Eyes ALGEBRA-! FTA, s DIRECTORATE OF DISTANCE EDUCATION SWAMI VIVEKANAND SUBHARTI UNIVERSITY Meerut (National Capital Region Delhi) REC: we ORECTO, fs * nouwo"®Units Page No. SS 1. GROUPS 1.27 1.1 Binary Operation 1 1.2 Algebraic Structure 2 13. Semi Group , 2 14 Group 2 L5 Order of a Group 3 1.6 Addition Modulo m 7 1.7 Additive Group of Integers Modulo m 7 1,8 Multiplication Modulo m 8 1.9 Multiplicative Group of Integers Modulo p, where p is a Prime Number g 1.10 Euler's g-function 7 9 1.11 Power of an Element B 1.12 General Properties of a Group 14 1.13. Order of an Element 7 1.14 Some Important Theorems 7 1.15 Homomorphism 20 1.16 Homomorphic Image of Identity and Inverse 2 1.17. Kernel of a Homomorphism 21 1.18 Isomorphism 23 1.19 Order of Isomorphic Image of an Element 2B 1.20. Kemel of an Isomorphism 24 1.21 Isomorphie Groups 24 2. SUBGROUPS 28-46 2.1 Complex 28 2.2 Subgroup 28 23 Tests fora Complex to be a Subgroup 29 24 — Product And Inverse of Complexes 32 25 Order of Product of Two Finite Subgroups 34 2.6 Construction of New Subgroups 35 27 Cosets 37 2.8 — Index ofa Subgroup in a Group 39 CONTENTS29 2.10 2.1L 2.12 213 3 32 33 34 35 Relation of Congruence Modulo a Subgroup Lagrange’s Theorem Order of Elements in « Finite Group Euler's Theorem Fermat's Theorem . NORMAL SUBGROUPS Introduction Normal Subgroup 7 Tests for a Subgroup io be a Normal Subgroup Quotient Group The Fundamental Theorem of Homomorphism CYCLIC GROUPS 41 42 Sa 5.2 53 54 55 5.6 57 58 59 5.10 5.1 5.12 5.13 61 6.2 63 64 65 Definition Properties of Cyclic Groups . PERMUTATION GROUPS : Permutation Theorem Permutation Group Cayley’s Theorem Permutations of Finite Sets Equality of Permutations Composition of Permutations in 2-Rowed Notation Cyclic Permutations ‘Transposition Disjoint Cycles ‘Theorem Even and Odd Pertnutations Alternating Group ._ GROUP AUTOMORPHISMS Introduction Automorphism Inner Automorphism Group of Automorphisms Characteristic Subgroup 39 42 43 45 47-55 47 47 51 56-61 56 ST 62-72SYLLABUS SEMESTER III Course I ‘Course Name: Algebra I ‘Course Code: BAMAT-301 Course Objectives: The objective of the course is to introduce the fundamental theory of groups and their homomorphisms. Symmetric groups and group of symmetries are also studied in detail. Unit 1: Binary operation on a set, Algebraic structure, Finite set:composition table for finite Sets, Definition of a group with examples and simple properties. Unit 2: Order of elements and groups, Subgroups, generators of groups, Cyclic groups, coset decomposition, Lagrange’s theorem and its applications, ,cyclic, even and odd permutations, permutation groups Unit 3: Homomorphism and Isomorphism. Cayley's theorem First fundamental theorm of homomorphism,second fundamental theorm of homomorphism, Third fundamental theorm of homomorphism, Normal subgroups. - Unit 4: Automorphism and inner automorphism, Automorphism groups and their computations. Nosmalizer and centre Course Learning Outcomes: The course will enable the students to 1, Recognize the mathematical objects that are groups, and classify them as abelian, cyclic and permutation groups, ete; 2. Link the fundaimental concepts of Groups and symmetrical figures; 3. Analyze the subgroups of cyclic groups; 4; Explain the significance of the notion of cosets, normal subgroups, and factor groups. Reference: 1. Gallian, Joseph. A. (2013). Contemporary Abstract Algebra (8th ed.). Cengage Learning India Private Limited, Delhi. Fourth impression, 2015. Additional Reading: 1. Rotman, Joseph J. (1995). An Introduction to The Theory of Groups (Ath ed.). Springer Verlag, New York.1. GROUPS NOTES teat Binary Operation Algebraic Structure Semi Group Group Order of a Group Addit Additive Group of Integers Modulo m Multiplication Modulo m : Multiplicative Group of Integers Modulo p, where p is a Prime Number Euler's 6-funetion Power of an Blement General Properties of a Group Order of an Blement Some Important Theorems Homomorphism * ‘Homomorphic Image of Identity and Inverse Kernel of a Homomorphism Isomorphism Order of Isomorphie Image of an Element Kernel of an Tsomorphism Isomorphie'Groups see 4.1. BINARY OPERATION Lot $ be a non-empty set. A function from S * $ into $ is called a binary operation on S. Thus, if‘ is a binary operation on §, then it means that ‘+"is a function from $ x S into S. : If (@, b) ¢ SxS, then the image #(a, b) of (, b) under the binary operation "is, written ab. For example : ( Ordinary addition is a binary operation on Z, because’ isa function from Zx LintoZ. We write + 4, )=4*7= lle Z (i Ordinary subtraction is nota binary operation