Dr.Siham J. Al-Sayyad.

Transcription

1 King Abdulaziz University Mathematics Department Abstract Algebra1 Math (342) Lecture Notes Dr. Siham. Dr.Siham J. Al-Sayyad. This notes has been typed as an appreciation gift from Norah H. Al-Shamrani To The Helpful & Kind teacher ; Dr.Siham /

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7 1Groups and Subgroup In this chapter we shall deal with mathematical systems which are defined by a prescribed list of properties. 1.1: Binary Operation: Diffinition1.1.1: A binary operation on a nonempty set S, is a function from the Cartesian product S S into S i.e, :S S S given by : (a,b)=a b. Thus a binary operation assigns to each ordered pair S S a uniquely element of the same set. Remark1.1.2: 1. We generally use symbols as {+,-,,.,,,U} to represent binary operation. 2. If is a binary operation on s then we say that s is closed under the operation or is a closed operation. 6

8 Abstract Algebra(I), Math (342) Chapter 1- Groups and Subgroups.. 3. If :s s T is a function such that T S then we say that is not a closed operation and if T S then is a closed operation. Example(1); Ordinary subtraction is clearly a binary operation on the set Z of integers, the subset Z + of +ve integers is not closed under subtraction. Difinition1.1.3: By a mathematical system or algebraic system, we shall mean a nonempty set of elements together without or more binary operations defined on this set. A mathematical system consisting of the set S and a single binary operation is denoted by the ordered pair (S, ), with two operation and will be denoted by (S,, ). Example(2): 1) (Z,+) is an algebraic system. 2) (Z +odd,-), ( Q, ) are not an algebraic systems. 3) (Q *, ),(R, ) and (Z,-) are algebraic systems. 7

9 Abstract Algebra(I), Math (342) Chapter 1- Groups and Subgroups. Example(3); The pair (S, ), where the set S={1,-1,I,-i} and the operation is ordinary multiplication, is a mathematical system provided one defines i 2 =-1. Example(4); If Z e and Z denote the even and odd integers, then (Z e,+, ) constitutes a mathematical system, while (Z,+, ) does not, since Z is not closed under addition (the sum of two odd integers is necessarily even). Definition1.1.4: A binary operation :S S S on the set S is; a) Commutative if x y:y x x,y S. So, x,y commutative or permutes. Associative if (x y) Z= x (y Z). x,y,z S. c) Left distributive: let be another binary operation on S, the is a left distributive over if; x (y z)=(x y) (x z) x,y,z S. D) Right distributive; is a right distributive over, if; (y z) x=(y * x) (z * x) x,y,z S 8

10 Abstract Algebra(I), Math (342) Chapter 1- Groups and Subgroups.. e) Distributive; if is both left and right distributive over, then is said to be distributive over. Example(5); Define as an Z, the set of integers, by taking a b=a+b+ab. Then for a,b,c Z we have a (b c)= a (b+c+bc) =a+ (b+c+bc)+a(b+c+bc) Also,(a b) c=(a+b+ab) c =a+b+ab+c+ (a+b+ab)c Thus a (b c)=(a b)c For a,b Z shown that a b=a+b+ab=b+a+ba=b a. thus is commutative. Example(6); Consider the operation defined on the set S={1,2,3}by the operation table; We note that, 2 (1 3)=2 3=2, whereas 9

11 Abstract Algebra(I), Math (342) Chapter 1- Groups and Subgroups.. (2 1) 3=3 3=1 that is 2 (1 3) (2 1) 3. The associative law thus fails to hold in the system(s, ) In tables if we need to define a binary operation on a set S we must be sure that; 1) exactly one element is assigned to each possible pair of element of S. 2) for each ordered pair of element, the elements assigned to it again in S. (2 3)=2 (3 2)=2 So is not commutative. theorem1.1.5: A mathematical system has at most one identity element. Proof: Suppose that (S, ) has identity elements e and e \ since e a=a for all a S then, e e \ = e \ but on the other hand e \ is also an identity so, e e \ : e thus, e= e e \ = e \ so, e= e \ then there is only one identity element. 10

12 Abstract Algebra(I), Math (342) Chapter 1- Groups and Subgroups.. Example(7); 1) Addition and multiplication of real numbers are commutative and associative operations. 2) Union and Intersection of sets are commutative and associative operations. 3) The algebraic systems (Z,+, ), (Q,+, ) and (U,+, ) are all algebraic system with operation ( ) is distributive over (+). 4) (P(A),,U), and (P(A),U, ) are algebraic systems with operation (U) is distributive over ( ) and ( ) is distributive over (U). 5) (Z,+,-) is a mathematical system, but the operation (-) is not distributive over the operation (+). Difinition1.1.5: A mathematical system (S, ) is said to have an identity element if there exist e S such that. Example(8); 1) In (R, +), Zero is the identity element. 2) In (P(A),U), is the identity element. 3) In (P(A), ) A (universal set) is the identity element. 4) (Z +,+) has no identity. x e= e x=x x S 11

13 Abstract Algebra(I), Math (342) Chapter 1- Groups and Subgroups.. Definition1.1.6: the operation denoted by x -1 if x x -1 = x -1 x=e Definition1.1.7: defined on S. Example(9); On the set Q of rational number define by a b=a+b+ab. Then (Q, ) is a commutative semi group. How? The identity is. If a Q such that a -\ then if b is the inverse of a we have; a b=0 Let (S, ) be a mathematical system with identity e S. Then two element x S is said to have an inverse under A semi group is a pair (S, ) consisting of a nonempty set S together with an associative (binary) operation a+ b+ab=0 a+b(1+a)=0 a(1+a)=-a -a b= 1+a Thus a -1 = = -a, a a Thus a a -1 = a -a = a + ( ) -a = a ( ) -a 1+a 1+a 1+a 12

14 Abstract Algebra(I), Math (342) Chapter 1- Groups and Subgroups. = a(1+a)-a-a -2 = 0 = a -1 a. 1+a There is a mathematical system, known as a group which display most of the properties we have discussed. Theorem1.1.8: Proof: In a semi group (S, ) with identity, each element has at most one inverse. Suppose that two inverses exist, let a S have two inverses a \ and a \\ by def. So that, a a \ =e= a \ a a a \\ =e= a \\ a a a \ = a a \\ multiplying both sides by a \ on the left we get by associativity, a \ (a a \ ) = a \ (a a \\ ) (a \ a) a \ = (a \ a) a \\ e a \ =e a \\ a \ = a \\ thus the element a cannot posses more than one inverse. 13

15 Abstract Algebra(I), Math (342) Chapter 1- Groups and Subgroups. 1.2: Groups: Definition1.2.1: A group is a system (G, ) consisting of a nonempty set G and a binary operation defined on G such that; G.1) G is closed under. G.2) The operation is associative. G.3) G contains an identity element e. G.4) Each element a G has an inverse a -1 G. Definition1.2.2: A group (G, ) is abelian if is commutative, that is a b=b a for all a,b G. Remark1.2.3: If it happen that the group operation satisfies the commutativity property, then (G, ) is called a commutative group or an abelian group. Example(10); Let a 0 be any real number. Define G={na n Z}. Then (G,+) is a commutative (abelian) group. The identity is 0, and the inverse of na is (na) = (-n)a G. 14

