Abstract Algebra Past Paper 20MAT21C1 PDF

M. Sc. (Mathematics) Semester-I ABSTRACT ALGEBRA Paper Code: 20MAT21C1 DIRECTORATE OF DISTANCE EDUCATION MAHARSHI DAYANAND UNIVERSITY, ROHTAK (A State...

M. Sc. (Mathematics) Semester-I ABSTRACT ALGEBRA Paper Code: 20MAT21C1 DIRECTORATE OF DISTANCE EDUCATION MAHARSHI DAYANAND UNIVERSITY, ROHTAK (A State University established under Haryana Act No. XXV of 1975) NAAC 'A+’ Grade Accredited University  Author Dr. Jagbir Singh Assistant Professor, Department of Mathematics Maharshi Dayanand University, Rohtak Copyright © 2021, Maharshi Dayanand University, ROHTAK All Rights Reserved. No part of this publication may be reproduced or stored in a retrieval system or transmitted in any form or by any means; electronic, mechanical, photocopying, recording or otherwise, without the written permission of the copyright holder. Maharshi Dayanand University ROHTAK – 124 001  MASTER OF SCIENCE (MATHEMATICS) First Semester Paper code: 20MAT21C1 Abstract Algebra M. Marks = 100 Term End Examination = 80 Assignment = 20 Time = 3 hrs Course Outcomes Students would be able to: CO1 Apply group theoretic reasoning to group actions. CO2 Learn properties and analysis of solvable & nilpotent groups, Noetherian & Artinian modules and rings. CO3 Apply Sylow's theorems to describe the structure of some finite groups and use the concepts of isomorphism and homomorphism for groups and rings. CO4 Use various canonical types of groups and rings- cyclic groups and groups of permutations, polynomial rings and modular rings. CO5 Analyze and illustrate examples of composition series, normal series, subnormal series. Section - I Conjugates and centralizers in Sn, p-groups, Group actions, Counting orbits. Sylow subgroups, Sylow theorems, Applications of Sylow theorems, Description of group of order p2 and pq, Survey of groups upto order 15. Section - II Normal and subnormal series, Solvable series, Derived series, Solvable groups, Solvability of Sn-the symmetric group of degree n ≥ 2, Central series, Nilpotent groups and their properties, Equivalent conditions for a finite group to be nilpotent, Upperandlower central series. Composition series, Zassenhaus lemma, Jordan-Holder theorem. Section - III Modules, Cyclic modules, Simple and semi-simple modules, Schur lemma, Free modules, Torsion modules, Torsion free modules, Torsion part of a module, Modules over principal ideal domain and its applications to finitely generated abelian groups. Section - IV Noetherian and Artinian modules, Modules of finite length, Noetherian and Artinian rings, Hilbert basis theorem. HomR(R,R), Opposite rings, Wedderburn – Artin theorem, Maschk theorem, Equivalent statement for left Artinian rings having non-zero nilpotent ideals. Radicals: Jacobson radical, Radical of an Artinian ring. Note :The question paper of each course will consist of five Sections. Each of the sections I to IV will contain two questions and the students shall be asked to attempt one question from each. Section-V shall be compulsory and will contain eight short answer type questions without any internal choice covering the entire syllabus. Books Recommended: 1. Luther, I.S., Passi, I.B.S., Algebra, Vol. I: Groups, Vol. III: Modules, Narosa Publishing House (Vol. I – 2013, Vol. III –2013). 2. Lanski, C. Concepts in Abstract Algebra, American Mathematical Society, First Indian Edition, 2010. 3. Sahai, V., Bist, V., Algebra, Narosa Publishing House, 1999. 4. Malik, D.S., Mordenson, J.N. and Sen, M.K., Fundamentals of Abstract Algebra, McGraw Hill, International Edition, 1997. 5. Bhattacharya, P.B., Jain, S.K. and Nagpaul, S.R., Basic Abstract Algebra (2nd Edition), Cambridge University Press, Indian Edition, 1997. 6. Musili, C., Introduction to Rings and Modules, Narosa Publication House, 1994. 7. Jacobson, N., Basic Algebra, Vol. I & II, W.H Freeman, 1980 (also published by Hindustan Publishing Company). 8. Artin, M., Algebra, Prentice-Hall of India, 1991. 9. Macdonald, I. D., The Theory of Groups, Clarendon Press, 1968.  Contents CHAPTER TITLE OF CHAPTER PAGE NO. 1 PRELIMINARIES 1-6 SECTION–1 2 THE SYLOW THEOREMS 7-30 SECTION–2 3 SUBNORMAL SERIES 31-50 4 NORMAL SERIES 51-65 5 COMPOSITION SERIES 66-76 SECTION–3 6 MODULES 77-100 SECTION–4 7 NOETHERIAN AND 101-137 ARTINIAN MODULES  1 PRELIMINARIES Structure 1.1. Introduction. 1.2. Definitions. 1.1. Introduction. This chapter contains definitions and results related to groups, cyclic group, subgroups, normal subgroups, permutation group, centre of a group, homomorphism and isomorphism. All of these results will be helpful throughout the further study of the course. 1.1.1. Objective. The objective of the study of these results is to understand the basic concepts and have an idea to apply them in further study of the course. 1.2. Definitions. 1.2.1. Cartesian Product of Two Sets. Let A and B be two non-empty sets. Then, the set of all distinct ordered pairs whose first co-ordinate is an element of A and whose second co-ordinate is an element of B is called cartesian product of A and B and is denoted by A×B. For example, let A = 1,2 , B = 4,5 , then A×B = 1,4  , 1,5 ,  2, 4  ,  2,5 and B×A =  4,1 ,  4, 2  ,  5,1 ,  5, 2. Thus, in general, A×B  B×A if A  B. Also, A×B =  if A or B or both of A and B are empty sets. 1.2.2. Function. Let A and B be two given non-empty sets. A correspondence denoted by f, which associates to each member of A a unique member of B is called a function. The function f from A to B is denoted by f : A B. 1.2.3. Binary Operation. A mapping f : S  S S is called a binary operation on the set S. 1.2.4. Algebraic Structure. A non-empty set S equipped with one or more binary operations is called an algebraic structure. Suppose ‘*’ is a binary operation on S. Then, (S,*) is called an algebraic structure. 1.2.5. Group. LetG be a non-empty set with a binary operation ‘*’. Then, G is called a group w.r.t. binary operation ‘*’ if following postulates are satisfied: (i) Associativity 2 Preliminaries (ii) Existence of Identity (iii) Existence of Inverse. 1.2.6. Abelian Group. A group G is called an Abelian group or commutative group if in addition to above postulates G also satisfies the commutative law. 1.2.7. Important Results. (i) The identity element in a group is unique. (ii) Every element in a group have a unique inverse. (iii) If a,b,c be elements of G such that ab = ac, then b = c (Left cancellation law) and ba = ac, then b = c (right cancellation law) (iv) If a  G , then  a -1   a. 1 (v) If a,b  G , then  ab   b 1a 1. 1 (vi) If G is an Abelian group, then for all a,b  G and any integer n , we have  ab   a n b n. n (vii) If every element of the group is its own inverse, then the group is Abelian. (viii) If a group has a finite number of elements, this number is called the order of the group and the group is called finite group. A group with an infinite number of elements is called an infinite group. (ix) If G is a group such that  ab   a nb n for three consecutive integers m and for all a,b  G , then G is n Abelian. 1.2.8. Subgroup. A non-empty subset H of a group G is said to be a subgroup of G if H itself is a group w.r.t. the same binary operation as in G. 1.2.9. Proper and Improper subgroups. The subgroups {e} and G itself are called improper subgroups of G. All other subgroups, other than {e} and G, are called proper subgroups of G. 1.2.10. Coset of a Subgroup. LetG be a group and H is any subgroup of G. Let ‘a’ be any element of G. Then, the set Ha = ha : h  H is called a right coset of H in G generated by ‘a’. A left coset aH can be defined in a similar way. Also, a subset is called a coset of H in G generated by ‘a’ if Ha = aH. 1.2.11. Normal Subgroup. A subgroup N of a group G is said to be a normal subgroup of G iff Na = aN for all a  G , that is, right and left cosets are same for every element of G. We denote a normal subgroup N of a group G by N  G. 1.2.12. Remark. (i) A subset H of a group G is a subgroup iff ab -1  H for all a, b  H. (ii) A finite subset H of a group G is a subgroup iff ab  H for all a, b  H. (iii) Let H and K be two subgroups of a group G. Then, the set Abstract Algebra 3 HK = x : x = hk where h  H, k  K is a subgroup of G iff HK = KH. (iv) If H is a subgroup of G then Hg = H = gH iff g  H. (v) Any two right(left) cosets of a subgroup are either disjoint or identical. (vi) If H is a finite subgroup of G. Then, o  H  = o  Ha  for all a  G. (vii) A group G  e which does not have any non-trivial normal subgroup is called a simple group. (viii) A subgroup H of a group G is normal iff g 1hg  H for every h  H , g  G. (ix) Every subgroup of an Abelian group is a normal subgroup. (x) Let H be a subgroup of G. The number of distinct right cosets of H in G is called the index of H in G and written as [G : H]. (xi) If [G : H] = 2, then H is normal in G. (xii) A subgroup H of a group G is a normal subgroup of G iff the product of two right cosets of H in G is again a right coset of H in G. (xiii) Every subgroup of a cyclic group is cyclic. (xiv) Order of a finite cyclic group is equal to the order of its generator. (xv) If the order of a group is a prime number, then the group is cyclic and hence Abelian. 