on N, because difference of wo natural numbers may not be a natural number. We have ~ (5, 8) =5-8=—3e N. SelfInstructonal Matérial Groups 1Algebra-1 1.2. ALGEBRAIC STRUCTURE Sr A non-empty set with one or more binary operations defined on it is called an NOTES algebraic structure. In an algebraic structure, we can combine two elements of it to obtain another element of the algebraie structure. For example, (Z, +), (Q, +), (R, -) are algebraic structures, 1.3. SEMI GROUP a Anon-empty set G with a binary operation, denoted by , is called a semi group if EX YS2=W@eyNeZ Vey260 Equivalently, an algebraic structure (G, *) is a semi group if ; E*Q*da(eey)az Vay,26G. Ifthe binary operation on the serui group G is denoted by ‘." then we have x.Q.2=(%.9).z Vu,y,zeG, SS 1.4. GROUP n ; 7 i TT Anon-empty set G with a binary operation, denoted by “’, is called a group if: @x.@.2=(.9).2 YxyzeG . i) There exists an element ¢ € G such that : ex WxeeG cis called an identity of G. ii) For xe G, there exists an, element y € G such that : K.ysesy.x ¥ is called an inverse of x. An inverse of x is written as x, If in addition, we have : xe (iu) s.y=9.8 VayeG, then G is called a commutative group, A commutative group is also called an abelian group after the famous Norwegian mathematician N.H. Abel. A group is called non-commutative or non-abelian if itis not commutative. If the binary operation of a group is written as“, then we write Ox+Q+Q=(e+y)+2 Vayy,2eG Gi) Ice Gixt (iid) For xe Gye G And for commutativity, xtysy+x Yu yeQ, Remark 1. The binary operation +’ used above has got nothing to do with the ordinary addition of numbers, In fact», +, . are just symbols representing binary operations. Remark 2, In order to show a non-empty set G with a given binary operation to be a group, we should verify all the three conditions given abo 2 Self-instructional Material1.5. ORDER OF A GROUP If the number of elements in a group is finite, the group is said to be a finite group, otherwise an infinite group. The number of elements in a finite group is called the order of the group. Ifa group Gis finite and has n elements, thén we write o(G) = n. Examples : : 1. @, #)is an abelian group, where‘#’ represents the usual addition of numbers. This is true because, we have @xty+Qs+)+z VuyzEZ Gi) Oe Zandx+0=x=0+x VreZ ii) For x € Z, we have-xe Zandx+ Cx) = (a)xtysytx VayeZ 2.@, 4, ®, #), ©, », ( @ =Q~ {0}, R°=RB - {0}, C°= C—O}: All these groups are infinite groups. 3. The set of all mn matrices whose elements are complex numbers forms an infinite abelian group with respect to matrix addition, 4, The set of all non-singular n x n square matrices whose elements are complex numbers forms an infinite non-abelian group with respect to matrix multiplication. 8. The set of all integers which are multiple of m (an integer) i.e., the set {0, £m, # 2m, + 8m, .....} forms an infinite abelian group w.r.t, usual addition. Note that 0 acts as the identity and inverso of Am is ~ Am. WORKING RULES TO SHOW THAT (G, +) ISA GROUP Step I. Take arbitrary a, b of G and show that a * b is a uniquely defined element of G. Step II. Take arbitrary a, b, ¢ of G and show that a * (b *c) and (a* b) * care equal. This establishes the associativity of. Step IIL. Find an element 'in G such that a*e=a=e*a forallae G. e works as the identity element. Step IV. Forac G, finda number b¢ G such thata*b=e=b+ a. bis the inverse of a. Step V. Take arbitrary a, b in G and show that a* the commutativity of *. cote =b* a. This establishes, SOLVED EXAMPLES Example 1. Show that the set Z is an abelian group with binary operation ‘* defined as. a* b=a+b+ 1 fora, be Z. Solution, Fora, be Z, the valueofa* bie, a+b+1isa uniquely determined integer. z.‘® is binary operation on Z. Associativity. Leta, b, e¢ Z at(bea= gebtetD= at(btct ye 1 a +btet+2 Self Instructional Material , (RS, .), (C2, .) are all abelian groups, where |, NOTES 3Algebra—I and (axb)tc=(a+b+ )ec=@+bt Iter +bter2 # a*(*d=(@rb)*e + 4 is associative, NOTES Existence ofidentity. —1e Z. Fora Z, a*CDZat+C)+i=a and . Cd+e=Citatiza a*)=a=Cl*a, vacZ 1 is an identity element. Existence of inverse, Letae Z. «.-a-2e Z. a*(-a~2)=a+(-a-2)+1=-1 (a-2)*a=¢a-2)+a41= a*Ca-2=-1=(a-9 a ~a~2is an inverse of a, ,+)isa group. Commutativity. For a, b < Z, atbaatbt+1=btatl=bea, « @,#) is an abelian group. Example 2. Show that the set of all positive rational numbers forms an abelian Stoup under the binary operation ‘+ defined by cote Solution. Let G be the set of all positive rational numbers. b Fora, be G, thevalue ofa bie, <> is a uniquely determined element of G. «. + is a binary operation on G. Associativity. Leta, b,ce G. . a-oroner(t)= 4 16 (@): ab) 4 abe and cr b)s0-() ee- 4 e: a*(b+e)=(a*b)*c -. + is associative. 