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A group may be defined as follows. 3. G has an identity element e. There is an e in G such that x * e 5 e * x 5 x for all x [ G. Groups ! Week 4 ition 3.1 ■4. G Group contains inverses. For each a [ G, there exists b [ G such that a * b 5 b * a 5 e. Definition 3.1 Group Suppose the binary operation p is defined for elements of the set G. Then G is a group with Thetophrase “with the respect to p” should be noted.hold: For example, the set Z of all integers respect p provided following four conditions is a group with respect to addition but not with respect to multiplication (it has no inverses 1. G is closed under p.61). That Similarly, is, x [ G and [G G5 imply in G. with respect to elements other than the yset {1, that 21}x is * yaisgroup Chapter 3 for Groups multiplication but notFor withallrespect 2. p is associative. x, y, z to in addition. G, x * (y *Inz)most 5 (xinstances, * y) * z. however, only one binary operation is under consideration, and we say simply that “G is a group.” If the binary 3. G has isanunspecified, identity element e. There is an e in Gnotation such that 5 ejuxtaposition all * e the * x 5 x for137 operation we adopt the multiplicative andxuse xy x [ G. to indicate the result of combining x and y. Keep in mind, though, that the binary operation 4. not G necessarily contains inverses. For each a [ G, there exists b [ G such that a * b 5 b * a 5 e. is multiplication. ition 3.2 tion 3.2 phraseGroup “with respect to p” should be noted. For example, the set Z of all integers ■ The Abelian Definition 3.2 Abelian is a group with respect toGroup addition but not with respect to multiplication (it has no inverses Letelements G be a group to p. Then the G isset called commutative group,with or anrespect abelianto† for otherwith thanrespect 61). Similarly, G 5a {1, 21} is a group group, if p is but commutative. That is, for all x, y in G.however, only one binary * y 5 y *Inxmost multiplication not with respect to xaddition. instances, operation is under consideration, and we say simply that “G is a group.” If the binary Remark is unspecified, we adopt the multiplicative notation and use the juxtaposition xy operation 1 result We can obtain some simple examples of groups bythat considering toExample indicate the of combining x and y. Keep in mind, though, the binaryappropriate operation subsets of the familiar number systems. is not necessarily multiplication. 1) The set of integers is a group under the OPERATION of addition: a. The set C of all complex numbers is an abelian group with respect to addition. ■We Abelian Group already integers undernumbers the OPERATION of addition are respect CLOSED, b. have The set Q ! seen {0} ofthat all the nonzero rational is an abelian group with to ASSOCIATIVE, have IDENTITY 0, and that any integer x has the INVERSE −x. Because multiplication. Let group with respect to p.satisfies Then Gall is called a commutative group, an abelian† the G setbeofa integers under addition four group PROPERTIES, it is or a group! + c. The of all positive real abelian with respect to multiplicagroup, if psetisRcommutative. That is,numbers x * y 5 yis*an x for all x,group y in G. tion, but it is not a group with isrespect addition (it hasitno additive identityalland no 2) The set {0,1,2} under addition not a to group, because does not satisfy of the group PROPERTIES: it does not have the CLOSURE PROPERTY (see the previous additive inverses). ■ lectures to see Therefore, the simple set {0,1,2} under of addition notconsidering a group! appropriate Example 1 why). We can obtain some examples groupsisby The examples some indication of the great variety there is in groups. subsets offollowing the familiar numbergive systems. (Notice also that this set is ASSOCIATIVE, and has an IDENTITY which is 0, but does not have theset INVERSE because and −2group are on not set!) a. The C all PROPERTY complex numbers is an−1 with respect to addition.mapping Example 2 ofRecall