Class Equation & Conjugacy in Groups

Transcription

1 Subject: ALEBRA - V Lesson: Class Equation & Conjugacy in roups Lesson Developer: Shweta andhi Department / College: Department of Mathematics, Miranda House, University of Delhi Institute of Lifelong Learning, University of Delhi 1

2 Table of Contents 1. Aims and Objectives 2. Introduction 3. Pre-requisites and Notations 4. Conjugate elements and conjugacy relation 5. Class Equation 6. Conjugacy in S n 7. Conjugacy in A n Vs. S n 8. Simple groups 9. Exercises 10. References Institute of Lifelong Learning, University of Delhi 2

3 1. Aim and Objectives In this chapter, we introduce the concept of conjugate elements and conjugacy relation in order to derive the class equation of a finite group. We will see that every group can be partitioned into disjoint conjugacy classes. Members of the same conjugacy classes of non-abelian group share many important features of their structure. After going through this chapter, we will be able to: Define conjugacy relation through group action. Study conjugacy relation as an equivalence relation. Derive class equation of finite groups. Understand the concept of partition of a positive integer n, cyclic decomposition type and conjugacy classes in S. n 2. Introduction Many extensive theories have been developed to know the structure of a group. One of them is counting principle. Counting principle is a powerful method to study various important concepts in Algebra. The arithmetic of conjugacy classes of a group like the number of conjugacy classes and conjugacy class sizes has helped in many ways to know the structure of a group. Normal subgroup is the union of some or all conjugacy classes of the group, therefore conjugacy classes help us to construct normal subgroups. roups are useful if we use them on other structures and group actions provide one such platform. Conjugacy is a very special type of group action which provides a lot of information about the group. All the elements of a conjugacy class are same" in some sense, for example geometrically reflection across any line through the origin (making an angle of θ with the positive x-axis) and reflection across the x-axis are same. Reflection across any line through the origin (making an angle of θ with the positive x-axis) can be viewed as rotation of the plane by an angle of θ to bring the line of reflection onto the x-axis followed by reflection across the x-axis and rotation of the plane by θ to return the line to its original position. In this chapter, we will explore the inner structure of a group, by partitioning the elements of group with the help of conjugacy relation among the elements of a group. Before giving formal definitions let us recall and discuss various concepts required in this chapter. 3. Prerequisites and Notations Definitions and Results Notations Cardinality of a finite set: Number of elements in a finite set is known as its cardinality. O() Institute of Lifelong Learning, University of Delhi 3

4 roup Action: Let be a group with identity element e and S be any set. Then an action of on S is a map from S S defined by ( g, s) g. s S, s S and g S Satisfying the following conditions: 1. g.( g. s) ( g g ). s, g, g, s S e. s s, s S. Matrix roup: Matrix roup is set of square matrices of order n over a field with non-zero determinant. L ( ) n F, where F is a field. 4. Conjugate elements and conjugacy relation Definition 4.1 (Conjugate elements): Let be a group and a, b. Then a is 1 said to be a conjugate of b in if there exists c such that a cbc Result 4.2: Relation of conjugacy is an equivalence relation. Notation: Symbol a ~ b will denote that a is conjugate to b. Value Addition Conjugacy can be studied via group action called conjugation. Conjugation is a 1 group action of on itself defined by : as ( g, a) gag. Two elements are said to be conjugate in if and only if they are in the same orbit of acting on itself by conjugation. Thus gets partitioned into equivalence classes or the orbits associated with group action conjugation. Definition 4.3 (Conjugacy class): Conjugacy class of any element a is denoted by cl(a) and is defined as : cl a gag g 1 ( ) { : }. Example 4.4 Let be quaternion group , 1, i, j, k, i, j, k : i j k ijk 1. It can be seen that cl(1) { g1 g 1 : g } {1}, cl g g g 1 ( 1) { ( 1) : } { 1} cl( i) { g( i) g 1 : g } { i, i} { g( i) g 1 : g }. Similarly cl j j, j and, cl k k k. Institute of Lifelong Learning, University of Delhi 4