7 Existence ofidentity. 4¢ G ax 4xa Forae G, as4= "=a and dea=—y n a*4=a=4*a VaeG 2. 41s an identity element. Existence ofinverse. Let aed. 2eq axis Bxa - ani 224 and Meqg= 224 : a. a / 4 Self-Instructional Material =, . izGroups 16 i 1. 7 isan inverse of a. (G, *) isa group. NOTES Commutativity. For a, b € G, asb=o =| #|F «. G, *) is an abelian group. amen soe anf IES YE ISS} is a multiplicative group. ‘Solution. We have A= {i li | io =| lea ait It can be easily verified that the usual multiplication is a binary operation on A. Let P,QReA P.Q.R)=(P.Q).R Lat the matrix [9 {be denoted by P.I=P=I.PWPeA Lis the identity element for A. =1 O}f-1 0 ‘We have fo ito 1 O}f1 0], f-1 a}f-1 0” ame [eal sheets lf 4] Each element in A is inverse of itself Every element in Ashas an inverse. Ais a group under multiplication. Example 4 Let G be thst of al 2 2 matrices | i -ba 2] seer ond ae real numbers, not both zero (i.e., a + b* # 0). Show that G is an infinite abelian group w.r.t. usual matrix multiplication. ® Solution, We have, o={| 5 Pf asbero? +0? 20 It can be easily verified that the usual multiplication is a binary operation on G. Let A.B, Ce G. ‘ A(BO) = (ABYC. ab Let =[S 2] whee a= 1.=0 Self-Instructional MaterialAlgebra —1 NOTES 1 0 7 = f=1+0= 1 (i QJ anaers 1+0=1#0 ForAe G,wehave Al=A=IA «. Lis an identity element for G. Let al ® ‘| ~b al ° | Al =a?+b?40 AO exists @, Ay =D, Ay =—B Ay = - adj. w=[§ Hi -[ a Also Wealsohave AA? A+ is an inverse of A. Gis an infinite group. fas} [ tm Let Atl al Belin 1&4 a bile m, b al-bm am+bl AB I | [aren aime] -b al|-m+ 7 1 mjfal 6 7 =[ la-mb tb+ma a wa-[ [SI ‘be nine. = AB=BA. .. Gis also abelian, Example 5. Let G={(a, b): 0#0; a,b € R} and let * be a binary operation on G defined by (a,b) + (c, €)= (ae, be +d) (a, b),(c, d)€ G. Show that (G, *) is anon-abelian group. Solution. We have G = {(a, b) :a#0; a, be R). Let (a, b), (ed) G. * (@,6)* ( d)= (ae, be+ d) ac#0 = ace0 ‘Also, the numbers ae and be + d are uniquely defined, <> + isa binary operation on G. 6 Self-Instrucrional MaterialAssociativity. Let (a, 6), (c,d), (e, ) € G. Grops (a,b) * (c,d) * @,f) = (a,b) * (ce, de +f) (a(ce), b(ce) + de + f) = (ace, bee + de+ fy (ac, be + d) * (e,f) (ace, (be + d) e+ f)= (ace, bee + de + f) NOTES i Also, ((a,b) * (,@) *@, is associative. Existence of identity element. (1, 0) € G. For @, bye G, (a,b) * (1, 0) = (a), BQ) + and (1, 0) * @, b)=(1@), 0@ + 8 n @,1)*(,0=@,H=(,0*@,b) V@beG 1, 0) is an identity element. Existence of inverse. Let (a, b) € G. . o a#0 * (fee | aoethet CG) CE) 9 and (2.-4)re »=(20(-2o+4]=a.0 7 G i 2) is an inverse of (a, 6). 2. (G,*) is a group. Non-commutativity. (2, 8), (1, 4) € G. 2,3) *, 4)= 20), 30) + and (4) * @, 3) =(1Q), 4@) + (2, 3) * (1, 4) #(1, 4) * @, 3). ‘We cannot have @H*@d=C4)*@b) v G,d),¢deG. + isnon-commutative. -. (G, *) is anon-abelian group. 1.6. ADDITION MODULO m Let {0, 1, 2, ...., m— 1} be the set of first m non-negative integers. The binary operation ‘@ is defined on this set’by a © b= r, where r is the least non-negative Femainder obtained by dividing a+ b by m. This binary operation @ is called ‘addition modulo m’. 1.7. ADDITIVE GROUP OF INTEGERS MODULO m ,m— Te , where 0S r ~ pla or plb-_ = Pes pis prime) Both are impossible because 1 plaor pin—m) Both are impossible because 1 1, o(n) as the number of positive integers less than n and relatively prime ton For example, ifn = 9, then @(9) = 6 because there are 6 numbers 1, 2, 4, 5,7, 8 which are less than 9 and relatively prime to 9. If p is a prime number, then we have 0) =p -1. Self-Instructional Material 9Algebra—1 NOTES SOLVED EXAMPLES Example 6. Let n> 1 bea positive integer. Let G be the set ofall positive integers less than n and relatively prime to n. Show that G is an abelian group of order o(n) under multiplication modulo n, Solution. Let G = {a :0 n does not divide ab Our supposition is wrong. O (@.0.b=@.a.¢ = e.b=0.c = b=e, IMPORTANT RESULTS Ina group, identity element is unique. Ina group, every element has unique inverse. If (G,.)is 0 group, then(w!!=a Vac G. If(G,.) is 0 group, then (a,b? =b1. a Va,be G. If(G,.) is a group anda. b =a. ¢, then =c. . IF G,.) is a group and b.a=c.a, thenb =e, rr erase 14 Self-nstructional MaterialSOLVED EXAMPLES Groups Example 9. if every element of a group is its own inverse, then show that the Broup is abelian, Solution. Let (G, .) be a group such that x?