from Chapter 1 that a abelian permutation ainsetthe A is a one-to-one fromThe A onto A and thatofS(A) set of all permutations on A. We havewith seenrespect that S(A) b. set Q ! {0} all denotes nonzerothe rational numbers is an abelian group to 3) The setwith of integers under subtraction is not a group, because it doesand notthat satisfy is closed respect to the binary operation + of mapping composition the multiplication. all of the group PROPERTIES: it does not have the ASSOCIATIVE PROPERTY (see the operation + is +associative. The identity mapping IA is an identity element: c. The set R of all real numbers is abelian group with to multiplicaprevious lectures to positive see why). Therefore, theanset of integers underrespect subtraction is not a f + IA 5tof 5 IA + f (it has no additive identity and no group! tion, but it is not a group with respect addition inverses). ■ for additive all f [ S(A), and each f [ S(A) has an inverse in S(A). Thus we may conclude from (Notice also that this set is CLOSED, but does not have an IDENTITY and therefore also results infollowing Chapter that S(A) PROPERTY.) is a group respectoftothe composition of mappings. The examples give some with indication great variety there is inHowever groups. does not have the 1INVERSE S(A) is not abelian since mapping composition is not a commutative operation. ■ Example from Chapterunder 1 that addition a permutation onaa set A is abecause one-to-one mapping 4) The set2 of Recall natural numbers is not group, it does not satisfy all of theWe group PROPERTIES: does notobtain have an theexplicit IDENTITY PROPERTY (see Example shall take A 5the {1,set 2,itof 3} ofthat S(A) . In from A onto A3and that S(A) denotes alland permutations on A. Weexample have seen S(A) previous totosee Therefore, off (1), natural numbers isthe closed withlectures respect binary + set of mapping composition andaddition thatthree theis order to define an element fthe ofwhy). S(A) , weoperation need tothe specify f (2), and f (3).under There are not a group! operation + is associative. identity mapping IA is anthere identity possible choices for f (1).The Since f is to be bijective, areelement: two choices for f (2) after f + IAASSOCIATIVE, 5 f 5 IA + f but does not have the INVERSE (Notice also that this set is CLOSED, † The term abelian is used innone honor of Nielsnegative Henrik Abel (1802–1829). A biographical of Abel appears on PROPERTY because are in the set.)wesketch for all f [ S(A), and each fof[the S(A) has annumbers inverse in S(A). Thus may conclude from the last page of this chapter. results in Chapter 1 that S(A) is a group with respect to composition of mappings. However S(A) is not abelian since mapping composition is not a commutative operation. ■1 Example 3 We shall take A 5 {1, 2, 3} and obtain an explicit example of S(A). In Groups ! Week 4 5) The set of rational numbers with the element 0 removed is a group under the OPERATION of multiplication: We have already seen that the set rational numbers with the element 0 removed under the OPERATION of multiplication is CLOSED, ASSOCIATIVE, have IDENTITY 1, and that any 1 integer x has the INVERSE . Because the set of rational numbers with the element 0 x removed under multiplication satisfies all four group PROPERTIES, it is a group! 6) The set of rational numbers (which contains 0) under multiplication is not a group, because it does not satisfy all of the group PROPERTIES: it does not have the INVERSE PROPERTY (see the previous lectures to see why). Therefore, the set rational numbers under multiplication is not a group! (Notice also that this set is CLOSED, ASSOCIATIVE, and has an IDENTITY which is 1.) 7) The set of rational numbers under division is not a group, because it does not satisfy all of the group PROPERTIES: it does not have the ASSOCIATIVE PROPERTY (see the previous lectures to see why). Therefore, the set of rational numbers under division is not a group! (Notice also that this set is not CLOSED because anything divided by 0 is not in the set, does not have an IDENTITY and therefore also does not have the INVERSE PROPERTY.) 