16 Abstract Algebra(I), Math (342) Chapter 1- Groups and Subgroups. Example(11); Define on Q + by g 1 g 2 g = 1 g 2 g 1,g 2 Q + Then (Q, ) is a group for is closed and if g 1,g 2, g 3 Q + then g (g 1 g 2 ) g 1 g 2 g 3 = g 1 g 2 g 3 3 = g g 1 (g 2 g 3 )= g 2 g 3 g 1 g 2 g 3 1 = Therefore (g 1 g 2 ) g 3 = g 1 (g 2 g 3 ) Now, let e Q + such that g 1 e= g 1= e g 1 g Thus g 2 e 1 e = = g 1 2 for g 1 G Hence g 1 e= 2 g 1 g 1-1 (g 1 e) = g 1-1 (2g 1 ) (g 1-1 g 1)e=2(g 1-1 g 1) e=2 ( h identity in Q + ). Finally, For any g Q +, we have

17 Abstract Algebra(I), Math (342) Chapter 1- Groups and Subgroups. g g -1 = e = 2 = g -1 g Thus, = 2 is the inverse of g Q + Test g g -1 =?? g g -1 4 g This shows that (Q +, ) is a group. Example(12); Let G = {(a,b):a,b U,a 0}. Define on G by (a,b) (c,d)=(ac,bc+d). Show that (G, ) is a non-commutative group. Solution; - Since a 0 ac 0 (ac, bc + d) G. i.e, is a closed operation. - is an associative operation, for [(a,b) (c,d)] (e,f) = (ac,bc+d) (e,f) = ((ac)e,(bc+d)e+f) = (a(ce),b(ce)+(de+f)) = (a,b) (ce,de+f) = (a,b) [(c,d) (e,f)] - (a,b) (e 1,e 2 ) = (a,b) = (e 1, e 2 ) (a,b) 16

18 Abstract Algebra(I), Math (342) Chapter 1- Groups and Subgroups. (ae,be 1 +e 2 ) = (a,b) ae 1 =a and be 1 + e 2 = b e 1 = 1 and e 2 =0 Thus the identity element is (1,0) G - (a,b) (a -1,b -1 ) = (a -1,b -1 ) (a,b) = (1,0) (aa -1,ba -1 +b -1 ) = (a -1 a,b -1 a+b) = (1,0) aa -1 = a -1 a = 1 and ba -1 +b -1 = b -1 a+b =0 a -1 = 1 and b -1 = - b a a So, the inverse of (a,b) is ( a 1, - ) b a G. - is not commutative for (1,2) (3,4) = (3,10) (3,4) (1,2) = (3,6) Example(13); Let be the Set of function F i, i=1,2,3,,6 from R {0,1} into it self where; f 1 (x)=x, f 2 (x)= 1 x, f 3 (x)=1-x, f 4 (x)= 1, f 5 (x)= x-1, f 6 (x)= x 1-x x x-1 Then G={f 1,f 2,f 3,f 4,f 5,f 6 } with the composition of function is a group. Solution: We see that (f 2 f 4 )(x) = f 2 (f 4 (x)) = f 2 ( ) 1 1-x 17

19 Abstract Algebra(I), Math (342) Chapter 1- Groups and Subgroups. = 1-x = f 3 (x) and for x R {0,1} (f 4 f 2 )(x) = f 4 (f 2 (x)) = f 4 ( ) 1 x = 1 = x = f 6 (x) for x R {0,1} 1-1 x-1 x Hence (G, ) is not a commutative group. Therefore we can get Theorem1.2.4: Proof: If a,b one given elements of the group G, then the equation ax=b and ya=b have a unique solution in G. Observe that x=a -1 b satisfies ax=b, for a(a -1 b)=(aa -1 )b=eb=b. This shows at least one solution exist in G. To see that the solution is unique, suppose there is another element x \ G for which ax \ =b 18

20 Abstract Algebra(I), Math (342) Chapter 1- Groups and Subgroups. Then ax \ =a(a -1 b) and so we may left cancel theorem is proved similarly. Corollary1.2.4 \ : In multiplication table for a group G, every element appears exactly once in each raw and column: this is called the group table of (G, ) clearly from the table (G, ) is a group. Theorem1.2.4 \ \ : Let (G, ) be a group, then ; 1. The identity element is unique. 2. The inverse of eachelement is unique. 3. If a b=c b, then a=c for a,b,c G (this is the right cancellation law). Also if a b=a c, then b=c. (the left cancellation law) Proof: 1. Let e and f be two identities in G, then for x,y G we have ; x e = x and f y=y. thus, f e=e for e G, f = f e, therefore f=e and the identity is unique. 2. Let x be an element in G with two inverses x 1, x 2, then x x 1 =x 1 x=e x x 2 =x 2 x=e 19

21 Abstract Algebra(I), Math (342) Chapter 1- Groups and Subgroups. Consider x 1 = x 1 e = x 1 (x x 2 ) = (x 1 x) x 2 = e x 2 = x 2 3. Let a b=b c, since (G, ) is a group, then every element has an inverse. i.e, b has inverse b -1 i.e, a (b b -1 )=c (b b -1 ) a e=c e a=c Similarly, we prove a b=a c b=c. Theorem1.2.5: Proof: If G is a group, then a) (a -1 ) -1 = a a G. b) (ab) -1 = b -1 a -1 a,b G. a) By definition, (a -1 ) -1 is the inverse of a -1 (a -1 ) -1 (a -1 )= a -1 (a -1 ) -1 = e But aa -1 = a -1 a = e, the element a is also the inverse of a -1, but the inverses of a group are unique and so (a -1 ) -1 = a. b) We need to show that (ab)(b -1 a -1 ) = (b -1 a -1 )(ab) = e (ab)(b -1 a -1 ) = ((ab)b -1 ) a -1 20

22 Abstract Algebra(I), Math (342) Chapter 1- Groups and Subgroups. Similarly (b -1 a -1 )(ab) = e Therefore, (ab) -1 = b -1 a -1 Example(14); = (a(bb -1 )) a -1 = (ae) a -1 = aa -1 = e. (Klein 4-group). Let G={e,a,b,c}. Define on G by ; e a = a e = a, e b = b e = b, e c = c e = c Also, cb=bc=a, ac=ca=b, c=ab=ba, a 2 =b 2 =c 2 =e. Thus G is a group which is abelian. This group is called the klein 4-group. We denote it by V or K 4 therefore K 4 =V={e,a,b,c}={e,a,b,ab=ba} Construct the group table. e a b c e e a b c a a e c b b b c e a c c b a e 21

23 Abstract Algebra(I), Math (342) Chapter 1- Groups and Subgroups. Definition1.2.6: Let G be a group. Then the nimber of the elements in G is called the order of G, and denoted by G. The group G is called finite if it has a finite number of elements, and called infinite if it has an infinite number of elements. Definition 1.2.7: Let G be a group. And let x G. We define the power of x as follows: i) x = e is the identity in G. ii) x n = xx.x, for n > 0, and n Z. Theorem1.2.8: Proof: iii) i) Excersize. n- times x -n = (x -1 ) n = x -1 x -1.x -1 for n Z +, x is +ve. n- times Let G be a group. And let x G. If m,n Z. Then: i) x n x -n = e. ii) x m x n = x m+n. iii) (x n ) -1 = x n = (x -1 ) n. iv) (x m ) n = x nm = (x n ) m. ii) x m x n = x m+n we have four cases. Case(1): if m or n is zero i.e, m=0, n 0 22