5 Recall that permutation group S n is the set of one-one onto maps on {1, 2,...,n}. Example 4.5 Let = S 3 that is, permutation group on three elements S3 { I, (12), (13), (23), (123), (132)}. Now we find the conjugacy classes of S 3 : cl( I) { I 1 : } { I } 1 cl(12) { (12) : } (1) (12) (13) (23) (123) (132) (12) 1 (12) (12) (23) (13) (23) (13) Therefore cl cl cl , 13, cl(123) { (123) : } (1) (12) (13) (23) (123) (132) (123) 1 (123) (132) (132) (132) (123) (123) Therefore cl(123) {(123), (132) cl 32 } 1. Example 4.6 When is an abelian group, conjugacy class of every element in the group is a singleton set. If is abelian group the action of on itself by conjugation is the trivial action 1 ( g, a) gag a. Recall that dihedral group D 2n is set of symmetries of regular polygon with n sides Example 4.7 Let D 8 be dihedral group i.e. set of symmetries of square. Thus it comprises of eight elements. { R, R, R, R, V, H, D, D } D D Institute of Lifelong Learning, University of Delhi 5

6 Where V, H, D and D are the reflection symmetries as shown above. R 0, R 90, R 180, R 270 is rotation about 0, 90, 180 and 270 respectively, then cl( R ) { R } Cl( R ) R (since R 0 and R 180 commute with every element). cl( R ) { gr g : g D } { R, R } Cl( R ) cl( V) { gvg : g D } { V, H} cl( H) 1 8 cl( D') { gd ' g : g D } { D', D"}. 1 8 Value Addition Conjugacy classes for the matrix group L ( 2 ) turn out to be very interesting. Recall that two matrices A & B are said to be similar if there exists a nonsingular matrix P such that A = PBP 1. Let A L ( 2 ). Now we find conjugacy classes of A. We have following three cases. 1) A has two distinct eigenvalues. In this case, A is diagonalizable and cl( A) { PAP : P L ( )}. Since A is 1 2 diagonalizable A is similar to diagonal matrix of the form where 1 and 2 are eigenvalues of A. As all the members of cl(a) are similar to matrix of the form , we have cl(a) consisting of matrices having eigenvalues 1 and 2. Corresponding to every pair { 1 2 } we have distinct conjugacy class comprising of matrices having eigenvalues 1 and 2. 2) A has repeated eigenvalues but A is a diagonalizable. In this case A is similar to matrix of the form 0 where is the eigenvalue of 0 0 A. Corresponding to every, we have distinct conjugacy class 0 comprising of diagonalizable matrices having single eigenvalue. 3) A is not diagonalizable In this case A has repeated eigenvalues otherwise A is diagonalizable. We can 1 block diagonalize A and therefore A is similar to matrix of the type, where 0 is the eigenvalue of A. Corresponding to we have conjugacy class comprising of matrices having single eigenvalue which are not diagonalizable. Institute of Lifelong Learning, University of Delhi 6

7 Recall that center Z() of a group comprises of those elements of the group which commutes with every element. We can see the following observations from our examples 1. Since conjugation is an equivalence relation we have cl( a ), where the union runs over mutually disjoint conjugacy classes. 2. If a Z( ) then cl( a) { a }. We define such an element as self conjugate element, central element or invariant element. Thus a Z( ) if and only if a is self conjugate element. In order to know various properties of a group, we study the cardinality of conjugacy classes, their relation with the order of the group and various other properties of conjugacy classes. Recall that centralizer of an element a denoted by C (a) is the set of those elements of which commute with a in. It can easily be seen that C (a)is a subgroup of and a Z( ) if and only if C a. In the context of cardinality of conjugacy classes, we state the following theorem which can be easily seen as a consequence of orbit stabilizer theorem by applying to action of by conjugation. Theorem 4.4 If is afinite group and o ( ) a then o( cl( a)) : C ( a) o( C ( a)), where : C ( a ) is the index of C (a) in. Number of conjugates of an element a in a group is the index of the centralizer of a. 5. Class Equation Let be a finite group and ~ be the conjugacy relation on. Then is the union of distinct conjugacy classes. Let the number of distinct conjugacy classes be m, and denote the conjugacy classes by cl( a1), cl( a2),..., cl( a m), where a1, a2,..., a m are representative of cl( a ), cl( a ),..., cl( a ), respectively. Therefore 1 2 m cl( a ) where a. i 1 i i 1 m i m o( ) o( cl( a )) where a ( cl( a ) are disjoint) i i i a m o( ) : C ( a) where a. i i 1 i Also, if ai Z( ) we get o( cl( ai )) 1 ( cl( ai ) { a i}). Therefore, we have o( ) ( O( Z( )) times) : C ( i a). (1) ai Z( ) Institute of Lifelong Learning, University of Delhi 7