8) The set of natural numbers under division is not a group, because it does not satisfy all of the group PROPERTIES: it does not have the IDENTITY PROPERTY (see the previous lectures to see why). Therefore, the set of natural numbers under division is not a group! (Notice that this set does not have the CLOSURE, ASSOCIATIVE or INVERSE PROPERTIES.) 9) The set of integers under multiplication is not a group, because it does not satisfy all of the group PROPERTIES: it does not have the INVERSE PROPERTY (see the previous lectures to see why). Therefore, the set of integers under multiplication is not a group! (Notice also that this set is CLOSED, ASSOCIATIVE, and has the IDENTITY ELEMENT 1.) Definition A permutation of a set is a bijection (one to one and onto) of this set onto itself, i.e., S ( A ) = { f f : A → A and f is bijective } . Example* 2 7 Permutations Permutations Groups 7 ! Week 4 The aim of this section is to discuss a large class of groups, called permutation groups. The aim ofThe thiscentral sectionresult is toofdiscuss a largeis class of groups, called permutation this section Cayley’s theorem, which states thatgroups. actually any group The centralis result of thisinsection is Cayley’s theorem, states that actually group example of a contained this class of group. Thus, which permutation groups give any a generic is contained in this class of group. Thus, permutation groups give a generic example of a group. group. 7.1 7.1 Permutations Homework*. Take A = {1,2, 3} and obtain an explicit example of S ( A ) . Permutations Definition 7.1. A permutation of a set is a bijection of this set onto itself. Definition 7.1. A permutation of a set is a bijection of this set onto itself. Note that Note that (i) if α : X → X is a permutation, than the inverse function α−1 : X → X is also a (i) if αpermutation; : X → X is a permutation, than the inverse function α−1 : X → X is also a permutation; (ii) if α : X → X and β : X → X are permutations, one can form a composition (ii) if αα :◦ X X X, andwhich β : Xis → permutations, one can form a composition β :→ X→ alsoXa are permutation. α ◦ β : X → X, which is also a permutation. Theorem 7.2. Let A be a nonempty set, and let S be the collection of all permutations of TheoremTheorem 7.2. Let A be nonempty set,the and let S let be the collection of all permutations of of A. Then G a nonempty set, and G ◦. be collection of all permutations A. Then SLet isAaabe group under composition A. Then Sis isa agroup groupunder underthe thecomposition composition ◦. Proof. Clearly, ◦ defines a binary operation on S. We need to check the three axioms G1, Proof. Clearly, ◦ defines a binary operation on S. We need to check the three axioms G1, G2, G3 of the group. G2, G3 of the group. (G1) We have seen in section 1 that composition is associative. (G1) We have seen in section 1 that composition is associative. (G2) Consider the mapping ι : A → A such that for all x ∈ A, ι(x) = x. Then for any (G2) Consider the mapping ι : A → A such that for all x ∈ A, ι(x) = x. Then for any α ∈ S one has α ◦ ι = ι ◦ α = α. Thus, ι is the identity element on S. α ∈ S one has α ◦ ι = ι ◦ α = α. Thus, ι is the identity element on S. −1 −1 −1 (G3) For any permutation α ∈ S, its −1 inverse α −1 satisfies α −1◦ α = α ◦ α (G3) For−1any permutation α ∈ S, its inverse α satisfies α ◦ α = α ◦ α = ι. Thus, = ι. Thus, is the of α with respect the group operation. α−1 is the α inverse of αinverse with respect to the grouptooperation. Example**. Construct the Cayley (multiplication) table for S(A) in Example*. Below we will denote composition of permutations α and β by αβ rather Notation Notation Below we will denote composition of permutations α and β by αβ rather than α ◦ β.than α ◦ β. 7.3.A be1.any Letset. A be anygroup set. of The of all permutations DefinitionDefinition 7.3. 