MMT-003 ALGEBRA
Indira Gandhi National Open University School of Sciences
Block
1 GROUP THEORY
Course Introduction 3 Block Introduction 5 Notations and Symbols 6 UNIT 1 Basic Group Theory – A Review 7 UNIT 2 Group Actions 31 UNIT 3 The Sylow Theorems 57
Course Design Committee*
Prof. Gurmeet Kaur Bakshi Prof. Parvati Shastri Deptt. of Mathmatics, Centre for Excellence in Basic Sciences Panjab University Mumbai University
Dr. Anuj Bishnoi Faculty members Deptt. of Mathematics, School of Sciences, IGNOU University of Delhi Dr. Deepika
Prof. K. N. Raghavan Prof. Poornima Mital Deptt. of Mathematics Prof. Parvin Sinclair Institute of Mathematical Sciences, Chennai Prof. Sujatha Varma Prof. Ravi Rao Dr. S. Venkataraman School of Mathematics Tata Institute of Fundamental Research Mumbai
* The course design is based on the recommendations of the Expert Committee for the programme M.Sc (Mathematics with Applications in Computer Science).
Block Preparation Team
Prof. B. Sury (Editor) Prof. Parvin Sinclair Deptt. of Mathematics School of Sciences Indian Statistical Institute, Bengaluru IGNOU Prof. K. N. Raghavan Deptt. of Maths Institute of Mathematical Sciences, Chennai Course Coordinator: Prof. Parvin Sinclair
Acknowledgement: To Sh. S. S. Chauhan for the preparation of the CRC of this block.
November, 2018 © Indira Gandhi National Open University ISBN-8]- All right reserved. No part of this work may be reproduced in any form, by mimeograph or any other means, without permission in writing from the Indira Gandhi National Open University. Further information on the Indira Gandhi National Open University courses, may be obtained from the University’s office at Maidan Garhi, New Delhi-110 068 and IGNOU website www.ignou.ac.in. Printed and published on behalf of the Indira Gandhi National Open University, New Delhi by 2 Prof. M. S. Nathawat, School of Sciences.
COURSE INTRODUCTION
Welcome to the next level of your study of algebra, the previous one being the course on Linear Algebra. As you know by now, abstract algebra is the single vehicle that encapsulates all the aspects of mathematical thinking — particularisation and generalisation, abstraction, brevity and clarity of expression, precision in communicating proofs. Through this course we aim to help you develop these abilities and processes further, while studying some abstract concepts in group, ring and field theory. Of course, you may not find this to be a “standard” algebra course. It has, actually, been specially designed to help you develop a better understanding of some of the application-oriented courses in the next two semesters of this Master’s programme. Therefore, you will find that the concepts and processes that you study in this will be closely tied to examples of their applications alongside.
Before getting further into the details of this course, here is a brief historical overview. The origins of group theory lie in the theory of substitutions, developed by the prolific 18th and 19th century mathematicians Lagrange and Galois, whose work you will keep meeting in this course. Interestingly, while most courses on algebra, including this one, begin with group theory, it is the theory of rings that began developing earlier.
Among the abstract notions of ring theory, the concrete example we first see is that of the ring of integers. Natural generalisations like the ring of integers modulo n and the ring of Gaussian integers occurred in Gauss's work and, simultaneously, rings of polynomials also arose. Initially, they were studied independently, with the former having applications to number theory and the latter giving tools to study geometric problems. Soon, it was realized that it would be beneficial to unify them under the same umbrella of abstract rings. The abstract notion of a commutative ring was first given by Fraenkel in 1914. However, it was the algebraist Emmy Noether, considered by many as the mother of modern algebra, who extended and abstracted the theories of polynomial rings to create abstract commutative ring theory in the 1920s. (A major part of ring theory is non-commutative ring theory, which we shall not really be dealing with in this course.)
You are familiar with the fields of real/complex numbers. In fact, field theory (especially finite fields) came up in Galois's work in the theory of equations. However, it is Weber who gave the first clear definition of an abstract field, in 1893. Later, in 1910, Ernst Steinitz (1871-1928) synthesised the knowledge of abstract field theory accumulated till then. He axiomatically studied the properties of fields and clearly defined many important field-theoretic concepts which you will be studying in Block 4 of this course. In modern times, finite fields have found many applications to areas like coding theory and cryptography. You will study some of these in this course, and details of the applications in the relevant courses in the coming semesters.
Now let us present how the course unfolds. There are two main paths through the land of groups. One emphasises the action of groups on sets, graphs and geometries. Here generators and relations play a central role and, in the finite case, counting arguments come to the fore. The other emphasises the action of groups on vector spaces and modules, both internal and external. Here linear algebra is the key tool. Sometimes these two paths intersect fruitfully. Often they go their divergent ways. The groups that act ‘linearly’ are particularly important and powerful. You will study groups from the first path’s point of view in Block 1, and in Block 2 you will get a flavour of the second path too.
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In Block 3, we focus on basics of ring theory, some of which will be a recalling of what you would have studied earlier. We also discuss congruences here, which has applications in different areas like coding theory and cryptography.
In Block 4, you will be studying fields, with the focus on finite fields and their applications. You will also study some Galois theory here.
Now, a word about our notation. Each unit is divided into sections, which may be further divided into sub-sections. These sections/sub-sections are numbered sequentially, as are the exercises and important equations, within a unit. Regarding the exercises in each unit, these have been placed at many points in a unit. They are meant for you to assess if you have understood the concepts and processes that have been discussed till that point. Therefore, you must attempt to solve all the exercises in the unit as you go along, before you look up the solution. Since the material in the different units is heavily interlinked, there will be cross- referencing. For this we will be using the notation Sec.x.y to mean Section y of Unit x. Further, the end of an example is shown by the sign ∗∗∗ and the end of a proof of a theorem/proposition/corollary is shown by the mark . Another compulsory component of this course is its assignment, which covers the whole course. Your academic counsellor at the study centre will evaluate it, and return it to you with detailed comments. Thus, the assignment is meant to be a teaching as well as an assessment aid. Further, you will not be allowed to take the exam of this course till you submit your assignment response. So please submit it well in time. The course material that we have sent you is self-sufficient. If you have a problem in understanding any portion, please ask your academic counsellor for help. In addition to these exercises, you may like to also look up some other books in the library of your study centre, and try to solve some exercises from these books too. Some useful books and websites are the following: 1) “Algebra”, M. Artin, Pearson 2) “Abstract Algebra”, D.S.Dummit and R.M.Foote, Wiley (Student Edition) 3) “Contemporary Abstract Algebra”, J. Gallian, Narosa 4) “Abstract Algebra: Theory and Applications”, T. W. Judson (freely downloadable e-book) 5) https://www.youtube.com/watch?v=ylAXYqgbp4M (on group theory) 6) http://www.jmilne.org/math/CourseNotes/GT.pdf 7) http://www.math.uconn.edu/~kconrad/blurbs/
Best wishes for an enjoyable time with algebra!
The Course Team
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BLOCK INTRODUCTION
There are basically two groups of people – those who act, and those who claim to act. The first group is less crowded! Paraphrasing Mark Twain
With this block you will resume your study of group theory. This area of mathematics has many applications. Be it a combinatorial problem like counting numbers of isomers of a chemical compound, or the problem of whether a polynomial in one variable admits its roots to be expressed in terms of the coefficients if we allow only algebraic operations and taking of square, cube and higher roots, finite groups play a crucial role. This is not surprising as the conceptual structure behind any kind of symmetry is that of a group.
To start with, in Unit 1 we will help you recall the basic concepts that you studied in your undergraduate studies. So, we will briefly review definitions and results about cosets, normal subgroups, quotient groups, etc.
Unit 2 is focused on group actions. This concept is the reason behind the importance of groups in such diverse areas of science and mathematics. Here you will study about ‘defining actions’, ‘natural actions’ and several other kinds of actions. Related to an action are the concepts of an ‘orbit’ and a ‘stabiliser’. You will study these objects, their properties and the relationship between them. You will also study about the Class Equation here.
In Unit 3 you will study how group actions can be used to prove some very fundamental results about finite groups. These results are due to the Norwegian mathematician, Ludwig Sylow. A considerable part of the unit is about applications of the three theorems of Sylow for studying properties of certain finite groups.
In the next block, we will continue the discussion on group theory, with more time being spent on infinite groups.
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NOTATIONS AND SYMBOLS
N the set of natural numbers Z Z*)( the set of integers (non-zero integers) Q Q*)( the set of rational numbers (non-zero rational numbers) R R* )( the set of real numbers (non-zero real numbers) CC*)( the set of complex numbers (non-zero complex numbers) « (or {}) the empty set ℘ )S( the power set of a set S
M ×nm F( ) the set of m × n matrices over the field F
n )F(GL the set of n × n invertible matrices over the field F |S| the cardinality of the set S
Zn the set of integers modulo n
D n2 the dihedral group of order 2n
Un the group of nth roots of unity ≤ ≤/ )( is a subgroup of (is a proper subgroup of) normal subgroup of (is not a normal subgroup of) H/G the set of left cosets of H in G s.t. such that w.r.t. with respect to ∴ therefore iff if and only if ⇒ ⇔)( implies (implies and is implied by) |G|),G(o order of a finite group G )g(o order of an element g in G |H:G| index of H in G ~ is isomorphic to
Sx the set of permutations on the elements of the set X GAut the group of automorphisms of a group G )G(Inn the group of inner automorphisms of a group G )G(Z the centre of a group G Gx the orbit of x under the action of a group G
G )x(Stab (or Gx ) the stabiliser of x under the action of a group G
Cg the conjugacy class of the element g of a group )g(Z the centraliser of the element g of a group Xg set of elements in X fixed by g in G XG set of elements in X fixed by every element of G H × K direct product of the groups / subgroups H and K
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UNIT 1 BASIC GROUP THEORY – A REVIEW Basic Group Theory – A Review
Structure Page Nos.
1.1 Introduction 7 Objectives 1.2 Groups 7 1.3 Subgroups 12 1.4 Lagrange’s Theorem 14 1.5 Normal Subgroups and Quotient Groups 17 1.6 Group Homomorphisms 19 1.7 Summary 23 1.8 Solutions / Answers 23
1.1 INTRODUCTION
With this unit you will be re-entering the world of abstract algebra. You will, of course, be delving deeper than you did in your undergraduate studies. For this purpose, in this unit we have put together those essential concepts that you have studied earlier on which the first two blocks of this course will take you deeper. Accordingly, you will find a quick overview of groups, subgroups, cosets and Lagrange’s theorem in Sec.1.2-1.4. Then, in Sec.1.5, we focus on normal subgroups and quotient groups. Following this, in Sec.1.6 we shall help you recall what a group homomorphism is, and the Fundamental Theorem of Homomorphism for groups. [If you need to re-look any other algebraic concept from your undergraduate studies, please refer to the titles suggested in the Course Introduction.]
The unit is peppered with exercises. Doing them will help you engage more with the concepts concerned. So, please do not go further till you have done every exercise as you come to it.
Let us now state the broad objectives of studying this unit.
Objectives After studying this unit, you should be able to: • recall what a group, a subgroup and a normal subgroup is; • state, prove and apply Lagrange’s theorem; • state, prove and apply the Fundamental Theorem of Homomorphism, and the other two isomorphism theorems, for groups; and • state, prove and apply Cayley’s theorem.
1.2 GROUPS
You would recall that a set is a well-defined collection of objects. So, for example, N Z Q R,,,, C are sets of numbers that you will recognise. You may also recall that the empty set is denoted by « or {}.
Is it necessary that the objects in a set have some common property, or follow a pattern? For instance, is the collection of all living humans and all living 7
Group Theory insects in India on January 1st, 2020, a set? This collection is well-defined, and hence is a set. For the same reason, {Indira Gandhi National Open University, N 7.5, } is a set.
Next, let us recall the definition, and some properties of a binary operation.
Definitions: 1) A binary operation on a set S is a function from ×SS to S . If ∗ is a binary operation on S, we also say that S is closed with respect
to ∗. We usually denote* 21 )s,s( by s * 21 .s 2) A binary operation * × → SSS : is called
i) associative if s( * )s * = ss * s( * 321321321 ∈∀ .Ss,s,s)s
ii) commutative if s * ss * 211221 ∈∀= .Ss,ss
For example, addition and multiplication are associative and commutative binary operations on C , the set of complex numbers. Also, the intersection and union of the subsets of a non-empty set S are binary operations on ℘ )S( , the power set of S , that is, the set of all subsets of S . Can you see why? Think about it and do the following exercises.
E1) Give an example, with justification, of a collection that is not a set.
E2) Check whether or not subtraction and division are binary operations on R , the set of real numbers. Further, for those operations that are binary, check whether or not they are associative and commutative.
E3) Show that ∩ and ∪ are closed on ℘ )X( , where X is a non-empty set. Are they commutative? Are they associative? Why, or why not?
E4) Define ∗ on R× R by ∗ = − + rssr)s,r( . Is ∗ a binary operation on R ? ‘w.r.t.’ stands for ‘with Is every subset of R closed w.r.t.∗ ? Give reasons for your answers. respect to’.
Now that you know what a set is and what a binary operation is, you would be ready to recall an hierarchy of algebraic objects.
Definitions:
In Unit 7 you shall study 1) A semigroup is a pair (S,∗ ), where S is a non-empty set and ∗ is an semigroups in detail. associative binary operation on S.
2) A monoid is a semigroup (S,∗ ), such that S has an identity element with respect to ∗ . 3) A group is a monoid (G,∗ ), such that every element of G has an inverse in G w.r.t. ∗ .
So, if we focus on the requirements for a group, we see that:
A group is a pair ∗ ),G( , where G is a non-empty set and ∗ is a binary operation on G for which the following properties hold:
G1) ∗ is associative; 8
G2) ∃ ∈ Ge such that ∀ ∈ ∗ = = ∗ xexex,Gx , that is, G contains an Basic Group Theory – A Review identity element e with respect to ∗ ; G3) for each ∈ ∃ ∈GyGx such that x ∗ y = e = y ∗ x , that is, every element in G has an inverse in G with respect to ∗ . (We denote the inverse of x by x −1 .)
Before going further, here are some related remarks.
Remarks: 1) We often drop the operation while referring to a group, if the operation is understood. Here, we will also denote ∗ ba by .ab 2) If the operation is addition, a −1 will be denoted by − a , and the identity will usually be denoted by .0 Similarly, the identity with respect to multiplication will usually be denoted by 1.
So, by definition, Z + ),( is a group, but Z − ),( is not. Also, N + ),( is a semigroup, but neither a monoid nor a group. Similarly, the set of whole numbers, W is a monoid w.r.t. + , but not a group.
Let us look at another example of a group.
Example 1: Consider −=±±= 1i},i,1{G . Show that G is a group with respect to multiplication. Solution: We will draw the Cayley table (named after the algebraist Arthur Cayley) for multiplication.
• 1 −1 i −i
1 1 −1 i −i
−1 −1 1 −i i
i i −i −1 1
−i −i i 1 −1
From the table, you can see that ⋅ is a binary operation on G which is associative. Also, ∈G1 such that = = ∀ ∈Gxxx.11.x , and for each ∈ ∃ ∈GyGx such that = 1y.x . Hence (G,)⋅ is a group. ***
The operation in Example 1 is also commutative. So, ⋅ ),G( is an example of an abelian group, named after the famous Norwegian algebraist Abel, who died very young.
Definition: A group ∗ ),G( , is called abelian, or commutative, if * is Fig.1: Niels Henrik Abel commutative. Otherwise ∗ ),G( , is called non-abelian. (1802-1829) 9
Group Theory You may recall that M ×nm R )( , the set of m× n matrices over R , is an abelian group with respect to matrix addition.
Some examples of non-abelian groups are:
i) Sn , the set of permutations on n objects (for n > 2 ), which is a group with respect to the composition of functions;
ii) n C )(GL , the set of non-singular n × n matrices over C , which is a group with respect to matrix multiplication.
The group in Example 1 also has another property that, for example, Z + ),( does not have. It is a set with only 4 elements. On the other hand, Z is an infinite set. This leads into the following definitions.
Definitions: Let ∗ ),G( be a group. Then i) if < ∞ ∗),G(,|G| is called a finite group. If G is an infinite set, then | X | denotes the number ∗),G( is called an infinite group. of elements in the set X, that is, the cardinality of ii) if = n|G| , then ∗ ),G( is a group of order n . We write = nGo . X. ( )
So, the group in Example 1 is a finite group, of order 4. Also, Z + ),( is an M C = − K1)1n(n!n for infinite group. Further, note that Sn is of order !n, and ×nm +)),(( and
n ∈ N . n C ⋅)),(GL( are infinite groups.
We now look at an important and oft-used finite abelian group.
Example 2 (Group of integers modulo n ): Consider the equivalence relation ~ on Z , given by ‘ b~a iff | − )ba(n ’, where ∈N.n This partitions Z into n
‘iff ’ stands for ‘if and equivalence classes, depending on the remainder left after dividing an integer K only if ’. by n . The set of these classes is Zn = − }1n,,1,0{ . (Thus, for a ∈Z , if
a = nq + r , with ≤ < nr0 , then a ∈ r .) Show that Zn + ),( is a finite abelian
group, where : Z Z Znnn +=+→×+ ba)b,a(: . Solution: Firstly, + is a well-defined operation since
= aa 1 and = bb 1 ⇒ − 1)aa(|n and − 1 )bb(|n
⇒ [( 1 ) ( −+− bbaan 1 )]
⇒ [( ) ( +−+ baban 11 )] ⇒ +=+ baba 11 .
