 has order p D Lemma 47 normal sylow psubgroups Let G be a nite group andp a prime Then the following are equivalent a G has a unique Sylow p subgroup b All Sylow p subgroups of G are normal in G c At least one Sylow p subgroup of G is normal in G d t E G l ltl pk for some k E N is a subgroup of G Proof agt 1 Let S E SylpG and g E G Then also 9S is a Sylow p subgroup of G and hence by assumption S 9S Thus S is normal in G bgt c Obvious C a Let S be a normal p subgroup and R any p subgroup By Sylow s theorem7 R 9S for some 9 E G Since S is normal in G we get R S and S is the unique Sylow p subgroup of G To remains to show that a and d are equivalent Let T t E G l ltl pk for some k E N By Lagrange s theorem S Q T for all S E SylpG a d Suppose S is the unique Sylow p subgroup of G Let t E T Then t is a p subgroup of G and so by Sylow s Theorem contained in some Sylow p subgroup of G Thus t E S and so S T d a Suppose T is a subgroup of G Let q be a prime with p 7 1 By de nition of T7 no element in T has order q Thus by 46 q does not diVide T Hence T is a p group Let S E SylpG Since S Q T7 the maximality of T implies S T So T is the unique Sylow p subgroup of G D Lemma 48 Image in SymSylG LetG be a nite group p aprime N CaSylpG and T SylpG Then a N SeSy1pGNGS39 b T 31 G and T is the unique Sylow p subgroup of G 4 SYLOW S THEOREM 13 c N o 5 T for all T e SylpG d The map SylpG a SylpGN 5 a 5NN is a well de ned bijection e Let 5 lSylpG Then GN is isomorphic to a subgroup of Symsp 51 l lGNl lGNl 1 221 Proof Let S E SylpG a is clear since 00S N0S b By a and sicne S N0S T g N Let A be a p subgroup of N By de nition of N A N0S Thus AS is a p subgroup of G and so by maximality of G A g S Since S was an arbitrary Sylpw p subgroup A g T Thus T is the unique maximal p subgroup of G and b holds c Notethat TgSandso TgNoS N Sisap subgroup ofGand soN SgA and N O S A d By bc and Sylows Theorem we compute lSNNl sNm Sl lSTl lGNlp W W 1 So 5NN e SylpGN Let 5 e SylpG with 5N 5N Then 5 5N NG5 and so 5 55 5 We conclue that the map in question is one one Now let AN E SylpGN and choose S E SylpA Since lAlp lGlp S E SylpG By what we proved lSNNl lGNlp lANl and so SNN AN So the map is also onto e By d 517 lSylpGNl and so 517 diVides lGNl Also GN is isomorphic to subgroup of SymSylpG E Symsp So by Lagrange lGNl diVides 5171 D De nition 49 A non tiuial group is called simple if it has no poper normal subgroups Lemma 410 Simple abelian groups Let G be a simple group If G is a abelian or a nite p group then G ZpZ for some prime p Proof If G is a nite p group then by a a homework problem ZG 7 e Since ZG is a normal subgroup of G and G is simple we conclude G ZG So G is abelian We may assume from now on that G is abelian Then every subgroup of G is normal and thus G has no proper subgroups Thus G g for all e 7 g E G In particular G E ZnZ for some n E N lfm 6 2 diVides n and 17 m 7 n then mZnZ is a proper subgroup of ZnZ Thus n is a prime D Lemma 411 The number of sylow p subgroups for simple groups LetG be a non cyclic simple group and p a prime diuisor of Then G is isomorphic to a subgroup of Symsp In particular lGl diuides spl 5 SYM5 AND SYM6 14 Proof Let S be a Sylow p subgroup of G Then lSl lGlp 34 1 Suppose that S 31 G Then as G is simple7 G S Thus by 410 G is cyclic7 a contradiction to the assumption Thus S is not normal in G Hence N0S lt G Let N 00SylpG By 48a7 N NaS and so N 34 G Since N is normal in G we conclude7 N 5 Thus by 48e G E Ge is isomorphic to a subgroup of Symsp So by Lagrange s theorem lGl divides l Symspl spl D 5 Sym5 and Sym6 In this section compute the action of K Sym5 on Sy15K and then construct an automorphism of G Sym6 which is not an inner automorphism Let F be a Sylow 5 subgroup of G and e 34 9 E G Then G is a ve cycle and so 9 dies not x 1 Hence GF1 e and lg1 lg E 5 Thus ther exists a unique 9 E F with 91 2 Then 9 12a7b76 with 171976 345 Also F 57979279727971 and there exists a unqie Hence there exists 24 elements of order 5 in K In particular7 l Sy15Kl 6 The following tables list the six Sylow 5 subgroups explicitly F1 F2 F3 F4 F5 F6 12345 12354 12435 12453 12534 12543 13524 13425 14523 14325 15423 15324 14253 15243 13254 15324 13245 14235 15432 14532 15342 13542 14352 13452 We now compute the action of 12 on Sy15K l 1 g 2 g 6 12gt12345 21345 13452 6 F6 So 12 4 F1 F6 Since 12 as order 2 this implies 12 4 F6 F1 12gt12354 21354 13542 6 F4 Thus 12 4 F2 F4 and 12 gtk F4 F2 Also 12gt12435 21435 14352 F5 and 12 gtk F2 F4 Let I be the homomorphism obtained from the action of K on Sy15K composed with natural isomorphism from SymE l 1 g i g 6 to G Sym6 Thus 5 SYM5 AND SYM6 15 FltIgtgi 9339 1 Then we have M12 162435 23gt12345 13245 6 F5 and 23gt12354 13254 6 F3 Since ltIgt12 is a triple 2 cycle and ltIgt12 and ltIgt23 are conjugate also ltIgt23 is a triple 2 cycle Thus lt1gt23152346 34gt12345 12435 6 F3 and 34gt12354 12453 6 F4 and so lt1gt34132456 45gt12345 12354 6 F2 and 45gt12435 12534 6 F5 and so ltIgt45 123546 We summarize these results in the following table l g l 12 l 23 34 l 45 l l 449 l 162435 l 152346 l 132456 l 123546 l 1 Now that we know ltIgtii l 1 for al 1 g i g 4 we are able to compute ltIgtg for any 9 E K As an example we compute ltIgtg for one 9 from each conjugacy class in K lt1gt123 lt1gt12 o 23 lt1gt1 lt1gt23 162435 o 152346 134256 2 lt1gt1234lt1gt123lt1gt34 134256o 132456 145236 35 2 AA lt1gt12345 lt1gt1234lt1gt45 1452 o 12 46 124653 lt1gt1234 lt1gt12lt1gt34 16 435 o 132456 1635 o 1356 1536 lt1gt12345 ltIgt123ltIgt56 134256 o 123546 154236 We summarize these results in the following table Let L ltIgtSym5 Since Sym5 12 23 34 45 we get that L lt1624357 1523467 1324567 123546gt

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