1. Let The all group permutations of A is calledofa A is called a on A and symmetric symmetric group on Agroup and denoted by denoted Sym(A). by Sym(A). 2. Anyofsubgroup Sym(A)Sym(A) (including Sym(A) itself) is called a group permutation group on 2. Any subgroup Sym(A) of (including itself) is called a permutation on the set A. the set A. 3. Iffinite A isset the{1, finite . . , n} of nthen elements, Sym(A) is commonly denoted 3. If A is the 2, . . .set , n}{1, of 2,n . elements, Sym(A)then is commonly denoted Sn .already We have already dealt with Sthe Sn . We have dealt with the groups n . groups Sn . 2 ···n. · · · (n − 1) · n. Proposition 7.4. of The is group n! = 1 S· n2 ·is· ·n! ··= (n 1−· 1) Proposition 7.4. The order theorder groupofSnthe Proof. TheProof. problem of problem computing number the of elements theproblem same as the problem The of the computing number in of Selements in Snasisthe n is the same of computing the numberthe of number different of ways the integers 1, 2,integers . . . , n can be. placed in the n of computing different ways the 1, 2, . . , n can be placed in the n Homework**. Construct the Cayley (multiplication) table for S(A) in Homework*. blanks indicated each(using integereach just integer once): just once): blanks(using indicated ! !" " Example 4 (Homework) 1 2 . . . n1 .2 . . . n . ... ... 25 25 3 igure 3.3 igure 3.3 igure 3.4 If the elements in the row headings are listed in the same order from top to bottom as the in all positions that are symmetrically placed relative to the diagonal from upper left to lower elements in the column headings are listed from left to right, it is also possible to use the table and inverses elements. element e is athe lefttable identity if and 3.2 onlyisifnot thesymmetric. row headed For by e right. In Example 3, theofgroup is notAnabelian since in Figure to check foratcommutativity. The operation is commutative if and only if equal elements appear 2 2 leftdend reads the! same as r the+column table. Similarly, e is a4 Groups Week example, g + the r 5 is in rowexactly 5, column 3, and g 5 sheadings is in rowin3,the column 5. in all positions are symmetrically from upperthe leftsame to lower rightthat identity if and only if theplaced columnrelative headed to bythe e atdiagonal the top reads exactly as the right. In Example 3, the group is not since table 3.2 is not For row in the the set table. it exists, the the inverse ofina Figure certain element asymmetric. can be found by Example 4 2headings Let G be of Ifabelian complex numbers given by G 5 {1, 21, i, 2i}, where 2 example, gsearching + r 5 dfor is in 5, column andheaded r + gby5asand is in rowin3,the column therow identity e in the3,row again column5.headed by a. i 5 !21 , and consider the operation of multiplication of complex numbers in G. The If the elements in the row headings are listed in the same order from top to bottom as the table in Figure thatset Gheadings is respect totomultiplication. elements inGthe arewith listed from left right, also use the table Example 4 3.3 Letshows becolumn the ofclosed complex numbers given by itGis5 {1,possible 21, i,to2i}, where Multiplication in G is associative and commutative, since multiplication has these propcheck for commutativity. The operation is commutative and only ifnumbers equal elements i 5 !21 , toand consider the operation of multiplication of ifcomplex in G.appear The erties in theinset of all complex numbers. We placed can observe from Figure 3.3 that 1 isleft thetoidenall positions that are symmetrically relative to the diagonal from upper lower table in Figure 3.3 shows that G is closed with respect to multiplication. tity elementright. andInthat all elements haveis not inverses. of 1table andin21 is its andFor i Example 3, the group abelianEach since the Figure 3.2own is notinverse, symmetric. Multiplication gin+Gr 2is5associative and commutative, sincesmultiplication has these propd is in Thus row 5,G column 3, and r2group + g 5 with is inrespect row 3, column 5. and 2i are example, inverses of each other. is