Next, you can check that + is associative on Zn , and commutative on Zn .
Here 0 is the additive identity and n − r is the inverse of r ∈Zn , since ==+− 0nrrn .
Thus, Zn + ),( is a group of order n , and is abelian. *** D is the group of n2 Now consider another important example of a finite group, which is non- symmetries of a regular n-gon. It is called a abelian. dihedral group. Example 3: Let D8 be the set of planar symmetries of a square. Show that 10 (D8, o ) is a finite non-abelian group.
Solution: Let x denote the reflection about a fixed central axis, y denote the Basic Group Theory – A Review rotation though π/ 2 in the positive direction about the centre of the square, and e denote the symmetry where the square does not move (see Fig.2).
Fig.2: Under x, the square is reflected in the y-axis. Under y, the square rotates about O through π/2 in the counter-clockwise direction. We can see that x ≠ e, x o x = e , that is, 2 = ex . (Here, we denote x o y by xy.) Similarly, y, y2 and y3 are not e, but y4 = e . Also = −1 xyxy . 2 4 −1 32 32 So 8 =>===<= }xy,xy,xy,y,y,y,x,e{}xyxy,ey,ex|y,x{D . Now, the Cayley table below shows that D8 is closed with respect to o , o is associative, e is the identity and every element has an inverse. However, xy ≠ yx .
o e x y y2 y3 xy xy2 xy3 e e x y y2 y3 xy xy2 xy3 x x e xy xy2 xy3 y y2 y3 y y xy3 y2 y3 e x xy xy2 y2 y2 xy2 y3 e y xy3 x xy y3 y3 xy e y y2 xy2 xy3 x xy xy y3 xy2 xy3 x e y y2 xy2 xy2 y2 xy3 x xy y3 e y xy3 xy3 y x xy xy2 y2 y3 e
To understand how the entries of the table have been written, take for example 325 2 −2 ===== −− 11 = −1 − == xyy )xy(y)xy(yy)xy(yxyyxyy)xy(yxyyx 321 . Similarly, you will get the other entries for y2 x, etc.
***
Why don’t you try some exercises now?
E5) Prove that, if ∗ ),G( is a group, then i) ∗ has a unique identity element in G ; ii) each ∈Gx has a unique inverse element with respect to ∗ ;
iii) −− 11 ∈∀= Gaa)a( ; 11
Group Theory iv) −−− 111 ∈∀∗=∗ Gb,aab)ba( ; v) ∗ = ∗ ⇒ = ∀ ∈ .Gc,b,acbcaba
E6) Check whether or not ∗ ),G( is a group, where GR* ×= R and R* = R }0{\ ∗ = + ).dbc,ac()d,c()b,a(
E7) Draw the Cayley table for Z7 ⋅ ),( , and hence decide whether it is a group or not.
E8) Give an example of a group ∗ ),G( , and ∈ Gb,a , such that −1 ≠∗∗ .baba
Let us now revisit subgroups and their properties.
1.3 SUBGROUPS
You know that Z and R are groups with respect to the same binary operation, and Z ⊆ R . This makes Z a subgroup of R , according to the definition below.
Definitions: 1) A subset H of a group ∗ ),G( is called a subgroup of ,Gdenoted by ≤ GH , if ∗ ),H( is a group. 2) If ≤ GH and H ⊂ G , then H is called a proper subgroup of .G This is / denoted by H < G , or < GH . / 3) }e{ is a subgroup of G, called the trivial subgroup of G , where e is the identity of G w.r.t. ∗.
For example, n ,A the set of even permutations in n ,S is a subgroup of Sn with o respect to . However, Zn , for n >1, is not a subgroup of Z , since Zn ⊄ Z .
Also, not every subset of a group is a subgroup. For instance, N ⊆ Z , but N is not a subgroup of Z + ),( . (Why?) So, we need some criteria to help us decide which subsets form subgroups. Do you recall the following criterion?
Theorem 1 (Subgroup Criterion): Let G be a group and H a non-empty subset of .G H is a subgroup of G iff −1 aHab ∈∀∈ .Hb,
Let us use this criterion in some examples.
So, by Theorem 1, < > ≤ .Ga Basic Group Theory – A Review ***
Example 5: Let G be a group and = ∈ = xxg|Gx{)G(Z ∀ ∈ }.Ggg Show )G( Z is called the centre that ≤ .G)G(Z of G . Solution: Since ∈ Z),G(Ze ≠ «.)G( Next, if ∈ )G(Zx , then = ∀ ∈ .Gggxxg Therefore, −1 −1 ∈∀= Gggxgx , so that −1 ∈ ).G(Zx Finally, if ∈ )G(Zy,x , then −1 = −1 )gy(xg)xy( =x(gy−1 ) = (xg)y−1 = (gx)y−1 −1 ∈∀= Gg)xy(g . Thus, −1 ∈ ).G(Zxy Hence, by Theorem 1, ≤ .G)G(Z ***
Why don’t you try some exercises now?
E9) Check whether or not the following subsets are subgroups of the group given alongside: i) 1 C =∈= }1|z|z{S of C* ⋅ ),( ; 0a ii) S= a ∈R of M2 R + )),(( ; a0 o iii) { n σ∈σ= oddisSO } of (Sn , ) , n≥ 3;
iv) n R ∈= n R = }1|A|)(GLA{)(SL of n R ⋅ )),(GL( , n≥ 1;
n * v) n C =∈= }1z|z{U of C ⋅ ),( .
E10) Let G be a cyclic group, that is, =< xG> for some ∈ Gx . Show that every subgroup of G is cyclic.
E11) Is every subgroup of an infinite group infinite? Give reasons for your answer.
E12) Show that every cyclic group is abelian, and the centre of a group is abelian.
E13) Is every abelian group cyclic? Why, or why not?
Let us now quickly recall the properties involving operations on subgroups. We state them as a theorem.
Theorem 2: Let H and K be subgroups of a group .G Then i) = ∈ ∈ }Kk,Hh|hk{HK is a subgroup of G iff = .K HHK ii) ∩ ≤ .GKH
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Group Theory iii) ∪ ≤ GKH iff ⊆ KH or ⊆ .HK
Note that if G is a group w.r.t. + , then HK should be seen as + .KH Further, recall that if G is abelian then HK = KH , so that ≤ .GH K
Let us apply Theorem 2 in some cases to regain a feel of its strength.
Example 6: Consider the dihedral group ji 2 5 −1 10 { ⋅⋅======⋅ yx yx,ey,ex,4,,1,0j,1,0iyxD }, and its cyclic subgroups = < xH> , and = < > .yK Find ∪ ∩ KH,KH and K.H Which of
these are subgroups of D10 ? Solution: Now, =∪ y,x,e{KH 432 };y,y,y, ∩ = };e{KH = 432 432 }xy,xy,xy,xy,y,y,y,y,x,e{HK , with all the elements being distinct, as you must check.
So, KH ≤∪ / D 10 , since, for example, ∉ ∪ KHxy though ∈ ∪ .KHy,x H ∩ K is the trivial subgroup.
= DH K10 because ⊆ KDH 10 and == 10 |D|10|HK| . Hence ≤ 10 .DHK ***
Try some exercises now.
E14) Consider the group of quaternions, Q8 = { ± 1, ± i, ± j, ± k}, where ,jiij,kij 2 ==−=−== kj1i 22 . i) Obtain all its cyclic subgroups. ii) Let =< iH> and =< kK> . Find H ∪ K , HK and H ∩ K .
Which of these are subgroups of Q8 ? Give reasons for your answers.
E15) Consider the subgroups = < 5H> and = < 9K> of Z . Check if H ∪ K and H + K are subgroups of Z or not.
We shall now focus on a very basic and very powerful idea in group theory.
1.4 LAGRANGE’S THEOREM
Let us begin this section with a definition.
A coset in a group G is a Definition: Let H be a subgroup of a group G, and let ∈ .Gx Then the subsets subset of G, not an of G , = |xh{xH ∈ }Hh and = ∈ }Hh|hx{Hx , are called a left coset and a element of G. right coset, respectively, of H in .G The element x is called a representative of these cosets.
Note that if the group operation is +, then the cosets of H in G represented by x are + = + ∈ }Hh|hx{Hx and + = + ∈ }Hh|xh{xH , respectively. 14
Here are some important properties about cosets now, which we shall state for Basic Group Theory – A Review right cosets. They hold analogously for left cosets.
Theorem 3: Let G be a group, ≤ GH and ∈ .Gy,x Then i) x ∈ Hx ; ii) x = ⇔ ∈ HxHH ; iii) −1 ∈⇔= .HxyHyHx
You can try and prove Theorem 3 yourself.
Now let us recall how distinct cosets are formed.
Theorem 4: Let G be a group and ≤ .GH The relation ~ on ,G defined by ‘ x ~ y iff −1 ∈ Hxy ’ is an equivalence relation. The equivalence classes are the distinct right cosets of H in .G
Thus, given any group G and its subgroup H ∪= HxG, . In fact, the distinct ∈Gx cosets partition G into sets of size o(H) (see E16).
An immediate result that follows from Theorem 4 is the following.
Corollary 1: If Hx and Hy are two right cosets of H in G ,x then = HyH or =∩ «.HyHx
Let us look at some examples of cosets of a subgroup in a group.
Example 7: Find all the right cosets of
Solution: The distinct cosets are obtained by taking the equivalence classes of the equivalence relation ‘ x ~ y iff − ∈ nyx Z ’, that is, ‘ x ~ y iff − )yx(|n ’.
Thus, these are the same classes as the elements of Zn , that is, Z Z + Z + K Z + − )1n(n,,2n,1n,n (ref. Example 2). We denote them by K − )1n(,,1,0 . ***
Example 8 (A geometric example): Let G = R 2 and = ∈R .}x|)0,x{(H Find the left cosets of H in .G
Solution: Let v = α β ∈ .G),( Then v + = + α β |),x{(H ∈R = R × β .}{}x Thus, v + H is the line y= β , parallel to the line representing ,Hthat is, the x-axis. Thus, all the left cosets of H are represented by the lines parallel to the x -axis (see Fig.3). Two parallel lines are either identical or non-intersecting (as two cosets of H are either equal or disjoint). So, H has infinitely many cosets in G. 15
Group Theory
Fig. 3: The x-axis represents H, L1 represents v + H , and the infinitely many other lines parallel to the x-axis represent the different cosets of H in G. ***
Here are some exercises now.
E16) Let ≤ GH . Show that there is a 1-to-1 correspondence between i) the elements of H and those of any right or left coset of H in G. ii) the set of left cosets and the set of right cosets of H in G.
E17) Find the distinct right cosets of A5 in S5 .
E16 (ii) tells us that the number of distinct cosets of a subgroup H in a group G is independent of whether we take all its left cosets or all its right cosets. Therefore, the following definition can be given.
Definition: Let H be a subgroup of a finite group G . The index of H in G is the number of distinct cosets of H in ,G and is denoted by |H:G| .
Z Z So, for example, = n|n:| and 55 = .2|A:S|
Here is an important remark about cosets.
Remark 3: For any ≤ GH and ∈ ∉ ,Hx,Gx you know that x ∈ xH and x ∈ Hx . So H xxH≠∩ « . However, it need not be true that = .H xxH For
example, consider ≤= 3.S)21(H Then = )}231(,e{H)31( and = )}.321(,e{)31(H So ≠ ).31(HH)31(
Now, let us state the important theorem about finite groups that we had been aiming to do. This is due to the French mathematician Lagrange. He made significant contributions to the fields of analysis, number theory, and mechanics. His treatise, Theorie des Functions Analytiques, laid some of the
Fig. 4: Joseph-Louis foundations of group theory, anticipating the work done by the French Largrange (1736-1813) mathematician Galois (1811-1832). (You will come across some of Galois´ work in the last block of this course.)
16
Theorem 5 (Lagrange): Let H be a subgroup of a finite group .G Then Basic Group Theory – A Review = |H:G|o(H)o(G) .That is, if a group is finite, then the order of a subgroup always divides the order of the group; and the index of a subgroup always divides the order of the group.
As you can see, this result has so many consequences. For one, we can apply this to show that a group of prime order has no non-trivial proper subgroup.
Now, let us recall what the order of an element of a group is.
Definition: Let G be a group and ∈ .Gx The order of x , denoted by o(x), is the order of the cyclic subgroup< x >.
Note that if < > = n|x| < ∞ , then n is the least positive integer such that n = ex . If
So, for instance, = 2))21((o in n ,S and ⋅⋅⋅ )n321((o = n) in n ,S ≥ 3n . Also * = = = )k(o)j(o4)i(o in Q8. And, for any ∈Z m,m is of infinite order.
Here are some exercises now.
E18) Let G be a finite group of order n and ∈ .Gg Show that | n)x(o , and The order of any n element of a finite = .ex group divides the order of the group.
E19) Find )4(o in Z18 and )4(o in Z5 .
n E20) Show that if = ,n)x(o where ∈ Gx , then m )x(o = , where )m,n( 1 ≤ m < n and )m,n( is the gcd of n and m.
E21) Show that if x is of infinite order in G , then so is m ∈∀ Z .}0{\mx
E22) Is the converse of Lagrange’s theorem true? That is, if G is a finite group and )G(o|m , will G have a subgroup of order m ? Give reasons for your answer. (You will study this question in detail in Unit 3.)
Let us now study a very special kind of subgroup.
1.5 NORMAL SUBGROUPS AND QUOTIENT GROUPS
Let us begin this section by considering Remark 3 in the context of certain subgroups.
Definition: A subgroup N of a group G is called a normal subgroup of Gif = ∀ ∈ ,GxxNNx and this is denoted by N G. 17
Group Theory Recall the following equivalent conditions:
i) N G ⇔ -1 NNgg ∈∀⊆ Gg ,
ii) N G ⇔ -1 NNgg ∈∀= Gg ..
Here, note that = xxNN does not mean yx = xy ∀ ∈ .Ny It just means that given ∈ ∃ ∈ Nz,Ny such that yx = xz.
You can check that every subgroup of an abelian group is normal.
However, the converse is not true, one example being 8 .Q
You would also recall that every subgroup of index 2 in a group is a normal
subgroup. Using this result, you know that An n >∀ 1nS .
So, can you recall any subgroup that is not normal? For instance, you have seen in Remark 3 that ≠ )21()31()31()21( . ‘ / ’ denotes ‘is not a normal subgroup of ’. So ( 21 ) / 3 .S
Try the following exercises now.
E23) Show that }e{ G and Z )G( G for any group .G
E24) a) Show that if H and K are normal subgroups of ,G then i) H ∩ K G ii) HK .G
b) Show that if H and K are subgroups of G such that H / G, K / G, then HK need not be a subgroup of G.
E25) Consider D8 in Example 3. Let = < > = < >.yK,xH Show that H / G and = .HKG
E26) Prove that the commutator subgroup of = { −− 11 ∈Gy,x|xyyx'G,G } , is normal in .G
So, what is special about a normal subgroup? To answer this, we look at all the cosets of a normal subgroup, and remind you of the following theorem.
Theorem 6: Let G be a group and H G . Let HG be the set of all (left) cosets of H in .G Define multiplication on the elements of HG by = ∀ ∈ .Gy,xxyHyH.xH Then ( ,HG⋅) is a group. The multiplicative identity is eH , that is, ,H and the inverse of xH is −1 .Hx
The group defined in Theorem 6 is called the quotient group of G by ,Hand is denoted by .HG Note that i) taking all the right cosets of H in G will give us an analogous group; 18
)G(o Basic Group Theory – A Review ii) if G is a finite group, then |H:G|)HG(o == ; )H(o iii) if H is not normal in G , then the cosets of H in G do not form a group. (Why?)
Let us consider a few examples. The first is closely related to Example 7.
Example 9: For ∈ N n,n Z ≤ Z and Z + ),( is abelian, so that nZ Z. Obtain the elements of Z nZ , and Z Z .)n(o Solution: Z nZ is a group with respect to addition, defined by + Z + + Z = + + n)sr()ns()nr( Z , i.e., +=+ .srsr The elements are the distinct cosets of nZ in Z , that is, K − .1n,,1,0 Thus, Z Z = .n)n(o ***
Example 10: Find the elements of 8 >< .iQ Solution: Let = < > = ±1{iH ± }i, . Since ∉ ≠ .HHj,Hj )Q(o Also HQ 8 == .2 Thus, = .}Hj,H{HQ 8 )H(o 8 ***
Here are some related exercises now.
E27) Obtain nn ,AS where ≥ .2n
E28) For any group ,G determine GG and .}e{G
E29) i) Show that if G is cyclic, then so is every quotient group of .G ii) If every non-trivial quotient group of G is cyclic, must G be cyclic? Give reasons for your answer.
E30) Show that if G is abelian, then so is ∀HHG .G
E31) If ∃ H G such that HG is abelian, then G is abelian. Is this statement true? Give reasons for your answer.
Let us now look at operation preserving functions between groups.
1.6 GROUP HOMOMORPHISMS
Let us begin this section by considering the function o n σ=σ⋅−→ )(sign)(f:)},1,1({),S(:f . You know that f is a well-defined function. It also preserves the group operation, that is, o ∈τσ∀τσ=τσ n.S,)(f).(f)(f
19
Group Theory This property makes it a group homomorphism, as you would recall. More formally, f satisfies the following definition.