=x Vive G. NOTES Let a, be G. + G,.) is abelian. Example 10, Let (G, .) bea group. Ifa, b € G, then solve the equation a. x. Solution. We have a.x.a=b. o.@.2.0.0 = @.o.2.@oy = x Example 11. //(G, .) is finite group, show.that there exists a positive integer N that > e.x.e=at.b.at such aN=e vaeG. Solution, Let G= {a,, ay, .....,a,} For a;€ G, consider a, a2, a8... Since all these elements are in G, which is finite, we must have a/ =a," for some J, m. Let t (lb) a. b) . [Using (2)] = amt} Uma) b= a" (a. BP) 2d = bn Lb ky bm asa. Ob) = bet a=(a.b%).b => "8. = bm. .a)=b".(a.b) = baa » Gis abelian, 7 Self Instructional Material 15Algebra—1* NOTES Example 13. Prove that if (G, 2 icon tion ron thn fore b¢ Gand ne Z, we have (a. by" =a". b> ~ Solution. Let a, b¢ G. Case In € N. We shail prove that ~ (@. br =ar. be ol) by using PMI. Letn= 1 ‘ . i LHS. of (1)=(@. bf =a.b and - RHS.of ()=a!.b'=a.b + () is true for n= 1. Let (1) be true for n=, * (a. b= at b aac / @) Let nek. at @. dM =@. HK@. b= @. ha: » Using 2) BY (b.'a) a, (6. b).a) (1) is true for n =k +1, ByPMI, (@.8) = Casell.n=0 (@.b)=e=e.e=a°.b° @.bo=0 5+ Case IIL. n is ve integer. Let n ‘ pen. > Now (By case 1) (, Gis abelian) ar besarebt. Combining, wa get EXERCISE B Let (G, ) be a group. If b. a= ¢. , then show that b =<. Ifin a group G, a? =o, ae G, show that a=e, Show that a group up to order 2is always abelian. Let (G, ) bea group. If «, be G, prove that G is abelian if and only if (a. 8)? =a". 6% If G, )is a group, then show that (a,b. a°!)*=a.b".a1 Vo, be Gandne Z sae Hints 4 (@.bPaa?.b? = @.8).(0.D=0.(2.b).b = 0..0).b=0.(@.8).6 = b.o=a.d Conversely, (2. b= (2.6). (2-8) =a. (ba) .b=a. (a. B).b. 5, First prove for n ¢ Nby using the principle of mathematical induction. 16 Self-Instructional Material1.13. ORDER OF AN ELEMENT Let (G, .) be a group with identity element e. For ae G, if there exists a smallest natural number m such that a” =e, then m is called the order of a and write o(a) = m. Ifno such natural number exists then we say that a is of infinite order. 1.14, SOME IMPORTANT THEOREMS Theorem 1, For a finite group, the order of every element is finite and cannot exceed the order of the group. Proof. Let (G, .) be a finite group. Let abe an arbitrary element of G. Cénsider the elements: a, a, a°, a, ‘These are all elements of G. Since Gis finite, all these elements cannot be distinct. a’ = a! for some natural numbers k and I. Letk>1.:. at. a!=a'=at=e ( k=h-D+D By definition of o(a), we have o(a) < k 1 + 0(@) is finite, ~ Let o(a) =n. If possible, let us suppose that n > o(@). Since o(a) = n, the elements : a, a, a°,......, a, a* (= @) are all distinct and these are all elements of G. This is impossible, because these are n distinct elements and G has less than n elements. Our supposition is wrong. -. 0(a) <0(G). Theorem 2, In a group, the order of an element and its inverse are same. Proof. Let (G, .) be a group. Let ae G. Case I. a is of infinite order ‘ atte YneN = @yt#e! VneN = a"#e VneN = (yee VneN = "is of infinite order. Case Il. a is of finite order “is also finite order. Let o(@)=m and o(a)=n o@=m > a"=e = (@")'=e'=> a™=e = ("=e nSm. o@)=n>(a'"=e> at Combining, we get m =n. i.e., o(a) = of). e = @yt=e'= at=e => msn IMPORTANT RESULTS 1. Forafinite group, the order of every element is finite and cannot exceed the order of the group. 2. Ina group, the order of an element and its inverse are same. Self-Instructional Material NOTES nAlgebra SOLVED EXAMPLES Example 14. In the group (Q’, .) find the order of elements of the group. Solution, In the group (Q°, .), the identity element is 1. oO o()=1 Also—1 € Q? and 1)?=1. = oCN=2 ~ Let x be any element of Q different from 1 and - 1. wel YneN © ‘The order of x is infinite. 0(1) == I, o@ 1) =2 and all other elements are of infinite order. Example 15. Let G=({0, 1, 2, 3, 4, 5}. Find the order of elements of the group G under the binary operation ‘addition modulo 6. Solution. We have G=0, 1, 2, 3, 4, 5}. Let ® denote the binary operation ‘addition modulo 6’. ‘The identity element is 0. -. 0(0)=1. Now (= 1, ()?=161=2, (= 18 (1)? =12=3, . «yt @3=4, (5=1@(1)'=16455, aye ot) @ @*=202=4, @P=20 @=204=0 02)=3. (9)! =3,@)?=3@3=0 0(8) =2. : a= (P=404=2, (= 40 (H*=402=0 o(4) =3, 7 @'=5, @=505=4, O=58 6 =504= =50 6% ©=5 0 6)'=5O2= 6 =50 6) 06) =6. Exaniple 16. Let G=(1, 2, 3, 4, 5, 6). Find the order of elements of the group G under the binary, operation ‘multiplication modulo 7”. Solution. We have G = {1, 2, 8, 4,5, 6). Let © denote the binary operation ‘multiplication modulo 7’, ‘The identity element is 1. -. o(1) =1. Now @)'=2, - @P=2@2=4, @3=20 @=204=1 02) =8. 3, @P=3 0 3=2, @3=8 © GP=802=6, 30 @=3 0 6=4, @S=3 0 GY =38.04=5, 30 @P=3O5=1, ?