24 Abstract Algebra(I), Math (342) Chapter 1- Groups and Subgroups. x m+n = x +n = x n = e.x n = x x n = x m x n Case(2): Suppose m,n > 0 x m+n = xx x = xx x x x = x m x n Case(3): Suppose m,n < 0, let m=-r, n=-s for r,s,z + So x m x n = x -r x -s = (x -1 ) r (x 1 ) s = (x -1 ) r+s = x -(r+s) = x -r-s = x m+n Case(4): Suppose that m,n have opposite signs. Let m < 0 and n > 0 with m+n 0. Then x -m x m+n = x -m+(m+n) = x n. Hence multiply both sides x m x m+n = x m x n. Similary for m > 0, n < 0, with m+n < 0. m+n < 0 - (m+n) > 0 we get x (m+n) x m = x m-n+m = x -n Consider n+m times x m+n = x m+n x -n x n scince x -n x n = 0 = x m+n (x m-n x m )x n = (x m+n x -m-n )x m x n = x m x n iii) x -n = (x n ) -1 = (x -1 ) n?? m- times n- times Abstract Algebra(I), Math (342) 23

25 Chapter 1- Groups and Subgroups. We consider three cases; Case1: if n=0 it s true. Case2: suppose n > 0 by definition. x -n = (x -1 ) n (use induction) For n=1, x -1 = (x 1 ) -1 = (x -1 ) 1 its true. For n=k, x -k = (x k ) -1 = (x -1 ) k. For n=k+1, x -(k+1) = (x -1 ) k+1 = (x k+1 ) -1??? i.e, x k+1 x -(k+1) = e x(x k x -k )x -1 = e So, x -k = (x k ) -1 x x k (x k ) -1 x -1 =e e=e. That is x -(k+1) = x so its true for n= k+1 Hence its true for all n > 0. Case(3): suppose n < 0, then m=-n > 0 x -n = x m = [(x m ) -1 ] -1 since (x -1 ) -1 = x. And = (x m ) -1 since m > 0 by case 2. 1 = (x n ) -1 since -m =-(-n) = n. x -n = [ (x -1 ) -1 ] -n = (x -1 ) -(-n) 2 = (x -1 ) n (k+1) -1 Abstract Algebra(I), Math (342) 24

26 Chapter 1- Groups and Subgroups. From eq1and2we get x -n = (x n ) -1 = (x -1 ) n iv) x mn = (x m ) n we will consider three cases: Case(1): if n=0 its true. Case(2): suppose n > 0 (use induction). for n=1 we have x m1 = (x m ) 1 for n=k assume its true for n=k so x mk = (x m ) k. for n=k+1 x m(k+1) = x mk+m = x mk x m = (x m ) k (x m ) =(x m ) k+1 Case(3): suppose n < 0 -n > 0 and so X mn = (x -(mn) ) -1 = (x m(-n) ) -1 = ((x m ) -n ) -1 = (x m ) n 1.3: Subgroup: Definition1.3.1: If a subset H of a group G is closed under the binary operation of G and if H with the induced operation from G itself a group, then H is a subgroup of G. Denoted H G or G H denote the H is a subgroup of G, and H<G or G > H shall mean H G but H G. Abstract Algebra(I), Math (342) 25

27 Chapter 1- Groups and Subgroups. Example (15); (z,+)<(r,+) but (Q +,+) is not a subgroup of (R,+). Even though Q + R. Every group G has a subgroup G of itself and {e}, where e is the identity elememt of G. Definition1.3.2: If G is a group, then the subgroup consisting of G itself is the improper subgroup of G. All other subgroup are proper subgroup. The subgroup {e} is the trivial subgroup of G. all other subgroup are nontrivial. Theorem1.3.4: Let G be a group and H G then H G ab -1 H a,b H Proof: suppose H G, a,b H b -1 H a,b -1 H ab -1 H (closure condition) suppose that H G which contains ab -1 H. i.e, ab -1 H whenever a,b H (i) b,b -1 H bb -1 H e H (ii) e,b H eb -1 H b -1 H (iii) a, b -1 H a(b -1 ) -1 H ab H Abstract Algebra(I), Math (342) 26

28 Chapter 1- Groups and Subgroups. Example(16): There are two different types of group structures of order 4. We describe them by their group tables. The group V is the klein 4-group, and the notation V comes from the German word Vier for four. The group Z 4 is isomorphic to the group U 4 = {1,I,-1,-i} of fourth roots of unity under multiplication. Namely, we can rewrite 1,i,-1, and i as i 0, i 1,i 2, and i 3, respectively. Renaming the elements of the table for U 4 under multiplication by their exponents 0,1,2, and 3 when expressed as a power of i, we obtain the group Z 4. Passing to the exponent has of course converted multiplication to addition. We add the exponent in the usual way and take the remainder when we divided by 4. For example, i 3 i 3 = i 6 = i 4 i 2 = 1 i 2 = i 2, so 3+3=2 in table The only nontrivial proper subgroup of Z 4 is {0,2}. Note that {0,3} is not a subgroup of Z 4, since {0,3} is not closed under +. For example, 3+3=2, and 2 {0,3}. However, the group V has three nontrivial proper subgroups, {e,a}, {e,b}, and {e,c}. Jere {e,a,b} is not a subgroup, since {e,a,b} is not closed under the operation of V.For example, ab=c, and c {e,a,b}. Abstract Algebra(I), Math (342) 27

29 Chapter 1- Groups and Subgroups. Z 4 : V: e a b c e e a b c a a e c b b b c e a c c b a e Example (17); Consider (Z -9,+ 9 ) the group of integers modulo 9. If H={0,3,6}, it will be found that (H,+ 9 ) is a subgroup of (Z -9,+ 9 ). The identity element of the subgroup is 0, while 3,6 are inverses of each others. Theorem1.3.5: x Let {H;} be a collection of subgroup of a group G then H i i=1 Proof: x - H i, since each H i contains the identity element of G. i=1 - a,b H i let a,b H i ab -1 H i H i G. Theorem1.3.6: Abstract Algebra(I), Math (342) x i=1 x i=1 x i=1 Let H 1,H 2 be subgroups of a group G. Then H 1 H 2 G H 1 H 2 H 2 H 1. 28

30 Chapter 1- Groups and Subgroups. Proof: suppose a,b H 1 H 2 a,b H 1 or a,b H 2 ab -1 H 1 or ab -1 H 2 ab -1 H 1 H 2 H 1 H 2 G. suppose to the contrary of this assertion were false i.e, H 1 H 2 and H 2 H 1, H 1 H 2 G. a H 1 -H 2 and b H 2 -H 1 where a H 2 and b H 1 a,b H 1 H 2 G ab H 1 H 2 ab H 1 or ab H 2. Closed If ab H 1 b = a -1 (ab) H 1 If ab H 2 a = (ab) b -1 H 2 Which is a contradiction. Corollary1.3.7: H 1 H 2 or H 2 H 1 A group G cannot be the union of two of its proper subgroups. Proof: where H 1 < G, H 2 < G. Let G = H 1 H 2, then by theorem above either H 1 H 2 or H 2 H 1 G=H 1 H 2 = H 1 or G= H 1 H 2 = H 2. Abstract Algebra(I), Math (342) Chapter 1- Groups and Subgroups. 29

31 Definition1.3.8: The center of a group G, denoted by cent G is the set Z(G) = cent G = { g G: xg=gx x G }. Theorem1.3.9: Proof: i) cent G, since ex=xe x G e cent G. ii) Example(16); Let a,b cent G. Then ax = xa and bx = xb x G. Consider (ab -1 )x = ab -1 xb b -1 = ab -1 bx b -1 ab -1 cent G. So, cent G G. = axb -1 = x(ab -1 ) The subgroup of the kelin 4 are V, {e},{e,a},{e,b},{e,c}. Example(17); Let G be a group. Then cent is a subgroup of G. i.e, cent G G. Let G be a group, a G. let H={x G:x=a n for n Z}. then H G. Abstract Algebra(I), Math (342) Chapter 1- Groups and Subgroups. Solution: 30