1.2.13. Cyclic group. A group G is said to be cyclic group generated by an element a  G if every g  G is such that g  at for some integer t. 1.2.14. Order of an element. Let G be a group and a  G and the composition being denoted by multiplication. By the order of an element a  G , we mean the least positive integer n, if exists, such that a n  e , the identity in G. 1.2.15. Results. (i) Let G be a finite group and a  G , then o  a  o  G . (ii) Let G be a finite group and a  G , then a o G   e. (iii) If a  G and o  a   n , then a t  e iff n / t (n divides t). n (iv) If o  a   n , then o  a k  . g.c.d.  n, k  1.2.16. Homomorphisms. If  G,. and  G ,* are two groups. A mapping f : G G is called a homomorphism, if f  x. y   f  x  * f  y  for all x, y  G. 4 Preliminaries 1.2.17. Results. If f is a homomorphism from the group G to the group G , then (i) f  e   e , where e and e are identities of G and G respectively. (ii) f  g 1    f  g   for all g  G. 1 (iii) it is called epimorphism, if it is onto. (iv) it is called monomorphism, if it is one – one. (v) it is called isomorphism, if it is one – one and onto. We write as G  G (vi) it is called endomorphism, if G  G. 1.2.18. Kernel of a Homomorphism. Let f : G G be a homomorphism. Then, the kernel of f is the  set Kerf  g  G : f  g   e, the identity element of G.  It should be noted that: (i) Kerf  G. (ii) f is monomorphism iff Kerf  e. (iii) A homomorphism from a simple group is eithe trivial or one-to-one. 1.2.19. Quotient Group. Let G be a group and H be a normal subgroup of G, then the set G/H (G mod H) of all cosets of H in G is a subgroup w.r.t. multiplication of cosets. Itis called quotient group or factor group of G by H. If a, b  G , then HaHb = Hab. The identity element of G/H is H. 1.2.20. Canonical Homomorphism. The mapping f : G G / H defined by f  g   Hg for all g  G is an onto homomorphism, where H be a normal subgroup of G. It is called natural or canonical homomorphism and Kerf  H. 1.2.21. Fundamental Theorem of Homomorphism. If G is homomorphic image of G under f (that is, f is onto), then G G. ker f 1.2.22. First Theorem of Isomorphism. Let f be a homomorphism of a group G onto a group G. Let     K is any normal subgroup of G and K  x  G : f  x   K  f 1 K. Then, K is normal subgroup of G containing ker f and G G. K K 1.2.23. Second Theorem of Isomorphism. Let H and K are subgroups of any group G, where H  G. Then, K  HK. H K H Abstract Algebra 5 1.2.24. Third Theorem of Isomorphism. Let G be any group and H , K be two normal subgroups of G G such that H  K. Then, G  H. K K H 1.2.25. Permutations. SupposeS is a finite set having n distinct elements. Then, a one-one mapping of S onto itself is called a permutation of degree n. Let S  a1 , a2 ,..., an  be a finite set having n distinct elements. If f : S S is a one-one onto mapping, then f is a permutation of degree n. Let f  a1   b1 , f  a2   b2 ,..., f  an   bn , where  a1 a2... an  a1 , a2 ,..., an   b1 , b2 ,..., bn . Then, f is written as f   .  b1 b2... bn  If S is a finite set of n distinct elements, then we have n distinct arrangements of these n elements. So there will be n distinct permutations of degree n. the set of all permutations of degree n is called symmetric set of permutations and is denoted by Pn or S n. 1.2.26. Product of Permutations. Product of two permutations f and g of degree n is given by first  a1 a2... an  carrying out the operation defined by g and then by f. It is denoted by fog. If f     b1 b2... bn   b1 b2... b n   a1 a2... an  and g   . Then, gof   .  c1 c2... cn   c1 c2... cn  1.2.27. Results. (i) The set Snof all permutations of n symbols is a finite group of order n w.r.t. product of permutations. (ii) This group is Abelian for n  2 and non-abelian for n  3.  a1 a2... ak ak 1... an  1.2.28. Cyclic Permutation. Let f   . It is cyclic of length k and can be  a2 a3... a1 ak 1... an  written as f   a1 a2... ak . 1.2.29. Transposition. A cyclic permutation of length 2 is called a transposition. 1.2.30. Inverse of a Cycle. Let f   a1 a2... ak  be a cycle of length k and degree n. Then, f 1   a1 a2... ak    a1 ak... a2 . 1 1.2.31. Disjoint Cycles. Two cycles are said to be disjoint if they have no object in common in their one-rowed representation. 1.2.32. Results. (i) Any two disjoint cycles commute with each other. 6 Preliminaries (ii) A permutation is said to be an “even permutation” if it can be expressed as a product of an even number of transpositions and is called “odd permutation” if it can be expressed as a product of odd number of transpositions. For example, (1 2 3 4 5 ) = (1 2)(1 3)(1 4)(1 5) which is product of even number of permutations and so is even permutations. (iii) Product of two even(odd) permutations is again an even permutation. (iv) The set of all permutations in Sn is a normal subgroup of Sn, is denoted by An and is called n alternating group of order n and has elements. 2 (v) The group An is simple for n = 1,2,3. But A4 is not simple. However, An is simple for n  5. 1.2.33. Centre of a group. Let G be a group then the centre of G is given by Z  G   C  G    x  G : xy  yx for all y  G. 1.2.34. Normalizer of a subgroup. Let G be any group and H be its subgroup. Then, normalizer of H in G is given by N  H    x  G : xH  Hx. N(H) is the largest subgroup of G in which H is normal. In particular, H  G iff N  H   G. 1.2.35. Result. (i) If o  G   p n for some prime p, then centre of G is non-trivial. (ii) If o  G   p 2 , where pis a prime, then G is abelian. Books Suggested: 1. Luther, I.S., Passi, I.B.S., Algebra, Vol. I: Groups, Vol. III: Modules, Narosa Publishing House (Vol. I – 2013, Vol. III –2013). 2. Sahai, V., Bist, V., Algebra, Narosa Publishing House, 1999. 3. Malik, D.S., Mordenson, J.N. and Sen, M.K., Fundamentals of Abstract Algebra, McGraw Hill, International Edition, 1997. 4. Bhattacharya, P.B., Jain, S.K. and Nagpaul, S.R., Basic Abstract Algebra (2nd Edition), Cambridge University Press, Indian Edition, 1997. 5. Artin, M., Algebra, Prentice-Hall of India, 1991.  2 THE SYLOW THEOREMS Structure 2.1.Introduction. 2.2.Conjugate of an element. 2.3.Commutator. 2.4.The Sylow Theorems. 2.5.Structure of Finite Abelian Groups. 2.6.Survey of Groups. 2.7.Check Your Progress. 2.8.Summary. 2.1. Introduction. This chapter contains many important results related to the p-groups, Sylow p- subgroups, equivalent classes of the Sylow subgroups, number of sylow p-subgroups. 2.1.1. Objective. The objective of these contents is to provide some important results to the reader like: (i) Conjugate of an element. (ii) Sylow First Theorem. (iii) Sylow Second Theorem. (iv) Sylow Third Theorem. (v) Survey of groups. 2.2. Conjugate of an element. Let G be any group and a, b  G, then a is called conjugate of b if there exists an element x  G such that a = x 1 bx. 2.2.1. Exercise. The relation of conjugacy is an equivalence relation. 