8 Definition 5.1 Then above equation where o() can be represented in terms of cardinality of Z() and conjugacy classes of a i not in Z() is called class equation of. Examples of class equation Example 5.1 If is an abelian group of order n then class equation of is n (n times). Since is an abelian group, conjugacy class of every element in the group is a singleton set. Example 5.2 Consider the Quaternion group Q {1, 1, i, j, k, i, j, k : i j k 1, ij k, jk i, ki j } Also ZQ ( ) {1, 1} and therefore, cl{1} {1}, cl { 1} { 1} 8 Also cl{ i} { i, i}, cl( j) { j, j}, cl( k) { k, k }. Hence the conjugacy classes are {1}, { 1}, { i, i}, { j, j}, { k, k }. Therefore, class equation is Example 5.3 Let = S 3, permutation group on three elements. Then S3 { I, (12), (13), (23), (123), (132)}. Also, Z( S ) { I } we have cl I I. 3 c ((12)) {(12), (13), (23)} C ((123)) {(123), (132)}. Therefore, class equation is Example 5.4 Let =D 8 be the dihedral group of order 8. Then we have, Z( D8) { R, R } cl( R ) { R, R } cl( R ) cl( V) { V, H} cl( H ) cl( D') { D', D"} cl( D "). Therefore, class equation is Remark: Two non-isomorphic groups can have same class equation. For example Q 8 and D 8 have the same class equation but they are not isomorphic. Institute of Lifelong Learning, University of Delhi 8

9 Applications Theorem 5.5 If is a group of order p n for some prime number p and n 1, then has a non- trivial center. Proof: Consider the class equation of O( ) ( O( Z( )) times) : C( a), az ( ) where the summation runs over exactly one element from every conjugacy class. If =Z() then clearly Z() is non- trivial. So assume Z(). Consider a Z( ). Then by the choice of a, there exists x satisfying xa ax. This shows C( a). Since C (a) is a subgroup of, by the application of Lagrange's theorem o( C( a)) o( ). m Thus order of C (a) is of the form of p (0 m n ) and therefore, : ( ) n m a p where n m 1 C. Thus p divides : C ( a) for all a Z( ) and we get p divides : C ( a). This give az ( ) shows p divides o( Z( )). Hence Z ( ) is non-trivial. In order to prove next theorem we prove a lemma. Lemma 5.6 If a Z( ) then Z ( ) Ø C (a) Ø Proof: It can be easily verified that p divides o( ) : C ( a) which az( ) Z( ) ( a ). Also as a C ( a) and given a Z( ), we have Z( ) Ø ( a ). Since a Z( ), therefore there exists x such C that x C( a ) and C ( ) a Ø proving the desired result. Theorem 5.7 Every group of order p 2 is abelian where p is a prime. Proof: Let be a group of order p 2 where p is a prime. Consider the class equation of o( ) ( O( Z( )) times) : C ( a ). By theorem 5.5 we have o(z()) 1. az ( ) By the application of Lagrange's theorem the possibility of order of Z() is p or p 2. C Institute of Lifelong Learning, University of Delhi 9