The word Definition: Let ∗11 ),G( and ∗ 22 ),G( be two groups, and φ → GG: 21 be a ‘homomorphism’ is derived from two Greek function. φ is called a group homomorphism if φ x *1 y = φ *2 φ )y() x()( . If φ words, ‘homos’ and is bijective, it is called a group isomorphism. ‘morphe’, meaning ‘same’ and ‘form’, respectively. Further, the image of φ, Im φ {φ }≤∈= GGx)x( 21 , and the kernel of φ ,
Ker φφφ { ∈= 1 φ )x( = eGx 2} G1 , where e2 is the identity in G2 w.r.t. ∗2 .
Here are some examples.
Example 11: Show that R R* =⋅→+ x(f:),(),(:f e) x is a homomorphism. Also check whether f is an isomorphism or not. Solution: Firstly, since = ⇒ = ,y(f)x(fyx f) is well-defined. Next, since + yxyx R =+∈∀= y, x)y(f).x(f)yx(f,y,xe.ee ∈∀ R . So f is a group homomorphism. Now, { R x }==∈= {01exfKer }, so f is 1-1 . Also, you can check that { x xefIm R}=∈= R+ , so that f is not surjective. Hence f is not an isomorphism. ***
* = )Adet(|A| for Example 12: Check whether 3 R → R = |A|)A( f:)(GL:f is a homomorphism −1 ∈ Mn R)(A . or not. If it is, then is it 1-1? If f is not a homomorphism, find )r(f , for a fixed ∈R* .r Solution: = = = .)B( f)A(f|B||A||AB|)AB(f So, f is a group homomorphism. R R ∈= 3 3 ≠== .}I{)(SL}1|A|)(GLA{fKer Hence f is not 1-1. ***
Example 13: Let → GG: f 21 be a group homomorphism. Give an example,
with justification, to show that fIm need not be normal in 2 .G
Solution: Consider 43 f:SS:f )( σ=σ→ . Then you can check that f is a well- defined group homomorphism.
fIm / S4 because, for example = ∉ m)43()41()31()41( .fI ***
Try some exercises now.
π is called the canonical map. E32) Let G be a group, H G and =π→π .gH)g(:HGG: Show that π is a surjective group homomorphism. Also find π .Ker E33) Let R → C = .r)r(i::i Show that i is a group homomorphism. Also obtain iIm and .iKer 20
Basic Group Theory – A Review E34) If → fGG: 21 and → GG: g 32 are two group homomorphisms, show that o fg is also a group homomorphism.
K −1n E35) Consider n ζζ= },,,1{U , where ζ is a primitive nth root of unity. Z r Define φ → nn φ )( ζ= .r:U: Check if φ is an isomorphism or not.
E36) Show that the homomorphic image of a cyclic group is cyclic. Is the
converse true? That is, if → fGG: 21 is an onto group homomorphism
such that G2 is cyclic, must G1 be cyclic? Give reasons for your answer.
Let us now recall some important theorems on homomorphisms and FTH is also referred to isomorphisms. The first such theorem is the basis on which the others depend. as the ‘First This is why it is called the Fundamental Theorem of Homomorphism, FTH in isomorphism theorem’. short. It is due to the algebraist Jordan.
Theorem 7 (Fundamental Theorem of Homomorphism): Let G1 and G2 be → two groups and φ GG: 21 be a group homomorphism. Then ‘ ~ ’ is the symbol of ‘is φ −~ ImK erG φ. isomorphic to’. . * Using Theorem 7, Example 12 tells us that 3 R 3 R)(SL)(GL ~ R . In fact, on R R ~ R* the same lines you can show that n n )(SL)(GL ≥∀ .2n Similarly, applying Theorem 7 to the signature homomorphism, you can see ~ that AS nn ≥∀− .2n}1,1{
Now for the other two isomorphism theorems.
Theorem 8 (Second Isomorphism Theorem): Let H and K be subgroups of a group G , with K normal in .G Then ∩ )KH( H and ∩ )KH(H ~ .KHK
Theorem 9 (Third Isomorphism Theorem): If H and K are normal subgroups of a group G and K ≤ H , then )KH()KG( ~ .HG
~ Using these results, you can conclude, for example, that Z nZ Zn , or that ~ Z 153 Z Z5 .
Here are some related exercises.
E37) Prove Theorem 7, and use it to prove the other two isomorphism theorems. Note that the set HK has |K||H| E38) Let H and K be subgroups of a finite group ,G with H .G Then cardinality even )K(o)H(o ∩ |KH| show that )HK(o = . ∩ )KH(o if H / G. 21
Group Theory E39) Let G be a group. Consider GAut , the set of isomorphisms from G to G (such isomorphisms are called automorphisms). i) Show that o ),GAut( is a group. −1 ii) For each ∈ Gg , show that g → g = xgg)x(f:GG:f is an
automorphism of .G ( fg is called an inner automorphism of G induced by ∈ .Gg ) iii) Show that GInn , the set of all inner automorphisms of G is a normal subgroup of .GAut
iv) Show that )G(ZG ~ .GInn
E40) Prove that any cyclic group is either isomorphic to Z + ),( or to Zn + ),( for some ∈N.n
We end this section with a result that tells us why permutation groups really form the basis of group theory. This theorem is due to the British mathematician Arthur Cayley, who proved it in 1854. We shall state it now and sketch its proof.
Theorem 10 (Cayley’s theorem): Every group is isomorphic to a group of permutations.
Outline of proof: Let G be a group and )G(S be the group of permutations on the elements of .G For each ∈ ,Ga define → = .ag)g(f:GG:f a a Then ⊆∈ ).G(S}Ga|f{ Fig. 5: Arthur Cayley a (1821-1895) made Next, define → = a .f)a(f:)G(SG:f important Then, you can check that f is a well-defined group homomorphism, contributions to many = }e{fKer and ≤ ).G(SfIm areas in Algebra. Applying the Fundamental Theorem of Homomorphism, we obtain the result.
Let us consider an application of what Cayley has shown.
2 2 Example 14: Consider the Klein 4-group, 4 = { == beaab,b,a,eK }. Find a
group of permutations to which K4 is isomorphic.
Solution: 4 = 4)K(o , so = S)G(S 4 here. Define 44 =φ→φ ,I)e(:SK: φ = φ = φ = φ φ )()( .ba)ab(),43()b(),21()a(
Here we are careful to define φ so that φ ∈∀= 4.Kx))x((o)x(o You can check that φ is a group isomorphism onto
φ = ≤= 44 .SV)}43()21(),43(),21(,e{Im ***
Why don’t you try some exercises now?
E41) Find a group of permutations to which Z is isomorphic. 22 5
E42) If a group-theoretic property holds for every finite permutation group, Basic Group Theory – A Review would it hold for every finite group? Give reasons for your answer.
With this we come to the end of this rather long discussion for reviewing some of what you have studied about groups in your undergraduate studies. Let us end with summarising the major points of this unit.
1.7 SUMMARY
In this unit, we helped you recall the following points. 1) The definition, and examples, of a group and a subgroup. 2) The statement, and applications, of Lagrange’s theorem. 3) The definition, and examples, of normal subgroups and quotient groups. 4) The definition, and examples, of group homomorphisms and isomorphisms. 5) The statement, proof and applications of the Fundamental Theorem of Homomorphism. 6) The statement, proof and applications of Cayley’s theorem for groups.
1.8 SOLUTIONS / ANSWERS
E1) There are infinitely many such examples. One is the collection of all clothes. This is not well-defined because neither do we know whose clothes (humans / animals / etc.) are included, nor is it clear what constitutes a clothing item. Will a fig leaf of any size be included?
R R R E2) Since 21 r,rrr 21 ∈∀∈− , subtraction is a binary operation on .
r1 R However, since ∈ only if 2 ≠ ,0r division is not a binary operation r2 on R. Further, you can check that, for example, − − ≠ − − 3)21()32(1 and − ≠ − .1221 So subtraction is neither associative nor commutative.
E3) Let S and T be subsets of .X Then ∩ TS and ∪ TS are also subsets of .X Hence both the operations are closed on ℘ .)X( You should check that they are both commutative and associative.
E4) Since * ∈R∀ ∈R,s,r)s,r( * is closed on R. However, consider the set of whole numbers, ⊆ R .W Then * = − ∉ .W1)1,0( So W is not closed w.r.t.*.
E5) i) Let e and e′ be identity elements of * in .G Then e* ′ = ,ee since e′ is an identity element. Also e* ′ = ′ ,ee since e is an identity element. Thus, = ′.ee
ii) Let y and z be inverses of x w.r.t.*, and let e be the identity element of * in .G Then x * y = y* x = e and x *z = z * x = e. Now = yy = ye x( = y()z )x = ez = .zz * * * * * * 23
Group Theory Hence the uniqueness.
−− 11 −− 11 iii) Since a * aa * ∈∀=== .Gaa)a(,ea
−− 11 iv) a( * )b * b( * = e)a , using associativity and the definition of −−− 111 inverse. Thus, from (ii) above, a( * = b)b * .a
−1 −1 v) a * = ab * ⇒ ac * a( * = a)b * a( * ⇒ = cb)c , using associativity and the definition of inverse.
E6) * is a binary operation on R*×R since for R* R ac,)d,c(),b,a( ∈×∈ R* and + ∈ R.dbc * is associative on G because )b,a(( * ))d,c( * = + )dbc,ac()f,e( * = + + )fdebce,ace()f,e( , and )b,a( * )d,c(( * = )b,a())f,e( * + = + + .)fdebce,ace()fde,ce( You can check that * has the identity .)0,1( −1 −1 The inverse of ∈ G)b,a( w.r.t.* is − .)ba,a( Thus, ,G(* ) is a group.
E7) • 0 1 2 3 4 5 6
0 0 0 0 0 0 0 0
1 0 1 2 3 4 5 6
2 0 2 4 6 1 3 5
3 0 3 6 2 5 1 4
4 0 4 1 5 2 6 3
5 0 5 3 1 6 4 2
6 0 6 5 4 3 2 1
The second row shows that 1 is the identity w.r.t. • but the first row Z shows that 0 has no inverse. Hence 7 • ),( is not a group.
E8) Consider the group in Example 3. Here, −1 oooo 3 ≠== .yyxyxxyx
E9) i) ∈S1 1 , so that 1 ≠ « .S
1 −1 z1 If 21 ∈Sz,z , then zz 21 == .1 z2 − 11 C*1 Thus, 21 ∈Szz . Hence ≤ .S ii) You can check that S≠ « and if ∈ SB,A , then − ∈ .SBA Thus, M R ≤ 2 .)(S iii) Consider = 3n . Here )21( and )31( are in O. But
= ∉ .O)321()31()21( So < |O 3.S R « 24 iv) You should check that n ≠ .)(SL
R −1 −1 R Basic Group Theory – A Review Further, if ∈ n )(SLB,A , then =1AB , so that ∈ LAB n .)(S R R Hence n ≤ n .)(GL)(SL
n v) Since ∈= n .U1,11 Hence n ≠ « .U −1 n n −n If ∈ Uz,z n21 , then 21 1 2 == 1zz)zz( , since • is commutative. C* Thus, n ≤ .U
E10) Let ≤ =< xGH> . If = }e{H , then it is trivially cyclic. If ≠ }e{H , let ∈ .Hh Then ∈ Gh , so that h = x m for some ∈Z ≠ .0m,m Now, if < 0m , then x −m ∈H also. So, without loss of generality we may assume m ∈∃ Hx with > .0m Let { N m ∈∈= Hxmminn }. Then you should prove that xHn ><= .
E11) No, for example ≤ Z + ),(}0{ , which is of order 1.
E12) Let =< > .xH Any two elements of H are of the form x m and n n,m,x ∈Z . Then + == mnnmnm .x.xxx.x So H is abelian.
Next, let G be any group and let 21 ∈ ).G(Zz,z
Then = zzzz 1221 , since 1 1 ∈∀= .Gxxzxz Hence )G(Z is abelian.
2 2 E13) Consider the Klein 4-group, 4 = == }.bea|ab,b,a,e{K Then K4 is
abelian since = .baab (Why?) But K4 is not cyclic.
E14) i) < > = < − >= − }1,1{1},1{1 < >=< − > = − − }1,1,i,i{ii < >=< − >= ± ± }j,1{jj < >=< − >= ± ± }k,1{kk
ii) ∪ = ± ± ± }k,i,1{KH . Check that this is not a subgroup of 8 .Q
=±+±±= Q}j,k,i,1{KH 8
≤±=∩ 8 .Q}1{KH < n > + < m > is the subgroup Z generated by gcd(n, m) in Z , E15) Since 5 and 9 are in H ∪ K , but − 95 is not, ∪ ≤ .|KH and < n > ∩ < m > is the + = < > + < > = < 195KH > = Z . Thus, + ≤ Z.KH subgroup generated by lcm n()m, in Z . E16) i) Define → = .hx)h(f:HxH:f Then you can check that f is a bijection. ii) Define =ψ∈→∈ψ −1 .Hx)Hx(:}Gg|gH{}Gx|Hx{: This is well-defined since −1 −1 =⇔∈⇔= −1 .HyHxHxyHyHx You can check that ψ is one-one and onto. Hence the result.
25
Group Theory o E17) The distinct cosets are A5 and 5 ).21(A
This is because any permutation in S5 is either odd or even. If it is even, o o it is in A5 . If σ is odd, then ∈σ A)21( 5 , say ∈α=σ 5 .A)21( o −1 o o Then ∈α=α=σ 5 ).21(A)21()21( o So, ∪= 555 ).21(AAS This partition is not unique, as you can see from the construction. For o instance, ∪= 555 )43(AAS also. Here, also note that o o −1 5 = 5 )43(A)21(A since ∈ 5.A)43()21(
E18) = < >)x(o)x(o , which divides )G(o by Lagrange’s theorem. ∴ .)G(o)x(o Let = m)x(o . Since = mrn,nm for some ∈Z.r Now m = ⇒ nmrrm === .exx)x(ex
Z E19) Since 18 == .18or9,6,3,2,1)4(o,18)(o Since ≠≠=⋅ .2)4(o,0842 Similarly, you can check that ≠ .6or3,1)4(o Now ==⋅ 03649 . Thus, 9 is the least natural number such that = 04.m . So = .9)4(o Z Similarly, check that )4(o in 5 is 5.
E20) Since = m )x(o,n)x(o must be finite, say r. Then mr = ⇒ .mrnex
Now let 1 === dnn,dmmandd)m,n( 1 , where 11 = .1)m,n(
Then 11 rdmdn , that is, 11 rmn . But 11 = .1)m,n(
So 1 .rn
m 1 dnmn 11 1nm Also === exx)x( . Thus, 1.nr n n Hence nr == , that is, m )x(o = . 1 )m,n( )m,n(
E21) Suppose m r)x(o ∈= Z , for some ∈Z∗ .m Then mr = ex , which means that )x(o is finite, a contradiction. Thus, what we assumed was wrong, that is, x m cannot be of finite order.
E22) Not necessarily. For example, 4 = 12)A(o , but A4 has no subgroup of order 6.
E23) You know that }e{ and )G(Z are subgroups of G. So, now we only need to check that they are normal in G. For any −1 ∈=∈ }.e{eegg,Gg Hence }e{ .G Next, let ∈ Gg and ∈ ).G(Zz Then −1 = −1 zgggzg , since zg = g z. = ∈ ).G(Zz 26
Thus, )G(Z .G Basic Group Theory – A Review
E24) a) i) Let x ∈ H ∩ K and ∈ .Gg Since H G and ∈ −1 ∈ .Hxgg,Hx Similarly, −1 ∈ .Kxgg Thus, −1 ∩∈ .KHxgg Hence H ∩ K .G
ii) Let ∈ ∈ .Gg,HKhk Then −1 −1 ∈∈ .Kkgg,Hhgg So −1 = −1 −1 ∈ HK)kgg()hgg(g)hk(g , and hence, KH .G
b) For example, )21( and )31( are not normal in 3 ,S and
E25) From the elements given in Example 3, you can see that = KHG . Note that −1 2 ∉= Hxyxyy . So H / G.
E26) Let −− 11 ∈ Gxyyx ′ and ∈ .Gg Then −−− 111 = −− 11 −− 11 −1 −1 )gyg()gxg()gyg()gxg(g)xyyx(g = −−−−− 11111 −1 ∈ ′.G)gyg()gxg()gyg()gxg( Therefore, for any ∈α ′ −1 ∈α ′ ∈∀ .GgGgg,G So G′ .G
E27) As in E17, you can show that = nnnn )}.21(A,A{AS
G E28) Since G ,1 G ==== }.e{}G{ G G G
Since G = G and }e{gg G ,Gg G ∈=∈∀∈= }.Gg|g{ }e{ }e{ }e{ Also note that for −1 =⇔∈⇔=∈ .hg}e{ghhg,Gh,g So G .G~ }e{ −
E29) i) Let =< gG> and ≤ GH . Then gHm ><= for some ∈Z.m Then G ⊇ 2 −1m =⋅⋅⋅ , say. H S}Hg,,Hg,Hg,H{ Also, any element of G is of the form t = t .)gH(Hg H If t ≤ m , then t ∈ .SHg If t > m , then t = mq + r , where r < m. So ⋅= g)g(g rqmt , hence rt ∈= .SHgHg Thus, G = , and =< HgS> , the cyclic group generated by H S .Hg
ii) This is not true. For instance, consider the Klein 4-group. It is not cyclic, but all its non-trivial quotient groups are cyclic, being of order 2.
27
Group Theory E30) Let ∈G , where ∈ .Gy,x Then x y = xy = yx = y x. Hence the y,x H result.
E31) This is not true. For instance, 8 iQ >< is abelian, since it is cyclic. But
Q8 is not abelian, as you know.