2.2.2. Equivalence Class. Let a  G, then equivalence class or conjugate class of ‘a’ is given by: Cl(a) = {x  G : a ~ x } = Set of all conjugates of ‘a’ = {g-1ag : g  G } 8 The Sylow Theorems Remark. Since the conjugacy relation ‘~’ is an equivalence relation on G, so G is union of all conjugate classes and any two conjugate class are either disjoint or identical. Keeping this in mind, we can say that o(G) =  o  Cl (a)  , where the sum runs over element a which is taken one each from each conjugate a class. Clearly, Cl (e) = {e} and Cl (a) = Cl (b) iff a ~ b. o(G ) Result. If G is a finite group and a  G, then o  Cl ( a )  = o  N (a)  2.2.3. Class Equation. Let G be a finite group and Z(G) denote the centre of G. Then, the equation o (G ) o(G) = o  Z (G )    o  N (a)  a where ‘a’ ranges over each conjugate class containing more than one element, is called class-equation. Another forms of class equation. o (G ) (i) o(G) =  o  N (a)  , where the sum runs over ‘a’ taken one from each conjugate class. a o (G ) (ii) o(G) = o(Z(G)) +  a Z ( G ) o  N (a)  , where the sum runs over ‘a’ taken one from each conjugate class. o(G ) (iii) o(G) = o(Z(G)) +  N (a)  G o  N (a )  , where the sum runs over ‘a’ taken one from each conjugate class. (iv) o(G) = o(Z(G)) +  a Z ( G ) [G : N (a )] , where the sum runs over ‘a’ taken one from each conjugate class. Results. 1. If o(G) = pn for some prime p then Z(G)  {e} that is, Z(G) is non-trivial, that is, o  Z (G )   1. 2. If o(G) = p2 for some prime p, then G is abelian. 3. A group of order p3 may not be abelian e.g. Q8 whose order is 23. 4. If G is a non-abelian group of order p3 for some prime p, then o  Z (G )  = p. 5. If Z is the centre of a group G such that G Z is cyclic, then show that G is abelian. 2.3. Commutator. Let G be any multiplicative group. The commutator of two elements x and y of G is the element x 1 y 1 xy  G. We denote it by [x, y]. 2.3.1. Proposition. G is abelian iff [x, y] = e  x, y  G. Abstract Algebra 9 Proof. If G is abelian then [x, y] = x 1y 1xy = x 1xy 1y = e.e = e. Conversely, let [x, y] = e  x, y  G.  x1y1xy = e  (yx)1. (xy) = e  xy = yx  x, y  G.  G is abelian. 2.3.2. Proposition. a  Z(G) iff [a, x] = e  x  G. Proof. Let a  Z(G) = centre of G Then [a, x] = a1x1ax = a1x1xa = a1a = e [Since a  Z(G)] Conversely. [a, x] = e  x  G  a1x1ax = e  ax = xa   G  a  Z(G). 2.3.3. Commutator Element. The element y of G is said to be a commutator element of G if  a, b  G such that [a, b] = y that is, a1 b1ab = y. e.g. Identity element is always a commutator element. Let us find the commutator elements of S3. We know that S3 =  I , (12) , (13) , (23) , (123) , (132) Now, [I, (12)] = I, Similarly [I, x] = I  x  S3. Now, [(12), I] = I, [(12), (12)] = I [(12),(13)] = (123), [(12), (23)] = (132) [(12), (123) = (132), [(12), (132)] = (123) So, (123) and (132) are also commutator elements of S3. We can show that I, (123), and (132) are the only commutator elements of S3. 2.3.4. Derived Subgroup. The subgroup of G generated by all the commutators of G is called the derived subgroup of G. We denote it by (G) and G that is, (G) = G = < [x, y] : x, y  G > For example, let G = S3, then (S3) = < [x, y] : x, y  S3 > = < I, (123), (132)> = {I, (123), (132)}. (G) is also known as first derived subgroup. 2.3.5. Exercises. i) Derived subgroup of a group G is a normal subgroup of G, that is, (G)  G. ii) A group G is abelian if and only if G = < e >. 10 The Sylow Theorems 2.3.6. nth Derived Subgroup. Let G be a group, for every non-negative integer n, define G ( n ) inductively as follows:   G 0  G , G ( n 1)  G ( n ) ,  the commutator subgroup of G ( n ). The G ( n ) is called nth commutator subgroup or nth derived subgroup of G.  G ( n 1)   G ( n )  = [ G ( n ) , G ( n ) ] = < [x, y] : x, y  G ( n ) >. 2.4. Sylow Theorems 2.4.1. Sylow’s First Theorem. Let p be a prime number such that p m o(G ) , where m is a positive integer. Then G has a subgroup of order p m. Proof. We shall prove the Theorem by induction on o(G ). If o(G ) = 1, then Theorem is trivially true. As our induction hypothesis, we assume that Theorem is true for all groups of order less than order of G. In other words, we have assumed that if G is a group such that o(G) < o(G ) and p k o(G) , for some integer k, then G has a subgroup of order p k. We shall prove the result for G. For this we consider two cases separately. Case I. If p m divides the order of a proper subgroup, say H, of G that is, p m o( H ) and o( H ) < o(G ). Then by induction hypothesis on H, we obtain that H (and hence G) has a subgroup of order p m. Case II. Let p m does not divide the order of any proper subgroup of G that is, p m / o( H ) , for all proper subgroup H of G. o(G ) We know that the class-equation for G is o(G ) = o( Z (G ))   N (a) G o( N (a )) (1) If N (a )  G , then N(a) is a proper subgroup of G, and so by hypothesis of this case, p m / o( N (a )) o(G ) Now, o(G ) =. o( N (a )) o( N (a )) Given that p m o(G ) , and if p m / o( N (a )) then by above expression, we obtain o(G ) p , whenever N(a)  G. o( N (a )) Abstract Algebra 11 o(G )  p  N ( a )  G o ( N ( a ))  o(G )  Also, p o (G ) , so p o(G )     p o ( Z (G )) [By (1)]  N ( a )  G o( N ( a ))  As Z(G) is abelian, so by Cauchy Theorem for finite abelian group, there exists an element a ( e)  Z(G) such that a p  e. Let K be the cyclic subgroup of G generated by a that is, K = < a > = {a, a 2 , a 3 ,..., a p } and o(K) = p. Now a  Z (G ) implies that K  ZG) and we know that a subgroup of Z(G) is always a normal subgroup of G and so K  G and so G K is well-defined. o(G ) o(G ) Now, o(G K )    o(G ) o( K ) p So, we can apply the induction hypothesis on G K. o(G ) Now, p m o(G )  p m1 p  p m 1 o(G K ) By induction hypothesis on G K for the divisor p m 1 , G K must have a subgroup, say, T of order p m 1 , that is, o(T) = p m 1. Now, we know that every subgroup of G K is of the form L K where L is a subgroup of G containing K. So, we must have T = L K , where L is a subgroup of G containing K. o( L )  o(T) = o( L K ) = o( K )  o(L) = o(T).o(K) = p m 1.p = p m Thus G has a subgroup L of order p m. Remark. Sylows first Theorem can also be stated in following ways: (i) If any power of prime divides the order of a group G, then G has a subgroup of order equal to that power of prime. (ii) If o(G ) = p k q , where p is a prime number and q is a positive integer such that gcd(p,q)=1, then G has subgroups of orders p, p 2 ,..., p k. 12 The Sylow Theorems Example of Sylow’s first Theorem. 2.4.2. Example. Let G be a group such that o(G ) = 9000. By Sylow first Theorem, find the order of subgroups which G certainly contains. Solution. First we do the prime factorization of 9000 and obtain o(G ) = 23. 32. 52 Here, 2, 3 and 5 are prime numbers so by Sylow’s first Theorem, G contains the subgroups of order 21, 22, 23, 31, 32, 51, 52 that is, 2, 4, 8, 3, 9, 5, 25. However, by Sylow’s first Theorem, nothing can be said about the existence of subgroups of order 6, 15, 10 etc. as they are not powers of a prime. 2.4.3. Sylow p-subgroup. Let p be a prime number such that p k divides o(G ) and p k 1 does not divide o(G ). Then a subgroup of order p k is called a Sylow p-subgroup of G. -OR- If o(G ) = p k q where p is a prime number and gcd(p, q) = 1, then a subgroup of order p k is called a Sylow p-subgroup of G. -OR- Sylow p -subgroup of a group G is a subgroup whose order is p k where k is the largest power of p such that p k divides o(G ). -OR- A subgroup of G is called a Sylow p-subgroup if its order is equal to the maximum power of p occurring in the order of the group. 