=40 4=2, @9=40 (4% =402=1 =5 @5=4, @°=5 © G=5 0 4=6, @F=5 0 G=502=3, 18 Self-Instructional MaterialGroups | @'*=606=1. Remark. Here note that the number 7 is primo. Example 17. /f (G, .) is a group, then show that ofa) = ofr!..a. x) Va, x G. Solution. We first show that @?.a.x)"=x1.a™.x VméN. NOTES Now | : yo A) Case I. a is of infinite “ . ate YneN => a’ex.e.x? VneN => lat xé¢xt.@e.e. x") x VneN = .a.a4e WneN. (By using (1)) => x! .a. xis‘of infinite order. s of finite order x7 a. xis also of finite order. 1 cand oft.a.x=n. ‘ = a™=e = am=x.e.e! = la x=e = (@.a.9™5e = msm otta.s)en = la s se 9 xtatxze = a"=x.e.x7 = =e = msn Combining, we get m=niee.,o(a)=o(r'.a.3). 7 eel 18. Let (G, .) be a group and a € G be of order m. Show that a" =e if ifm/n. | ond oul ifm Solution. (0) = m. 7 mis the smallest natural number such that a” = e. Let m/n. :. n= km for some ke N a" = alm = (an) @ “Conversely, let o"=e, +. m2 m * By division algorithm, 3 integers q, r, where 0 e.atze u = r=0 (: 0o(@)=mand0 G' by (x) = {. 7 ‘ ae - Show that 9: G > G' isa homomorphism. Solution, Let x, y € G. There are four possibilities Case I.x>0,y>0 ay > Oand o(xy) = 1 Oy) = $62) O60) Case II x>0,y<0 ay 0 . syOand 6 @y)=1 OGy) = O07) 90) Vn ye G = $:GG’is a homomorphism. and_ (x) 66) = (1) (1) = 1. and 4(x) 69) =(@) C1) =-1. and (x) $0) and 94) 6)=CDCD=1 Example 21, Let G be the group of al reat 2x 2 matrices (% ) such tha ad ~ be # Ounder matrix multiplication. Let G’ be the group ofall non-zero real numbers under multiplication. Define 6: GG’ by o(: 3) Sad - be, Show that 9 is a homomorphism of G onto G’. Solution. Let (° ) and (% 3) dein. a) . . yo) Bf ‘aa’ +be’ ab’ + bd’ [Jee | 2) lee 23) = (aa’ + be’) (cb + da’) - (ca + de’) (ab’ + ba’) : = a0reb’ + aa'dd! + be'eb’ + be’ dd’ — ea’ab’ — ca’bat — de’ab’ de'bd’ add’ + be'cb’ ~ ca’bd’ ~ de’ab’ =a'd"(ad- be) - Ye'(ad~ be) = (ad ~be) (a'd' - We) “fa b) fa’ bP , no(t Jae) «+ ¢ is a homomorphism from G into G’, gisonto.LetmeG. * ameOandmeR CC ‘jes (: m(Q)=00) = m0) and ° (3 ‘) = m(1) -0() = m. 22° Self-Instructional MaterialEvery element in G has a pre-image in G. 6 is onto. 9:G> Gis an onto homomorphism. —_ 41.18. ISOMORPHISM A mapping ¢ from a group G into a group G’ is called an isomorphism of G into eit: (@ $s a homomorphism i.e., (ab) = $(@) 46) Va, be G (i) ¢ is one-one. Remark. If : GG’ is a mapping and we write ¢(ab) = ¢(a) (0), then it should be clearly understood that @ abis that element of the group G which is obtained by applying the binary operation of G on the ordered pair (@, 2) GxG. * G 4) 4(8) is that eloment of the group G’ which is obtained by applying the binary operation of G’ on the ordered pair (9(a), 6(8)) © @” x G’. eee 4.19. ORDER OF ISOMORPHIC IMAGE OF AN ELEMENT Sees ‘Theorem. If is an isomorphism of group G into group G’, then oa) =0((@)) Vae G. Proof, $ ; G—+ G’ is an isomorphism of G into G’. Let e and e’ be the identity elements of the groups G and G’ respectively. Leta ¢ G. Case I. o(a) is finite. Let o(a) =>: at=e = o")=9@ = = (a) $(0)..... ntimes =e => => o(f(@)) Sn Let ‘omsn = (a) G(0)..... m times = & = g(a”) = oe) (eis one-one) nsm Cr o@)=n) Manic, 0(6(a)) = ofa). Case II. o(a) is infinite. If possible, let 0(9(0)) be finite. Let 0($(a)) = m. = @@)"=¢ = 9(a) 6(@) m times = & = oa m times) = = 9a") = 9) = ame (+ @ is one-one) (a) < m, which is impossible. -. 0(9(a)) cannot be finite. 0(¢(a)) is also infinite, -. a and 6(a) have same order. SelfInsructional Material 23 NOTESAlgebra 1 1.20. KERNEL OF AN ISOMORPHISM “Theorem. Let } be a homomorphism from a group G into group G’. The NOTES * | homomorphism ¢ is an isomorphism of G into G if and only if ker = te). Proof. 6 : GG’ is.a homomorphism of group C:into group G’. Let 6 be an isomorphism of G into C’. We have o(¢) =e’. ce kerd. Let a be any other element in ker ¢. 4) = o@) . (+ ¢ is one-one) ker $ = {e}. “Conversely, let ker 9 = {e}. We shall show that $ is one-one. Let fora, be G, we have 9(a) = $(). = 9) OY = 90) (OO) = 9a) Ub) =e = = able kero = 7 = ab") b=eb = ae )=b = ae=b = a=b, +. 9 is one-oné, +. bis an isomorphism of G into G’. 1.21. ISOMORPHIC GROUPS ‘Two groups G and G’ are called isomorphic groups if there exists an isomorphisin of G onto G’. If groups G and G’ are isomorphic groups then we write. G=G. Remark. If groups G and G’ are isomorphic then there may. exist one or more than one isomorphism from G onto G’. IMPORTANT RESULTS 7 1 |. If is a homomorphism of group G into group G’, then 9(@) = &. : If is ahomomorphism of group G into group G’, then o@)=@H vaeG. }. If} is an isomorphism of a group G into ‘group G’, then ofa) =0(0(a) Vac G. . Let: GG’ bea homomorphism. The homomorphism ¢ is an isomorphism of G into Gif and only if ker 6 = (¢. SOLVED EXAMPLES Example 22, Let G be any group and a be any fixed element in G. Define a map 9: GG by the formula (3) = axer!, V xe G. Prove that 6 is an isomorphism of G onto itself. Solution. For x€ G, (2) = axa-"is a unique element of G. 4 is well defined. 