⇒ E32) π is well-defined because = = 2121 HgHggg , that is, 1 π=π 2 )g()g( .
Further, 212121 ππ===π 21 ).g()g(gggg)gg( Now, any element of G is g for some ∈ Gg , which is π )g( . So, π H is surjective. π = ∈ π = = ∈ ∈ = .H}Hg|Gg{}H)g(|Gg{Ker
E33) 12121 +=+=+ 2 ).r(i)r(irr)rr(i You can show that iIm= R and = }.0{iKer
E34) For ∈ Gy,x , o = = ))y(f)x(f(g))xy(f(g)xy()fg( = = oo ).y()fg()x()fg())y(f(g))x(f(g o Thus, fg is a group homomorphism from G1 to 3 .G
Z E35) To check that φ is well-defined, let r = s in n . Then − )sr(|n , say − = .nt)sr( Then − ntsr =ζ=ζ 1, that is ζ=ζ sr . Hence φ is well-defined.
+ 21 rrrr 21 Now, 21 φφ 21 )rr()rr( =ζ⋅ζ=ζ=+=+ φ⋅φ rr 21 )()( , so that φ is a group homomorphism. Z r Z Next, Ker φ n ∈==ζ∈= n | = .}0{}rn|r{}1|r{ r r Finally, any element of Un is ζ , where ≤ ≤ −1nr0 and =ζ φ ),r( Z where ∈ n .r So φ is onto. Hence φ is a group isomomorphism.
E36) Let → GG:f 21 be an onto group homomorphism, where 1 ><= .gG m m Then, any element of G2 is of the form = )]g(f[)g(f for some Z m∈ . Thus, 2 ><= .)g(fG
For the converse, consider →π / 333 .ASS: Then π is an onto
homomorphism and / AS 33 is cyclic, but S3 is not.
E37) Outline of proof of FTH: Define ψ : G → Im φ=ψφ )g()g(: . Ker φ Check that ψ is well-defined, ψ is a group homomorphism, ψ is surjective, ψ is − .11 To prove Theorem 8, define HK =λ→λ Check that λ is H: K .h)h(: well-defined, a group homomorphism, and surjective. Now ∩=∈∈==∈=λ .KH}Kh|Hh{}1h|Hh{Ker Now apply Theorem 7 to get the result.
28
To prove Theorem 9, define G G =µ→µ Basic Group Theory – A Review : K H .gH)gK(: Show that µ is well-defined, a group homomorphism, surjective and =µ H . Now apply Theorem 7 to get the result. Ker K
E38) By Theorem 8, HK ~− K H ∩ KH . )HK(o )K(o Thus = . Hence the result. )H(o ∩ )KH(o
E39) i) Firstly, since G≠ «, the identity map,∈ .GAutI So ≠ « .GAut
Next, if 1 →ψ GG: and 2 →ψ GG: are in Aut G, then o o 21 →ψψ GG: is also an isomorphism. Thus, is closed on Aut G. Thirdly, o is associative, since it is associative in general. Fourthly, → = g)g(I:GG:I is the identity w.r.t. o . Finally, −1 →ψ GG: is the inverse of ψ → .GG: Thus, o ),GAut( is a group.
ii) You can check that fg is 1-1, onto and a group homomorphism, for each ∈ .Gg iii) Firstly, since G≠ « , Inn G≠ « . (Why?) Next, for ∈ ,Gg,g show that o ∈= .GInnfff 21 21 gggg 12 −1 Also, for 1 ∈ g −1 ∈= .GInnf)f(,Gg 1 g1 So ≤ .GAutGInn Now, for ψ∈ GAut and ∈ Gg , show that −1 oo ∈=ψψ GInnff . Hence GInn .GAut g ψ−1 )g(
iv) Define → = −1 .f)g(f:GInnG:f g Check that f is well-defined. o Now 21 −1 1 −− 1 −1 −1 ==== 21 ).g(f)g(fffff)gg(f 21 2 1 1 gggg)gg( 2 Thus f is a group homomorphism. Also, check that f is onto. Finally, ∈==∈= −1 ∈∀= }Gxxgxg|Gg{}ff|Gg{fKer g−1 e = ∈ = ∀ ∈ }Gxxggx|Gg{ = ).G(Z Now use FTH to get the result.
E40) Let =< gG> be a cyclic group. Define Z =ψ→ψ m .g)m(:G: Now ψ is a well-defined group homomorphism, which is surjective. If G is an infinite cyclic group, then ψ = ψ ⇒ = nm)n()m( , that is, ψ = .}0{Ker And hence, by FTH, Z−~ < > .g n m If G is finite, of order n, then = eg and if = eg for m∈Z , then m|n . So, in this case, Z m ==∈=ψ Z.n}eg|m{Ker 29
Group Theory Thus, by FTH, Z ~ ><− nZ .g Since Z ~ Z g, −><− ~Z . nZ n n
Z E41) Define 55 =ψ→ψ )54321()1(:S: and =ψ I)0( , and extend ψ to be a homomorphism, that is, =∀ψ=ψ .5,4,3,2m)1(m)m( Then, the way ψ is defined, it is a group homomorphism which is 1-1,
and <=ψ ≤> 5.S)54321(Im
E42) Yes, it would. This is because, by Cayley’s theorem, the finite group is isomorphic to a finite permutation group; and being isomorphic they must satisfy the same properties.
30
UNIT 2 GROUP ACTIONS Group Actions
Structure Page Nos.
2.1 Introduction 31 Objectives 2.2 What is a Group Action? 32 2.3 Orbits and Stabilisers 34 Orbits Stabilisers The Necklace Problem 2.4 The Class Equation 45 2.5 Summary 49 2.6 Solutions / Answers 49
2.1 INTRODUCTION
Symmetry is everywhere around us. It underlies beauty and structure. Its study is, therefore, important. The theory of group actions, which we will be concerned with in this unit, is indeed the study of symmetry.
The unit begins with examples of group actions, in Sec.2.2. Here we also talk about the “group of symmetries” of any object, and how it acts naturally on the object. In the next unit, you will study how the actions of a group on itself are very useful in unravelling the internal structure of the group.
In Sec.2.3, we introduce the notions of orbits, stabilisers, and invariant subsets. These are the nuts and bolts of the engine of group actions. The result about the structure of an orbit is simple to state and prove, but fundamental. Combining it with Lagrange’s theorem, which you have studied earlier, leads to the conclusion that the cardinality of any orbit of a finite group divides the order of the group. In this section, you will also study the Orbit Counting Lemma, and one of its applications, in detail.
The Class Equation, which relates the orders of conjugacy classes in a finite group, is introduced in Sec.2.4. This is an important tool for studying finite A group is simple if groups, as you will see in this section and in the next unit. Here you will also it has no proper non- trivial normal see how it is used to prove that the alternating group, n ,A is simple for ≥ .5n subgroup.
Let us now list the precise learning expectations around which this unit is created.
Objectives
After studying this unit, you should be able to: • define, and give examples, of a group action; • define orbits and stabilisers, and work them out in given situations; • state, and apply, the results about the structure of an orbit, the Orbit Counting Theorem and the Class Equation; and
• prove that An is simple, for ≥ 5n . 31
Group Theory 2.2 WHAT IS A GROUP ACTION?
In this section we focus on what it means for a group to act on a set, and give some examples of group actions. Let us begin with considering the binary operation + : R × R → R.
Here + = + sr)s,r( . Note that + = ∀rr)r,0( ∈R , and R 321321321 321321 ∈∀++=++=++=++ .r,r,r)r),r,r((r)rr()rr(r))r,r(,r( So, according to the definition we now give, R acts on itself through addition.
Definition: Let G be a group and X a set. We say that G acts on X, or that there is an action of G on X, or that X is a G-set, if there exists a function ϕ × → XXG: satisfying the following axioms: i) ϕ = x)x,e( for every x in X , where e denotes the identity element of the group G ; ii) (Associative law) ϕ ϕ = ϕ ∀ ∈Xx)x),gh(())x,h(,g( and ∀ ∈ .Gh,g
Usually ϕ )x,g( is denoted by g.x . Then the defining axioms become i) = ∀ ∈Xxxx.e , and ii) = ∀ ∈Xxx.gh)x.h(.g and ∈ .Gh,g
In the example above of + : R × R → R , you saw how a group can act on itself via the binary operation that makes it a group. This is called the regular action of R on itself. In general, ⋅ × → ⋅ = gh)h,g(:GGG: is called the left regular action of G on itself. Similarly, =⋅→×⋅ hg)h,g(:GGG: −1 is called the right regular action of G on itself.
Let us now consider some examples where ≠ .XG
Example 1: Let X be a non-empty set. Let Sx be the group of its symmetries,
that is, the 1-to-1 maps from X to .X Define a map x →× XXS by a )x(g)x,g( for ∈Sg x and ∈ .Xx Show that this defines an action of Sx on
.X (We call this the defining action of x ,S because X helped define Sx in the first place.) Solution: In this case, a = ,x)x(I)x,e( and a o o = = ∀ ∈Xx)x,gf()x()gf())x(g(f))x,g(,f( and ∈ x .Sg,f Thus, we see that the given map is an action of G on .X ***
Remark 1: A special case of Example 1 is very basic and important. Let n denote a fixed positive integer and X the set = .}n,...,1{]n[ In this case we get a the defining action of Sn on [n] as ∈σ∀σσ S)x()x,( n and ∈[ ].nx
Now, for any group G, consider ψ × → ψ ϕ = ϕ .)g()g,(:GGGAut: Then ψ is an action of GAut on G. A special case of this is discussed in the next example. 32
R R2 R Group Actions Example 2 (The action of GL2 )( on ): Show that the group 2 )(GL acts on the set of column matrices of size 2×1 with real entries. r r 2 RRR 2 1 a 1 Solution: Consider the map ϕ 2 )(GL: →× A: , A . r2 r2 r r r r 1 a 1 1 1 Since ,I and ϕ B,A ϕ= ,AB , this map is an action. r2 r2 r2 r2 ***
R Following more generally from Example 2, the group n L)(G acts by usual matrix multiplication on the set Rn of column matrices of size n ×1 with real entries.
Here are some more examples of group actions.
Example 3 (The conjugation action of G on itself): Let G be a group. Define a map × → GGG by a −1.gxg)x,g( Show that this defines an action of G on itself. (This is referred to as the conjugation action of G on itself.) Solution: Since −1 = xexe and −1 = −− 11 ,g,xg)hxh(g)gh(x)gh( ∈∀ ,Gh the given map is an action. ***
Example 4 (Action of G on left cosets of a subgroup): Let G be a group and H be a subgroup of G . Denote the set of all (left) cosets of H in G by .H/G Show that G acts on .H/G Solution: Consider the map × → H/G)H/G(G given by a .gxH)xH,g( You can check that this defines an action of G on ,H/G the set of left cosets of .H ***
Example 5 (The trivial action): For any group G and any non-empty set X, consider the map defined by g.x = x for all g in G and all x in .X Show that this defines an action. Solution: You can check that ⋅ = ∀ ∈Gxxxe and ⋅ ⋅ = ⋅ x)gh()xh(g ∀ ∈ ∈ .Xx,Gh,g ***
Example(Restricting 6 the action) : Let H be a subgroup of a group .G Show that any G-set X can also be considered as an H-set by just restricting the action to .H Solution: Consider the action × → x,g(:XXG a .x.g) Define the map ψ × → ψ = .x.h)x,h(:XXH: Then ψ is an action of H on X, which is the restricted action. Thus, the action map × → XXH is the composition of the inclusion × ⊆ × XGXH followed by the action map × → .XXG ***
33
Group Theory Some very interesting actions follow from Example 6. For instance, taking R R R2 = 2 )(SLH in Example 2, you get an action of L2 )(S on .
Try some related exercises now.
M E1) Show that n )F(GL acts on n )F( by conjugation, where F is a field.
E2) Let G be a group and let H be a proper normal subgroup of .G Give two distinct G-actions on .HG
E3) Let X be a non-empty set. Does there exist a group acting on X ? If yes, will such a group be unique? Give reasons for your answers.
E4) Give an example, with justification, of a map × → XXG which is not an action.
We now look at ways in which a G-set decomposes into particular types of subsets that help us study the G-set.
2.3 ORBITS AND STABILISERS
In this section, we will introduce some basic sets related to a group action: orbits, stabilisers and invariant subsets. These are very useful for getting a complete picture of a given action. This section includes a structure theorem for orbits. While the theorem is quite simple to state and prove, its importance cannot be overstated. It is used time and again in what follows. For instance, you will see how it is combined with Lagrange’s theorem to derive the important fact that the order of any orbit of a finite group divides the order of the group. We end the section with a detailed application of the Orbit Counting Theorem, also known as the Cauchy-Frobenius Lemma.
Throughout this section, G denotes a group and X a G-set.
2.3.1 Orbits Let us define a relation ~ on the set X by ‘ x ~ y iff there exists g in G such that gx = y ’. This relation is: • reflexive: since ex = x , where e denotes the identity element of ∀ ∈ ;Xxx~x,G • symmetric: If gx = y , then −1 = ,xyg that is, ⇒ ∀ ∈ ;Xy,xx~yy~x
• transitive: If gx = y and = zhy , then = ,zx)hg( that is, x ~ y and ⇒ ∀ ∈ .Xz,y,xz~xz~y So, ~ is an equivalence relation.
The equivalence classes under this equivalence relation are called the G-orbits in X, or simply orbits, if G and X are clear from the context. As you know, being equivalence classes, the orbits form a partition of X. Hence, we have proved the following theorem.
Theorem 1: The G-set X is the disjoint union of its orbits. 34
Note that, for ∈ ,Xx the orbit containing x is ∈⋅= .}Ggxg{Gx So Group Actions ∪= .GxX ∈Xx
Let us consider an example.
Example 7: Consider the action in Example 2. Obtain all the orbits of R2 under this action. Solution: We will show that there are only two disjoint orbits. So, let )y,x( be any non-zero element of R2 . We will show that it belongs to the orbit 1 1 R of , that is, A ∈ 2 .)(GLA 0 0 There are two possibilities: i) )y,x( and )0,1( are linearly dependent, say = ≠ .0a),0,1(a)y,x( 0a 1 x 0a R Then = and ∈ 2 .)(GL a0 0 y a0 1x R ii) ( y,x) and ( 0,1) are linearly independent. Then ∈ 2 )(GL and 0y 1x 1 x = . 0y 0 y x 1 Thus, in either case, is in the orbit of . So, there is only one orbit y 0 consisting of all the non-zero elements of R2 .
The other orbit is {( )}.0,0 ***
Example 7 leads us to the following definition.
Definition: An action of a group G on a set X is called transitive if = ,GxX for some ∈ .Xx In other words, the action is transitive if for ∈ ∃ ∈GgXy,x such that y = gx.
R R2 So the action in Example 7, of 2 ),(GL is transitive on .)}0,0{(\ Now consider another important example of orbits.
Example 8: Obtain the orbits for the conjugacy action of a group G on itself.
Solution: →× a h,g(:GGG ghg) −1 is the action. So, for ∈ ,Gx = −1 ∈ .}Gg|gxg{x.G So ∪= −1 ∈ |gxg{G .}Gg ∈Gx ***
Example 8 leads us to the following definition.
Definition: Let G be a group and ∈ .Gx The conjugacy class of x in G is −1 Cx = ∈ }Gg|gxg{ . 35
Group Theory So, for example, in the group R2 + ),,( R2 0,1( ) −+= =∈ .)}0,1{(})y,x(|)y,x()0,1()y,x{(C In fact, more generally we have the following remark.
−1 Remark 2: If ∈ Zz ),G( then z = },z{C since ∈∀= .Ggzgzg Conversely, if
for x =∈ }x{C,Gx , then ∈ .)G(Zx
Now for an example with a geometric flavour!
Example 9: Let G be the group of all rigid motions of a plane. Show that G acts on ,S the set of triangles in the plane. Also obtain the orbit of a triangle under this action. Solution: The elements of G are rotations about a point in the plane or reflections about an axis in the plane. The identity, ,I is the motion that makes no change in the plane. Thus, for any ∈ a s)s,I(,Ss and
21 = 21 .)s,gg())s,g(,g(
For any ∈Ss Gs, will consist of all the triangles in the plane obtained by applying the rigid motions to s. Hence, Gs is the set of triangles that are congruent to s. ***
Try an exercise now.
E5) Prove the following: i) The whole of G is a single orbit for the regular action of G on itself. ii) The action of G on HG in Example 4 is transitive. iii) Let H be a subgroup of a group .G Restrict to H the regular action of G on itself. The H-orbits for this action are precisely the right cosets of .H
iv) Sn acts transitively on .}n,...,2,1{
Now, in Example 6 you saw how a G-set becomes an H-set for ≤ .GH Here is a somewhat similar concept.
Definition: A subset Y of a G-set X is said to be G-invariant yif ∈ Yg ∀ ∈ ∈ ,Yy,Gg that is, if Y is a G-set in its own right. Hence, a subset Y of X is G-invariant if and only if it is a disjoint union of G-orbits.
Let us consider an example.
Example 10: Show that if X is a G-set, then each orbit of X is G-invariant. Solution: Let ∈ .Xx Then, for any ∈ ⋅ ⋅ = ⋅ ∈ .Gxxgh)xh(g,Gg So Gx is G-invariant. *** 36
Let us consider some properties of G-invariant subsets. Group Actions
Theorem 2: The intersections and unions of G-invariant subsets are G-invariant.