an abelian to multiplication. erties in the set of all complex numbers. We can observe from Figure 3.3 that 1 is the identity3element and21 that all have Each of 1 and its {1, own inverse, i 4 elements Let the setinverses. of complex numbers given21 by is G5 21, i, 2i}, and where 1Example i G be2i i 5 !21 , and consider the operation of multiplication of complex numbers in G. The and 2i are inverses of each other. Thus G is an abelian group with respect to multiplication. in Figure 3.3 1 1table 21 i shows2ithat G is closed with respect to multiplication. Multiplication 3 1 21 i in G is 2iassociative and commutative, since multiplication has these prop21 21erties in1the set2i i numbers. We can observe from Figure 3.3 that 1 is the idenof all complex 1 1tity element 21 and that i 2i have inverses. Each of 1 and 21 is its own inverse, and i i i 2i 21 all elements 1 and 2i are inverses of each other. Thus G is an abelian group with respect to multiplication. 21 1i 2i 2i 21 2i 1 21i 21 i 3 2i 1 21 i 1i 2i 21 i 2i 2i 2i 1 21corollary of Theorem 2.28 that the set Example 51 It isi 1an immediate Example 5 (Homework) 21 21 1 2i i i i 2i 2i 2i i Zn 5 5 304, 314, 324, c, 3n 2 14 6 21 1 1 21 Example 5 classes It is anmodulo immediate corollary of Theorem 2.28 respect that thetosetaddition. of congruence n forms an abelian group with ■ ■ ■ ■ Z 5 5 304, 314, 324, c, 3n 2 14 6 Example 6 Let G 5 {e, a, b,n c} with multiplication as defined by the table in Figure 3.4. of congruence classes modulo n forms an abelian group with respect to addition. ■ ■ Figure 3.3 be ca e Example 5 It is an immediate corollary of Theorem 2.28 that the set Example 6 Let G 5 {e, a, b, c} with multiplication as defined by the table in Figure 3.4. b c Zn 5 5 304, 314, 324, c, 3n 2 14 6 Example 6 Let G 5 Example 6 (Homework) {e, a, b, c} with multiplication as defined by the table in Figure 3.4. ■ e e of congruence a b cclasses modulo n forms an abelian group with respect to addition. a? ae ba cb ec ? be ca a ca eb eb ac ac b e b ? e a e e a b c c b b c e a a b c e From the table, wea observe the following: c c be ba c b e a igure 3.4 1. G is closed under this multiplication. c c e a b ■ Figure From the table, we observe 2. e 3.4 is the identity element.the following: From the table, wemultiplication. observe the following: 1. G is closed under this 2. e is the 1.identity element. G is closed under this multiplication. 2. e is the identity element. 4 Groups 3. Each 3.1 Definition of a Group 141 3.1 Definition of a Group 141 Week 4 of e and b is its own inverse,! and c and a are inverses of each other. 3.1 Definition of a Group 141 This of multiplication commutative. 3.4. Each e and b is itsisown inverse, and c and a are inverses of each other. This multiplication is also associative, but we shall not verify it here because it is a labori4. This multiplication 3. Each of e is andcommutative. b is its own inverse, and c and a are inverses of each other. ous task. It follows that G is an abelian group. ■ 3.2 Properties of Group Elements 145 4. This is multiplication is commutative. This multiplication also associative, but we shall not verify it here because it is a laborious task. It follows thattable G is in an abelian group. ■ This is also associative, but we shall not verify it here because it isset a laboriExample 7 multiplication The Figure 3.5 defines a binary operation p on the S5 33. Let G be the set of all matrices in M3(R) that have the form ous task. It follows that G is an abelian group. ■ {A, B, C, D}. 1 a b Example 7 The table in Figure 3.5 defines a binary operation p on the set S 5 Example 7 The table C 0 in 1Figure c S 3.5 defines a binary operation p on the set S 5 {A, B, C, D}. * A {A, B B, C, C D}. D 0 0 1 realAnumbers B a, b, and c. Prove or disprove that G is a group with respect *A for arbitrary AB BC * C A DB C D to multiplication. AB ProveBCor disprove CD BACthe set A ABBthat A GBin Exercise 32 is a group with respect to addition. 