32 i) H since e=a H. ii) Consider If x,y H x=a n, y = a m for m,n Z. xy = a n a m = a n+m H where n+m Z. iii) Let x= a n H, n Z. Then x -1 = a -n H where n Z. Thus H G. This subgroup is called subgroup generated by a and is denoted by a. Definition1.3.10: A subgroup H of a group G is a proper subgroup if H G, its denoted by H < G. Theorem1.3.11: Let H be a nonempty subset of a group G. Then H G. H G (i) H is closed (ab H a,b H). (ii) a -1 H a H. Proof: If H G its travial that (i)(ii) satisfied suppose that (i) (ii) holds, we want to prove that H G i.e H is a group itself. Abstract Algebra(I), Math (342) Chapter 1- Groups and Subgroups. 1) (i) holds, then the closure condition is satisfied > 31

33 2) Since its associative so a (b c) = (a b) c a,b,c H 3) H the at least one element say a H. but by (ii) a -1 H a a -1 H (by(i)).i.e, e H. 4) Condition (ii) means that every element has inverses thus H G. Problems Set I 1. Let S be a set containing more than one element. If the operation is defined by a b = b a,b S. verify that (S, ) 32

34 is non-commutative semi-group which admits no identity. 2. Determine whether the binary operation defined below is commutative and whether x is associative. a) defined on Q by a b=ab+1. b) defined on Z + by a b=a b. 3. Mark each of the following true or fals: a) If is any binary operation on any set S. then a b =a for all a S. b) A binary operation on a set S is commutative if thereexist a,b S such that a b = b a. c) A binary operation on a set S assigns more than one element of S to some order pair of elements of S. 4. In the following determine whether the binary operation gives a group structure on the given set. If no group results give the fail axioms. 1) let be defined on 2Z={2n n Z} by a b = a + b. 2) let be defined on R + by a b = ab. 5. In the following determine whether the system (G, ) are commutative group for those system fail to be so indicate which axions are not satisfied. a) G=Z, a b = 0 b) G=Z Z, (a,b) (c,d) = (a+c,b+d). c) G=R R, (a,b) (c,d) = (ac+bd,ad+bc). d) G is a commutative group iff (ab) n = a n b n a,b G. 6. Let G be the set of mapping f i (i=1,2,3,4) from R-{0} into itself where ; f 1 (x)=x, f 2 (x)= Æ, f 3 (x)=-x, f 4 (x)= -Æ for each x R-{0}. verify that the system (G, ) forms a commutative group. 7. For any group G, prove that the following conditions are equivalent. a) G is commutative. b) (ab) -1 = a -1 b -1 a,b G. c) (ab) 2 = a 2 b 2 a,b G. 33

35 8. For each of the following sets H, establish that (H, ) is a subgroup of (G, ). a) G={1,-1,I,-i}, H={1,-1} where i 2 = -1. b) G=Q-{0}, H={2 n, n Z}. 9. If H is a subgroup of the group G and K is a subgroup of H, verify that K is a subgroup of G. 10. Let G be a group and a G define c(a)={x G: xax -1 =a} show that c(a) is a subgroup of G, knows as the centralizer of in G. 11. Assume that H,K are subgroup of the abelian group G let HK={g G: g=hk, h H, k K}. prove that HK<G. Two Important 2Groups 34

36 this section is devoted to an examination of two important and frequently used group: the group of integers moduo n and the group of permutative of elements of a set, the so called symmetric group. 2.1: Group of integers modulo n: Definition 2.1.1: Let n be a fixed +ve integer. Two integers a and b are said to be congruent modulo n written. a b (mod n) a-b is divisible by n. i.e, a-b=qn for q Z. if a-b is not divisible by n, we say that a is incongruent to b modulo n and is written a b (mod n) Example(1); Clearly is an equivelance relation to find equivelance classes we see. Abstract Algebra(I), Math (342) Chapter 2- Two Important Groups. Definition2.1.2: For an arbitrary integer a, let [a] aradenote the set of all integer congruent to a modulo n. [a] = a = {x Z : x a (mod n)} = {x Z : x a+kn, for k Z)} We call [a] = a the congruence class modulo n, determined by a and refer to a as representation of this class. 35

37 Example(2); By way of illustration, suppose that we dealing with congruence modulo 4. then {x Z x 0 mod 4} [0]= {x Z x =4k, k Z} = {,-8,-4,0,4,8,12, } {x Z x 1 mod 4} [1]= {x Z x =4k+1, k Z} = {,-7,-3,1,5,9, } {x Z x 2 mod 4} [2]= {x Z x =4k+2, k Z} = {,-6,-2,2,6,10, } {x Z x 3 mod 4} Abstract Algebra(I), Math (342) Chapter 2- Two Important Groups. [3]= {x Z x =4k+3, k Z} = {,-5,-1,3,7,11, } Remark2.1.3: 1) Every integer lies in one of these 4 classes. 2) Integers in the same congruence class are incongruent modulo4. While integers in different classes are incongruent modulo 4. To return to the general case of congruence modulo n, 36

38 Theorem 2.1.4: Let n be a +ve integer and Z n be as defied above then; Definition2.1.5: Abstract Algebra(I), Math (342) Chapter 2- Two Important Groups. Example(3); 1) For each [a] Z n, [a]. 2) If [a] Z n and b [a], then [b]=[a]. 3) For any [a],[b] Z n, where [a] [b], [a] [b] =. 4) { [a] a Z n }=Z. A binary operation + n may be defined on Z n as follows [a] + n [b] = [a+b] [a],[b] Z n. Thus, [a] + n [b] = [a+b] if a+b < n. We consider congruence modulo 7 [3] + 7 [6] = [ 3+6-7]=[2] [2] + 7 [4] = [ 2+4] =[6] [ a+b-n] if a+b n. 37

39 We are now in a position to prove one of the principal theorem. Theorem2.1.6: For each positive integer n, the mathematical system (Z n,+ n ) forms a commutative group, known as the group of integers modulo n. Proof: 1) ([a 1 ],[b 1 ]) = ([a],[b]) [a 1 ] + n [b 1 ] = [a] + n [b]?? Let ([a 1 ],[b 1 ]) = ([a],[b]) [a 1 ]=[a] and [b 1 ]=[b]. a 1 -a = qn and b 1 -b = q \ n, q,q \ Z. a 1 -a + b 1 -b = qn + q \ n = (q +q \ )n = q \\ n, q \\ Z. a 1 +b 1 a+b (mod n) [a 1 +b 1 ] = [a+b] [a 1 ] + n [b 1 ] = [a] + n [b]. so + n is well defined. Abstract Algebra(I), Math (342) Chapter 2- Two Important Groups. 2) (Z n,+ n ) is closed. Since (Z n,+ n ) is closed, (a,b) Z Z : a+b Z. a+b [r] where [r] Z n. [a+b] = [r] Z n. 38

40 ([a],[b]) Z n Z n : [a] + n [b] Z n. 3) + n is commutative: [a], [b] Z n [a]+ n [b] =[a+b] = [b+a] = [b] + n [a]. 4) + n is associative: [a], [b], [c] Z n ([a]+ n [b])+ n [c]=[a+b]+ n [c] =[(a+b)+c] =[a+(b+c)] =[a]+ n [b+c] =[a]+ n ([b]+ n [c]) 5) [e] Z n. Consider [a]+ n [e] [a+e] = [a] a+e-a=qn, q Z. e=qn [e] = [qn] =[0] Z n. Abstract Algebra(I), Math (342) Chapter 2- Two Important Groups. 6) Inverse: Let [b] = [a] -1 be the inverses of [a], then [a]+ n [b]=[0] [a+b] = [0] a+b 0 (mod n) 39