Proof: Let X be a G-set and Y1 and Y2 be two G-invariant subsets of .X
For ∈ Gg and 21 ∈∩∈ Ygy,YYy 1 and ∈ 2 ,Ygy since ∈Yy 1 and ∈ 2 .Yy Thus
∩∈ 21 .YYgy Hence ∩ YY 21 is G-invariant.
Let ∪∈ 21 .YYy If ∈ 1 ,Yy then for ∪⊆∈∈ 211 .YYYgy,Gg
Similarly, if ∈Yy 2 , ∪∈ 21 .YYgy Hence, ∪ YY 21 is G-invariant.
This theorem is of great use while studying G-sets, as you will see in Theorem 3.
Try an exercise now.
E6) Prove the following: i) A subgroup H of a group G is normal if and only if it is invariant under the conjugacy action of G on itself. ii) Restrict to a subgroup H the regular action of G on itself. A subset of G is H-invariant under this action if and only if it is the union of right cosets of .H iii) Let X be a G-set. For the induced action of G on X× X (that is, × × → × a )gy,gx())y,x(,g(:XXXXG ), the “diagonal” {( ) ∈ Xxx,x } is an invariant subset.
We now look at another set that comes into play when a group acts on a set.
2.3.2 Stabilisers If G is a group acting on a set X, there are two concepts that come into play, which we shall now define.
Definitions: Let G be a group and X a G-set. 1) For g in G and x in ,X we say that g fixes x, or that x is fixed by g, if g.x = x. Further, Xg { =∈= xx.gXx } is the subset consisting of all elements of X that are fixed by g. 2) Given an element x in X, we define the stabiliser of x in G by
G (x)Stab { =∈= }.xgxGg This is also denoted by x ,G and Stab(x) (if G is clear from the context).
g Note that G )x(Stab is a subgroup of G and X is a subset of X.
There is a nice relationship between the stabilisers of points that belong to the same orbit, which we now give.
Proposition 1: For g in G and for x in ,X = −1.g)x(Stabg)gx(Stab 37
Group Theory Proof: ∈ )g x(Stabh ⇔ = gxh gx ⇔ −1 = xx)hgg( −1 ∈⇔ x(Stabhgg ) ∈⇔ )x(Stabgh g −1 Hence the result.
Let us consider an example of the use of this relationship.
R R2 Example 11: For the action of L2 )(G on by matrix multiplication, find 1 the stabiliser of the vector . Hence describe the stabilisers of the other non- 0 zero vectors in R2 . Also find the stabiliser of zero in R2 . ba ba 1 1 R Solution: Let = .)0,1(v Then )v(Stab = ∈ L2 )(G = dc dc 0 0 x1 = R ≠∈ .0yandy,x y0 By Example 7, we know the action is transitive for all non-zero vectors. So, 1 α R2 R given any other ∈βα ∈∃ 2 )(GLA)},0,0{(\),( s.t. A = . Then, by 0 β Proposition 1, =βα −1.A)v(baStA)),((Stab 0 0 R R Also, ∈= 2 A)(GLA))0,0((Stab = = 2 .)(GL 0 0 ***
Why not try and solve some related exercises now?
E7) Prove the following results: i) For the regular action of G on itself, the stabiliser of any element of G is the trivial subgroup, .}e{
The centraliser of ∈ Gg is ii) For the conjugation action of G on itself, the stabiliser of any { =∈= gxxgGx)g(Z }. element of G is its centraliser. iii) For the action of G on the coset space HG of a subgroup, the stabiliser of a coset xH is the conjugate xHx −1 of H by x.
a ~ E8) We have a natural group homomorphism S −1n Sn given by ϕ ϕ , where we define ϕ~ by ϕ~ = ϕ )j()j( for < nj and ϕ~ = .n)n( Consider the
defining action on ]n[ of the symmetric group n .S Show that the stabiliser of n is precisely the image of the group homomorphism given above.
E9) Let a group G act on a set .X Show that there is a group homomorphism a →ϕ SG: x given by g ϕg , where g =ϕ x(.x.g) Further, show that
38
Group Actions Ker ∩=ϕ )x(Stab . Hence show that →ϕ G ,SG: where G acts on itself ∈Xx by the regular action, is 1-1. (Note: This is precisely Cayley’s Theorem, which you have studied in Unit 1.)
Now let us see why G-orbits are so important. As you have seen in Theorem 1, the G-set X is the disjoint union of its orbits. Moreover, from Example 10 you know that each orbit is G-invariant and can be considered as a G-set in its own right. Thus, to understand the action of G on ,X it would be enough to understand the action of G on each and every orbit. We could then “stitch together” the G-action on orbits to obtain the G-action on .X We shall now show how this can be done.
Theorem 3 (Structure Theorem for Orbits): There is a natural bijection between the orbit Gx of x ∈X and the set of left cosets (or right cosets) of
G )x(Stab in .G
Proof: Consider ,)x(StabG the set of left cosets of )x(Stab in ,G and ϕ ϕ→ = gx))x(Stabg(:Gx)x(StabG: . Firstly, ϕ is well-defined (since = ⇒ −1 ∈ )x(Stabgh)x(Stabh)x(Stabg ⇒ −1 = ⇒ = ).hxgxxgxh Also, you should check that ϕ is a bijection. Hence the result.
From Theorem 3, we immediately have the following corollary. ‘Corollary’ is explained Corollary 1: Let G be a finite group, X a G-set and ∈ .Xx Then the in Remark 4. cardinality of the orbit of Gx,x , equals )x(StabG . In particular, it divides = |(x)Stab:G||Gx| .G G
Proof: By Theorem 3, you know that )x(StabG and Gx have the same cardinality. Also, by Lagrange’s theorem (Theorem 5 of Unit 1), you know that = )x(StabG)x(StabG . Therefore, = )x(StabGGx . Thus, )G(o|Gx| .
Remark 3: Both, the G-set X and the element x in Corollary 1, are arbitrary. The only assumption made on G is that it is finite. So the conclusion is that the cardinality of any orbit of any finite group G, for any action whatsoever, divides the order of G.
Corollary 2: Let G be a finite group and ∈ .Gg Then |C| = |G| . g |)g(Z|
In particular, g ).G(o|C|
Proof: Consider the action of G on itself by conjugation. The orbits for this action are the conjugacy classes in .G By E7, you know that the stabiliser of g for the conjugacy action is precisely ),g(Z the centraliser of g. Hence the result. 39
Group Theory Before we go further, here is a general remark.
Remark 4: The term ‘lemma’ is used for a minor result that is a stepping stone for proving a major result, which is the ‘theorem’. The same lemma could be used in the proof of more than one theorem, to avoid using the same argument in their proofs. However, some lemmas, like Zorn’s lemma or the Cauchy-Frobenius Lemma, are so widely applicable that they are as important, or even more important, than some theorems. Such lemmas are very few, and are singled out by giving them such names.
The term ‘proposition’ is used for a result that is proved, but which is not as important as a result that is given the status of a ‘theorem’. The term ‘corollary’ is used for a result that immediately follows from a Fig. 1: The English theorem. mathematician, William Burnside (1852-1927), Now we come to a very important result that follows from Corollary 1. Though was one of the pioneers in it is commonly known as Burnside’s Lemma, Burnside attributed this result to the study of groups and their actions. the mathematician Frobenius (1849-1917), a pioneer in the theory of groups. This result was also known as early as 1845 to the famous mathematician Cauchy (1789-1857).
Theorem 4 (Orbit Counting Theorem): Let G be a finite group and X a The Orbit Counting 1 Theorem is also finite G-set. Then the number of G-orbits in X = ∑ g .X called the Cauchy- )G(o ∈Gg Frobenius Lemma. (The right hand side of the equation may be interpreted as the number of points fixed “on an average” by a group element.)
Proof: Imagine a matrix M whose rows are indexed by the elements of X and columns are indexed by the elements of .G There are X rows and G columns in .M If x in X is the row index and g in G the column index of an entry, we set the entry to be 1 if g fixes x (that is, g⋅x = x ), and to be 0 otherwise.
[For instance, the matrix M for the defining action of the symmetric group S3 on = }3,2,1{X is given below. Here, for instance, the entry in ))31(,2( is 1, since ⋅ = 22)31( , and so on.]
S3 X identity (1 2) (1 3) (2 3) (1 2 3) (1 3 2) 1 1 0 0 1 0 0 2 1 0 1 0 0 0 3 1 1 0 0 0 0
The simple and elegant idea behind the proof of the lemma runs as follows. Imagine that we want to compute the sum of the entries of the matrix .M One way to do this would be to first take the sum of the entries along each row, and then sum the row-sums. Another way would be to first sum the entries along each column, and then sum the column-sums. The theorem is obtained by equating the results obtained by following these different ways.
Let us first consider the row-sums. Fix x in X and consider the row indexed by x. The entry of this row in the column corresponding to g in G is 1 if and 40
only if g belongs to the stabiliser of x . Thus the row-sum of the row indexed Group Actions by x is )x(Stab , and the sum of the row-sums is ∑ .)x(Stab ∈Xx Turning to the column-sums, you can see that Xg is the sum of the entries in the column indexed by g in .G So the sum of the column-sums is ∑ g .X ∈Gg Equating the sum of the row-sums of M to the sum of its column-sums, we obtain: ∑ = ∑ g .X)x(Stab …(1) ∈Xx ∈Gg Recall from Corollary 1, that = .GxG)x(Stab Substituting this value in (1), and dividing both sides by ),G(o we obtain 1 1 = ∑∑ g .X …(2) ∈Xx Gx )G(o ∈Gg
Let 1 X,. ..,Xk be the G-orbits of .X Note that X is the disjoint union of the
isX and that = XGx i for x in i .X We now re-arrange the left hand side of (2) and show that it equals the number k of orbits in ,X as follows: 1 1 1 1 = ∑ ∑∑ = ∑ ∑ = ∑ ∑ ∑ == .k11 ∈Xx Gx ≤≤ki1 ∈ Xx i Gx ≤≤ki1 ∈ Xx i Xi ≤≤ ki1 Xi ∈Xx i ≤≤ ki1 1 Hence, k = ∑ g .X …(3) )G(o ∈Gg
Try some exercises related to this theorem now.
E10) Let X be a finite set with at least two elements. Suppose that a group G acts transitively on .X Show that there exists at least one element g in G that does not fix any element of .X
E11) Suppose C is a conjugacy class in a group G such that C is finite but ≠ .1C Show that there exists an element g in G that does not commute with any element of .C
E12) Suppose H is a proper subgroup of finite index of a group .G Show that G is not a union of conjugates of .H
E13) Show that the hypothesis of finiteness of X in E10 is necessary.
E14) Show that the hypothesis of finiteness of C is necessary in E11. (You may assume the fact that there are infinite groups with exactly two conjugacy classes.)
We now apply the Orbit Counting Theorem to an interesting question of counting the number of objects in a finite set that are considered non- equivalent.
2.3.3 The Necklace Problem
Imagine that you are in the business of necklace making. The necklaces are 41
Group Theory made by stringing together round beads that come in two colours, black and white. You would then naturally be interested in the following question: given a large supply of beads of both colours, how many distinct necklaces with a given fixed number of beads, let us say 8, can you make?
Your instinctive reaction to the problem may be, “Why not just enumerate – draw pictures of all the possible necklaces and count them?” Indeed, this strategy of enumeration can be carried out for necklaces with eight beads, although some care and time are needed to ensure that every possible necklace is counted, and counted exactly once. However, for necklaces with larger numbers of beads the enumeration method quickly becomes impractical: the number of necklaces with 10, 15, and 20 beads are 78, 1224, and 27012 respectively, as you will see. You will also see that generalising the process takes us very naturally to using the Orbit Counting Theorem for finding the answer, that is, the number of different kinds of necklaces possible, given the same shaped beads in two colours.
So let us start doing some enumeration. And while we are at it, why not start with necklaces with fewer than 8 beads? With a single bead, there are just two necklaces – the bead could be either black or white.
With two beads, there are 3 necklaces – either both beads are black, or one is black and another white, or both are white.
How many necklaces can we make with 3 beads? There are 4 of them, and they are depicted in Fig.2. Note that we have placed the beads at the vertices of a fixed equilateral triangle inscribed in a circle.
Fig. 2: All possible necklaces with three beads
How many necklaces with 4 beads? There are 6, as depicted in Fig.3. The beads are placed at the vertices of a fixed square inscribed in a circle.
Fig. 3: All necklaces with four beads
Now try an exercise.
E15) Enumerate all necklaces with 5 beads each of which could be either black or white. Do the same for necklaces with 6 beads, and for necklaces with 7 beads.
Let us now move on to our originally stated problem of counting necklaces with 8 beads. What kind of a picture should we draw to depict a necklace with 42
8 beads? As in the earlier cases, we will analogously place the beads at the Group Actions vertices of a fixed regular octagon inscribed in a circle, as in Fig.4.
Fig. 4: Some necklaces with eight beads
Now consider the regular octagon whose vertices are the positions of the eight beads in any of the three necklaces in Fig.4. Let us think about colouring the vertices of this octagon with two colours, black and white. How many distinct colourings are there? The eight vertices can be coloured independently of one another, and there are two choices of colours for every vertex. Thus, the total number of colourings is 8 = 2562 .
So, there is a relation between colourings of the octagon and necklaces with eight beads. Indeed, we can represent any colouring as a picture similar to the ones in Fig.4 and we can think of the resulting picture as depicting a necklace. But, if you think about it, you will see that it is possible for different colourings to depict the same necklace. For instance, the four pictures in Fig.5 below are distinct as colourings but the same as necklaces.
Fig. 5: Four different colourings that determine the same necklace
So, the key to the solution of our problem lies in understanding precisely the relation between colourings (of the octagon) and necklaces (with eight beads). So let us try and analyse why the four pictures in Fig.5 are the same, as necklaces.
If we turn the first picture through an angle of 135o in the counter clockwise direction about the centre of the circle, we obtain the second picture. Turning now to how the third picture is related to the first, it is the reflection of the first picture along the vertical diameter of the circle (see Fig.6(a)).
Fig. 6: Flipping about or, equivalently, reflecting along, a diameter. 43
Group Theory The fourth picture is related to the first similarly as the third is to the first, except that the line of reflection, is now the diameter that makes an angle of 25.2 o with the vertical (see Fig.6(b)). Whereas the vertical diameter passes through two opposite vertices, this diameter passes through the midpoints of an opposite pair of edges of the octagon.
The discussion above may be summarised as follows: While there are 8 = 2562 distinct colourings of the octagon, many different colourings turn out to be the same, as necklaces. In fact, two colourings result in the same necklace if and only if they are related to each other by a symmetry of the octagon, where such a symmetry is one of the following: • rotation by an angle that is a multiple of 45o : there are eight such rotations, including the one by 0o (which does nothing to the colouring); • reflection about a diameter joining two opposite vertices: there are four such reflections, corresponding to the four such diameters; • reflection along a diameter through the midpoints of a pair of opposite edges: there are four such reflections, corresponding to the four such pairs of opposite edges.
D16 acts on colourings of As you may recall, these sixteen symmetries form D16 , the dihedral group of the octagon. order 16.
For any two colourings C and D of the octagon, let us write ‘ D~C if C and D result in the same necklace’, that is, if they are related by a symmetry of the
octagon. Since the symmetries form a group (namely, D16 ), the relation defined by ~ is an equivalence. The number of distinct equivalence classes equals the number of necklaces. And our original problem of counting the number of distinct necklaces is the same as counting the number of equivalence classes.
To summarise: The number of distinct necklaces with 8 beads equals the number of equivalence classes of colourings of the octagon, which are the
orbits under the action of D16 .
It turns out that the Orbit Counting Theorem may be applied to count these equivalence classes, which we now do.
Let G denote the dihedral group 16 .D Let X be the set of colourings of the vertices of the octagon with two colours, black and white. The cardinality of X is 8 = 2562 . We want to compute the right hand side of the equation in Theorem 4 for the action of G on .X For this, we need to compute Xg , for every g in .G How do we do this?
We first enumerate the elements of .G Let r denote the rotation by an angle of 45° (say counter clockwise). Let s denote the reflection in the line joining a pair of opposite vertices. Then 2 76543 765432 16 = }.s r,sr,sr,sr,sr,sr,sr,s,r,r,r,r,r,r,r,e{D Now, let us fix g in .G Consider the action of the group < g > on the vertices of the octagon. Observe that a colouring is fixed by g if and only if it is fixed by every element of < g > . Further, observe that for a colouring to be fixed by 44
< g > , it is necessary and sufficient that vertices in the same < g > -orbit must Group Actions be assigned the same colour. Thus, Xg equals )g(m ,2 where )g(m is the number of orbits for the action of < g > . The values of )g(m and Xg are listed in the following table for all the elements of .G Note that two elements that are conjugate have the same number of elements of G fixing them – indeed, if ′ = −1 ,hghg then there is a bijection between Xg and g′ ,X given by a ⋅ .xhx So, it is enough to compute Xg for just one element from every conjugacy class.
Elements )g(m Xg e 8 28 r,r7 1 21 r,r 53 1 21 r,r 62 2 22 r4 4 24 sr,s r,sr,s642 5 25 sr,s r,sr,sr 753 4 24
Plugging the values of Xg from the table into the equation in the Cauchy- Frobenius Lemma, we obtain: The number of G-orbits in X 1 8 1 1 2 4 5 4 =⋅+⋅+⋅+⋅+⋅+⋅+⋅= .30)24242122222221( 16 Thus, there are precisely 30 different necklaces that one can make, each of eight beads of two colours.