10 If o(z()) = p 2 then we are done. If a o( Z( )) p, then there exists 5.6 we get Z( ) Ø C ( a) Ø. such that a Z( ) and from the Lemma Also C (a) is a subgroup of, therefore o(c (a)) divides p 2. Thus 2 o( C( a)) 1, p or p. Since Z( ) Ø C ( a) Ø, we have p o( C ( )) 2 a p which is impossible. Thus we get o( Z( )) p. This proves the result. The above result can also be proved from the following more generalized result and Theorem 5.5. Theorem 5.8 If is a finite group of order n p, then n1 o( Z( )) p. n1 Proof: Suppose o( Z( )) p, then o p. Z ( ) Since every group of prime order is cyclic, we get Z ( ) is cyclic. n This implies is abelian and o( Z( )) o( ) p, which is contradiction to our assumption and therefore we get n1 o( Z( )) p. Now we prove another very important consequence of class equation. Theorem 5.9 Let be non-abelian group of order p 3, where p is prime then (a) o( Z( )) p (b) a Z( ) o( C ( )) 2 a p. 2 (c) The number of distinct conjugacy classes of is p p 1. Proof: (a) Let be a non-abelian group of order p 3. By the application of Lagrange's theorem, we get Case I. o( Z( )) 1. o( Z( )) 1, p, p 2 or p 3. This case is not possible because by Theorem 5.5, has non-trivial center. Case II. o( Z( )) p 3 This case is not possible because is non-abelian. Case III. o( Z( )) p 2 Since o( Z( )) p 2, Z( ) Ø and therefore there exists a such that a Z( ). By Lemma 5.6 we have Z( ) Ø C ( a) Ø and this 2 3 gives p o( C( a)) p. Since C( a ) is a subgroup of, o( C( a )) divides p 3 we get a contradiction. Institute of Lifelong Learning, University of Delhi 10

11 Thus this case is not possible and o( Z( )) p. Part (b) can be proved by an argument given in case III. of part (a). If a Z( ) we have Z( ) Ø C ( a) Ø and o( C ( )) 2 a p. For part (c) consider the class equation of group o( ) ( O( Z( )) times) : C ( a ). Let the number of distinct conjugacy classes be k. az ( a) Number of conjugacy classes corresponding to elements in the center is o( Z( )) p. Therefore the number of classes corresponding to elements not in center is k p and its cardinality is p. 3 Thus by substituting these numbers in class equation we get p p ( k p) p 2 which implies k p p 1. Note: The class equation of Q 8 and D 8 is which is non-abelian groups of order 2 3. Now we prove few important consequences of class equation related to normal subgroups. Theorem 5.10 If N is a normal subgroup of a group then either a conjugacy class C is contained in N or N does not have any member of the conjugacy class C. Proof: Let N be a formal subgroup of a group. Let C be any conjugacy class of. We have following two cases. Case I. We are done. Case II. N C. N C. Let x N C then cl(x) = C. Now we show that C N. Let y C and since y cl( x ), we have y gxg g 1 for some. Also as x N and N is a normal subgroup of, 1 gxg N which gives y N. Corollary 5.11 Every normal subgroup of a group is union of some or all conjugacy classes of that group. Now we give another important theorem which reveals few properties of normal subgroups. Theorem 5.12 If is a group of order p n for some prime number p and n 1 and N is a proper non trivial normal subgroup of then N contains an element of Z() other than the identity element of the group. Proof: Consider the class equation of Institute of Lifelong Learning, University of Delhi 11

12 o( ) ( O( Z( )) times) : C( a ). az( ) Let N be a nontrivial proper normal subgroup of. From the Theorem 5.10, we have Therefore, cl( x) N or cl( x) N for every x. o( N) o( Z( ) N) o( cl( a )). az ( ) an Since p o(n) and also p o(cl(a)) for every a Z( ), a N, we have p o( Z( ) N ). Therefore, we get Z( ) N { e }. Now we provide one more application of class equation. Example 5.13 Find all the possible class equations of a group of order 10. Hence find the possible orders of normal subgroups of such a group. Let be a group of order 10. We have following two cases: Case I. is abelian. Then class equation is times and all subgroups of are normal. Caser II: is non-abelian Since o( Z( )) 10, we have o( Z( )) 1, 2 or 5. If If o( ) o( Z( )) 2 then o o( Z( )) Z( ) = 5. o( ) o( Z( )) 5 then o 2. o( Z( )) Z( ) In both the cases Z ( ) is of prime order and hence cyclic. This shows is abelian which is contradiction to case taken. Thus o(z()) = 1, and 10 =1+k 1 +k k n, where k1 k2... k n 9 and each k i divides 10. Thus, for any i we have ki 2 or 5. Now all k i s cannot be 2 and similarly all k i s cannot be 5. Thus atleast one k i is 2 and k j = 5. Therefore class equation is 10 = Now we find possible orders of normal subgroups. By Corollary 5.11 the possible order of normal subgroups is 1, 1+2, 1+5, 1+2+2, or Institute of Lifelong Learning, University of Delhi 12