41 a+b-0=qn, q Z. b=qn-a, q Z. b= n-a where q=1 Thus the inverse of [a] is [n-a]. Therefore (Z n,+ n ) is a commutative group. Definition 2.1.7: Let n be a fixed +ve integer. Define Z n, by [a] [b] = [ab] for [a],[b] Z n Theorem 2.1.8: Proof: Home work. Example(4); (Z n, ) is a commutative semi-group with identity. 1) (Z 5,+ 5 ) 2) (Z 3, ) 3) (Z 5, ) 4)( Z 4, ). Abstract Algebra(I), Math (342) Chapter 2- Two Important Groups. 1) (Z 5,+ 5 ) + n [0] [1] [2] [3] [4] [0] [1]

42 [2] [3] [4] ) (Z 3, ) ) (Z 5, ) Abstract Algebra(I), Math (342) Chapter 2- Two Important Groups. 4)( Z 4, )

43 :Symmetric group of order n: Definition 2.2.1: Let G be a non-empty set, A permutative on a set Gis a function from G into itself which is both 1-1 and onto. S G is the set of all permutative on G. Take G = N = { 1,2,,n}. then the set of all permutative on N will be denoted by S n. Example(5); Suppose that G={1,2,3,4,5} σ is the permutation given by σ = Such that σ(1)=4, σ(2)=2, and so an. Abstract Algebra(I), Math (342) Chapter 2- Two Important Groups. Theorem 2.2.2: The set S G is a group under function composition this group is called symmetric group on the set G. Proof: 42

44 - Since composition of two permutations of G yields a permutation of G, so S G is closed under permutation multiplication. - Since function composition is associative, is associatie. - The identity of S G is the permutation id=i:g G defined by i(a) a G. - For a permutation σ the inverse σ -1 is the permutation that reverses the direction of the mapping σ. i.e. σ -1 (a)=a \ A such that a a=σ(a \ ). The existence of one such element a \, since σ is both 1-1 and onto, for each a A, we have: And also, i(a) = a = σ (a \ ) = σ(σ -1 (a)) = (σσ -1 ) (a). i(a \ ) = a \ = σ -1 (a) = σ -1 (σ (a \ )) Abstract Algebra(I), Math (342) Chapter 2- Two Important Groups. = (σ -1 σ) (a \ ). So σ -1 σ and σσ -1 are both the permutation i. Thus (S G, ) is a group.# From now on S n will be called the symmetric group of order n. Note that S n has n! elements where n! = n(n-1)(n-i)

45 Since the injective function σ can end number 1 to any n- element of {1,2,,n}:σ(2) can then be any one of the element of this set except σ(1). So, there are (n-1) choices for σ(2), σ(3) can be any element except σ(1), σ(2). So, there are (n-2) choices for σ(3) and so on thus there are precisely ; n(n-1)(n-2) 3.2.1=n! possible injective function from N to itself. Example(6); An intresting example for us is the group S 3 of 3!=6 elements. σ 1 = σ 4 = σ 2 = σ 5 = σ 3 = σ 6 = Abstract Algebra(I), Math (342) Chapter 2- Two Important Groups. The multiplication table for S 3 is: 44

46 Note: That this group is note abelian. It is our firt finite ex. We have seen that any group of at most 4 elements is abelian, also we will see that a group of 5 element is also abelian. Thus S 3 has minimal order for any non-abelian group. Example(7); Let us form the Dihedral group D 4 of permutations corresponding to the ways that two copies of a square with vertices 1,2,3 and 4 can be placed, one covering the other. D 4 will then be the group of symmetric of the square, it is also eight permutations involved. Abstract Algebra(I), Math (342) Chapter 2- Two Important Groups. ρ 0 = ρ 4 =

47 ρ 1 = ρ 5 = ρ 2 = ρ 6 = ρ 3 = ρ 7 = Make the table and prove that the table is group. Remark 2.2.3: Any permutation α S n may be described as a set of ordered pairs. α= { (1,α(1), (1,α(2)), (n,α(n))} Or as we saw before α = n α(1) α(2) α(3) α(n) The columns may be rearranged without affecting the nature of the fun. For ex: the following two symbols both represent the same element in S 4. Abstract Algebra(I), Math (342) Chapter 2- Two Important Groups or

48 Permutation, being functions, may be multiplied under the operation of functional composition. The resulting produce is not just a function, its also a permutation of N. Thus for α, β S n. α β = 1 2 n 1 2 n α(1) α(2) α(n) β (1) β (2) β (n) = β (1) β (2) β (n) 1 2 n α (β (1)) α (β (2)) α (β (n)) β(1) β(2) β(n) = 1 2 n α (β (1)) α (β (2)) α (β (n)) What we have done is to rearrange the column of first left permutation until its top row is the same as the bottom row of the second (right) permutation, the product α β is then the permutation whose top row is the top row of the second factor and whose bottom row is the bottom row of the first factor.s Example(8); In example(6) we have S 3 where the permutation are: Abstract Algebra(I), Math (342) Chapter 2- Two Important Groups. α 1 = α 4 =

49 α 2 = α 5 = α 3 = α 6 = α 4 α 6 = = = = α α 6 α 4 = = s = = α α 4 α 6 α 6 α 4 is not commutative. Abstract Algebra(I), Math (342) Chapter 2- Two Important Groups. Theorem 2.2.4: 48

50 The system (S n, ) forms a group, known as the symmetric group on n symbols which is non-commutative for n 3. Note that: id = The identity element on (S n, ) is the permutation While the multiplicative inverse of any permutation α S n is described by; In the following technical definition, we introduce a special type of permutation called a cycle. Definition 2.2.5: Lemma 2.2.6: A cycle of length k has order k. Abstract Algebra(I), Math (342) Chapter 2- Two Important Groups. Example(9); id=i= α α(1) α(2) α(n) 1 2 n An element α S n is a k-cycle or cycle of length k if there exist a set {i 1,i 2,, i k } of distinct integers such that α(i 1 ) = i 2, α(i 2 )= i 3,, α(i k-1 ) = i k, α(i k ) = i 1 While α(i) = i 1 for i { i 1,i 2,, i k } 49

51 Let α= = ( 2 5 3) This is a cycle permutation of length 3. Note that: (2 5 3) = (5 3 2) = (3 2 5). Definition 2.2.7: Note: All cycles of length one (1-cycle) are equal to the identity permutation. Definition 2.2.8: Example(9); Consider the element Let α= = (1 3 7)(2 5)(6 8) Abstract Algebra(I), Math (342) Chapter 2- Two Important Groups. Example(10); A cycle of length 2 is called a transposition. 1) two cycles α = ( a 1, a 2,,a n ) and β = ( b 1, b 2,,b n ) of a set i.e, N are called disjoint if they have no integers in common. {a 1, a 2,,a n } { b 1, b 2,,b n }= 2) A set of cycles is disjoint if any two distinct cycles are disjoint. 50

52 Consider the element. Let N={1,2,3,4,5} α= β= , α= (1 2) (3 4), β = ( ) Example(11); In S 7, let α= ( ) and β= ( ) then α β. The product is defined using mapping composition. α= β= α β= = ( )(2 3). βοα= ( )( ) =(1 3)( ). α β βοα Note: The product of two cycles need not be a cycle. Lemma 2.2.9: Let α S n have its cycle decomposition into disjoint cycles of length m 1,,m k. then the order of α is the least common multiple of m 1,,m k. Abstract Algebra(I), Math (342) Chapter 2- Two Important Groups. Note: Disjointness is very important in the above lemma. 51