24 Self-Instructional Material| is a homomorphism. Let x,, x, ¢ G. Groups : 94,3) = ole) 1 = ax, lar, a = (ax, 07) (ax, 1) = Oy) 0) : Cee) = 04) 06) Ve G NOTES | +. 618 a homomorphism. 44s one-one. Let x, x, € Gand $(e,) = Ox). | = oxo = ax, 7 = eax, o) a= a7 (ax, oa = @'ax, (r'0)=(e1 a) x, (00) - = me Hak. * Cr) =0;) 4+, Gis one-one $is onto. Let xe G. -. c'xae Gand aa? xa).a* (aa) x (aa!) = exe =x. For ve G,3a7 sae G: 6" x0)=x is onto. | +. @is an isomorphism of G onto itself. ox a) Example 23. Let G be the additive group of real numbers and G’, the multiplicative group of positive real numbers. If: G— Gis a mapping defined by OG) =e V xe G, then show that 6 is an isomorphism of G onto G. Solution. For x € G, 6(2) = e*is a unique positive real number. 4 is well defined. is a homomorphism. Let x,, x, € G. Ge, + xy) = et = OF 0™ = O(x,) O16) O(a; +x) = 00) 00) Vay eG 6s a homomorphism $s one-one. Let x, 2 € Gand o(x,) = 6). : = “eh =e" = log e* = log e*# | = x, log e= x, loge = 4S 964, = O¢62) = 458, + disone-one. is onto. Let ye G’ [ Let log xe Randy=e. . o@=yandxeG. + gisonto. 9:G>G’ is an isomorphism. Example 24, Show that the mapping x -> x of @ group G onto itself is an + isomorphism if and only ifG is abelian, x being any arbitrary element of G. Solution. We have $: G— G defined by 6) = x for xe G. 4s well defined because inverse of an element of a group is unique. Let G be abelian. Fors yeG, dm) =Goytsytxt = xix? = 00) 00) * $y) =9@) 00) VayeG ++ @ is ahomomorphism, Self-Instructional Material 25Algebra NOTES Let $(%) = 96) for x, ye G. . = sisyl =) yey + 9s one-one Let ze G. «21 ¢ Gand oe") = (@1) =z, “+ ¢is onto. +. $:G— Gis an isomorphism. Conversely, let ¢: G > G be an isomorphism. For, ye G, (xy) = 0G) 00) = tery? 3s fr Ptsety yt = oy (ey => . ay = yx = Gis abelian. WORKING RULES TO SHOW TWO GIVEN GROUPS TO BE ISOMORPHIC Rule I. Designate the groups by G and G’. Also observe the binary operations of the groups. Rule II. Construct a mapping ¢ from G into G’ keeping in mind that itis to be one-one, onto and a homomorphism. Rule IIL, An isomorphism is to preserve identities, inverses and orders. These facts may be used while constructing the mapping 6. Rule IV. Show that the mapping $ is well defined. Rule V. Show that ¢ is one-one, onto and a homomorphism. 26 SelfInstnictional Material EXERCISE D “1 Let, and G+) bo groups. Define a mapping ):G-»G'by 6@)=¢ Ve G, where «is the identity clement of the group G’. Show that is a homomorphism from G into G’, 2 Let G be the group of all complex number under usual addition. Define ¢ : G > G by 9@) = ZV ze G. Show that 6 is an isomorphism from G onto G. 3. Show that the mapping /: C > R such that f(x + i) = xis a homomorphism of the additive group of complex numbers onto additive group of real numbers. Also find ker / 4, Let Cy and R, be the multiplicative groups of non-zero complex numbers and of non-zero real numbers respectively. Define 9: Cy» Ry by 6(@)= | z | for ze C,. Show that ois ‘homomorphism of C, into R, and gis not onto. Also find ker 6. 5. Let G be the multiplicative group of all n m non-singular matrices with elements as real numbers and let G’ be the multiplicative group of all non-zoro real numbers. Show that the mapping @ : GG’ defined by $(A) = det A for A € G is a homomorphism of G onto G’. Also find ker 6 6. IfG be the multiplicative group of positive real numbers and G’, the additive group of real numbers. Show that the groups G and G’ are isomorphic. 7 Show that tho st Gof all matics ofthe tspe [2 4], whore ae R, 2 O form a multiplication group and is isomorphic to the group G’ of non-zero real numbers under multiplication. 8 If 3s an isomorphism from group G onto group G’, then show that ¢-1is an isomorphism from G’ onto G.= Answers kar f= tiy:y¢ RB Pre-images of negative real numbers are not defined.- ker g={e:|2|=1,2€ Oj) ker 9 is tho sot of all n x n non-singular matrices with elements as real numbers and whose determinant is 1. Hints . gee a) e# = 96) ¥ 80). k 00. 