34. B BCCthat B GAin Exercise 33 is a group with respect to addition. BC ProveCAor disprove DB ADDthe set 35. C C A D B theCpower D set p(A) was defined in Section 1.1 by p(A) 5 36. A, Dset D CDFor an AA arbitrary BB {X 0 X 8 A}, and addition in p(A) was defined by D A B D D D 3.5A B D D X 1 Y 5 (X c Y) 2 (X d Y) ■ Figure gure 3.5 5 (X 2 Y) c (Y 2 X). From the table, we make the following observations: From the table, we make the following observations: Prove that we p(A) is p. a group with respectobservations: to this operation of addition. 1. a.Sthe is closed under From table, make the following 1. S is closed under p. b. If A has n distinct elements, state the order of p(A). igure 3.5 c. 1.1, #7c @ c. 1.1, #7c @ 2. C is an identity element. 1. S is closed 2. under C is an p. identity element. 37. outnot the have elements of p(A) since for theDX set 5 A 5C{a, b,no c}, solution. and construct an addition 3. Write D does an inverse has 3. D does not have an inverse since DX 5 C has no solution. 2. C is an identity element. table for p(A) using addition as defined in Exercise 36. Thus S Ais5not a group respect to p.p(A) Thus Sc}. is Prove not ainverse group with respect to is p.Ca has 3. D not anwith since 5 nowith solution. 38. Letdoes {a, b,have or disprove thatDX group respect to the operation of union. Thus S is not a group with respect to p. Finite Group, Infinite Group, Order Group c. 1.1, #7c @ 39. A 53.3 {a, b,Finite c}. Prove or disprove that p(A) isOrder aof group with respect to the operation Definition 3.3 ■ Group, Infinite Group, ofof aaGroup ition 3.3 Definition ■ Let Finite Group, Infinite Group, Order a Group tion 3.3 3.2 ■ ■ ■ of intersection. If a group G has a finite number of elements, G is called a finite group, or a group of finite group GGroup, has a finite number of elements, a finite group, or a group of finite ■If a Finite Infinite Group, OrderGofis acalled Group order. The number of elements in G is called the order of G and is denoted by either order. Theo(G) number of elements in G is called the order of G and is denoted by either or 0 G 0 . If G does not have a finite number of elements, G is called an infinite group. IfProperties a group a does finite number of elements, G isofcalled a finite group, group ofgroup. finite o(G) or 0 GG0 .has If G not have a finite number elements, G is calledorana infinite of Group Elements order. The number of elements in G is called the order of G and is denoted by either 8notdefinition Inhave Example the grouprecorded o(G) orconsequences 0 G 0 .Example If G does a finite number of elements, G is called an infinite group. Several of the a3,group in Theorem Example 8 In Example*, the ofgroup G are = {e, σ } has order 3.4. Example 8 In Example 3, the group G 5 5e, r, r2, s, g, d6 o (G ) = 2 . In Example 5, o ( Z n ) = n. The set Z of all integers is a group2 under addition, and this is an exaple of an G 5 5e, r, r s,n.g,The d6set Z of all integers is a group under adhas o(G) 5 6.3,Inthe Example 5, o(Z n) ,5 8oforder In Example rategy ■ Example Parts a group. and bdition, the next are group statements aboutgroup. uniqueness, andinfinite they can infinite and thistheorem is an example of an infinite If A is an set,be then S(A) furnishes ) 5 n. The set Z of all integers is a group under adhas order o(G) 5 6. In Example 5, o(Z proved by the standard type of uniqueness proof: Assume that two such quantities exist, 2 n an example of an infinite group. ■ G 5 5e, r, r , s, g, d6 and thenand prove be equal.of an infinite group. If A is an infinite set, then S(A) furnishes dition, this istwo antoexample Properties ofthe Group Elements an example of 5 an 6. infinite group. 5, o(Zn) 5 n. The set Z of all integers is a group under ad■ has order o(G) In Example dition, and3.4 this is an example of an infinite group. If A is an infinite set, then S(A) furnishes Exercises Properties of Group Elements orem 3.4 Theorem ■ Properties of Group Elements an example of an infinite group. ■ 3.1 True or False Exercises 3.1 Exercises 3.1 Let G be a group with respect to a binary operation that is written as multiplication. Label each of the following statements as either true or false. a. The e in Gelement is unique. 