The method above can be applied easily to necklaces with not just eight beads, but any number of beads (see E16). The Cauchy-Frobenius Lemma is, similarly, useful for counting the number of benzene molecules, or other such molecules.
Try this exercise now.
E16) How many distinct necklaces of 12 beads can you make by stringing together beads of two colours? Repeat the problem with 12 replaced by p , where p is an odd prime, and obtain a formula for the number of distinct necklaces.
Let us now use what we have discussed about G-sets to study an important consequence.
2.4 THE CLASS EQUATION
Let G be any group and X a G-set. As we have seen, the set X is the disjoint union of its G-orbits. Now, let us separate the orbits into two classes: singleton 45
Group Theory orbits and non-singleton orbits. As you can check, an element x of X forms an orbit by itself if and only if it is fixed by every element of the group G, that is, gx = x for all g in ,G that is, iff g ∈∀∈ .GgXx This leads us to the following definition.
Definition: An element x in a G-set X is called a fixed point (for the action of G on X ) if = ∀ ∈ .Ggxgx The set of all fixed points is denoted by XG , i.e., G { ∈∀=∈= }.GgxgxXxX
Since XG is the union of all the singleton orbits, Theorem 1 tells us that X is the disjoint union of XG and the disjoint non-singleton G-orbits of X.
In the case when X is finite, we obtain the following immediately as a consequence: XX G += the sum of the cardinalities of all disjoint non-singleton orbits Let us now see what this means when a finite group G is acting upon itself by conjugation. So here = .GX What are the orbits in this case? And what is XG ? As you have already seen, the orbits are precisely the conjugacy classes, and here G = ),G(ZX the centre of G. Thus, we have the following:
The Class Equation: If G is a finite group, and = K n,, 1i,C are all the g i distinct non-singleton conjugacy classes in ,G then n
o(G) o(Z(G))+= ∑{ gg >1|C| |C| } i i =1i
The Class Equation is a very useful device for understanding the structure of groups. We will now apply this equation to derive a basic fact about “p-groups”. In the next unit you will see many more applications. But first, what is a p-group?
Definitions: Let p be a prime. A group whose order is a positive power of p is called a p-group.
Further, if H is a subgroup of a group G such that H is a p-group, then H is called a p-subgroup of .G
3 For instance, D8 is a 2-group, as 8 = ,2D and S3 is not a p-group for any
prime p, but )321( is a 3-subgroup of 3 .S
Now we shall prove the following result.
Theorem 5: The centre of a p-group is non-trivial, that is, its order is n ,p for some ≥ .1n
Proof: Let G be a p-group and n ≥= .1n,pG Consider the Class Equation of
+= ∑{ gg >1|C| |C|))G(Z(o)G(o,G }. ∈Gg
Since the order of any conjugacy class divides g g >∈∀ .1|C|,Gg|C|p|,G| 46
Thus, the Class Equation tells us that .|)G(Z|p Group Actions Now, ≥ ,1)G(Z since the centre always contains the identity element. Also, by Lagrange’s theorem, )G(Z divides .G So we conclude that )G(Z must be a positive power of p. So, )G(Z has more than one element, that is, ≠ .}e{)G(Z
A corollary of this result is the following. Corollary 3: Every group of order p2 is abelian, where p is a prime.
Proof: Let G be a group of order 2 .p By Theorem 5, )G(Z(o = p) or 2 .p If )G(Z(o = 2 G,p) is abelian. If = p))G(Z(o , then ∃ ∈ )G(Zz s.t. = z)G(Z .
Also G = p , and hence ∃ ∈ Gx s.t. G x >=< . )G(Z )G(Z Then x and z together generate the group .G Also xz = zx, because z belongs to the centre. Thus, again, G is abelian.
Does this corollary lead you to expect all p-groups to be abelian? Well, what about Q8 ? As you know, Q8 is a 2-group but it is non-abelian.
Try some exercises now.
E17) Let G be a p-group and let X be a G-set. Use the Class Equation to show that ≡ G .)p(modX|X| Hence prove that if G is a p-group and X is a G-set such that = 1)p|,X(| , then there is a fixed point for the action of .G
E18) Let G be a group. Find G ,X 7 when: i) G acts on itself by conjugation; ii) G acts on itself via the regular action.
Now, when you think about it, how many groups can you list which are simple? It is not easy to do this. But, if we apply the Class Equation, we can show one type of group is simple. Let us start with stating some results that we will need for doing this.
Lemma 1: i) Let G be a group and N .G If ∈ Nx , then −1 x = ⊆∈ .N}Ggxgg{C Thus ,∪= x .CN ∈Nx ii) If G is a finite group, then = ∑ x ∈ ,Nx|C|{N the Cx are all distinct}.
Lemma 2: A5 is simple.
The proof of Lemma 1 is left to you (see E19). You will study the proof of Lemma 2 in the next unit, as it is an application of the Sylow theorems.
Now, assuming these results, let us prove the following theorem. 47
Group Theory Theorem 6: An is simple for ≥ .5n
Proof: We will prove this by induction on n. For = ,5n the result holds by
Lemma 2. Now, assume it is true for = − ≥ 6m,1mn , and let = m .AG Suppose ∃ H ≠ }I{H, G or ,G where I is the identity permutation. For ≤ ≤ ,mi1 let )i(Stab be the stabiliser of i w.r.t. the natural action of G on K ~ 1{ }m,, . Then i(Stab ≤ G) and )i(Stab −1m =∀− .m,...,1iA So, )i(Stab is simple ∀ = .m,...,1i
Step 1: We first prove that σ ≠ i(i) ∀σ ∈H σ ≠ I,. Suppose ∃ ∈ ∩ j(StabH.t.s}m,...,1{j ≠ }I{) , then ∩ = ),j(Stab)j(StabH since )j(Stab is simple. So, ≤ H)j(Stab , and hence, −1 ∈ρ∀≤ρ=ρρ .GH))j((Stab)j(Stab K Next, we note that for every ,G τττ=ρ∈ρ r21 , where r is an even number
and each τi is a transposition.
Now, 21 ∈ττ G and since 21 ∈ττ≥ tab,6m )j(S for some j. r Since ρ is a product of such elements, G is generated by ,1(Stab K ,) 2 tab)m(S , i.e.,=< K >≤ ,H)m(Stab,),1(StabG that is, = ,HG a contradiction. Thus, ∩ = ∀ = .m,...,1i}I{)i(StabH Hence σ ≠ ∀σ∈ σ ≠ .I,Hi)i(
Step 2: Here we prove that ∀∀∀σ ∈H, the decomposition of σσσ into disjoint cycles only has transpositions.
Suppose ∃σ ∈ H such that σ has a cycle of length more than 2 in its cycle
decomposition. Then =σ 21321 ...... )bb(...)aaa(
Since ∈α∃≥ 6n A, n such that (,a)a(,a)a( ≠α=α=α 332211 .a)a −1 Then 1 2 3 1 ααα=ασα=β 2 ..))b()b((...))a(aa( ....
Thus 1 =σ=β∈βσ ()a(,H, a)a 21 and 2 33 σ=≠α=β (a)a()a( 2 ).a Thus, β ≠ σ −1 but =βσ 11 ,a)a( a contradiction to Step 1. Hence the statement in Step 2 is proved.
Step 3: Here we shall prove that= {I}H .
Suppose =σ≠ 654321 ∈H...)aa()aa()aa(and}I{H . −1 Let =α 5321 ∈ .G)aa()aa( Then =ασα=β 634521 ...)aa ()aa()aa( −1 So β ≠ σ β σ∈H,, and =βσ 11 ,a)a( a contradiction. Hence σ ≠ I cannot be in
.H Hence = }.I{H Hence, An is simple.
Why don’t you try some related exercises now?
E19) Prove Lemma 1.
E20) Which of the following can form the RHS of a Class Equation of a group of order 10? Give reasons for your answers. (i) + + + + 52111 , (ii) + + + 5221 ,
48 (iii) + + + 4321 , (iv) 1+1+ ⋅⋅⋅ +1.
01 Group Actions Z E21) Determine the cardinality of the conjugacy class of in 52 ).(GL 20 Z 2 [Hint: 52 =−−= 80)15()15(5))(GL(o .4 ]
E22) Can = + + + + 331119 be a Class Equation? Why, or why not?
We end this discussion on G-actions, orbits and stabilisers now. Let us summarise the discussion we have had in this unit.
2.5 SUMMARY
In this unit, you have studied the following points.
1. A group G acts on a set X if there is a function ϕ × → XXG: such that i) ϕ = x)x,e( , and ii) ϕ ϕ ′ = ϕ ′ ∀ ∈ ′∈ .Gg,g,Xx)x,gg())x,g(,g(
2. Several example of actions of a group G on various sets, including G acting on itself by the regular action and by conjugation.
3. If X is a G-set, then the orbit of x∈X is = ⋅ ∈ },Gg|xg{Gx
the stabiliser of x∈X is G =⋅∈= },xxg|Gg{)x(Stab Xg is the set of elements in X that are fixed by ∈ ,Gg XG , the set of fixed points of the action on = ∈ = ∀ ∈ }.Ggxgx|Xx{X
4. If X is a G-set, then X is the disjoint union of its G-orbits.
5. The proof, and applications, of the structure theorem for orbits, which
states that if X is a G-set and x ∈X , then = / G )).x(Stab(o)G(o|Gx|
6. The proof, and applications, of the Orbit Counting Theorem, which states that if G is a finite group and X is a finite G-set, then the number of 1 distinct G-orbits in X = ∑ g .|X| )G(o ∈Gg
7. Applications of the Class Equation, which is n K += ∑ gg > }1|C| |C{|))G(Z(o)G(o , where g = )n,, 1i(C are all the i i i =1i
distinct conjugacy classes in G for which g > |.1|C
8. An is simple ∀ ≤ 5n .
2.6 SOLUTIONS / ANSWERS
M M −1 E1) Define n n →×ψ n =ψ ABA) B,A(:)F()F()F(GL: . M Then ∈∀=ψ n ),F(BB)B,I( and 49
Group Theory −− 11 −1 =ψψ = ∈∀ψ= n )F(G LB,A)C,AB()AB(C)AB(A)BCB(A))C,B(,A( M and ∈ n ).F(C
E2) Consider G G =ϕ→×ϕ gx and G: H H )x,g(: G G =ψ→×ψ −1 . Check that ϕ and ψ are G-actions. G: H H gxg) x,g(: Next, they are distinct because ϕ = ψ ⇔ ∈ .Hg)x,g()x,g( So ∃ α∈ H\G s.t. ϕ α ≠ ψ α ).x,()x,(
E3) Take = X .SG Then G acts on X (see Example 1). Any proper subgroup of G will also act on .X So the group acting on X is not unique.
E4) There are several such examples. For instance, consider Z N →× N a n,z(: n) |z| . Now − a − 3 a = 3)3 ()3,2())3,3(,2( 623 And ≠→+− 61 .33)3,32(
E5) i) × → → gh)h,g(:GGG is the regular action. For any ∈ ,Gg we will show that = .GgG Firstly, ⊆ .GGg Next, for any ∈ ,Gh −1 ∈= ,Ggghgh so that ⊆ gG .G Hence = .GgG ii) Refer to Example 4. Then ⋅ = ∀ ∈ }GggxH{xHG Now, for any G =∈ −1 ⋅∈ gH H .xHGxH)gx(gH, Hence the result. iii) The H-orbit of ∈ Gg under this action would be ∈ = ,Hg}Hh|hg{ the right coset of H containing g. Conversely, any right coset is of the form ∈ .Gx,Hx Now = ∈ },Hh|hx{Hx which is the H-orbit of ∈ .Gx Hence the result. a iv) Consider n σσ→× ),i()i,(:]n[]n[S where = 2,1{]n[ ⋅⋅⋅ }.n,,
Now n n ⊆∈σσ=⋅ ].n[}S|)i({iS
Next, let ∈ ≠ .ij],n[j Then ∈=τ ji( S) n and τ = ,j)i( so that
n ⋅∈ .iSj
Thus, n ⋅= .iS]n[
E6) i) Let H .G Then −1 gg ∈∀⊆ ,GgHH i.e., H is G-invariant. Conversely, let H be G-invariant, that is, −1 ∈∀⊆ ⇒ HGgHHgg .G ii) Let ⊆ .GS S is H-invariant iff × a a hs)s,h(:SSH is an action iff ∪= .HsS ∈Ss iii) Let = ∈ }.Xx|)x,x{(D Then a ∈ .D)gx,gx())x,x(,g( Hence D is G-invariant.
E7) i) For ∈ ,Gx G ==∈==∈= }.e{}eg|Gg{}xgx|Gg{)x(Stab 50
−1 Group Actions ii) For ∈ ,Gx G ∈= ==∈== ).x(Z}xggx|Gg{}xgxg|Gg{)x(Stab
−1 iii) For ∈Gx , G ∈==∈= ∈ }Hg xx|Gg{}xHgxH|Gg{)xH(Stab −1 =∈∈= −1.xHx}xHxg|Gg{
~ E8) Let − n1n )(:SS: ϕ=ϕα→α . Then α is a 1-1 group homomorphism. =σ∈σ= }n) n(|S{)n(Stab Sn n α∈σ ⇒ =σ n)n(Im ⇒ ∈σ tab)n(S , so that ⊆α ).n(S tabIm Sn Sn Conversely, let ∈σ tab).n(S Define ∈ϕ S by Sn −1n ϕ = σ ∀ = ⋅⋅⋅ −1n,,1i)i()i( . Then σ =ϕ~ ∈Imα , that is, α⊆ .I m)n(Stab Sn Hence =α tabIm ).n(S Sn
E9) gh ∈∀ϕ=ϕ ,g)gh( .Gh o Now gh ϕϕ==ϕ ).x()h() g(ghx)x( ∴ϕ = ϕ o ϕ ∀ ∈ .Gh,g)h()g()gh(
g x|Gg{}I|Gg{Ker ∈∀=∈==ϕ∈=ϕ }Xxxg = ∈ ∈ ∀ ∈ }Xx)x(Stabg|Gg{ ∩= ).x(Stab ∈Xx
Since G =ϕ∈∀= }e{Ker,Gg}e{)g(Stab if = GX , by E7(i).
Thus →ϕ SG: G is a 1-1 homomorphism.
E10) We may assume that the group G is finite because, using E9, we may
replace G by its image in Sx under the group homomorphism → SG x defining the action of G on .X Now, let us apply the Orbit Counting Theorem. The left hand side is 1 since the action is transitive. Looking at the right hand side, we conclude that ∑ g = ).G(oX ∈Gg Suppose every element of G fixes some element of .X Then g ≥1X for every ∈ .Gg Moreover, for the identity element e of the group, we have Xe = X, so that e ≥ .2X Hence, ∑ g > ,)G(oX a contradiction. Hence ∈Gg ∃ ∈Gg s.t. g = ,0|X| that is, ≠ ∀ ∈ .Xxxgx
E11) Consider the conjugation action of G on .C This is transitive. So, by E10, ∃ ∈ Gg s.t. −1 ∈∀≠ Cxxxgg , i.e., ≠ ggx ∀ ∈ .Cxx
E12) Put = HGX , the set of left cosets of H. By E5, you know that the action is transitive. So, by E10, there exists g in G that does not fix any left coset ,xH that is, ≠ .xHgxH This means −1 ≠ ,HgxHx i.e., −1 xx ∉ ,Hg i.e., −1 ∈∀∉ .GxxHxg
C C2 E13) Consider the action of = 2 )(GLG on . Consider the set X of lines 2 z through the origin in C , i.e., X λ= ∈λ∈ CC .,w,z Then X is w infinite. There is an induced action of G on .X You can check that this 51
Group Theory action is transitive. However, every element of G fixes some line because every element of G has an eigenvector.
E14) Let G be an infinite group with exactly two conjugacy classes. }e{ is a conjugacy class by itself. Let C be the other conjugacy class (consisting of all elements of G other then e ). Now, each element g of G fixes some element of C under the conjugacy action because if g = e , then of course, every element of C is fixed; if g ≠ e, then g belongs to ,C and g fixes g.
E15) Enumeration of necklaces with 5 beads: If the number of black beads is 0, 1, 4 or 5, then there is a single necklace. There are two necklaces with 2 black beads – one in which the two black beads are together, and another in which they are separated (by a single white bead in one side and two white beads in the other). By the same reasoning, there are two necklaces with 2 white beads, or what amounts to the same, with 3 black beads. Thus, the total number of necklaces with 5 beads, counting by ascending order of number of black beads, is 1+1+2+2+1+1=8.
Enumeration of necklaces with 6 beads: If the number of black beads is 0, 1, 5 or 6, then there is a single necklace. There are three necklaces with 2 black beads – one in which the two black beads are together, a second in which they are separated by a single white bead on one side and three white beads on the other, and a third in which they are separated by two white beads on either side. By the same reasoning, there are three necklaces with 2 white beads, or what amounts to the same, with 4 black beads. There are three necklaces with 3 black beads – one in which the black beads are together, another in which two of the black beads are together and the third is separated (from the other two by white beads in between), and a third in which no two black beads are together. Thus, the total number of necklaces with 6 beads, counting by ascending order of number of black beads, is 1+1+3+3+3+1+1=13.
