BA AB =Abelian GroupEveryoneMultiplication table of C2vgroupvovo'2Cvovo'vovovo'vo'2VC2C2C2Cvo'vovo'vo2C2Cvovo'3C23Cvo''vovo'3C23Cvo''3C23C3Cvo''vovo'23C23C3Cvo'vo''vovovo'vo''vo3C23Cvo'vo''vo'vo23C3Cvo''vo''vovo'3C23C3VCBA AB =There is oneNon-AbelianGroupSubgroup Multiplication table of C3vgroupvo'' vovo '3C Definition of an abstract group Multiplication table for a finite group Cyclic groups Subgroups II. II. Definitions and theorems Definitions and theorems of group theoryof group theoryAB B A IIII--1. Definition of an abstract group 1. Definition of an abstract group Abstract Group:GGroup elements:G=(A, B, C, D, )Multiplication rule: Finite Group( )hA A A G , , ,2 1h is called order of GAbelian GroupAny two elements in G:BA AB =Non-Abelian GroupIf there are two elements in G:BA AB =To form an abstract groupThere are very strict mathematical laws to be satisfied1. Closure If AG ,B G AB Ge e e2. Associative operation (AB)C=A(BC)3. Unique identity element: E GEA=AE=Ae4. Unique inverse element: A-1GA-1A=AA-1=EeE CC EE C CC E EC E Cv v vv v vv vv vv v V222 222 2o o oo o oo oo oo o' '''''ExampleOnce multiplication table is established, rules 1, 3, and4 are easily checked.But rule 2 is not in the table|||||.|

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\|'ECCCECCEvvvvvvv22222oooooooAssociative operationE C CC E CC C EC E C CE C C CC C E EC C E Cv v v vv v v vv v v vv v vv v vv v vv v v V3232333233232323 3 323 323 3 3o o o oo o o oo o o oo o oo o oo o oo o o' ' ' ' '' ' ' '' ' '' ' '' ' '' ' '' ' 'Check associative rule for C3vgroup by calculatingprovevA o =(((((((((

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\|=1 00 1E(1) Prove that 3 2 1A A A Edonot form a Group(2) ( )3 2 1 3 2 1 iA A A E iA A A E G =forms a GroupCalculate multiplication table.(3)Is group in (2) Abelian group?GTAchem2-2 Prove that ( )1 1 1 1 11 = A B C X Y XY ABC D ABC =1 1 1 = A B DC E1 1 = B DC A1 = DC AB( )11= D ABC1 1 1 1 = A B C DExampleIIII--2. Multiplication table for a finite group2. Multiplication table for a finite grouph 1 - h 4 3 2 1AA A AAA h1 - h4321AAAAAAGh h 1 - h h 4 h 3 h 2 h 1 hh 4 1 - h 4 4 4 3 4 2 3 1 4h 3 1 - h 3 4 3 3 3 2 3 1 3h 2 1 - h 2 4 2 3 2 2 2 1 2h 1 1 - h 1 4 1 3 1 2 1 1 1A A A A A A A A A A A AA A A A A A A A A A A AA A A A A A A A A A A AA A A A A A A A A A A AA A A A A A A A A A A Ah 1 - h 1 - h 1 - h 4 1 - h 3 1 - h 2 1 - h 1 1 - hA A A A A A A A A A A A Arrange (element of column) * (element of row)E A1 =h 4 1 - h 4 4 4 3 4 2 4 1 4A A A A A A A A A A A A (1) Each row in the table lists each element in G once and only oncej i j 4 i 4A A A A A A = =wrong(2) No two rows may be identicalh j 1 - h j 4 j 3 j 2 j 1 jh i 1 - h i 4 i 3 i 2 i 1 iA A A A A A A A A A A AA A A A A A A A A A A Aj i 3 j 3 iA A A A A A = =wrong(3) Each row is a rearranged list of the group elementsh-order G; there are only h different elementsThe above conclusion is held for column as wellExample: Order h = 1E A1 =Example: Order h = 2A EAEAA E=AEAAOKwrongEGE A AA AAA = = 1 1( ) B A E G =Example: Order h = 3Based on definition of Group=BAEA2Case 1Case 2B A EBAEAGCase 1BwrongB A EBAEGCase 2EEOKBE AB A EBB AB A EExample: Order h = 4C B A ECBAECBAC B A E=CBEA2GStaring from the second rowCase 1Case 2Case 3C B A ECBAECBE AC B A EG1Case 1B CBCB C CC B BB C E A AC B A E EC B A E G1aE AA EB C CC B BB C E A AC B A E EC B A E G1bA EE AAbelianOKOKC B A ECBAECBB AC B A EG2Case 2E CECE C CC B BE C B A AC B A E EC B A E G2B AA EAbelianOKC B A ECBAECBC AC B A EG3Case 3B C CE B BB E C A AC B A E EC B A E G3E AA CB EBEAbelianOKB C CC B BB C E A AC B A E EC B A E G1aE AA E( ) 24G G1a =Page 11E C CC B BE C B A AC B A E EC B A E G2B AA E( ) 14G G2 =Page 11Only two different Groups SummaryG2 G3 G1b = =We can proveProveC B A E' ' ' 'CBAE'''CBAC B A E'''' ' ' 'C B A ECBAECBAC B A EG3E C B' ' 'B E CB A EA E C' ' '' ' 'E A BA C EC B A ECBAECBAC B A EE C BB A EA E CG2B CA BC AE E''''If we put Same G3Prove that G1b group is the same as G2 groupE C CC B BE C B A AC B A E EC B