1. element The identity in a group G is its own inverse. True oridentity False 21 !1 b. For each x2.[ If G,Gthe inverse x in G unique. is an abelian group,isthen " x true for all in G. Label each of the following statements asxeither orxfalse. 21 21 c. For each x [ G, (x ) 5 x. True or False 1. The identity element in a group G is its own inverse. Label each of the following statements as either true or false. 2. If G is an abelian group, then x!1 " x for all x in G. 1. The identity element in a group G is its own inverse. 2. If G is an abelian group, then x!1 " x for all x in G. 5 Then Chapter 3 Groups Groups y 5 ey ! since e is an identity 5 (zx)y since zx 5 e d. Reverse order law. For any and y inbyG, (xy)21 5 y21x21. 5xz(xy) associativity Week 4 e. Cancellation laws. If a, x,5and in G, z(e)y aresince xy then 5 e either of the equations ax 5 ay or xa 5 ya implies that x 5 y. 5z since e is an identity. Thus y 5 and this justifies x21the as others the unique inverse ofTo x in G. part b, let Wez, prove parts b andthed notation and leave as exercises. prove queness Proof ¿ q) ⇒ r x [ G, Weand shall use part b in theof proof ofzpart Specifically, we shall suppose that each y and is and.inverse of x. That is, use the fact that the in21 21 21 verse (xy) is unique. This means that in order to show that y x 5 (xy) 21, we need only xy 5 yxx21and 5 e 5calculations zx. to verify that (xy) (y21x21 ) 5 e 5e 5 (y21 ) (xy).xzThese are straightforward: Then (y21x21)(xy) 5 y21(x21x)y 5 y21ey 5 y21y 5 e y 5 ey and 5 (zx)y since zx 5 e (xy)(y21x21) 5 x(yy21)x21 5 xex21 5 xx21 5 e. 5 z(xy) by associativity z(e)x21 insince xy 5 e order law (xy) 21 5 y21x21 is crucial The order of the factors y215and the reverse 5 z where since e is2anx21 identity. in a nonabelian group. An example (xy) 21 y21 is requested in Exercise 5 at the end ofy this Thus 5 z,section. and this justifies the notation x21 as the unique inverse of x in G. Part e of Theorem 3.4the implies thatpart in d. theSpecifically, table for a finite group element G We shall use part b in proof of we shall useG, theno fact that theofinappears in the same row, and of G that appears twice in the 21 21 same column. verse (xy)twice is unique. This means thatno in element order to show y21x21 5 (xy) , we need only These results can be21 extended to the statement in the following strategy box. The proof of 21 21 21 to verify that (xy) (y x ) 5 e 5 (y x ) (xy). These calculations are straightforward: this fact is requested in Exercise 10. (y21x21)(xy) 5 y21(x21x)y 5 y21ey 5 y21y 5 e q) ⇒ r ■ and In the multiplication table for a group G, each element of G appears exactly once in each ategy 21 once 21column. row and also appears (xy)(y exactly each x21) 5inx(yy )x21 5 xex21 5 xx21 5 e. Remark. em 3.5 tegy since e is an identity Although our of and a group a standard one, alternative becrucial made. The order of thedefinition factors y21 x21 inisthe reverse order law (xy) 21 forms 5 y21can x21 is One of these isgroup. given in next theorem. in a nonabelian Anthe example where (xy) 21 2 x21y21 is requested in Exercise 5 at the end of this section. e 3.5 of Theorem 3.4Conditions implies in athe table for a finite group G, no element of G Theorem Equivalent for Group ■ Part Equivalent Conditions for athat Group appears twice in the same row, and no element of G appears twice in the same column. Let G results be a nonempty set that is under aninassociative binary operation called multiThese can be extended to closed the statement the following strategy box. The proof of plication. Then G is a group if and only if the equations ax 5 b and ya 5 b have solutions this fact is requested in Exercise 10. x and y in G for all choices of a and b in G. ■ In the multiplication table for a group G, each element of G appears exactly once in