Enumeration of necklaces with 7 beads: If the number of black beads is 0, 1, 6 or 7, then there is a single necklace. There are three necklaces with 2 black beads (and three more with 5 black beads, by the symmetry between black and white) – one in which the two black beads are together, a second in which they are separated by a single white bead on one side and four white beads on the other, and a third in which they are separated by two white beads on one side and three on the other. There are four necklaces with 3 black beads (and four more with 4 black beads, by the symmetry between black and white) – one in which the black beads are all together, a second in which two of them are together and are separated from the third by a single white bead on one side and three white beads on another, a third in which two black beads are together and are separated from the third by two white beads on either side, and a fourth in which the black beads are separated from each other. Thus the total number of necklaces with 7 beads, counting by ascending order of number of black beads, is 1+1+3+4+4+3+1+1=18.
E16) We follow the argument given for the case of 8 beads. Let G be the
group of symmetries of the regular 12-gon: it is the dihedral group 24 ,D 52
and has cardinality 24. Let X be the set of colourings of the vertices of Group Actions the regular 12-gon by two colours, black and white. We have = 12.2X We first enumerate the elements of .G Let r denote the rotation by an angle of 30° in the counter-clockwise direction. Let s denote the reflection in the line joining a pair of opposite vertices. Then the 24 elements of D24 are as follows: ,r,r,r,r,r,r,r,r,r,r,e 111 098765432 ,r 111 098765432 .s r,sr,sr,sr,sr,sr,sr,sr,sr,sr,sr,s Now, we have the following table:
Elements )g(m Xg e 12 212 ,r,rr,r 1175 1 21 r,r 102 2 22 r,r93 3 23 r,r 84 4 24 r6 6 26 r,sr,sr,sr,s sr,s 108642 7 27 sr,s r,sr,sr,sr,sr 119753 6 26
Plugging in the values of Xg from this table into the right hand side of (3), we obtain: The number of G-orbits in X 1 12 1 2 3 64 7 6 =⋅+⋅+⋅+⋅+⋅+⋅+⋅+⋅= .224)2626212222222421( 24 Thus, the answer is 224 for the case of 12 beads.
For p in place of 12, we follow the same procedure. There are p2 elements in the group D p2 of symmetries of the regular p-gon: let r denote the rotation by an angle of )p/360(° (say counter-clockwise). Let s denote the reflection in the line joining a vertex to the midpoint of the side opposite to it. Then the elements of D p2 are as follows: −1p2 r,...,sr,sr,s,r,...,r,r,e −1p2 .s We have the following table:
Elements )g(m Xg
e p 2p
r,. ..,r −1p 1 21
sr,. ..,s −1p ( + ) 2/1p 2( + ) 2/1p
Plugging values of Xg from this table into the right hand side of (3), we obtain: 53
Group Theory The number of G -orbits in X 1 −1p −12 p 1 + 2/)1p( )2p2)1p(21( =⋅+⋅−+⋅= − 2/)1p ( ++ 12 p2 p Thus, we have a formula for the number of distinct necklaces with p beads.
E17) G += ∑ > }1|G x||Gx{|XX ∈Xx Now ∈∀ Xx|Gx|p for which x| > .1|G So − G .|)X||X(|p Thus, ≡ G ).p(mod|X||X| Next, suppose = .1)p,|X|( Then ≡/ ).p(mod0|X| So G ≡ / ).p(mod0|X| So G ≠ .0|X| So ∃ ∈ Xx s.t. = ∀ ∈ .Ggxgx
E18) i) G ∈= −1 ∈∀= }Ggxxgg|Gx{G = ).x(Z ii) G ∈∀=∈= }Ggxgx|Gx{G = « , unless = }.e{G
E19) i) Since N −1 gg,G ∈∀∈ GgNx and ∈ .Nx
Thus, x ⊆ NC ∀ ∈ .Nx
So x ⊆∪ .NC ∈Nx
Also, for y ∪⊆∈∈ x .CCy,Ny ∈Nx
Hence ∪= x .CN ∈Nx ii) This follows immediately from (i) above.
E20) i) Here = ,3))G(Z(o but = 10)G(o and .)G(o))G(Z(o Hence this partition cannot form a Class Equation. ii) Here = 1))G(Z(o and there are three conjugacy classes, of sizes 2, 2 and 5. This is the Class Equation of 2 5 −1 10 } >===<= xyxy,ey,ex|y,x{D . Here == 432 = 4 = 32 }.y,y{C},y,y{C},xy,xy,xy,xy,x{C},e{C e x y y2 iii) This is not possible since this shows a conjugacy class of order 3, and 3/| 10. iv) Here = .10))G(Z(o Hence = ),G(ZG i.e., G is abelian. Check that ~ Z G − 10.
E21) By Corollary 2 (of Theorem 3), the required number is 01 Z 52 )A(Z))(GL(o , where A = . 20 Z Now, you can check that ∈= 52 = }XAAX)(GLX{)A(Z
54
0a Group Actions Z = 5 ,a,b,a ≠∈ 0b b0 Z* Z* So 5 5 =×= .1 6|||)A(Z| Thus, the required cardinality is 480 = 16 .30
E22) Any group of order 9 must be abelian. Hence each conjugacy class will have only one element. Therefore, this equation is not a valid Class Equation.
55
UNIT 3 THE SYLOW THEOREMS The Sylow Theorems
Structure Page Nos.
3.1 Introduction 57 Objectives 3.2 Direct Product of Groups 58 External Direct Product Internal Direct Product 3.3 Existence of Subgroups of Given Orders 62 3.4 Applications of Sylow’s Theorems 70 3.5 Summary 73 3.6 Solutions / Answers 74
3.1 INTRODUCTION
In this unit, we use the notions and results about group actions developed in the previous unit to derive some basic theorems about finite groups and their structures.
To be able to describe, and analyse, the structure of a finite group, we need the concept of a direct product of groups. In Sec.3.2, you will study this concept.
In Sec.3.3, we look at when the converse of Lagrange’s theorem holds, namely, given a factor k of ),G(o where G is finite, does there exist a subgroup of G of order ?k As you will see, the answer is “no” in general. Further, in Sec.3.3 and Sec.3.4, you will see that there are several special cases when the answer is “yes”. To reach this stage, you will be banking on very important theorems of group theory, which are due to the mathematician Sylow. In Sec.3.3, you will study the proofs of three theorems due to him. In Sec.3.4, you will be studying several applications of these theorems.
As noted above, you will be closely looking at several theorems and techniques for analysing the structure of finite groups. Therefore, after studying this unit, come back to this point and make sure you have achieved the following objectives of studying this unit.
Objectives After studying this unit, you should be able to: • construct the direct product of a finite number of groups; • check if a group is the direct product of its subgroups; • give examples to show when the converse of Lagrange’s theorem holds, and that it doesn’t hold in general; • state, and prove, all three Sylow theorems; and • apply Sylow’s theorems to prove the existence of subgroups of a specified order, and to analyse the structure of some finite groups.
57
Group Theory 3.2 DIRECT PRODUCT OF GROUPS
In this section you will study a very important method of constructing new groups by using given groups as building blocks. We will first look at one way in which two groups can be combined to form a third group. Then we will discuss a similar way in which two subgroups of a group can be combined to form another subgroup.
3.2.1 External Direct Product
In this sub-section, you will study one way of constructing a new group from two or more groups that are given.
Let ∗11 ),G( and ∗22 ),G( be two groups. Consider their Cartesian product
21 =×= 1 ∈∈ 2}.Gy,Gx|)y,x{(GGG Can we define a binary operation on
G by using the operations on G1 and 2 ?G Let us try the obvious method, namely, componentwise multiplication. That is, we define the operation ∗ on
G by 1 2 1 b,Gc,a)db,ca()d,c()b,a( ∈∈∀∗∗=∗ 2.Gd, The way we have defined ∗ ensures that it is a binary operation. To check that ∗),G( is a group, you need to solve the following exercise.
E1) Show that the binary operation ∗ on G is associative. Find its identity element and the inverse of any element )y,x( in .G
So, by solving E1, you have proved that ×= GGG 21 is a group with respect to
∗ . We call G the external direct product of ∗11 ),G( and ∗22 .),G(
For example, R2 is the external direct product of R with itself.
Another example is the direct product Z R∗ ⋅×+ ),(),( in which the operation is given by ∗ = + )xy,nm()y,n()x,m( for Z ∈∈ R∗.y,x,n,m
Remark 1: The groups forming the direct product do not need to have the Z same properties or algebraic structure. For instance, S3 × is a well-defined Z Z direct product of 3 o ),S( and + ),,( where S3 is non-abelian and is abelian, in fact, cyclic.
We can define the external direct product of 3, 4 or more groups along the same lines.
Definition: Let ∗∗ 2211 K ∗nn ),G(,),,G(),,G( be n groups. Their external
direct product is the group ∗ ),,G( where 21 K×××= GGGG n and
21 K ∗ 21n K n ∗∗= 222111 K ∈∀∗ iiinnn .Gy,x)yx,,yx,yx()y,,y,y()x,,x,x(
Thus, Rn is the external direct product of n copies of R ∀ ∈ N .n,
We would like to make a remark about notation now.
58
The Sylow Theorems Remark 2: Henceforth, we will assume that all the operations 1 K,,, ∗∗∗ n are multiplication, unless mentioned otherwise. Thus, the operation on
21 K×××= GGGG n will be given by
1 K 1n K = 2211n K ∈∀ iiinn .Gb,a)ba,,ba,ba()b,,b).(a,,a(
Now, you know that given two sets A and × ≠ × .ABBA,B So, you may Z Z wonder if, for example, ×S3 and S3 × are different groups. It turns out they are essentially the same, which is what the following exercise says.
~ E2) Show that GG ×× 1221 ,GG for any two groups G1 and 2 .G
Because of E2, we can speak of the direct product of 2 (or n ) groups without bothering about the order in which their products are taken.
Now, let G be the external direct product × 21 .GG Consider the projection map 11211 =π→×π .x)y,x(:GGG:
You should check that π1 is a surjective group homomorphism. Also,
1 121 =π×∈=π 1}e)y,x(|GG)y,x{(Ker
= 1 ×=∈ 212 .G}e{}Gy|)y,e{(
×∴ G}e { 21 × 21 .GG So, by the Fundamental Theorem of Homomorphism ~ × × )G}e({)GG( 12121 .G
You can, similarly, prove that × 21 }e{G ×GG 21 and ~ ×× 2121 })e{G()GG( 2 .G
In the following exercises you can prove some general facts about external direct products of groups.
E3) Show that × GG 21 is the product of its normal subgroups ×= 21 }e{GH
and ×= 21 .G}e{K Also show that 21 21 =×∩× 21 )}.e,e{()G}e({})e{G(
E4) Prove that 21 1 ×=× 2 ),G(Z)G(Z)GG(Z where )G(Z denotes the centre of .G
E5) Let A and B be cyclic groups of order m and n, respectively, where = .1)n,m( Prove that A× B is cyclic of order mn. Z Z Z Z Z [Hint: Define :f nm ++=×→ ).nr,mr()r(f: Then apply the Z Z ~ Z Fundamental Theorem of Homomorphism to show that × mnnm . ]
E6) Show that if ≤ GH 11 and ≤ 22 ,GH then ×≤× 2121 .GGHH Further, if
H1 G1 and H2 2 ,G is × HH 21 × 21 ?GG Give reasons for your answer.
E7) Give an example, with justification, to show that not every subgroup of
the direct product × GG 21 of the groups G1 and G2 is of the form 59
Group Theory × 21 ,HH where H1 is a subgroup of G1 and H2 is a subgroup of 2 .G
So, far we have discussed the construction of × GG 21 from two groups G1
and 2 .G Let us now see under which conditions we can express a group as a direct product of its subgroups.
3.2.2 Internal Direct Product
Let us begin by recalling that if H and K are normal subgroups of a group G, then HK is a normal subgroup of .G We are interested in the case when HK is the whole of .G Related to this, consider the following definition.
Definition: Let H and K be normal subgroups of a group .G We call G the internal direct product of H and K if = HKG and ∩ = }.e{KH This fact is denoted by = ×KHG .
Let us consider some examples.
Example 1: Show that the Klein 4-group, 4 ,K is an internal direct product of its subgroups.
2 2 Solution: 4 = },ab,b,a,e{K where == eb,ea and = .baab
Let = aH and = .bK Then, H and K are normal in K4 (as K4 is
abelian), and ∩ = }.e{KH Also, 4 = .HKK
4 ×=∴ KHK . ~ Z ~ Z ~ Z Z Here, note that H 2 and K 2 . K ×∴ 224 . ***
Z Example 2: Show that 10 is the internal direct product of its subgroups = }5,0{H and = }.8,6,4,2,0{K Z Z Solution: Note that 10 is abelian, so that H and K are normal in 10 . Further, Z Z i) 10 += ,KH since any element of 10 is the sum of an element of H and an element of K (note that +∈−= KH451 ), and ii) =∩ }.0{KH Z Hence 10 ×= .KH ***
Now, can an external direct product also be an internal direct product? Well,
go back to E3. What does it say? It says that the external product of × GG 21 is
the internal product 21 ××× 21 ).G}e({})e{G(
Consider the following remark now.
Remark 3: Let H and K be normal subgroups of a group .G Then the internal direct product of H and K is isomorphic to the external direct product of H and .K Therefore, when we talk of an internal direct product of 60
subgroups we can drop the word internal, and just say ‘direct product of The Sylow Theorems subgroups’.
Let us now extend the definition of the internal direct product of two subgroups to that of several subgroups.
Definition: A group G is the internal direct product of its normal subgroups
21 K H,,H,H n if i) = K n21 ,HHHG and ii) ∩ K +− K n1i1i1i =∀= K .n,,1i}e{HHHHH
For example, look at the group G generated by },c,b,a{ where 2 === cbea 22 and = = = .cbbc,caac,baab This is the internal direct ~ Z Z Z product of b,a and .c That is, G ×× 222 .
Now you may ask: can every group be written as an internal direct product of two or more of its proper normal subgroups? To answer this, consider Z . Suppose Z = × ,KH where K,H are non-zero subgroups of Z. So, = mH and = nK for some non-zero ∈Z .n,m Then ∈ ∩ .KHmn But if H× K is a direct product, ∩ = }.0{KH So, we reach a contradiction. Therefore, Z can’t be written as an internal direct product of two subgroups.
By the same reasoning, we can say that Z can’t be expressed as Z 21 K××× HHH n , where i =∀≤ K n,,2,1iH for any > .1n
Also, you know that there are some groups that do not have any proper normal subgroup except }.e{ So, they can certainly not be written as such a direct product!
So, how do we find out whether a group can be written as a direct product of its subgroups? The answer is given by the following theorem.
Theorem 1: Let a group G be the internal direct product of its subgroups H and .K Then i) each ∈Gx can be uniquely expressed as = ,hkx where ∈ ∈ ;Kk,Hh and ii) = ∀ ∈ ∈ .Kk,Hhkhhk
Proof: i) We know that = HKG and ∩ = }.e{KH Therefore, if ∈ ,Gx then
= ,hkx for some ∈ ∈ .Kk,Hh Now suppose = khx 11 also, where −1 −1 1 ∈ Hh and 1 ∈ .Kk Then = 11 .khhk 1 =∴ 1 .kkhh −1 −1 −1 −1 Now 1 ∈ .Hhh Also, since 1 1 1 ∈∈= .Khh,Kkkhh −1 −1 1 =∩∈∴ }.e{KHhh 1 =∴ ,ehh which implies that = 1 .hh Similarly,
1 = .kk Thus, the representation of x as the product of an element of H and an element of K is unique. ii) The best way to show that two elements x and y commute is to show 61
Group Theory that their commutator −− 11 xyyx is identity. So, let h ∈ H and k ∈ K and consider −− 11 .hkkh Since K −− 11 ∈ .Khkh,G ∴ −− 11 ∈ .Khkkh By similar reasoning, −− 11 ∈ .Hhkkh ∴ −− 11 =∩∈ }.e{KHhkkh ∴ −− 11 = ,ehkkh that is, = .khhk
What Theorem 1 says is that if a group does not have normal subgroups for which (i) and (ii) are satisfied, then G cannot be an internal direct product of its subgroups.
Try the following exercise now.
E8) Let H and K be normal subgroups of G which satisfy (i) of Theorem 1. Then show that = × .KHG
Now let us look at the relationship between internal direct products and quotient groups.
Theorem 2: Let H and K be normal subgroups of a group G such that = × .KHG Then HG ~ K and KG ~ .H
Proof: We will use the second isomorphism theorem to prove this result. Now = HKG and ∩ = }.e{KH Therefore, = HHKHG ~ =∩ }e{K)KH(K ~ .K You can similarly prove that KG ~ .H
We now give a result which immediately follows from Theorem 2, and which will be used later in this unit.
Corollary 1: Let G be a finite group and H and K be its subgroups such that = × .KHG Then = ).K(o)H(o)G(o
We leave the proof of Corollary 1 to you, and give you some related exercises.
E9) Let G be a finite group such that 1 K××= n ,HHG where
Hi ∀ = K .n,,1iG Prove, by mathematical induction, that n = ∏ i ,)H(o)G(o for ≥ .1n =1i
And now let us discuss some basic results about the structure of a finite group.
3.3 EXISTENCE OF SUBGROUPS OF GIVEN ORDERS
As already mentioned in Sec.3.1, this section is centred around the following question: Is the converse of Lagrange’s theorem true? That is, given a factor k of the order of a finite group ,Gdoes there exist a subgroup in G of order 62
?k Let us begin with an example to show that the answer is “no” in general. The Sylow Theorems
Example 3: Show that the alternating group A4 has no subgroup of order 6. Solution: You will see how this can be proved by contradiction. Now
( 4 ) = .12Ao Suppose H is a subgroup of A4 of order 6. Being a subgroup of index 2 in4 ,A H is normal in 4 .A Also ∈∀= 4 .H\Ag2)gH(o Thus, 2 2 2 ∈∀∈ 4.AgHg So ∈= H)231()321( and ∈= .H)321()231( In the same way all the 3-cycles in A4 are in .H So ≥ ,8)H(o a contradiction.