13 Now 1+2, 1+5, cannot be the order of subgroups by Lagrange's theorem. Thus possible normal subgroups are of order 1, 1+2+2, Example 5.14 Write class equation of group of order 2p where p is an odd prime. Let be a group of order 2p, then by Lagrange's theorem o( Z( )) 1,2, p or 2p Case I: is abelian, then o(z()) = 2p and the class equation of is 2p (2 ptimes ) Case II: is non-abelian o( Z( )) 1, 2 or p But o( Z( )) cannot be 2 or p otherwise number. o p Z ( ) or 2 which is a prime This shows that is abelian, which is a contradiction to the case taken. Thus Z ( ) is trivial. Also all the classes cl(a) for a Z( ) cannot be of order 2 because order of the group is even, therefore we have a conjugacy class of cardinality p and remaining classes are cardinality of 2. Thus 2p 1 p 2k, where k is the number of conjugacy classes of order 2. This shows that p 1 k. 2 Therefore the class equation of is 1 2p 1 p times. p 2 Value addition If is a group with trivial center then the center of the group Aut() (set of automorphisms from to ) is trivial. 6. Conjugacy in S n If is a permutation group then we have a simple rule to find conjugacy classes. We now introduce various definitions and concepts required to obtain class equation of S n. Definition 6.1 (Partition of a positive integer): Let n be a positive integer. A sequence of positive integers n1, n2,... n k is said to be a partition of n if 1) n1 n2... n k 2) n1 n2... nk n. Institute of Lifelong Learning, University of Delhi 13

14 We denote the partition of n by { n1, n2,... n k }. For example {1,2,3}, {3, 3}, {2, 2, 2} are partitions of 6. It can easily be seen that a positive integer n can have several partition. Let p(n) denote the number of partitions of n. It can be computed that p(1) = 1, p(2) = 2, p(3) = 3, p(4) = 5, p(5) = 7 and p(6) = 11. Definition 6.2 (Cycle decomposition type): Every member f of permutation group S n can be expressed as a product of disjoint cycles of lengths n1, n2,..., n k (including 1-cycles) with n1 n2... n and k n1 n2... nk n. Then the finite sequence { n1, n2,..., n k } is called cycle decomposition type of f. For example (123) S 7 has cycle decomposition type {1, 1, 1, 1, 3} and (12)(45) S 5 has cycle decomposition type {1, 2, 2}. Hence each cylcle type determines a partition on n. Relation between conjugacy and similarity Now we prove a result which will be used to prove an important theorem which relates conjugacy and similarity. Result 6.3 Let f S n be a permutation which can be represented as product of disjoint cycles as ( a a... a )( b b... b )...( c c... c ) then for any 1 2 n1 1 2 n2 1 2 n3 g a1 g a2 g an g b1 g b2 g bn g c1 g c2 g c n ( ) ( )... ( ) ( ) ( )... ( )... ( ) ( )... ( ) g S, n gfg is That is, all conjugates of an element have same cycle decomposition type. Proof: Let f be any permutation in S n and a, b {1, 2,..., n} such that f(a) = b. Now consider gfg 1 ( g ( a )) gf ( a ) g ( b ). Therefore, shows 1 gfg has same cycle decomposition type as f. 1 gfg maps g( a) to ( ) g b which Theorem 6.4 Two elements of S n are conjugates in S n if and only they have the same cycle decomposition type. Proof: Let f and g be two permutations in S n which are conjugates. Then h S. 1 g hfh for some n By the Result 6.3, g and f have same cycle decomposition type. Conversely, suppose f and g have same cycle decomposition type say { n 1, n 2,... n k }, where n1 n2... nk n. f ( x x... x )( y y... y )...( z z... z ) 1 2 n1 1 2 n2 1 2 n k g ( p p... p )( q q... q )...( r r... r ). 1 2 n1 1 2 n2 1 2 n k Also n1 n2... n k n and x i, y i and z i exhausts all of {1,2,3,...,n}. Consider an element of S n as follows: Institute of Lifelong Learning, University of Delhi 14