2.4.4. Example. Find the order of different Sylow p–subgroups for G where (i) o(G ) = 45 (ii) o(G ) = 1125. Solution. (i) o(G ) = 45 = 3251. Then, G has Sylow 3–subgroups and Sylow 5–subgroups. A sylow 3subgroup is that whose order is 32, that is, 9 and a sylow 5-subgroup is that whose order is 51 = 5. (ii) o(G ) = 1125 = 3253. In this case, a sylow 3-subgroup is that whose order is 9 and a sylow 5-subgroup is that whose order is 125. Note. By above example, it is clear that in different groups Sylow p-subgroup may have different orders for some fixed prime p. 2.4.5. Example. If H is a Sylow p-subgroup of G, then prove that x 1 Hx is also a sylow p-subgroup of G for any x  G. Abstract Algebra 13 Solution. Let p n o(G ) and p n 1 / o(G ). As H is a Sylow p -subgroup of G, we have o( H )  p n. Let H = {h1 , h2 , h3 ,..., h p n } , then for any x  G , we have x 1Hx  {x 1h1 x , x 1h2 x ,..., x 1hpn x} (1) First we prove that x 1 Hx is a subgroup of G. For this let x 1h1 x and x 1h2 x be any two arbitrary element. Then ( x 1 h1 x ) ( x 1 h2 x ) 1 = x 1 h1 x x 1h21 ( x 1 ) 1 = x 1h1h21 x  x 1 Hx Since h1h21  H as H is a subgroup  Thus, x 1 Hx is a subgroup. Secondly, we prove that o( x 1 Hx)  o( H ). For this it is sufficient to prove that all elements in (1) are distinct. Let if, possible x 1h1 x  x 1h2 x , where h1  h2  xx 1h1 xx 1  xx 1h2 xx 1  h1  h2 , which is a contradiction. Hence, o( x 1 Hx)  o( H ) , that is, o( x 1 Hx)  p n. Thus, x 1 Hx is a Sylow p-subgroup of G. 2.4.6. p group. Let p be a prime number. A group G is said to be a p – group if order of every element of G is some power of p. For example, Q8  {1,  1, i ,  i , j ,  j , k ,  k } The group of quaternions is a 2–group because o(1)  20 , o(1)  21 , o( i ,  i , j ,  j , k ,  k )  22 , that is, order of every element of Q8 is a some power of 2. 2.4.7. Theorem. A finite group G is a p-group iff o(G ) = pn for some integer n. Proof. Suppose G is a p-group. We shall prove that o(G ) = p n for some integer n  1. For this, it is sufficient to prove that p is the only prime dividing o(G ). Let, if possible, q (  p ) be any other prime such that q o(G ). By Cauchy Theorem, there exists an element a (  e)  G such that o( a )  q 14 The Sylow Theorems Since a  G and G is a p-group, so o(a )  p r for some r  1. Thus p r  q  p q , which is a contradiction since a prime can never divide other prime. Hence p is the only prime dividing o(G ) , so o(G ) = pn for some n. Conversely, Suppose o(G ) = pn. Let a  G be any element, then o ( a ) o (G )  o( a ) p n  o(a)  p r for some r. Thus order of every element of G is some power of p. Hence G is a p-group. Remark. Now we introduce the concept of Double Cosets which will be very useful in proving the Sylow’s second and third Theorem. 2.4.8. Double Coset. Let H and K be two subgroups of a group G and x  G be any element. Then the set H x K = {hxk : h  H , k  K } is called a double coset. 2.4.9. Double Coset Decomposition. If H and K are two subgroups of a group G then prove that (i) any two double cosets are either disjoint or identical (ii) G is the union of all distinct double cosets that is, G =  H xK xG where union runs over x taken one from each double coset. Proof. We define a relation ~ for any two elements x and y of G as x ~ y iff x = hyk for some h  H and k  K. First we prove that this relation is an equivalence relation. (i) Reflexivity: Clearly x ~ x as x = e x e, where e  H, e  K. (ii) Symmetry: Let x ~ y  x = h y k for some h  H, k  K  h 1 x k 1  h 1 h y k k 1  y  h 1 x k 1 where h 1  H , k 1  K  y ~ x. (iii) Transitivity: Let x ~ y and y ~ z  x  hyk and y  hzk  for some h, h  H and k , k   K  x = hhzk k Clearly hh  H and k k  K , as H and K are subgroups and so x ~ z. Hence ~ is an equivalence relation of G, so this relation partitions the group G into equivalence classes and so we can write Abstract Algebra 15 G=  cl ( x) xG (1) where union runs over x taken one from each conjugate class. Then, cl ( x) = { y  G : y ~ x} = { y  G : y  hxk for some h  H , k  K } = {hxk : h  H , k  K}= H x K  cl ( x) = H x K (2) Thus equivalence class of any element comes out to be a double coset. Also we know that any two equivalence classes are either disjoint or identical. Thus we obtain that any two double cosets are either disjoint or identical, which proves (i). Using (2) in (1), we obtain G=  H xK xG where union runs over x taken one from each double coset which proves (ii). This is called double coset decomposition of G by H and K. 2.4.10. Lemma. Let H and K be finite subgroups of a group G and x  G then o( H ) o( K ) o( H x K ) =. o ( H  x K x 1 ) Proof. We define a mapping  : H x K H x K x 1 by setting  (hxk )  hxkx 1 for h  H and k  K. We prove that  is well-defined, one-one and onto. (i)  is well-defined: Let h1 xk1  h2 xk2  h1 xk1 x 1  h2 xk2 x 1   (h1 xk1 )   (h2 xk2 ) So,  is well-defined. (ii)  is one-one. Let  (h1 xk1 )   (h2 xk2 )  h1 xk1 x 1  h2 xk2 x 1  h1 xk1  h2 xk2 16 The Sylow Theorems So,  is one-one. (iii)  is onto. Let hxkx 1  HxKx 1 be any element then clearly hxk  HxK and  (hxk )  hxkx 1  hxk is pre-image of hxkx 1 under . So,  is onto. Hence, there exists a one-to-one correspondence between H x K and HxK x 1 and so their orders must be same that is, o( HxK )  o( HxKx 1 ) (1) Now we know that if K is a subgroup of G then xKx 1 is also a subgroup of G of the same order, that is, o(K) = o( xKx 1 ). Also, we know a result that if A and B are two finite subgroups of G, then o( A) o( B) o( AB)  o( A  B ) Putting A = H and B = xKx 1 in above, we obtain 1o( H ) o( xKx 1 ) o( HxKx )  o( H  xKx 1 ) o( H ) o( K )  o( HxKx 1 )  [Since o( K )  o( xKx 1 )] (2) o( H  xKx 1 ) By (1) and (2), we obtain o( H ) o( K ) o( HxK )  o( H  xKx 1 ) 2.4.11. Sylow’s Second Theorem. Any two Sylow p-subgroups of a finite group G are conjugates in G. Proof. Let H and K be two Sylow p-subgroups of G. Let n be the highest power of p such that p n o(G ) that is, p n 1 / o(G ) (1) Then, o( H )  o( K )  p n We have to show that H and K are conjugate in G that is, H = xKx 1 for some x  G Let, if possible this is false that is, H  xKx 1 for all x  G.  H  xKx 1 is a subgroup of H which is properly contained in H that is, H  xKx 1  H (2) Abstract Algebra 17 Now, by Lagrange’s Theorem, o( H  xKx 1 ) o( H )  p n  o( H  xKx 1 ) = pm for some m  n. But in view of (2) clearly m  n , so o( H  xKx 1 ) = pm, where m < n. By above Lemma, we have o( H ) o( K ) pn. pn o( HxK )    p 2nm o( H  xKx 1 ) pm  p n 1 n  m 1  p n 1. p n  m 1  p n 1 divides o( HxK )  p n 1 divides  o( HxK ) (3) xG Now, by double coset decomposition, we know that G=  H x K , where H x K are mutually disjoint. xG  o(G) =  o( HxK ) xG (4) By (3) and (4), we have p n 1 divides o(G), which is a contradiction to (1). Hence H = xKx 1 for some x  G that is, H and K are conjugates in G. 2.4.12. Lemma. Let P be a Sylow p -subgroup of a group G, then the number n p of Sylow p-subgroups o(G ) of G is equal to. o( N ( P)) o(G ) Proof. We know that o(cl ( P))  (1) o( N ( P)) Now, cl ( P ) contains all subgroups which are conjugate to P. But by Sylow second Theorem, all sylow p-subgroups are conjugate to each other and hence cl ( P ) contains all Sylow p-subgroups of G. Hence, number of Sylow p-subgroups = n p = o(cl ( P )) (2) o(G ) By (1) and (2), np =. o( N ( P)) 2.4.13. Sylow’s Third Theorem. The number n p of Sylow p-subgroups of a finite group G is given by n p = 1 + kp such that 1  kp o (G ) , and k is a non-negative integer. 18 The Sylow Theorems Proof. Let P be a Sylow p-subgroup of G. Let n be the highest power of p such that p n o(G ) that is, p n 1 / o(G ). By double coset decomposition of G, we know that G=  H x K , where union runs over x taken one from each double coset. xG  o(G) =  o( HxK ) where sum runs over x taken one from each double coset. xG Taking H = K = P in above, we get o(G) =  o( PxP) xG  o(G) =  xN ( P ) o( PxP )   xN ( P ) o( PxP ) (1) We take up two sums in (1) one by one. If x  N(P) then xPx 1 = P  xP  Px  PxP  PPx  PxP  Px Since P is a subgroup , so PP = P    xN ( P ) PxP  xN ( P ) Px (2) Now P is a subgroup of N(P) and so Px is a right coset of P in N(P). Further we know that union of all distinct right cosets of a subgroup is equal to the group, so we get  xN ( P ) P x  N ( P) Using this in (2), we get  xN ( P ) P x P  N ( P)   xN ( P ) o( PxP ) = o( N ( P )) (3) Again, if x  N ( P ) then xPx 1  P  P  xPx 1 is a subgroup of P properly contained in P, that is, o( P  xPx 1 )  o( P )  p n Also by Lagrange’s Theorem o( P  xPx 1 ) o( P)  p n Abstract Algebra 19  o( P  xPx 1 )  p m with m < n. Now we know that, o( P ) o( P ) pn. pn o( PxP )  1   p 2nm o( P  xPx ) p m = p n 1 n m1 = p n 1. p n  m 1  p n 1 divides o(P x P) whenever x  N ( P ) that is, p n 1  xN ( P ) o( PxP )   xN ( P ) o( PxP ) = p n 1 t for some integer t (4) Using (3) and (4) in (1), we obtain o(G) = o(N(P)) + p n 1 t o(G ) p n 1t  =1+ (5) o( N ( P)) o( N ( P )) o(G ) As N(P) is a subgroup of G, by Lagrange’s Theorem, o(N(P)) divides o(G) and so o( N ( P)) is an integer. p n 1t So, by (5), we obtain that is an integer. o( N ( P )) Now, P is a subgroup of N(P), so by Lagrange’s Theorem o ( P ) o ( N ( P ))  p n o( N ( P))  o( N ( P))  p n r for some integer r. p n 1t p n 1t t Thus, we obtain that  n  p is an integer. o( N ( P )) p r r t p n 1t  is an integer, say k  = kp r o( N ( P )) Using this in (5), we have o(G ) = 1 + kp o( N ( P)) o(G ) By above Lemma, the number n p of Sylow p-subgroups is given by n p =. o( N ( P)) 20 The Sylow Theorems o(G ) Hence, n p = = 1 + kp o( N ( P)) Finally, o(G) = o(N(P)).