53 Example(12); If α= (1 2)(1 3) = (1 3 2) The order of α=3 which the order of (1 2)= 2 and (1 3)= 2 L.c.m =2. Example(13); The order of αοβ = ( )(2 3) is the least common multiple of 4,2 = 4. Note: Disjoint cycles commute, while nondisjoint cycles don t commute. Example(14); Let α=(4 5), β=(1 2 3) S 5 αοβ= (4 5)(1 2 3) = βοα= (1 2 3)(4 5) = (1 2 3)(4 5). Theorem : Every permutation α i (identity permutation) in S n can be written uniquely as a product of disjoint cycles. Remark : 1) Every transposition is its own inverse (ij)(ij)=(i)(j)=i=id. 2) Every cycle can be expressed as a product of transpositions i.e, α= (a 1, a 2,a 3,a k )=( a 1 a k )( a 1 a k-1 ) ( a 1 a 2 ). Abstract Algebra(I), Math (342) Chapter 2- Two Important Groups. Let α S 4 s.t. α= ( )=( 1 4)( 1 3)( 1 2). 3) thus, every permutation in S n can be expressed as a product of transpositrions. 4) If α= (a 1, a 2,,a k ) α -1 = (a k, a k-1,,a 1 ). 52

54 Example(15); Let α=(1 2 3) S 3 and β=(1 6)(2 5 3) S 6. Then α,β can be factorized as follows: α=(1 2 3) = (1 2)(2 3). or = (1 3)(1 2). β = (1 6)(2 5 3) = (1 6)(2 5)(5 3). or = (1 6)(2 3)(2 5). Definition : 1) A permutation is even if it can be written as a product of an Abstract Algebra(I), Math (342) Chapter 2- Two Important Groups. even number of transposition. 2) A permutation is odd if it can be written as product of an odd number of transpositions. 53

55 Theorem : No permutation in S n can be expressed as a product of an even number of transposition and as a product of an odd number of transpositions. Example(16); The permutation ( ) (2 1 5) in S 6 can be written as ( ) (2 1 5) = (1 6)(1 5)(1 4)(2 5)(2 1) which has 5 transpositions so its an odd permutation. Example(17); Let α= ( )(2 4)(3 4 5) S 6 α= (1 5)(1 2)(1 3)(2 4)(3 5)(3 4) It has 6 transposition so α is an even permutation. Remark : a) The product of any two even or any odd permutations is even. b) The product of an odd permutation and an even permutation is odd permutation. Abstract Algebra(I), Math (342) Chapter 2- Two Important Groups. c) The inverse of an even permutation is an even permutation, and inverse of an odd permutation is an odd permutation. d) The identity permutation is even permutation. 54

56 Definition : 1) For n 2 let A n denote the subset of S n consisting of all even permutation. A n = {α S n : α is an even permutation}. 2) For n α 2 let n denote the subset of S n consisting of all odd permutation. B n = {ρ S n : ρ is ann odd permutation} Theorem : For n 2, thus collection of all even permutations (A n, ο) forms a group of order n! which is called the alternating group of n- 2 letters. Proof: Since the identity permutation is even id A n. The other condition are easy straight forward. Now, let S n =A n B n such that A n B n =. Therefore, S n = A n +B n = n! Abstract Algebra(I), Math (342) Chapter 2- Two Important Groups. For a fixed transposition, define the function. f :A n n by f (α) = οα where α A n thus οα B n. Moreover, 55

57 i) f is 1-1 for let f (α) = f (µ). οα= οµ since S n -1 S n. -1 ο οα = -1 ο οµ α=µ ii) f is onto since ρ B n δ A n S.t. f (δ)=ρ οδ = ρ. δ = -1 ο ρ A n. i.e, f ( -1 ο ρ)= ρ Hence we have a corresponding between A n and B n. Thus A n = B n = n!. 2 Problems Set II 56

58 1. If a b modn and m n. Prove that a b mod m. 2. Construct the operation tables for the groups (Z 8, ) and (Z 7, ). 3. Show that the system ({ 0, 4, 8, 12}), ) mod 16 forms a group. 4. Express the following permutations as. (a) Products of disjoint cycles. (b) Products of transpositions. (c) Decide whether these permutations are even or odd. (d) Find the order of each permutation. α= β= , γ= From the set G = {f,f 2,f 3,f 4,f 5,f 6 } where f is the permutation. f = Prove that the pair (G,ο) is a commutative group. 57

59 6. Compute gfg -1 for each pair f,g of the following: (a) f = (1243) g=(132) (b) f = (1356) g=(2546) (c) f = (135)(24) g=(25)(34) 7. Which of the following functions is a permutation: (a) f 1 : R R defined by f 1 (x) = x+2. (b) f 1 : R R defined by f 2 (x) = x. 8. List all element of S 3 and then find the alternating group A 3 written in cycle notation. 9. Show that the cycle ( ) may be written as a product of 3 cycles. 10. If a b mob n and x is any integer, then a+x b+x modn and ax ba mod n. 11. If a b mob n and c d mod n, then a+c b+d mod n and ac bd mod n. 12. If ax ay mod n and (a,n) = 1, then x y mod n. 13. An element a Z n has a multiplicative inverse iff (a,n) are relatively prime. 14. Calculate αβ,βα,α -1,β -1,(αβ) -1,(βα) -1,β -1 α -1 and α -1 β -1 where α= β= , Construct the operation table of (S 3,ο). 58

60 16. What is the center of S 3 and S 4. Cyclic 3Groups 59

61 A central goal of group theory is to classify all groups i.e, to see what kinds of groups those are. One would like to start with the easiest groups. It turns out that these are the cyclic groups, those groups that are just the set of powers of some one element. 3.1: Cyclic groups: Definition 3.1.1: A group G is called cyclic if there exist at least one element x G such that G ={x n n Z} if the operation is ( ). Example(1); (Z 1,+ 1 ) is the trivial group {0} consisting of just an identity element So, (Z 1,+ 1 ) = <0>. ={nx n Z} if the operation is (+). X is called the generator for G. The cyclic group generated by x is denoted by <x>. Abstract Algebra(I), Math (342) Chapter 3- Cyclic Groups. 60

62 Example(2); If n 2, then (Z n,+ n ) = < 1>. For the powers 0, 1, 1 + n 1, 1 + n 1 + n 1,, 1 + n 1 + n 1 + n + n 1. n-times Z n = < 1 > = {n. 1 : n Z} = { 0, 1, 2,, n-1} Example(3); Let G ={1,-1,I,-i} where i=-1. then (G, ) is a group, Moreover it s a cyclic group generated by i,-i for (i) 1 =i (-i) 1 =-i (i) 2 =-1 (-i) 2 =-1 (i) 3 =-i (-i) 3 =i (i) 4 =1 (-i) 4 =1 Therefore G = <i> = <-i>. Example(4); Infinite cyclic. (Z,+) is a cyclic group generated by 1,-1 for Z = <1> = { n.1 : n Z} And = {,-2.1,-1.1,0.1,1.1,2.1,3.1, } = {,-2,-1,0,1,2,3, } Z = <-1> = { n(-1) : n Z} = {,-1,-2,-1.-1,-1.0,-1.1,-1,2, } = {,2, 1,0,-1,-2, } = {,-2,-1,0,1,2, } Abstract Algebra(I), Math (342) Chapter 3- Cyclic Groups. 61