A E G2B AA EB C CC B BB C E A AC B A E EC B A E G1bA EE AGTAchem2-3 What about order h = 5Classification from cycle-group method will help a lot { }h 1 - h 4 3 2 1AA A AAE, A G = =It must have h n in whichA anyforE Ainis =ProveE A E, A E, A , E, A E, A1 h h n 2= = = = =+ ( )E AA Am - nm n= > = m nContradiction For any group (order h) GE A =1h s 1Suppose there is A that satisfies This is order-h group, so it can not have h+1 different elementsn-m h sIIII--3. Cyclic group3. Cyclic groupOrder-hgroup has h different elements h 1 - h 4 3 2 1AA A AAE, A =E A A, A A A A , A A, Ah2 h1 - h2 1 - h32 422 3 2= = = = = We call this group as cyclic groupExample: Order h = 2A EAEAA EE AA,2=EGOnly two choices Example: Order h = 3 B A E3 2AA A B A2=E A3=2A A E A2=B A A 2Can not form groupB A EBAEBE AB A EGOKIf we take B as example, we have the same conclusionX is one of A, B, and C Example: Order h = 4 C B AEE XX XX4 3 2= Y E X X X3 2=Z Y E X X2=C B A ECBAECAEBEB AC B A EG OK always Can not form groupMay be OK Z YE XX2=C B A ECBAECBE AC B A EGOnly choiceBCBCOnly choiceA EE AE AA ELeft two choices Same as OKnewCreate only one tableOKE X X XX4 3 2={ }g h 2 1g 1 - g 3 2A A AE X XXXX= = GOrder-hG:any{ } E X XXXX Ag 1 - g 3 2 1i= eNote: all elements are different in GSuppose n -1iX A =g m n ifE X Xm n= + ={ } E X XXXX X Ag 1 - g 3 2 mi= e = { } E X XXXX Ag 1 - g 3 2i= e ProvemiX A ={ }g h 2 1g 1 - g 3 2A A AE X XXXX= For Order-hG:= == =j i E A Ai i E AAj i2iiOnly two choicesas unique inverse Exercise 2.3 (page 16)Order h = 5{ } DC B AE G =X is one of A, B, C , DE X X X XX5 4 3 2=This is always OK.Make multiplication table(1) Z Y E X XX3 2= Z Y T E XX2=Y E X X XX4 3 2= (2)(3)(4)Prove (2), (3) and (4) are not OK to form a GroupGTAchem2-4 E C CC B BE C B A AC B A E EC B A E G2B AA EB C CC B BB C E A AC B A E EC B A E G1aE AA EIIII--4. Subgroup4. SubgroupExample of order-h = 4 Four subgroups: (E)(E,A)(E,B)(E,C)Two subgroups: (E)(E,B)Cyclic group E X XX X4 3 2=(E) trivial subgroupOrder-hGroup: G(A)() { }h 1 - h 4 3 2 1AA A AAE, A A G = =If there are g elements in G that form a subgroup: G(B)The order of any subgroup g must be divisor of the group hg 3 2 1B B BE B =We call that G(B) is subgroup of G(A) Order-h Order-gh/g = k must be an integerProve: () { }h 1 - h 4 3 2AA A AAE, A G =( ) { }g 3 2B B BE B G =(1) If( ) B G A eany( ) B G ABi e( ) B G A B B A B AB If-1i k k ie = = contradiction(2) Ifk iB B If =k iAB AB =k i k iB B AB AB If = =contradiction{ }g 3 2AB AB ABAE (3) so g different elements(4) Take ( ) { }g 3 2AB AB ABAE A and B G A e'e'{ }g 3 2B A B A B AE A' ' ' 'the other g different elementsm-1i k k iAB B AB A AB B A If = =' ='{ }g 3 2AB AB ABAE A e'Contradiction kg = h, k must be an integerF D C B A EFDCBAED E AC BFE FB A C DB A EF D CA C D E FBC B F D E AF D C B A E(2) Take(2)6GE A A2=ExampleA E C E B EOne order-3 subgroup k=6/3=2(3) TakeE DF D F DD3 2= = =F D EThree order-2 subgroup k=6/2=3(1) Take E(4) Take F(2)6GisOrder 6-Group has multiplication table ( )D C B A E F FC B A E F D DB A E F D C CA E F D C B BE F D C B A AF D C B A E EF D C B A E G16Find out how many subgroupsfrom the above table(1)6G isGTAchem2-5 ExampleShow that there can be only one group of order h, where h is a prime number { }hA A A E A G , , , ,3 2 1 = =If h is prime number ()Only one{ } E X X X X Gh h= =, , ,1 2Exercise 2.2 (page 16)Why can we not have a group in whichE B A = =2 2Prove that this exercise is wrongGive an example where group does haveE B A = =2 2GTAchem2-6 Exercise 2.4 (page 16)Exercise 2.6 (page 16) Show that for any cyclic group, { } E X X X X Gh h= =, , , ,1 2there must be one subgroup corresponding to each integral divisor of the order h. Give an example.Ifnm=h, prove that there are() { } E A A A A A Gn n= =, , , ,1 2() { } E B B B B B Gm m= =, , , ,1 2( ) { } E X X X X G B Ah h= = e, , , , ,1 2GTAchem2-7 X is one of A, B, C , D , FExample: Order h = 6 FDC B AEE X X XXXX6 5 4 3 2= Z Y T E XXX3 2=Z Y T SE XX2=Y E X XXXX5 4 3 2=OKNot OK Z Y E X XXX4 3 2=Not OK