0 10. Ifke G’, then 9(4) =k, whoreA=| 0 0 1 ‘Try 9:G-G" defined by $(2) =logx Yxe G. . Define ¢ aoaryo([s (J-s 2B SelfInstructional MaterialRemark 1. In practice we gonerally write a. b as ab. In other words, ab denotes the olomont of the group G when the binary operation '’ of G is.applied on the ordered pair Gb). If binary operation on G is +, then ab represents a « 6. Remark 2. If a non-empty sot H of group G is a group under some other binary ‘operation on H and not with the binary operation of G, then H cannot be said a subgroup of. Ilustrations. (i) Z, Q, R are all subgroups of the group (C, +). Gi) @, BP are subgroups of the group (C°, .). Here Q*, R°, C® represents the set of non-zero rational numbers, non-zero real numbers and non-zero complex numbers respectively. Git) Let G = (1, +1,7,- and H=41,-1}. Since G and H are both groups under usual multiplication, we can say that H is a subgroup of (G, (iv) The set of even integers is a subgroup of the additive group of integers. (0) The set of positive real numbers is a subgroup of multiplicative group of non- zero real numbers. ‘Theorem. Let H be a subgroup of a group G. Then: (@ The identity element of the group H is the same as that of the group G. (ii) The inverse of any element of the group H is the same as the inverse of the element regarding as an element of G. (ii) The order of any element of His the same as the order of the element regarding ‘as an element of the group G. Proo€. (i) Let eande’ be the identity elements of the groups H and G respectively. « Forae H, ae=a=ea - + Alsoae Himplies a G. (and) > By cancellation law = =e. : (ii) Let e be the identity element of the subgroup H and the group G. Let a € H. Let b be the inverse of @ in H. | ab=e=ba Al) Let c be the inverse of a in G. 7 ac= (2) (@)and@) = ab=ace = baa (ii) Let ce H. Let m and n be the order of @ as an element of H and G respectively. 2 a and at=e ofa)=m => msn o@)=n = nsm Combining, we get. = m=n. TESTS FOR A COMPLEX TO BE A SUBGROUP Acomplex of a group may or may not be @ subgroup of the group. The direct method for testing a complex to be a subgroup is to verify all the properties of a group for the given complex wz.t. the induced composition of the group. In the theorems to follow, we shall establish some"results for testing a complex to be a subgroup. Subgroups NOTES Self-Instnuctional Material 28Algebra~1 Theorem 1. Let G be a group with binary operation denoted multiplicatively. _ Let H be a non-empty subset (i.e., a complex) of G. H is a subgroup of G if and only if: @abeH VabeH @joteH VaeH. Proof Lat Hea subgroup of G. Lato be H. Since the binary operation of G NOTES is also a binary operation on H, we have ab < H. Let ae H. Since H is a group in itself, we have a ¢ H. Conversely, let ab ¢ H, at ¢ H whenever a, be H. Since ab HV a,b € H, the binary operation of G is also a binary operation on H. Alternatively, we say that H is closed w.r:t. binary operation of G.- Associativity. Let a,b, ce H. abceG. + abe) =(abye. Existence of identity element. Let a € H. By second condition, 1¢ H. . By first condition, aoe Hie,ee H Letbe HH. 2. beG = be=b=eb cis an identity element of H. Existence of inverse. Leta.e H. .. By second condition, a! € H. His a group w.r-t: binary operation of G, His a subgroup of G. 7 ‘Theorem 2. Let G be a group with binary operation denoted multiplicatively. Let H be a non-empty subset (i.e, a complex) of G. H is a subgroup of G if and only if ote H Wa,beH, Proof, Lot H be a subgroup of G. Leta, b€ HI Since His a group in itself, wehave b* € H, Also, H is closed w.r.t. operation of G, we have ab“ € H. 7 Conversely, let-ab-! ¢ H whenever a, b € H. Associativity. Let a, b, ce H. abceG., + albe)=albe). . Existence of identity element, Let a€ H. By the given condition, aa"! Hite, e¢ H. Lethe H.beG be=b= eb e is the identity element of H. Existence of inverse. Leta ¢ H. By the given condition, ear! « Hie, ote H, Closure property. Let a, b ¢ H. Inverse of b ie, 1 € H, By the given condition, a(b-!y! € Hie, abe H. His closed w.r.t. operation of @. -. His. a group wrt. operation of G. Hisa subgroup of G. Remark. If the binary operation of G is denoted additively then the statement of the above theorem would take the following form : Let G be a. group with binary operation denoted additively. Let H be # non-empty subset (i.e, a complex) of G. H is a subgroup of G if and only if a-beH VabeH 30 SelfInstructional Material‘Theorem 3. Let G be a group with binary operation denoted multiplicatively. Let H be a non-empty finite subset (i.e, a finite complex) of G. H is a subgroup of G if and only ifabe Ha, be H. Proof. Let H be a subgroup of G. Let a, b ¢ H. Since the binary operation of G is also a binary operation on H, we have ab ¢ H. 7 Conversely, let ab € H whenever a, b « H. ‘The binary operation of G is also a binary operation on H. Associativity. Let a, b, ce H. a,b,ceG + albe)=(ab)e. Existence of identity element. Let ae H. a? =aae H, a= a, a, a°, a, ..... all lie in H. Since H is finite, there must be repetitions in the above elements. Let a! = a” for some integers J, m, where |> m. = dam=a™a™ = ah™=a = abmae 7 ecH Letbe HbeG be=b=eb eis the identity element of H. Existence of inverse. We have a/-" =¢, whore 1> m. ‘ where 1m 2 1 aan where !