63 Example(5); From Ex(2) elements other than T may be also generators. To illustrate this, consider the particular case ; Z 6 = { 0, 1, 2, 3, 4, 5}. 5 is a generator of Z 6 Z 6 = < 5> = { n. 5: n Z} = {0. 5, 1. 5, 2. 5, 3. 5, 4. 5, }. = { 0, 1, 2, 3, 4, 5}. The cyclic subgroup generated by the other elements of Z 6 under addition are. < 0> = { n. 0: n Z} = { 0} < 2> = { n. 2: n Z} = { 0, 2, 4} < 3> = { n. 3: n Z} = { 0, 3} < 4> = { n. 4: n Z} = { 0, 4, 2} Thus 1 and 5 are the only elements that are generators of the entire group. Example(6): Consider (3Z,+) this is a group of the multiple of 3 then this group is cyclic generated by 3 for 3Z = <3> = {n.3: n Z} = {,- 3,0,3,6, }. Abstract Algebra(I), Math (342) Chapter 3- Cyclic Groups. 62

64 In general nz= <n>, foe n Z, which is called the group of the multiple of n. Definition 3.1.2: Example(6); In Ex(5) 0( 5) = 6, 0( 2) = 3, 0( 3) = 2, 0( 4) = 3. In Ex(4) every non-zero element in Z is of infinite order. Theorem 3.1.3: Proof: i) let G be a group. The order of g G is the smallest +ve ii) integer n such that g n = e (ng = e if the operation is (+)). We denote it by 0(g) = n > 0. If there is no such integer we say that g has infinite order denoted by 0(g) =. Let G be a group and let g G then i) H : { g n : n Z} G. ii) H is the smallest subgroup of G containing g i.e, if K G containing g H K. 1) H since g ο = e H. H is closed under the group operation for if g r,g s H g r g s = g r + s H., r,s Z. Abstract Algebra(I), Math (342) Chapter 3- Cyclic Groups. 63

65 If y = g r H (g r ) -1 = g -r H, where r Z. Hence H contains inverses of all its elements H G. iii) Suppose H = <g> = {g n : n Z} And K G s.t. g K g g K. Inductively, its true that g g. g = g n K for n Z. that is x H s.t x = g n x = g n K. therefore H K i.e, H is the smallest subgroup of G containing g. Definition 3.1.4: Let G be a group and g G, then H ={g n : n Z } is a subgroup of G. generated by g. we write H = <g> = { g n n Z}. If G is a group such that for x G, G = {x m m Z}, then G is a cyclic group denoted by G = <x> Theorem 3.1.5: Proof: Every Cyclic group abelian. Let G = {x n n Z} = <x> be a cyclic group generated by x. Abstract Algebra(I), Math (342) Chapter 3- Cyclic Groups. 64

66 Let a,b G, then a=x r and b=x s. thus ab= x r x s = x s+r = x s x r =ba since s + r = r + s for r,s Z. Therefore, G is abelian. Theorem 3.1.6: (Division Algorthim for Z) If a,b Z s.t. b>0, then a unique integer q, r s.t. a=qb+r 0 r<b. Theorem 3.1.7: A subgroup of a cyclic group is cyclic. Proof: Let G be a cyclic group generated by a, and let H be a subgroup of G. if H={e}, then theorem is trivially true, for {e} is a cyclic subgroup generated by the identity element. Suppose that H {e}. If a m H a -m H since H G. H must contain a +ve powers of a. suppose n is the least +ve integer such that a n H. We have to show that any element in H is a power of a n, i.e, H = <a n > = {(a n ) t : t Z}?? Abstract Algebra(I), Math (342) Chapter 3- Cyclic Groups. 65

67 Let a k be an arbitrary element in H. since n is the least +ve integer such that a n H, then we can divide k by n and by division algorithm we get, K = qn + r Consider, a r = a k-qn = a k a -qn = a k (a n ) -q where 0 r < n a k,a n H a r H (since n is the least +ve integer Such that a n H, r<n). r = 0 k = qn i.e, k is a multiple of n. a k = ( a n ) q <a n > H = <a n >. Theorem 3.1.8: Let a be an element in a group G. If n is the least +ve integer such that a n = e, i.e, 0(a) = n. Then Proof: i) <a> has order n, and <a> = {a 0,a 1,a 2,,a n-1 } i.e, 0(<a>) ii) iii) = 0 (a). If a m = e n/m. If a k = a s k s (mod n). i) first we show that e,a,a z, a n-1 are all distinct. second, we show that any power of a is equal to these elements. Abstract Algebra(I), Math (342) Chapter 3- Cyclic Groups. 66

68 First, suppose a i = a j, 0 i < n, 0 j < n, i j. a i a -j = a j a -j = e a i-j = e 0 i-j < n. But n is the least +ve integer s.t. a n = e i-j = 0 i = j. Thus <a> contains the n-distinct elements e, a, a t,, a n-1. Second, let a k be any arbitrary element. By division algorithm q,r s.t. K= nq + r where 0 r < n, r = {0,1,2,,n-1} consider a k = a nq+r = a nq a r = (a n ) q a r = e q a r = ea r =a r. It follows that <a> = { e,a,a 2,, a n-1 } and <a> has order n. ii) ( ) suppose that n/m m = nq, q Z. Consider a m = a nq = (a n ) q = e q = e. ( ) suppose that a m = e, then by division algorithm q, r s.t. m = nq + r. 0 r < n, q Z then but a m = e a r = e s.t. r < n. r = 0 m = nq n/m. a m = a nq+r = ( a n ) q = e q a r = e q a r = a r. iii) let a k = a s a k a -s = a s a -s = a k-s = e. By (ii) n/k-s k-s = nq, q Z. k S ( mod n). Abstract Algebra(I), Math (342) 67

69 Chapter 3- Cyclic Groups. let k s ( mod n) k-s is divisible by n k-s =qn k = nq + s., q Z Consider a k = a nq+s = ( a n ) q a s = e q a s = e a s = a s Definition 3.1.9: The order of an element a (0(a)) of the group G is the order of the subgroup generated by a that is 0 (a) = 0 (<a>). Theorem : Let G be a finite cyclic group of order n (i.e, G = <a>, G = n) with generator a G. Then; 1) if d=(m,n) is the greatest common divisor for the integer m,n, then the subgroup generated by a d i.e, <a d > = <a m > or <da> = <ma> if its (+ve). 2) The distinct subgroups of G are those subgroups <a d > (<da> if the operation (+)) where d is a +ve divisor of n. 3) If (m,n) = 1, i.e, m,n are relatively prime, then a n is a generator of G. i.e, G = <a m > if the operation is ( ). <ma> if the operation is(+). Theorem provides a systemic way to obtain all subgroups of a cyclic group of order n. Abstract Algebra(I), Math (342) 68

70 Chapter 3- Cyclic Groups. Example(7); Let G = <a> be a cyclic group of order 12 : a) find all distinct subgroup of G. b) find all generators of G. solution: a) the divisors of 12 are 1,2,3,4,6 and 12 by th: , the distinct subgroups of G are <a> = G. <a 2 > = { a 2, a 4, a 6, a 8, a 10, a 12 = e} <a 3 > = { a 3, a 6, a 4, a 12 = e} <a 4 > = {a 4, a 8,e} <a 6 > = { a 6, e} <a 12 > = {e} b) By th 3.10 (3) all generators of G are (5,12) = 1 G = <a> = <a 5 > (7,12) = 1 G = <a> = <a 7 > (11,12) = 1 G = <a> = <a 11 > Note: theorem 3.10.(1) makes it easy to determine which subgroup is generated by each element of the group (8,12) = 4 <a 4 > = <a 8 > Abstract Algebra(I), Math (342) 69