15 x1x2... x y 1 1y2... y z 2 1z2... z p1 p2... pn q 1 1q2... qn r 2 1r2... rn k n n nk. maps xi to pi, yi to qi and zi to ri. Now consider 1 f ( ( x )... ( x )( ( y )... ( y ))( ( z )... ( z ) 1 n1 1 n2 1 n k (since f 1 maps ( x ) to ( xi 1) and similarly others). i Thus we get 1 f ( p p... p )( q q... q )...( r r... r ) which proves f and g are conjugate. 1 2 n1 1 2 n2 1 2 n k Corollary 6.5 There is a one-one correspondence between the collection of conjugacy classes of S n and the partitions of n. These results help us to find Normalizer of an r-cycle C Sn( ) where is any cyclic permutation in S n. Before stating a generalized result we prove a particular case in the form of following example. Example 6.6 Let (123) S 7, find C S7(123). To find C S7(123), firstly we find cardinality of C S7(123). From Theorem 4.4, os ( 7) o( CS7(123)) where cl(123) is the set of conjugates of (123). o( cl(123)) Also from Theorem 6.4, Conjugates of (123) in S 7 are precisely the 3-cycles. Three numbers out of the set {1,2,3,...,7} can be chosen in 7 C 3 ways and we can form (31)!=2 different 3-cycles out of these. Thus we have 7 C 3 (3 1)! many 3-cycles in S 7 which are conjugate to (123). Therefore, cardinality of centralizer of (123) is ( 7(123)) OCS 7 C 3 7! 7!(7 3)!3! (3 1)! 7!2! = 4! 3 = 72. Also, (123) commutes powers of (123). Every element S 7 which keeps 1,2 and 3 fixed that commutes with (123). Number of such permutations are 24. Thus we get that all the permutations which commute with (123) are of the form (123) i, where i = 0, 1, 2 and is a permutation which keeps 1, 2, and 3 fixed. There are 34!=72 permutations. Therefore we have, CS 7(123) {(123) i : i 1,2,3 and is a permutation which keeps {1,2,3} fixed}. We can generalize the above result and the proof follows similarly. Theorem 6.7 If is an r-cycle, then O( C Sn ())= r(n r)! and Institute of Lifelong Learning, University of Delhi 15

16 CSn( ) { i : Sn r, i 0,1,2,..., r 1}. Now we do a similar exercise by taking product of disjoint cycle. Example 6.8 Find all the permutations in S n which commute with = (12)(34). First, we find the cardinality of We know oc ( Sn(12)(34)) C Sn ((12)(34)). os ( n). o( cl(12)(34)) Cardinality of conjugacy class of (12)(34) is the number of permutations having same cyclic decomposition type as (12)(34). Let = (12)(34). Now first place in can be chosen in n ways, second place in (n1) ways, third place in (n-2) ways and fourth place in (n3) ways. Thus we have n( n 1)( n 2)( n 3) ways to choose any having same cyclic decomposition type as. Any such can be represented in 8 ways that is (12)(34) has 8 representations as follows: (12)(34),(21)(34),(21)(43),(12)(43),(34)(21),(43)(12),(43)(21) and (34)(12). Thus by the counting principle n( n 1)( n 2)( n 3) has 8 repetitions and therefore the distinct permutations which have the same cyclic decomposition type as (12)(34) are n( n 1)( n 2)( n 3) o(( cl (12)(34))) in number. 8 This gives OC ( Sn(12)(34)) 8 n!. n( n 1)( n 2)( n 3) Since disjoint cycles commute, (12)(34) commute with all the permutations which fixes 1, 2, 3 and 4. Number of such permutations in S 4 in (n 4)!. Number of such is (n 4)! (Equivalent to S n4 ). Now let be any permutation which fixes 5,6,7,...,n. Then commutes with if and only if 1 1 (12)(34) (12)(34) ( (1) (2)) ((3) (4)) (12)(34). Now (12)(34) on right hand side can be written in eight ways and therefore we have following eight options for. i) (1) 1 (2) 2 (3) 3 (4) 4 i.e. is identity element. ii) (1) 2 (2) 1 (3) 3 (4) 4 i.e. is (12) (1) 2 (2) 1 (3) 4 (4) 3 i.e. is (12)(34) iii) iv) (1) 1 (2) 2 (3) 4 (4) 3 i.e. is (34) v) (1) 3 (3) 1 (2) 4 (4) 2 i.e. is (13)(24) Institute of Lifelong Learning, University of Delhi 16