-m—120 ‘Also abl qa ghmitl a ghmae ; atm a+ is the inverse of a. His a grqup w.r.t. binary operation of G. His'a subgroup of G. = S} is closed w.rt, multiplication and is a finite subset Remark 1. ‘The set H = of B His a subgroup of the group (R, ). Remark 2. In the above theorem, tho subset H must be a finite non-empty subset of ©, otherwise the result may not hold good. For example, (Z, +) is a group and N is a non- ‘ompty subset of Z and is also closed under addition. Here N is not a subgroup of (Z, +) because N is not finite. ‘Theorem 4. I/¢ is a homomorphism of a group (G, +) into a group (G’, *’. Prove that 9(G) is a subgroup of the group (G’, *?). Proof. We have 4@) = (0 : g€ G Let ¢ and e’ be the identity elements of the groups G and G respectively s =e = ¢e KG) Let 9(@), 0(6) € @(G). (a) # @ DY" = Oa) * O67) (a+b) € 4). (abeG > atbteQ) 6(G) is a subgroup of the group (G’, #9. | | ‘Self Instructional Material Subgroups NOTES 3Algebra 1 NOTES SOLVED EXAMPLES Example 1. Let G be a group with‘ binary operation denoted multiplicatively. For ae G, lel H={a": ne a Show that His a subgroup of G. Solution. We have H={a": ne Z}. H#6, because a! =ae on eta, Be H. a=a™,B=a" for some m, ne Z. of! =a" (ay! =a" a*= a" € H because m-ne Z, ofeH Va,BeH. ~ Hisasubgroup of G . Example 2. Let G be a group with binary operation denoted multiplicatively, ‘The set the G:hx=xh V xe G) is called the centre of the group G. Show that the centre of G is a subgroup of G. Solution. Let H= {he G: he ah V¥ xe G} Let h, h’ € H. * he=xh VxeG =) WWe=xh’ VeeG (2) LetyeG . (hh)y=A(h’y) =hOh’) (Using @) = (hy)h! = GA)h = y(hh). (Using (1)) (hh)y=yhh) Wye @ he H. =, dy D4 = ony = yt sinly = Ity=yht hh, © Hwhenever h, h’ eH, His a subgroup of G. __ Bxample 8. Let G.be a group with binary spefation denoted multiptisatively. Lat H be a subgroup of G. For x €-G; define xtfe! = {shxclzhe },.Show that xls" is a subgroup of G. i Solution. We have xHx" = {xhx? : he H}. Let xh), xh’xt © xHxt. Now — (xha)(xh'sx!) eclsyhc = theh'x y a(hh’) x ¢ xB? (he Also Ghar = (ett = aha e Ht Wiel) Gch Noch’), (chet) ext ll is a subgroup of G. 2.4, PRODUCT AND INVERSE OF COMPLEXES Let H and K be any two complexes of a group G with binary operation denoted multiplicatively. ‘The product IK of complexes H and K is defined as HK ={hk:he H, ke K). 32 Selfsnstructional MaterialSince H and K are non-empty, HK is also non-empty. ‘Subgroups Also, hit ¢ G for every he H, ke K, because G is a group. HKG. HK is a complex of group G. NOTES For the complex H, we define the inverse (H~) of H as, 1 heH. Since H is non-empty, H™ is also non-empty. Abo hte G Vhe H, because G is a group. 2. H+ is also a complex of group G. . ‘Theorem 1, Let H and K be two subgroups of a group G with inbinary operation, denoted multiplicatively. The complex HK of the group G is a subgroup of G ifand only if HK =KH. . Proof. H and K are subgroups of G. Let the complex HK of G be a subgroup ofG. Letre HK. +. ete BK Let x1 =k, where he H, ke K Now x= (elyts (hb)? =I ite KH, Ea HK cKH Now, let-kh KH, where hk ¢ K and he H. > eK iteH = Wkte HK = @hyte HK = (Gh ¢ HK = khe HK. 7 KHcHK Combining, we get HK= KH. Conversely, let | HK =KH. Let ik We HK: GENIE) = HUN = MR" WY k= ah) © HK. (kh. ¢ KH = hh’ e HK = kh’ = hl for he H, ke K) The binary-operation of G is also a binary operation on HK. Associativity: Let a, b, ce HK. abceG .. albe)=(abye. Existence of identity element. ¢¢ Hand ee K. . eee HK ie, ee HK LetaeHK. . aeG 2 ae=a=e0 e's the identity element of HK Existence of inverse. Let hk HK, where he Hand ke K. ht © KH = e HK. ( HK=KI) Inverse of elements exists in HK. HK is a subgroup of G. Remark 1. [f G is an abelian group and H, K are subgroups of G, then the complex HIK of G is always a subgroup of G. This happons because HK end KH are equal sets when Gis abelian. Romarlk 2. If tho binary operation of group G is denoied by ‘then the statement of the above theorem takes the following form : ‘Let H and K be two subgroups of a group (G, +). The complex H + K of the group (G, *) isa subgroup of (G, +) if and only if H* K= K+ H. '" Seypsnstructional Material 33