(1 + kp) implies that 1  kp o (G ) Thus number of Sylow p-subgroups is 1+ kp such that 1  kp o (G ). 2.4.14. Corollary. Show that a Sylow p-subgroup of a finite group G is unique iff it is normal. Proof. Condition is necessary: Suppose H be a unique Sylow p-subgroup of G. Let p n o(G ) and p n 1 / o(G ) , then Clearly, o(H) = pn Let x  G be any arbitrary element, then we know that x 1 Hx is also a Sylow p-subgroup. Since H is the only Sylow p-subgroup of G, therefore x 1 Hx  H for all x  G  Hx  xH for all x  G  H is a normal subgroup of G. Condition is sufficient : Let H be a Sylow p-subgroup of G such that H  G. We shall prove that H is unique. Suppose K be any other Sylow p -subgroup of G. Then, by Sylow second Theorem, H and K must be conjugate in G that is, K = x 1 Hx for some x  G  K = x 1 xH [Since H  G]  K=H Hence H is unique Sylow p-subgroup of G. 2.4.15. Simple Group. A simple group is one having no proper normal subgroup. Remark. To show that a finite group G of certain order is not simple, obtain a unique Sylow p - subgroup G for some prime p. Then, it becomes normal and obviously H is proper, which shows that G is not simple. 2.4.16. Example. Show that a group of order 28 is not simple. -OR- Let o(G ) = 28, then show that group G has a normal subgroup of order 7. Solution. We have o(G ) = 28 = 2271. By Sylow first Theorem, G has Sylow 2 – subgroups each of order 4 and Sylow 7 – subgroups each of order 7. By Sylow third Theorem, the number n7 of Sylow 7 – subgroups is given by 1 + 7k such that Abstract Algebra 21 1  7 k o (G )  1  7 k 28  1  7 k 22.7  1  7k 4 [Since (1 + 7k, 7) = 1]  k=0 Thus, n7 = 1 that is, there is unique Sylow 7 -subgroup say H and o(H) = 7 But we know that “a Sylow p -subgroup is unique iff it is normal”. Thus H is a normal subgroup of order 7. Obviously H is proper. Hence G is not simple. 2.4.17. Exercise. 1. Let G be a group of order 52.7.11, then G has how many (i) Sylow 5–subgroups (ii) Sylow 7–subgroups (iii) Sylow 11–subgroups. Check whether G is simple or not. 2. Show that a group of order 40 is not simple. -OR- Show that a group of order 40 has a normal subgroup of order 5. 3. Show that a group of order 20499 is not simple. -OR- Show that a group of order 20499 has a normal subgroup of order 11. 4. Show that a group of order 56 is not simple. 2.4.18. Proposition. Let G be a finite group such that o(G ) = p n , where p is a prime. Prove that any subgroup of order p n 1 is a normal subgroup of G. Proof. We shall prove the result by induction on n. For n = 1, G is a group of order p and the only subgroup of order p n 1 that is, of order p11  p 0  1 is {e}. The identity subgroup {e} is obviously a normal subgroup of G. Thus the result is true for n = 1. As our induction hypothesis, we assume that result is true for all groups of order p m , where m < n. Let H be a subgroup of G of order p n 1. We shall prove that H is normal in G. Now, H  N ( H )  G and so by Lagrange’s Theorem, 22 The Sylow Theorems o ( H ) o ( N ( H )) and o ( N ( H )) o (G ) that is, p n 1 o( N ( H )) and o( N ( H )) p n  o( N ( H ))  p n 1 or p n If o(N(H)) = pn, then o( N ( H ))  o(G )  N(H) = G  H is normal in G, which is what we want to prove. Now, we finish our proof by showing that o(N(H)) = p n 1 is impossible. Let, if possible, o( N ( H )) = p n 1 , then as o( H ) = p n 1 and H  N ( H ) , we get H = N(H) (1) Now, o(G ) = pn, we know by class equation, that o( Z (G ))  1 (2) By Lagrange’s Theorem, o( Z (G )) o(G )  p n  o(Z(G)) = ps, 0  s  n But if s = 0, then o(Z(G)) = 1, which is a contradiction by (1). Hence, o(Z(G)) = ps, s > 0  p o ( Z (G )) So, by Cauchy Theorem for finite groups, there exists an element a ( e)  Z (G ) such that o( a )  p. Let K be the cyclic subgroup of G generated by ‘a’ that is, K = < a > = {a, a 2 , a 3 ,........., a p  e} As ‘a’ belongs to centre, every element x  G commutes with a and all its powers, so Kx  xK for all x  G  K is a normal subgroup of G. Hence G K is well-defined and o (G ) p n o(G K )    p n 1 , where n  1  n o( K ) p o( H ) p n 1 Also, o( H K )    p n2 o( K ) p So, by induction hypothesis, H K must be a normal subgroup of G K  H is a normal subgroup of G  N(H) = G (3) By (1) and (3) we obtain, H = G, which is absurd. Hence o(H) = p n 1 is not possible. 2.4.19. Example. Show that no group of order 108 is simple. Abstract Algebra 23 -OR- Let G be a group of order 108. Show that G has a normal subgroup of order 27 or 9. Solution. We have o(G ) = 108 = 22.33. By Sylow third Theorem, the number n3 of Sylow 3–subgroups is given by 1 + 3k such that 1  3k o(G )  22.33  1  3k 4 [Since (1  3k ,33 )  1]  k = 0 or 1  n3 = 1 + 3.0 or 1 + 3.1  n3 = 1 or 4 We consider the two cases separately. Case (i). n3 = 1, that is, G has a unique Sylow 3 -subgroup, say H. Since H is unique, it must be normal and o(H) = 33 = 27. Thus G has a normal subgroup of order 27 in this case and hence G is not simple. Case (ii). n3 = 4, that is, G has four Sylow 3 – subgroups each of order 27. Let H and K be any two distinct sylow 3 – subgroups. We claim that o( H  K )  9 and H  K is a normal subgroup of G. Clearly, H  K  H , and so by Lagrange’s Theorem. o ( H  K ) o ( H )  27  o( H  K ) = 1 or 3 or 9 or 27. If o( H  K ) = 27 then since o(H) = o(K) = 27, we obtain H = K, which is a contradiction. Hence o( H  K )  27. o( H ) o( K ) 27.27 If o ( H  K ) = 1 or 3 then o ( H K ) =   108  o(G ) , which is not possible. o( H  K ) 1 or 3 Hence o( H  K )  1, 3 and so o ( H  K ) = 9. We now show that H  K is normal in G. For this we shall prove that N ( H  K ) = G. Now, we know a that, if o(H) = p n 1 and o(G) = p n then H is a normal subgroup of G. Using this, we conclude that H  K is a normal subgroup of both H and K as o( H  K ) = 32 and o(H) = o(K) = 33. Let x  H be any element, then ( H  K ) x  x( H  K ) [Since (H  K )  H ]  x  N ( H  K ) , normalizer of H  K.  H  N (H  K ) 24 The Sylow Theorems Similarly, K  N ( H  K )  HK  N ( H  K ) o( H ).o( K ) 27.27  o  N ( H  K )   o ( HK ) = =  81 o( H  K ) 9  o  N ( H  K )   81 (1) On the other hand, N ( H  K ) is a subgroup of G so by Lagrange’s Theorem o  N ( H  K )  o (G ) , that is, o  N ( H  K )  108 (2) Both (1) and (2) are possible only when o  N ( H  K )  = 108 = o(G)  o  N ( H  K )  = o(G)  N (H  K ) = G  H  K is normal in G. [Since N(H) = G iff H  G ] Hence G is not simple. 2.4.20. Theorem. Let o(G ) = pq, where p and q are distinct primes, p < q and p / q  1 , then show that G is cyclic. Proof. By Sylow third Theorem, the number n p of Sylow p-subgroups is given by 1 + kp such that 1  kp o (G )  pq  1  kp q [Since (1  kp , p ) = 1]  1+ kp = 1 or 1 + kp = q [Since q is a prime] If 1 + kp = q, then kp  q  1  p q  1 , which is a contradiction. Hence n p  1  kp  1. Thus G has a unique Sylow p-subgroup, say, H of order p. Also since H is unique, it must be normal. Thus we obtained o ( H )  p and H  G (1) Again, by Sylow third Theorem, the number nq of Sylow q-subgroups is given by 1  k q such that 1  k q o (G )  pq  1  k q p [Since (1  k q , q ) = 1]  1  k q  1 or 1  k q  p If 1  k q  p then we get q  p , which is a contradiction. Abstract Algebra 25 Hence nq  1  k q  1. Thus G has a unique Sylow q-subgroup, say, K of order q. Also since K is unique it must be normal. Thus we obtained o ( K )  q and K  G (2) Now, we know that a group of prime order is always cyclic and here H and K both are of prime orders, so they must be cyclic. Let H = < a > and K = < b > then o(H) = o(a) and o(K) = o(b) (3) Using (1) and (2) in (3), we get o(a) = p and o(b) = q (4) Now, we prove that H  K = {e}. Let x  H  K be any element. Then x  H and x  K  o ( x ) o ( H ) and o ( x ) o ( K )  o ( x ) p and o ( x ) q  o( x ) gcd( p , q )  o( x)  1  x = e for all x  H  K  H  K = {e} (5) Now, we prove that ab = ba. For this consider the element a 1b 1ab. We see that a 1b 1ab  a 1 (b 1ab)  H , because H  G , so that b 1ab  H and also a 1  H. Again, a 1b 1ab = (a 1b 1a)b  K , because K  G , so that a 1b 1a  K and also b  K. Hence, we get a 1b 1ab  H  K  a 1b 1ab = e [By (5)]  baa 1b 1ab  ba.e  ab  ba Lastly, by (3), we see that gcd(o( a ), o(b))  gcd( p, q )  1 We know that, if a, b  G such that ab = ba and  o( a ), o(b)  = 1 then o( ab)  o( a ).o(b). Therefore, o( ab)  o(a ) o(b)  pq  o(G)  G contains an element ab of order pq  G = < ab > 26 The Sylow Theorems  G is cyclic. Remark. Due to the above result, we can say that groups of order 15, 33, 35, 65, 51 etc. are cyclic. 