each em 3.5 row and also appears exactly once in each column. Although our definition of a group is a standard one, alternative forms can be made. One of these is given in the next theorem. ■ Equivalent Conditions for a Group Let G be a nonempty set that is closed under an associative binary operation called multiplication. Then G is a group if and only if the equations ax 5 b and ya 5 b have solutions x and y in G for all choices of a and b in G. 6 tion ax 5 a has a solution x 5 u in G. We shall show that u is a right identity for every element in G. To do this, let b be arbitrary in G. With z a solution to ya 5 b, we have za 5 b and Groups ! Week 4 bu 5 (za)u 5 z(au) 5 za 5 b. Thus u is a right identity for every element in G. In a similar fashion, there exists an element v in G such that vb 5 b for all b in G. Then vu 5 v, since u is a right identity, and vu 5 u, since v is a left identity. That is, the element e 5 u 5 v is an identity element for G. Now for any a in G, let x be a solution to ax 5 e, and let y be a solution to ya 5 e. Combining these equations, we have x 5 ex 5 yax 5 ye 5 y, and x 5 y is an inverse for a. This proves that G is a group. Remark. In a group G, the associative property can be extended to products involving moreInthan three factors. For example, if a1be ,a2extended ,a3 , and ato are elements G, than then 4 products a group G, the associative property can involvingofmore applications condition 2 inifDefinition three factors.ofFor example, a1, a2, a3,3.1 andyield a4 are elements of G, then applications of con- dition 2 in Definition 3.1 yield 3a1(a2 a3)4 a4 5 3(a1 a2)a34 a4 and (a1 a2)(a3 a4) 5 3(a1 a2)a34 a4. These equalities suggest (but do not completely prove) that regardless of how symbols of grouping are introduced in a product a1a2a3a4, the resulting expression can be Chapter 3reduced Groupsto Chapter 3 Groups 3.6 Product ition 3.6 Definition ■ Product NotationNotation 3(a1 a2)a34 a4. ition 3.6 ■ Product Notation Let n be a positive integer, n $ 2. For elements a1, a2, c, an in a group G, the expression rem 3.7 em 3.7 omplete nduction omplete nduction Withc these observations in mind, we make the following definition. a1a2 a is defined recursively by Let n be a npositive integer, n $ 2. For elements a1, a2, c, an in a group G, the expression c ak ak11 a1a2 can is definedarecursively by 5 (a1 a2 c ak)ak11 for k $ 1. 1 a2 a1 a2 c ak ak11 5 (a1 a2 c ak)ak11 for k $ 1. We can now prove the following generalization of the associative property. We can now prove the following generalization of the associative property. Generalized Associative Law Theorem 3.7 Generalized Associative Law ■ Associative Law Let nGeneralized $ 2 be a positive integer, and let a1, a2, c, an denote elements of a group G. For ■ any positive integer m such that 1 # m , n, Let n $ 2 be a positive integer, and let a1, a2, c, an denote elements of a group G. For c (a1 athat am)(a any positive integer m such 1# m m11 , n,c an) 5 a1 a2 c an. 2 (a1 a2 c am)(am11 c an) 5 a1 a2 c an. Proof For n $ 2, let Pn denote the statement of the theorem. With n 5 2, the only possible value for m is m 5 1, and P2 asserts the trivial equality Proof For n $ 2, let Pn denote the statement of the theorem. With n 5 2, the only possi(a1)(a 5 a1 aequality ble value for m is m 5 1, and P2 asserts the2)trivial 2. 7 Assume now that Pk is true: For any positive (a1)(a2) integer 5 a1 a2.m such that 1 # m , k, c c (a1 a2For ampositive )(am11 c ak) 5masuch Assume now that Pk is true: any integer 1 a2 thata1k. # m , k, Groups ! Week 4 Example 1 M m×n ( R ) is an abelian group with respect to addidtion. This is an example of another infinite group. Example M m×n ( ) , M m×n ( ) , M m×n ( k ) , M m×n ( ) is a group with respect to addition. Example The nonzero elements of M n ( R ) do not form a group with respect to multiplication. Example 2 The invertible elements of M n ( ) , M n ( ) , M n ( k ) , M n ( ) , M n ( R ) form a group G with respect to matrix multiplication. 8