17 vi) (1) 4 (4) 1 (2) 3 (3) 2 i.e. is (14)(23) vii) (1) 3 (3) 2 (2) 4 (4) 1 i.e. is (1324) viii) (1) 4 (4) 2 (2) 3 (3) 1 i.e. is (1423). Thus can be (12),(34),(12)(34),(13)(24),(14)(23),(1423),(1324) which commutes with. Also if 1 and 2 commutes with then 1 2 also commutes with. Thus we have Therefore, we have 8( n 4)! permutations which commutes with. C Sn(12)(34) { (12), (34), (13)(24), (12)(34), (14)(23) (1324), (1423), : is a permutation which fixes {1,2,3,4}. 7. Conjugacy in A n Vs Conjugacy in S n. Recall that A n is the alternating group on n symbols consisting of even permutations in S n. Result 7.1 Two conjugate permutations in A n have same cycle decomposition type but converse may not be true. Converse is only true in S n. Proof: Consider two permutations f and g in A 5, as (12354) and (12345) respectively. Since f and g are 5-cycles, they are even permutations and same cycle decomposition type. If f and g are conjugate in A 5 there exists A 5 such that g 1 f ( (1) (2) (3) (4) (5)) (12354) Also (12354) can be written in 5-ways, therefore we have following 5 cases. Case I. (1) 1 (2) 2 (3) 3 (4) 5 (5) 4 (45) is odd permutation. Case II. (1) 2 (2) 3 (3) 5 (4) 4 (5) 1 (1235) is odd permutation. Case III. (1) 3 (2) 5 (3) 4 (4) 1 (5) 2 (134)(25) is odd permutation. Case IV. (1) 5 (2) 4 (3) 1 (4) 2 (5) 3 (153)(24) Institute of Lifelong Learning, University of Delhi 17

18 is odd permutation. Case V. (1) 4 (2) 1 (3) 2 (4) 3 (5) 5 (1432) is odd permutation. Thus can never be even permutation. Therefore f and g are not conjugates in A 5. Exercise 7.2 Show that any two 3-cycles in A n are conjugate in A n for n 5. Also show that the result does not hold for n = 4. We assert that every 3-cycle is conjugate to (123) in A n, then result follows by transitivity. Let A be a 3-cycle then is conjugate to (123) in S n by Theorem 6.4, therefore If (123) 1, for some S n., the result follows. If n An Now consider ((45) ) ((45) ) 1 (45) (45) (45)(123)(45) (123). n 1 A then consider (45), then A n. Thus and (123) are conjugate in A n and this proves the result. Consider two 3-cycles (234) and (243) in S 4. Since they are conjugate in S 4 there exists S 4 such that (234) 1 =(243). This implies ( (2) (3) (4)) (243). Now (243) can be written in 3-ways and therefore has 3 possibilities and none of these lies in A 4. Thus (234) and (243) are not conjugate in A 4. Exercise 7.3 Find all permutations in A 5 which are conjugate to 1) (12345) 2) (12)(34). For part (1), we have to find C Sn (12345). We know that C Sn (12345) = {(12345) i, i = 0, 1, 2, 3, 4: is a permutation which keeps {1,2,3,4,5} fixed}. Thus, CSn(12345) {(12345) i, i 0,1,2,3, 4}. Thus cardinality of normalizer of ( ) is 5 and OA ( ) 60 O( cl(12345)) OC ( Sn(12345)) 5 In order to find C Sn ((12)(34)), we know that by theorem 6.8, eight permutations namely (12)(34), (12)(34), (13)(24), (14)(23), (1324), (1423) commute with Institute of Lifelong Learning, University of Delhi 18