2.4.21. Exercise. 1. Show that a group of order 15 always cyclic. 2. Let G be a group of order 231, then show that (i) G is not simple (ii) Sylow 11 -subgroup of G is contained in the centre of G. 2.4.22. Theorem. Let P be a Sylow p-subgroup of G and let x  N ( P ) be an element such that o( x)  p r. Then show that x  P. Proof. Since P is given to be a Sylow p-subgroup of G and let p n o(G ) but p n 1 / o(G ) (1) Then, clearly o(P) = pn. We know that P  N ( P ) , so N ( P ) P is well-defined. r r As x  N ( P ) , Px  N ( P ) P and ( Px ) p  P.x p  P.e [Since o( x)  p r ] r  ( Px ) p = P = Identity of N ( P ) P  o( Px) p r  o( Px )  p s for some s  0 If s = 0, then o( Px)  p s  p 0  1  Px  P  x  P , which is required to prove. Now we finish the proof by showing that s > 0 is impossible. Let, if possible, s > 0 and let H be the cyclic subgroup of N ( P ) P generated by Px that is, H   Px  then o( H )  o( Px )  p s (2) Since H is a subgroup of N ( P ) P , it must be of the form H = H P where H is a subgroup of N(P) containing P. Now, o( H ) = ps [By (2)]  o( H P )  p s o( H )   ps  o( H )  p n  s , s  0 o( P ) As H is a subgroup of N(P) and N(P) is a subgroup of G, so H is a subgroup of G and by Lagrange’s Theorem, o( H ) o(G ) Abstract Algebra 27  p n  s o(G ) , which is a contradiction by (1), as s > 0. Hence s > 0 is not possible and in case s = 0, we have already shown that x  P. 2.5.Structure of Finite Abelian Groups. If a group is direct product of some of its subgroups, then the structure of the group can be determined by determining the structures of subgroups appearing in the direct product. This simplifies our work as determination of structure of a big group is broken into determination of structures of comparatively smaller groups. Let us call the subgroups appearing in the direct product as “building blocks”. Now the procedure will be more simple if these building blocks are taken to be cyclic subgroups since cyclic groups are always easy to deal with. Now a natural question arise “Is it always possible to write a group as the direct product of its cyclic subgroups”. The answer is no, in general. However, luckily, it is possible for finite abelian groups, due to Fundamental Structure Theorem for finite abelian groups. Before the formal statement of this Theorem, let us study another Theorem in this regard. 2.5.1. Theorem. Prove that a finite abelian group is direct product of its Sylow subgroups. Proof. Let G be a finite abelian group and o(G ) = p1n1 p2n2... prnr , where p1 , p2 ,..., pr are distinct primes and ni  1 for all i. Since internal direct product is always isomorphic to external direct product, we shall prove that G is internal direct product of its Sylow subgroups. Let H1 , H 2 ,..., H r be the Sylow subgroups of G such that o( H1 )  p1n1 , o( H 2 )  p2n2 ,..., o( H r )  prnr To show that G is internal direct product of H1 , H 2 ,..., H r we have to prove following three things. (i) Each Hi is normal in G. (ii) H i  H1 H 2...H i 1 H i 1...H r  {e} for any i. (iii) G = H1 H 2...H r Let us prove all these one by one. (i) Since G is abelian, so its every subgroup is normal. Hence each Hi is normal in G. (ii) Let x  H i  H1 H 2...H i 1 H i 1...H r be any arbitrary element. Then x  H i and x  H1 H 2...H i 1 H i 1...H r  x  h1h2...hi 1hi 1...hr where h j  H j for all j  i 28 The Sylow Theorems n pj j = e for j  i n As h j  H j and o ( H j )  p j j , so (h j ) (1) Now, let t  p1n1 p2 n2... pi 1ni1 pi 1ni1... pr nr , then for j  i , we have remaining factors  pj j  n =  h j   n n n 1 n 1 n (h j )  (h j ) t p1 1 p2 2... pi1i p i 1 i... p r r Since p nj j appears in t      = [e] remaining factors = e. Thus, h1t  h2t ...  hit1  hit1 ...  hr  e (2) xt   h1h2...hi 1hi 1...hr  t Now, = h1t h2t...hit1hit1...hrt = e. n n n i 1 n i 1 nr  o( x ) t  o ( x ) p1 1 p2 2... pi 1 pi 1... pr (3) Since x  H i  o( x) o( H i )  o( x ) pini  o( x )  pimi , 0  mi  ni (4) Putting value of (4) in (3), we get n n n i 1 n i 1 nr pimi p1 1 p2 2... pi 1 pi 1... pr  mi  0 , since pi does not appear on R.H.S.. So by (4), we have o( x )  pimi  pi0  1  x=e Thus, H i  H1 H 2...H i 1 H i 1...H r  {e} for any i, which proves (ii). (iii) We know that a result that o( A) o( B) If A and B are two finite subgroups then o( AB)  (5) o( A  B ) Using this result for A  H1 and B  H 2 H 3...H r , we get o( H1 )o( H 2 H 3...H r ) o( H1 H 2...H r )  (6) o( H1  H 2 H 3...H r ) Taking i = 1 in (ii), proved above, we have H1  H 2 H 3...H r  {e}  o( H1  H 2 H 3...H r )  1 Using this in (6) o( H1 ) o( H 2 ) o( H 3 H 4...H r ) o( H1 H 2...H r )  o( H1 )o( H 2 H 3...H r )  (7) o( H 2  H 3 H 4...H r ) Now, H 2  H 3 H 4...H r  H 2  H1 H 3 H 4...H r  {e} [By (ii) for i = 2]  H 2  H 3 H 4...H r  {e}  o( H 2  H 3 H 4...H r )  1 Abstract Algebra 29 Using this in (7), we get o( H1 H 2...H r )  o( H1 ).o( H 2 )o( H 3 H 4...H r ) Continuing in this way, we obtain o( H1 H 2...H r )  o( H1 ).o( H 2 )...o( H r )  p1n1 p2n2.......... prnr  O (G )  G = H1 H 2.......H r , which proves (iii). Thus G is the internal direct product of its Sylow subgroups. 2.6. SURVEY OF GROUPS. In our previous work we have obtained a complete description of number and nature of a finite abelian group. Unfortunately, there is no such general result for finite non-abelian groups. The Sylow Theorems and Cauchy Theorem (which is itself is a particular case of Sylow first Theorem) are a powerful tool in finding the number and nature of finite non-abelian groups. However, to keep our study within the scope of the book, we shall study the groups of orders 6 and 8 only. 2.6.1. Example. Find all non-abelian groups of order 6. Solution. Let G be a non-abelian group such that o(G ) = 6. Now 3 and 2 are prime numbers dividing o(G ) so by Cauchy Theorem, there exist two non-identity element a and b in G such that o( a )  3 and o(b)  2. Let H = < a > be the cyclic subgroup of G generated by a that is, H = {e, a, a 2 } and o( H ) = 3. o(G ) 6 Now, index of H in G =  2 o( H ) 3 Since every subgroup of index 2 is normal, H is normal in G. If b  H , then o (b ) o ( H )  2 3 , which is a contradiction. Hence b  H As index of H in G is 2, so there are two distinct right coset of H in G and clearly these are H and Hb. Then G = H  Hb = {e, a, a 2 , b, ab, a 2b}. Since H  G , b 1ab  H = {e, a, a 2 }  b 1ab  e or a or a2 If b 1ab  e , then bb 1abb 1  beb 1 that is, a = e, a contradiction. If b 1ab  a , then ab = ba  G is abelian, a contradiction. So, b 1ab  a 2  a 1. Hence there is only one non-abelian group of order 6 given by G  {e, a, a 2 , b, ab, a 2b} where a 3  b 2  e and b 1ab  a 1 2.6.2. Exercise. 30 The Sylow Theorems 1. Find all non-isomorphic abelian groups of order 6. 2. Find all non-isomorphic groups of order 6. 3. Find all non-isomorphic non-abelian groups of order 8. 4. Find all non-isomorphic groups of order 8. 2.7. Check Your Progress. 1. For any group G, G G  is always abelian. 2. If G is a group, then G ( n ) G ( n 1) is always abelian. 3. Show that a group of order 30 is not simple. 