71 Chapter 3- Cyclic Groups. (9,12) = 3 <a 3 > = <a 9 > (5,12) = 1 <a 2 > = <a 10 > Example(8); Let G = <a> be a cyclic group of order 10, then by theorem 3.10(3) G = <a> = <a 3 > = <a 7 > = <a 9 >. Example(9); 1) consider the group (Z 7,+ 7 ), then Z 7 is a cyclic group generated by [1]= 1. since Z 7 = 7, then by th. 3.10(3), every element of Z 7 except [0], generators Z 7 under addition. Z 7 = <ma> = <m 1>, where m is relatively prime with 7.i.e, Z 7 = < 1> = <2 1> = <3 1> = <4 1> = <5 1> = <6 1> Z 7 =< 1> =< 2> =< 3> = < 4> = < 5> = < 6>. 2) Consider (Z 7, ). Then this cyclic group generated by 3, for < 3> = {( 3) n : n Z}. = { 3 1, 3 2,, 3 6 }. = { 3, 2, 6, 4, 5, 1}. Hence Z 7 = < 3> of order 6. then by th.3.10(3) the other generator is Z 7 = < 3> = < 3 5 > since (5,6) = 1 but 3 5 = 5, So, Z 7 = < 3> = < 5>. Abstract Algebra(I), Math (342) 70

72 Chapter 3- Cyclic Groups. Example(10); Consider the cyclic group (Z 8,+ 8 ). i) find all generated of Z 8. ii) Find all subgrouop of Z 8. Solution: i) (Z 8,+ 8 ) is a cyclic group generated by 1. Z 8 = 8 n, then all numbers which are relatively prime with 8 are 3, 5, 7. so, by th.3.10(3). The generators of Z 8 other than 1 are Z 8 = < 1> = <3 1> = <5 1> = <7 1>. = < 1> = < 3> = < 5> = < 7>. ii) The divisors of 8 are 1,2,4,8 so by theorem 3.10(2), we get <1 1> = Z 8. <2 1> = < 2> = {n 2: n Z} = { 0, 2, 4, 6} <4 1> = < 4> = {n 4: n Z} = { 0, 4} <8 1> = < 8> = {n 8: n Z} = { 0} Exercise; As examples before find all generators and all distinct groups for the following cyclic group: (Z 12,+ 12 ), (Z 16,+ 16 ). Abstract Algebra(I), Math (342) 71

73 Chapter 3- Cyclic Groups. Example(11); Consider the group (Z 5, ) a) show that Z 5 is cyclic. b) Find all generators. c) Find all distinct subgroups. Solution: a) < 2> = {( 2) n : n Z} = { 2, 4, 3, 1}. Therefore Z 5 is a cyclic group generated by 2 i.e, Z 5 = < 2> and Z 5 = 4. b) By th.3.10(3) is the only number which is relatively prime with 4 therefore Z 5 = < 2> =<( 2) 3 > = < 3>. c) The divisors of 4 are 1,2,4. therefore the distinct subgroup are < 2 1 > = < 2> = Z 5 < 2 2 > = < 4> = {( 4) n : n Z} = { 4, 1} < 2 1 > = < 2> = {( 1) n : n Z} = { 1}. Do the same as in the example for a) (Z 13, ) b) (Z 17, ) Abstract Algebra(I), Math (342) 72

74 Chapter 3- Cyclic Groups. Example(12); Find the cyclic group generated by by σ = ( ) S 4. Solution: <σ> = <( ) > = {( ) n : n Z} = {σ 1,σ 2,σ 3,σ 4 } σ 2 = ( ) ( ) = (1 4) (2 3) σ 3 = ( ) ( ) ( ) = (1 4) (2 3) ( ) = ( ). σ 4 = ( ) ( ) = (1) (2) (3) (4) = i = id <σ> = {i,( ),(1 4)(2 3),( )} Note that 0(σ) = 4 = 0 (<σ>). Example(13); Find the order of α=( ) S 5. Solution: We want to find the smallest +ve integer n s.t. α n = i = id. α 2 = ( ) ( ) = ( ). α 3 = ( ) ( ) = ( ). α 4 = ( ) ( ) = ( ). α 5 = ( ) ( ) = (1) (2) (3) (4) (5) = i = id. 0(α)=5 the order of subgroup generated by α=5. Abstract Algebra(I), Math (342) 73

75 Chapter 3- Cyclic Groups. Example(14); Let α = (1 2 3) S 6, τ = ( ) S 6. find the cyclic subgroup generated by α -1 τ 2. i.e, <α -1 τ 2 >. Solution: α -1 = (1 3 2), τ 2 =(1 4)(2 5) and σ = α -1 τ 2 = ( ) σ 2 = ( ) ( ) = ( ). σ 3 = ( ) ( ) = ( ). σ 4 = ( ) ( ) = ( ). σ 5 = ( ) ( ) = (1) (2) (3) (4) (5) = i = id. <σ> = {i,σ,σ 2,σ 3,σ 4 }. Note: 0(σ) = 0(<σ>) and the 0(σ) = lenghth of the cycle σ. Theorem : Let G =<a> be an infinte cyclic group i.e, 0(a) =. Then 1) If n m Z a n a m. 2) a, a -1 are the only generators of G. Proof: 1) Let 0(a) = a k e for any +ve integer k. Suppose n m Z, s.t a n = a m. then a n a -m = a m a -m = e. Abstract Algebra(I), Math (342) 74

76 Chapter 3- Cyclic Groups. a n-m =e if n-m>0 or a m-n =e if m-n>0. In either cases we get that the element a hasfinite order. Therefore n m Z a n a m. 2) We want to prove that if G = <a> has infinite order then G = <a -1 >?? By the way of contradiction, suppose that G = <a m > where m ±1. a = (a m ) k = a mk, k Z a mk-1 =e, mk-1 Z. therefore the element a has finite order. Thus the only generators are a,a -1. Example(15); In (Z,+), the only generators are 1,-1, i.e, Z = <1> = <-1>. (Z e,+), has two generators 2,-2, i.e, Z e = <2> = <-2>. 75

77 Problems Set III 1. Consider the cyclic group (Z 15, ) and (Z 18, ) i) find all generators of Z 15, Z 18. ii) find all subgroups of Z 15, Z Consider the cyclic group (Z 17, ) i) find all generators. ii) find all distinct subgroups. 3. Suppose G = <a> is a cyclic group of order 24. list all generators. 4. Find the order of each of the elements 4, 5, 6 in the group ( Z 7, ). 5. Find the order of each of the elements 2, 3, 4 in the group ( Z 8, ). 6. Let G be a group a,x G. Suppose b = x -1 ax. Prove the following: (i) b n = x - 1a n x n Z +. ( use induction). (ii) If 0(a) = n 0(bab -1 ) = n. 7. Find the elements of ( Z 12, ) of order a) 2, b) 3, c) 4, d) 6, e) Show that the group (S, ), where S is the subset 76

78 { 1, 3, 5, 7} of Z 8 is not cyclic. 9. Prove that the group (H, ), where H is the subset H= { 2, 4, 6, 8} of Z 10 is cyclic. Find the generators. 10. Let G be a group, and x,y G. Use induction to show that (xy) n = y -1 (yx) n y 11. Find the number of elements in the indicared cyclic group: a. The cyclic subgroup of Z 30 generated by 25. b. The cyclic subgroup of Z 42 generated by