19 (12)(34) in S 5. Thus OC ( Sn(12)(34)) 4 in A 5 (since only four of them belong to 60 A 5 ), therefore we get O( cl(12)(34)) 15 in A 5, which is same as in S 5. Thus 4 conjugacy class of (12)(34) remain the same in S and 5 A 5 Exercise 7.4 Determine the class equation of A 5. Clearly cl(i) = {I}, where I is identity permutation. By theorem 7.2 cl ((123)) {3-cycles}. Number of 3-cycles can be calculated by counting principle. 3-cycle is of the type (abc) where a can be chosen in 5-ways, b in 4-ways and c in 3-ways. Thus 543 are possible ways. But every 3-cycle can be written in 3-ways. For example (123) = (231) = (312), therefore Hence o( cl (123)) are distinct 3-cycles. Conjugates of (12)(34) are the permutations with cyclic decomposition type same as (12)(34) Thus there are fifteen such permutations chosen in 15 ways. 8 Similarly, as discussed previously cl(12345) consists of twelve 5-cycles conjugate to (12345) and cl(12354) consists of twelve 5-cycles conjugate to (12354). This gives a total of 60 elements and therefore class equation of A 5 is 60 = Simple groups Definition 8.1 A group is said to be simple if it does not have a normal subgroup other than {e} where e is the identity element of the group and the group itself. Example of Non-simple group Example 8.2 Every non-trivial abelian group of composite order is not simple. Example 8.3 Permutation group S 3 on {1,2,3} is not-simple as {I, (123),(132)} is its normal subgroup. Now using the concept of class equation and conjugacy, we prove A 5 is simple. Theorem 8.4 A 5 is simple. We know the class equation of A 5 is 60 = Any normal subgroup of A 5 is union of some or all conjugacy classes of A 5. Therefore possible orders of normal subgroups are 1, 1+12, 1+15, 1+20, , , , , , , , Institute of Lifelong Learning, University of Delhi 19

20 Out of these only 1 and divide 60. Thus it has only two normal subgroups {I} and A 5 itself. Exercises 1. Show that conjugacy is an equivalence relation? 2. Let be a group and a, b. If a is conjugates to b then o( a) o( b ) If is a group of odd order and x, show that x is not in cl(x)? 4. Let be a finite group of order n. Then largest possible cardinality of any conjugate class is n, where [n] denotes the greatest integer Find class equation of non-abelian group of order Let be a finite group and a,b find the probability that a and b commutes. 7. Let A n be a permutation show that cl() in S n remains as a single conjugacy class in A n or it breaks into two conjugacy classes in A n of equal size. 8. Show that A 3 is simple whereas A 4 is not? 9. Find all conjugacy classes of S Find the class equation of Let be a group of all symmetries of the square then the number of conjugate classes is (a) 4 (b) 5 (c) 6 (d) Determine which of the following cannot be class equation (a) 10 = (b) 4 = (c) 8 = (d) 6 = Let be a group of order 49. Then (a) is abelian (b) is cyclic (c) is non-abelian (d) Centre of has order In the permutation group S 6, the number of elements conjugate to (123) (456) is (a) 80 (b) 120 (c) 60 (d) Let (12) (3 4 5) and ( ) be permutations is S 6. Which of the following statement is true: a) and commute b) and are conjugate in S 6 c) is trivial where is subgroup generated by. d) are isomorphic to each other. Institute of Lifelong Learning, University of Delhi 20

21 REFERENCES 1. allian, Joseph A., "Contemporary Abstract Algebra", Narosa Publishing House, New Delhi, Indian Reprint Herstein, I.N., "Topics in Algebra", Wiley India Edition Publications, Third Reprint, Dummit, David S.; Foote, Richard M., "Abstract Algebra", John Wiley and Sons (Asia) Pvt. Ltd., Singapore, Kapur J.N., Kalra, K.R. "Modern Algebra". 5. Khanna, Vijay K.; Bhambri, S.K., "A Course in Abstract Algebra", Vikas Publishing House Pvt. Ltd., New Delhi, Singh, Surjeet; Zameeruddin, Qazi, "Modern Algebra" Vikas Publishing House Pvt. Ltd., New Delhi, Seventh Revised Edition, Institute of Lifelong Learning, University of Delhi 21