2.8. Summary. In this chapter, we discussed about commutator elements, Sylow’s theorems which is an important part of group theory. Also observed that for a finite group and a prime p dividing its order, if pm is the largest power of p dividing order of group, then the group must have subgroups of orders p0, p1,..., pm. However, we have no idea about the number of subgroups of orders pi for i = 1, 2, …, m – 1, but for i = 0, it is 1 and for i = m, it can be decided by Sylow’s third theorem. Books Suggested: 1. Luther, I.S., Passi, I.B.S., Algebra, Vol. I: Groups, Vol. III: Modules, Narosa Publishing House (Vol. I – 2013, Vol. III –2013). 2. Lanski, C. Concepts in Abstract Algebra, American Mathematical Society, First Indian Edition, 2010. 3. Sahai, V., Bist, V., Algebra, Narosa Publishing House, 1999. 4. Malik, D.S., Mordenson, J.N. and Sen, M.K., Fundamentals of Abstract Algebra, McGraw Hill, International Edition, 1997. 5. Bhattacharya, P.B., Jain, S.K. and Nagpaul, S.R., Basic Abstract Algebra (2nd Edition), Cambridge University Press, Indian Edition, 1997. 6. Musili, C., Introduction to Rings and Modules, Narosa Publication House, 1994. 7. Jacobson, N., Basic Algebra, Vol. I & II, W.H Freeman, 1980 (also published by Hindustan Publishing Company). 8. Artin, M., Algebra, Prentice-Hall of India, 1991. 9. Macdonald, I. D., The Theory of Groups, Clarendon Press, 1968.  3 SUBNORMAL SERIES Structure 3.1. Introduction. 3.2. Subnormal Series 3.3. Solvable Group 3.4. p-group 3.5. Commutator Element 3.6. Lower Central Series 3.7. Upper Central Series 3.8. Check Your Progress 3.9. Summary 3.10. Exercise 3.1. Introduction. This chapter contains definition of subnormal series and its examples. Definition and important properties related to that of a solvable group are discussed. One important result in this direction is that Sn is not solvable for n > 4. Also it is proved that every p-group is solvable. 3.1.1. Objective. The objective of these contents is to provide some important results to the reader like: (i) Every subgroup of a solvable group is solvable. (ii) Every factor group of a solvable group is solvable. (iii) Converse result of these results. (iv) Every p-group is solvable. (v) Sn is not solvable for n > 4. (vi) A group is solvable iff nth derived subgroup is solvable. 3.1.2. Keywords. Subnormal Series, Solvable Groups, Abelian Groups, Order of Group, Quotient Groups. 32 Subnormal Series 3.2. Subnormal Series. A sequence of subgroups G  G0  G1  G2 ...  Gr  e  of a group G is called a subnormal series of G if Gi 1  Gi for 0  i  r  1. 3.2.1. Refinement of a Series. Let G be a group and G  G0  G1  G2 ...  Gr  e  be a subnormal series for G. A subnormal series G  G0'  G1'  G2' ...  Gs'  e  ' ' '  is called a refinement of the former series if G0 , G1 , G2 ,..., Gr   G0 , G1 , G2 ,..., Gs. ' ' '  The refinement is said to be a proper refinement if G0 , G1 , G2 ,..., Gr   G0 , G1 , G2 ,..., Gs. For ' '  example, let G = S3, then G  G0  G1  I  is a subnormal series for G. Now consider the series G  G0  G1  A3  G2  I . We note that A3  S3 and  I   A3. So, this series is also a subnormal series for G. Also, S3 ,  I   S3 , A3 ,  I . Hence this series is a proper refinement of the last one. 3.2.2. Length of a Series.Consider a subnormal series G  G0  G1  G2 ...  Gr  e  (1) Then, it is possible for some i, Gi = Gi+1 in (1). The number of distinct members of (1) different from (1) is called the length of the series (1). Due to this definition, the length of series (1) is r, if all Gi’s are distinct. The subnormal series (1) is said to be redundant if for some i = 0,1,2,…,r-1; Gi = Gi+1, otherwise it is said to be irredundant. One can always construct an irredundant series from a redundant one by deleting Gi whenever for some i, Gi+1 = Gi. So, if (1) is irredundant, then length of (1) is r. 3.2.3. Factors of a Series. Let G be a group and G  G0  G1  G2 ...  Gr  e  is a subnormal Gi series for G. Then, Gi 1 is called factor group or quotient factor group of the series. 3.3. Solvable Group. A group G is said to be solvable if there exists a sequence of subgroups G  G0  G1  G2 ...  Gr  e  such that Abstract Algebra 33 (i) Gi 1  Gi , 0  i  r  1 Gi (ii) Gi 1 is abelian, 0  i  r  1. 3.3.1. Example. Every abelian group G is solvable. The series G  e  is a subnormal series, that is  e   G and its only factor group is G  e  , which being isomorphic to G, is abelian. 3.3.2. Results. (i) A subset H of a group G is a subgroup iff ab  H for all a, b  H. -1 1 (ii) A subgroup H of a group G is normal iff g hg  H for every h  H , g  G. (iii) Let G be a group and H be a normal subgroup of G, then the set G/H (G mod H) of all cosets of H in G is a subgroup w.r.t. multiplication of cosets. It is called quotient group or factor group of G by H. If a, b  G , then HaHb = Hab. The identity element of G/H is H.   (iv) If  G,. and G,* are two groups. A mapping f : G G is called a homomorphism, if f  x. y   f  x  * f  y  for all x, y  G. Also, it is called isomorphism, if it is one – one and onto. We write as G  G (v) Fundamental Theorem of Homomorphism. If G is homomorphic image of G under f (that is, f is onto) , then G ker f  G. If f is not onto then G ker f  f  G. 3.3.3. Proposition. Every subgroup of a solvable group is solvable. Proof. Let G be a solvable group and let G  G0  G1  G2 ...  Gr  e  be a solvable series for G such that (i) Gi 1  Gi , 0  i  r  1 Gi (ii) Gi 1 is abelian, 0  i  r  1. Let H be any subgroup of G and let Hi  H  Gi , 0  i  r , then H  H 0  H1  H 2 ...  H r  e  ---(1) is a sequence of subgroups of H. For it, since e  H i for 0  i  r. Therefore, Hi   for all 0  i  r. Let a, b  Hi  H  Gi which implies a, b  H and a, b  Gi. 34 Subnormal Series 1 1 1 Since H and Gi are subgroups, therefore, ab  H and ab  Gi and so ab  H  Gi  H i. Hence Hi is a subgroup of H, 0  i  r. Now, we claim that the series (1) is a solvable series for H. First we prove that H i 1  Hi , 0  i  r  1. Let h  H i 1 and k  Hi. Then h  Hi 1  H  Gi 1  h  H and h  Gi 1. Similarly, k  Hi  H  Gi  k  H and k  Gi. 1 Since Gi 1  Gi , thus h  Gi 1 , k  Gi  k hk  Gi 1 and h, k  H  k hk  H. Therefore, 1 k 1hk  H  Gi 1  H i 1 and so Hi 1  Hi , 0  i  r  1. Hi Now, we shall prove that , 0  i  r  1 , are abelian. Hi 1 Gi To prove this we define a mapping f : Hi Gi 1 by considering f  x   xGi 1 , where x  Hi  H  Gi. We shall prove that f is well defined and a group homomorphism. Let x, y  Hi such that x  y  xy  e  Gi 1. 1 So, Gi 1 xy 1  Gi 1  Gi 1 x  Gi 1 y  xGi 1  yGi 1  f  x   f  y . Therefore, f is a well-defined mapping. Again, let x, y  Hi , then f  xy   xyGi 1  xGi 1. yGi 1  f  x  f  y  So, f is a group homomorphism. Thus, by fundamental theorem of group homomorphism, we have Hi Gi  f  Hi   ker f Gi1. (1a)   We shall prove that Kerf  Hi 1 , where Kerf  x  Hi : f  x  Gi1. For this, we have x ker f  f  x  Gi1  xGi1  Gi1  x Gi1 Now, x  Hi  H Gi  x  H. So, x  H, x Gi1  x  H Gi1  Hi1 Abstract Algebra 35  ker f  H i 1 (2) Let y  Hi1  H Gi1  y  H and y Gi1. So, f  y  yGi1  Gi1  y ker f  Hi1  ker f (3) By (2) and (3), Hi1  ker f. Putting this value in (1a), we obtain Hi Gi  f  Hi   Hi1 Gi1. Gi Gi Since Gi1 is abelian and f  Hi  is a subgroup of Gi1 , so f  Hi  is also abelian. Hi Therefore, Hi1 is also abelian. Hence H is solvable. 3.3.4. Canonical Homomorphism. The mapping f : G G / H defined by f  g   Hg for all g  G is an onto homomorphism, where H is a normal subgroup of G. It is called natural or canonical homomorphism and Kerf  H. 3.3.5. Proposition. Every quotient group of a solvable group is solvable. -OR- Let G be a solvable group and H is a normal subgroup of G, then G/H is also solvable. Proof. Let G be a solvable group and let G  G0  G1  G2 ...  Gr  e  be a solvable series for G, then Gi 1  Gi , 0  i  r  1 (i) Gi (ii) Gi 1 is abelian, 0  i  r  1. Let q : G G H be a canonical homomorphism, that is, q  x   xH for all x  G Since Gi 1  Gi so q  Gi 1   q  Gi . Now consider the series G H  q  G0   q  G1   q  G2  ...  q  Gr   H . 36 Subnormal Series We claim that this series is a solvable series for G H. Let x q  Gi1  and y q  Gi  , then there exist  Gi1 and  Gi such that x  q   and y  q   . Now  Gi1,  Gi and Gi1  Gi , so  1 Gi1.     Therefore, q    q  Gi1   q  q   q     q  Gi1  1 1  q    q   q     q  Gi 1   y 1 xy  q  Gi 1  and hence q  Gi1   q  Gi . 1 q  Gi  Now we shall prove that q  Gi1  is abelian. q  Gi  Let ,   q  Gi1  be any two arbitrary elements. Then,   q  Gi1  and    q  Gi1  for some , q  Gi . Also, , q  Gi     q  ' and   q   ' for some  ',  ' Gi. Therefore,   q  Gi1   q  ' q  Gi1   q  ' Gi1  and   q   ' Gi1 . Then,   q 'Gi1  q  'Gi1   q 'Gi1 'Gi1   q  'Gi1 'Gi1   q  'Gi1  q 'Gi1   . q  Gi  Therefore, q  Gi1  is abelian. Hence G/H is solvable. 3.3.6. Proposition. Let G be a group and H be a normal subgroup of G. If H and G/H both are solvable, then G is also a solvable group. Proof. Since H is solvable, so let H  H 0  H1  H 2 ...  H n  e  be a solvable series for H. Therefore, (i) Hi 1  Hi , 0  i  n  1 Hi (ii) Hi 1 is abelian, 0  i  n  1. Now, since G/H is solvable, so let G H  G0 H  G1 H  G2 H ...  Gm H  H  be a solvable series for G/H. Therefore, Abstract Algebra 37 (i) Gi 1 H  Gi H, 0  i  m  1 Gi H (ii) Gi 1 H is abelian, 0  i  m  1. Now consider the series G  G0  G1  G2 ...  Gm  H  H 0  H1  H 2 ...  H n  e  We claim that this is a solvable series for G, that is, we are to show that (i) Gi 1  Gi , 0  i  m 1 and H j 1  H j , 0  j  n 1 Gi Hj (ii) Gi 1 is abelian, 0  i  m  1 and H j 1 is abelian, 0  j  n  1. (i) It is clear that H j 1  H j , 0  j  n 1. So, we are to show that Gi 1  Gi , 0  i  m  1. Let x  Gi and y  Gi 1. Then, xH  Gi H and yH  Gi 1 H. Due to solvability of G/H, we have  xH   yH  xH   Gi 1 1 Gi 1 H  Gi H  H  x1 yxH  Gi 1 H  x1 yx  Gi 1 Thus, Gi 1  Gi for every i. Hj (ii) Again, it is clear that H j 1 is abelian, 0  j  n  1. Gi So, we have only to prove that Gi 1 is abelian, 0  i  m  1. Due to Third Theorem of Isomorphism, Gi H Gi Gi   Gi 1 H Gi 1 Gi 1 is abelian. Hence G is solvable. 3.4. p-group. A finite group whose order is pn for some integer n  1, where p is a prime, is called a p- group. 3.4.1. Cyclic group. A group G is said to be cyclic group generated by an element a  G if every g  G is such that g  at for some integer t. We denote it by G =. Remark. If the order of a group is a prime number, then the group is cyclic and every cyclic group is abelian and every abelian group is solvable. 38 Subnormal Series 3.4.2. Center of a group. Let G be a group then the center of G is given by Z  G   C  G    x  G : xy  yx for all y  G. 3.4.3. Corollary. Every finite p-group is solvable. Proof. Let G be a finite p-group and let o  G   p for some n  1. n If n=1, then o  G   p so G is cyclic. Hence G is abelian and so solvable. So let n > 1 and suppose as our induction hypothesis that result is true for all p-groups with order prwherer < n Now, if G is abelian then result is again true, so let G is non-abelian. Then, Z(G), the center of G, is non- trivial by class equation. But we know that Z (G)  G. Now, o  G   p n  o  Z  G    p s for some s  n. Since Z (G) is non-trivial, so o  Z (G)   1  s  1. o G  Then, o  G Z (G)    pns  p n. o  Z G  So, by induction hypothesis, G Z (G) is solvable. Now, Z (G) is abelian, so Z (G) is also solvable. Then, by above proposition, G is solvable. Remark. Let H and K be two groups. Then, the direct product of these groups is the group HxK   h, k  : h  H , k  K . Also, if K  K ', then X K  X K ' and K  HxK. 3.4.3. Corollary. Direct product of two solvable groups is again solvable. Proof. Let H and K be two solvable groups and X = HxK. We know that K  HxK  X. Define a mapping f : X H by setting f  x   f  h, k   h for all x  X. It is easy to show that f is well-defined. To show that f is a homoporphism. Let x, y  X , then x   h, k  and y   h1 , k1  for some h, h1  H ; k , k1  K. Therefore, f  x, y   f   h, k  ,  h , k    f  hh , kk   hh  f  x . f  y  1 1 1 1 1 Hence, f is a homomorphism. Abstract Algebra 39 To show that f is onto. Let h  H , then for every k  K , we have  h, k   X and f  h, k   h. Hence f is onto. Thus, by fundamental theorem of homomorphism, we have X Kerf  H. ---(*) Now, we claim that ker f  K. Let k  K , we define K '   e, k  : e  H , k  K. We shall show that ker f  K '. For this, let x  ker f  f  x   e, where e  H. But x   h, k  for some h  H , k  K. Therefore, f  h, k   e he  x   e, k   K '  ker f  K ' ---(1) Let x  K '  x   e, k  for some k  K. Thus, f  x   e  x  ker f  K '  ker f ---(2) By (1) and (2), we obtain K '  ker f   e, k  : e  H , k  K  Now, by (*), we obtain X K' H ---(**) We claim that K  K '  e xK , where e  H. Define a mapping  : K K ' , by setting   k    e, k  for all k  K. To show that  is a homomorphism. Let x, y  K , then   x, y    e, xy    e, x  e, y     x    y  To show that  is one-one. Let k1 , k2  K such that   k1     k2    e, k1    e, k2   k1  k2. 40 Subnormal Series To show that  is onto. Let  e, k   K ' , then for k  K ,   k    e, k  Hence  is an isomorphism. Therefore, K  K '. Thus, by (**), since K  K ' , so X K  X K '  X K  H. Since H is solvable, so X/K is solvable. Also subgroup K of X is solvable. Hence X = HxK is solvable. 3.4.4. Remark. (i) Let H and K be two subgroups of a group G. Then, HK is a subgroup of G iff HK = KH.If H  G , then HK = KH and so HK is a subgroup of G. (ii) Second Theorem of Isomorphism. Let H and K are subgroups of any group G, where H  G. Then, K H  K  HK H. 3.4.5. Corollary. Let H and K are solvable subgroups of G and H  G , then HK is also solvable. Proof. Since H  G  HK  KH , therefore HK is a subgroup of G. Now, by second theorem of isomorphism, we have K  HK H K H. Now, K is solvable, so K H  K is solvable, since factor group of a solvable group is solvable. So HK K H being isomorphic to H  K is solvable.But H is given to be solvable, so HK is also solvable, by above proposition. 3.4.6. Sylow p-subgroup.Let G be a finite group and p, a prime number, such that p | o G and k p k 1 | o  G . Then, any subgroup of G of order pk is called a Sylow p-subgroup of G, where o(G) = pkq, q is an integer. 3.4.7. Sylow’s First Theorem. Let G be a finite group of order pkq, k  1 , where p is a prime number and q is a positive integer such that g.c.d.(p, q) = 1. Then, for each i, 0  i  k , G has atleast one subgroup of order pi. 3.4.8. Sylow’s Third Theorem. Number of Sylow p-subgroups is of the form 1+mp, where p is a prime and m is non-negative integer such that 1  mp | o  G . 3.4.9. Result. (i) If o  G   p 2 , where pis a prime, then G is abelian. (ii) If G has only one Sylow subgroup of order pi, then that subgroup will be normal in G. Abstract Algebra 41 3.4.10. Corollary. Every group of order pq is solvable, where p, q are prime numbers not necessarily distinct. Proof. Let o  G   pq. If p  q , then o  G   p 2 and G is abelian and hence solvable.So, let us assume that p  q. Then, by Sylow’s theorem, G has Sylow p-subgroups each of order p and number of sylow p-subgroups is of the form 1+mp such that 1  mp | o  G .That is, 1  mp | pq  1  mp | q  m  0. Hence, G has unique Sylow p-subgroup, say H, and o  H   p. Also, we know that unique Sylow p- subgroup is always normal, so H  G. o G  Now, o  H   p and o  G H    q. oH  Thus, H and G/H are both subgroups of prime order, so they are cyclic and hence abelian and in turn solvable. Therefore, by above proposition, G is solvable. 3.4.11. Corollary. Every group of order p2q is solvable, where p, q are prime numbers not necessarily distinct. Proof. Let o(G) = p2q , we consider the following three cases Case (i) p = q, Case (ii) p > q, and Case (iii) p < q. Case (i). If p = q, then o(G) = p3 and we know that a finite p-group is solvable hence G is solvable. Case (ii). If p > q then by Sylow theorems, G has Sylow p-subgroups each of order p2 and number of these subgroups is 1+ mp such that 1  mp o( G) , m is non-negative integer. Since gcd (p , q) = 1, so 1  mp p2 q , implies, 1  mp q.  m = 0, as p > q. Thus, G has unique Sylow p-subgroup, say H, and o(H) = p2. Also, we know that unique Sylow p- subgroups is always normal so H  G. o( G) Now, o(H) = p2 and o  G H  =  q. o( H ) Now, H is abelian, since a group of order p2 is always abelian and so H is solvable. Again, G H , being of prime order, is cyclic hence abelian and therefore solvable. Now, H and G H both are solvable, so by above proposition, G is also solvable. Case (iii). If p < q then by Sylow theorems, G has Sylow p-subgroups each of order p2 and number of these subgroups is np = 1+ mp such that 1  mp o( G) , m is non-negative integer. Since gcd (p , q) = 1, so 1  mp p2 q , implies 1  mp q , so np = 1+ mp = 1 or q. 42