Hence, A4 has no subgroup of order 6. ***
In the example above, the group A4 is not abelian. So, your next question may be that if G is a finite abelian group, does the converse of Lagrange’s theorem hold true? In some cases it does, as the following theorem tells us.
Theorem 3: If G is a finite abelian group, and p is a prime dividing ),G(o then G has a subgroup of order p.
Proof: We first observe that it suffices to prove the following claim (under the given hypothesis): There exists in G an element g whose order m is divisible by p. In fact, if this claim is proved, then for this g, the element gp/m has order p, and the group generated by gp/m has order p, thus proving Theorem 3.
So let us prove the claim, which we shall call a lemma.
Lemma 1: If G is a finite abelian group such that ,|G|p then ∃ ∈ Gg s.t. ).g(op
Proof: Let us prove this by induction on |G| . If = ,1|G| then the statement is true vacuously since there is no p that divides ).G(o
Now suppose that > .1|G| Let h be any element of G other than identity and let n be its order. Note that n > 1 and that n divides .|G| If p divides n, then the lemma holds – just choose g to be h. If p does not divide n, consider the quotient group .hG Its order is ,n|G| which is divisible by p since |G|p and /| .np Now since < ,|G|n|G| we may apply the induction hypothesis to hG to conclude that there exists an element in hG whose order k is divisible by p. Let g be the pre-image in G of such an element in hG (under the natural epimorphism → hGG ). Let m be the order of g. Denoting hg by g , we have ( )m m == egg , where e denotes the identity element of .G Since e is the identity element of the quotient ,hG it follows that the order k of g divides m, and hence | .mp
Thus, Lemma 1 is proved, and so is Theorem 3. 63
Group Theory Let us now consider the next step in showing that the converse of Lagrange’s theorem holds for abelian groups.
Proposition 1: Let N be a finite normal subgroup of order n of a finite group G and let →π NGG: be the natural epimorphism. If K is a subgroup of order k of NGthen π−1(K) is a subgroup of G of order kn, that is, −1 =π .|N||K||)K(|
Proof: You know that the inverse image of a subgroup under a −1 homomorphism is a subgroup. So ≤π G)K( . Let g,...,gk1 be elements of G −1 such that = { 1 k }.Ng,...,NgK Then 1 ∪⋅⋅⋅∪=π k .NgNg)K( Since each coset −1 i Ng has n elements, and any two of them are disjoint, it follows that π )K( consists of kn elements.
Now you will see how useful Proposition 1 is for proving that the converse of Lagrange’s theorem holds for abelian groups.
Theorem 4: If G is a finite abelian group and k divides ,G then G has a subgroup of order .k
Proof: This proof is by induction on .k For ∈N ,n let )n(P be the statement, ‘if G is a finite abelian group and |G|n , then G has a subgroup of order n ’. If = ,1n then you know that }e{ is a subgroup of G of order 1, so that )1(P is true. Now suppose that k > 1 and that )n(P is true ∀ < .kn Now we shall show that )k(P is true. Let p be a prime divisor of .k ByTheorem 3, there exists a subgroup N of G of order p. Note that, as G is abelian, N .G Now, the quotient group NG has order ,p|G| which is divisible by .pk Since < ,kpk the induction hypothesis applies. Therefore, there exists a subgroup of NGof order .pk Its pre-image in G (under the natural epimorphism → NGG ) is a subgroup of G of order = kppk (by Proposition 1). So, )k(P is true.
Hence, )n(P is true ∀ ≥ .1n
So far we have looked at abelian groups. We now move to the first of the three famous theorems proved by the Norwegian mathematician Ludwig Sylow in 1872. This is one of the most important theorems in finite group theory. It tells us about the p-subgroups of any finite group, abelian or not.
Theorem 5 (Sylow’s first theorem): Let G be a finite group and p be a prime such that pr divides |G| for some integer ≥ .1r Then there exists a Fig. 1: L. Sylow r (1832-1918) subgroup of order p in .G
Proof: Let = r .mp|G| We shall proceed by induction on .|G| Let = r .pq If = ,1m then = q|G| so that the result clearly holds – G itself is a subgroup of the desired order. 64
So assume that > .1m Now there are two cases. The Sylow Theorems
Case 1: Suppose there exists a proper subgroup H of G such that q divides |H| . Write = ′ .mq|H| Then, < .|G||H| So, by the induction hypothesis, H (and so also G ) has a subgroup of order = r .pq
Case 2: Suppose there is no proper subgroup H such that = pq r divides .|H|
Now, since p divides ,|G| p divides ,|H||G| where ≤/ .GH So, |H:G|p for every proper subgroup H of .G So, by the Structure Theorem for Orbits, every orbit of G (and, in particular, every conjugacy class of G ) that is not a singleton has cardinality divisible by p. So, it follows from the Class Equation (see Sec.2.4, Unit 2) that .|)G(Z|p So, by Theorem 3, there exists a subgroup N of order p in ).G(Z Being contained in N),G(Z is normal in .G Consider the quotient group .NG Its order |N||G| is −1r .mp So, by the induction hypothesis, there exists a subgroup K of order p −1r in .NG It follows, from Proposition 1, that the pre-image of K under the natural epimorphism → NGG is a subgroup in G of order = r .pq
This important theorem has an immediate corollary, a result that was actually proved much earlier by the mathematician Cauchy, and is a generalisation of Theorem 3.
Corollary 2 (Cauchy’s theorem): If G is a finite group and p is a prime dividing ,|G| then G has an element of order p.
Proof: By Sylow’s theorem, G has a subgroup H of order p. Since |H| is prime, H is cyclic, say = .xH Then ∈ Gx s.t. = .p)x(o
Here are some examples of the use of Sylow’s first theorem and Cauchy’s theorem.
Example 4: Does a group of order 750 have at least 5 distinct proper non- trivial subgroups? Give reasons for your answer. Solution: Yes. To give the reason, let G be a group of order ××= 3 .532750 By Sylow’s first theorem, G has subgroups of order .125,25,5,3,2 All these must necessarily be distinct since they are of different orders. ***
Example 5: If G is a group of order ,4 what can its structure be? Solution: Firstly, being of order 2 G,2 is abelian. Next, since = ,4|G| by Sylow’s first theorem it has subgroups of order 2 and ,4 and at least one element of order 2 (by Cauchy’s theorem). ~ Z If G has an element of order ,4 then G is cyclic and G 4. If G has no element of order ,4 then it has three elements of order ,2 say, .c,b,a Since ∈ ∈ .Gab,Gb,a Also ≠ ≠ .bab,aab So = .cab As G is abelian, a b,G .G Also =∩ }.e{ba 65
Group Theory So ×= baG is the Klein 4-group. ***
Here is a related exercise for you to try now. While solving it, consider all the information you have about the group.
N E10) For which values of n ∈ must S5 have subgroups of order n, and why?
Sylow’s first theorem tells us about subgroups of G of order 2 K r ,p,,p,p where r .|G|p This leads us to the following definition.
Definition: Let G be a finite group and p be a prime such that = rmp|G| , with r ≥ 1 and = .1)m,p( A subgroup H of G such that = p|H| r is called a Sylow p-subgroup of .G
For instance, if = ,1400|G| then G will have a Sylow 2-subgroup (of order 8), a Sylow 5-subgroup (of order 25), and a Sylow 7-subgroup (of order 7). Also, G will have no Sylow 3-subgroup, since /| .|G|3
Remark 4: It follows from Sylow’s first theorem, as a special case, that G has Sylow p-subgroups.
Let us consider some examples of Sylow p-subgroups in some groups of small order. You can return to these examples later, and look at them again in the light of Sylow’s second and third theorems, which you will study later in this section.
Example 6: Exhibit a Sylow p-subgroup of S3 for each prime p dividing
3 .|S| Solution: The group has order = × .326 So it has a Sylow 2-subgroup of order ,2for instance, { ( )}.21,e Note that )}31(,e{ and )}32(,e{ are also
Sylow 2-subgroup of 3 .S It also has a Sylow 3-subgroup of order ,3 for instance, { ( ) ( 231,321,e )}. ***
Example 7: Obtain a Sylow p-subgroup of 4 ,S for each prime p dividing
4 ).S(o 3 Solution: ( 4 ) ×== .3224So So S4 has a Sylow 3-subgroup of order 3, for instance, { ( ) ( 231,321,e )}. It also has a Sylow 2-subgroup of order 8, for instance, { }.)3241(),4231(),32()41(),42()31(,)43()21(),43(),21(,e ***
Example 8: Does A4 have a Sylow 2-subgroup? A Sylow 3-subgroup? A Sylow 5-subgroup? Give reasons for your answers. 66
2 The Sylow Theorems Solution: ( 4 ) ×== .3212Ao So A4 has a Sylow 2-subgroup and a Sylow
3-subgroup. However, since /| A),A(o5 44 has no Sylow 5-subgroup. ***
Try an exercise now.
E11) Is every Sylow p-subgroup of Sn also a Sylow p-subgroup of +1n ?S Give reasons for your answer.
E12) Consider the direct product ×= GGG 21 of finite groups G1 and 2 ,G
where p divides 1 )G(o as well as 2 ).G(o Prove that the Sylow
p-subgroups of G are precisely those of the form × 21 ,PP where P1 and
P2 are Sylow p-subgroups of G1 and 2 ,G respectively.
In Example 6 you have seen that a group can have several Sylow p-subgroups for a given p. Are these subgroups related, apart from the fact that they have the same order? Let’s see.
Let X denote the set of all Sylow p-subgroups of a finite group .G There is a “conjugation action” of G on ,X defined as follows: For g in G and P in ,X define ⋅ Pg to be the “conjugate of P by g ”, namely, −1 .gPg Note that this is a well-defined action since the conjugate of a Sylow p-subgroup is also a Sylow p-subgroup. (Why?) This leads us to the next theorem of Sylow.
Theorem 6 (Sylow’s second theorem): Let G be a finite group and p a prime. Let P be a p-subgroup of G and P0 be a Sylow p-subgroup of G.
Then P is contained in a conjugate of 0 .P In particular, any two Sylow p-subgroups are conjugates.
Proof: Let C denote the set of cosets, 0 .P/G Restrict to P the natural action of G on .C Since P0 is a Sylow p-subgroup, it follows that (= 0 |P||G||C| ) is coprime to p. Since P is a p-group, we may invoke the result in E17 of Unit 2 to obtain ≡ P )p(mod|C||C| , where CP is the set of points of C fixed by P. Since ≡/ ),p(mod0|C| P ≠ .0|C| This means that P fixes some coset −1 0.gP So, ⊆ 0G )gP(StabP , that is, P is contained in 0 .ggP
For the proof of the second assertion, let Q be a Sylow p-subgroup of .G By −1 −1 the first part, there exists g in G such that ⊆ 0 .ggPQ But Q and 0ggP −1 have the same order. So, = 0 .ggPQ
Theorem 6 has several important consequences. Consider three of them, which we give as corollaries to the theorem.
Corollary 3: A Sylow p-subgroup of a finite group G is unique if and only if it is normal. 67
Group Theory Proof: First, assume that the Sylow p-subgroup of G is unique, and is 0 .P −1 −1 Then 0ggP is another subgroup of the same order. So 0 0 ∈∀= .GgPggP
Thus, P0 .G The converse holds by Sylow’s second theorem.
Corollary 4: Sylow p-subgroups of abelian groups are unique.
Proof: Since every subgroup is normal in an abelian group, Corollary 3 gives the result.
For a subgroup ,H let G (HN ) denote the normaliser of ,H that is, ∈= −1 = }.HgHg|Gg{)H(N Observe that (HN ) is a subgroup of G and H NG(H) G G
that H is a normal subgroup of G (HN ). The next corollary to Theorem 6 is related to this.
Corollary 5: Let P and Q be Sylow p-subgroups of .G If P normalises ,Q
that is, ⊆ G ( ),QNP then = .QP
Proof: Note that P is a Sylow p-subgroup of G ( ),QN and Q is a normal
Sylow p-subgroup of G ( ),QN and hence is unique. Thus, = .QP
Try these exercises now.
E13) Show that ( GG ( )) = G (PNPNN ) for any Sylow p-subgroup P of .G
E14) Let G be a finite group and pd divide ,|G| where p is a prime. (We are not assuming that pd is the highest power of p that divides .|G| ) Do any two subgroups of G of order pd have to be conjugates in G? Give reasons for your answers.
We now turn to the question of how many Sylow p-subgroups a group has.
In Example 6, you saw that S3 has at least 3 Sylow-2 subgroups. Can it have more? Let’s see.
Theorem 7 (Sylow’s third theorem): Let G be a group, p a prime and = r ,mp|G| with r ≥1 and m coprime to p. Let n be the number of Sylow p-subgroups of .G Then n divides m and ≡ ).p(mod1n
Proof: Let X denote the set of all Sylow p-subgroups of ,G so that = .|X|n Consider the action of G by conjugation on .X By Sylow’s second theorem, this action is transitive. It follows, from the Structure theorem for orbits, that ~ X − G )P(StabG as G-sets, where P is an arbitrary point in .X ~ Note that the stabiliser of P is G ( ).PN Thus X − G .)P(NG
In particular, the cardinality of X is G = G .|)P(N||G||)P(NG| And, since
⊆ G ( ),PNP it follows that G |)P(N||G| divides = .m|P||G| We conclude that |X| divides m.
68
For the second assertion, we restrict the action of G on X to a Sylow The Sylow Theorems p-subgroup P of .G Since P is a p-group, we may invoke the result in E17 of Unit 2 to conclude that ≡ p ),p(mod|X||X| where XP denotes the set of points of X fixed by all elements of .P Now, P −1 ∈∀=∈= },XQQpQp|Pp{X that is, XP is precisely the subset of Sylow p-subgroups normalised by .P By Corollary 5, there is precisely one such subgroup, namely, P itself. Thus, P =1|X| , and hence ≡ ).p(mod1|X|
Let us look at the tremendous utility of this result now.
Example 9: Obtain all the Sylow p-subgroups of the dihedral group 10 .D
Solution: The dihedral group 10 ,D of all symmetries of the regular pentagon, has order = × .5210 It has Sylow 2-subgroups and Sylow 5-subgroups. By Theorem 7, the number of Sylow 5-subgroups is )5(mod1 and divides .2
Thus, D10 has a unique Sylow 5-subgroup, namely, the subgroup consisting of all its rotations (by angles that are integer multiples of π 5/2 ), which is normal. The number of Sylow 2-subgroups is )2(mod1 and divides .5 Hence, it is 1 or 5 . If it were ,1 this subgroup would have been normal, and then D10 would Z Z have been isomorphic to × 52 , and hence abelian, which it is not. So D10 has five Sylow 2-subgroups, each of which is generated by an element of order 2, that is, a reflection. ***
Example 10: Check whether or not a group of order 300 is simple. Solution: Let G be a group with 2 ××== 2.532300|G| The number of Sylow 5-subgroups is )5(mod1 and divides 12, i.e., it is 1 or 6. If it is 1, then the Sylow 5-subgroup is normal in .G
If the number is 6, then all 6 are conjugates in .G Call them 21 K 6.P,,P,P
Consider the action of G on = 1 K 6}P,,P{X by conjugation. Using E9 of
Unit 2, you can get KerG ~ ≤ϕϕ 6.SIm So KerG ϕ divides = .720!6 Now since /| ≠ϕ .}e{Ker,720300 Hence, Kerϕ is a non-trivial normal subgroup of .G Thus, in either case, G is not simple. ***
Try some related exercises now.
E15) Show that a group of order 28 is not simple. Further, how many elements of order 7 does such a group have?
E16) Obtain all the Sylow p-subgroups of the dihedral group 12 .D
Z Z E17) Obtain all the Sylow p-subgroups of the group × 3.S)2(
E18) [This exercise outlines another proof of the existence of Sylow 69
Group Theory p-subgroups, using group actions. The proof can be modified to obtain the more general result in Theorem 5. This proof does not rely on induction.] Let G be a finite group and p a prime such that = r mp|G| , with r ≥ 1 and m coprime to p. Prove the following: r mp n i) is coprime to p. n r r )r,n(C,C, p r ii) The regular action of G on itself induces an action of G on the denote the same number. power set of .G The set ,Xof subgroups of cardinality pr of G, form a G-invariant subset for this action. r mp iii) |X|= and there is a G-orbit in X of cardinality not r p divisible by p. iv) Let Y be a G-orbit of X such that |Y| is coprime to p. Let H be the stabiliser of an element in .Y Then r .p|H| [Hint: see E6 (ii) of Unit 2.] v) |H| is divisible by r .p vi) H is a Sylow p-subgroup of .G
There are two reasons for E19) Let G be a subgroup of a group ,K and let p be a prime dividing insistence on first principles. The first is to encourage you ).G(o Let Q be a Sylow p-subgroup of .K Prove the following from to review the technique by first principles, that is, using only the results about group actions in the which the results on group previous unit and none of the results of this unit: actions are applied in this unit. The second is that the (i) G has a Sylow p-subgroup. results of this exercise form a basis for yet another (ii) Any p-subgroup of G is contained in a conjugate of Q in .K independent proof of Sylow’s theorems. (iii) Any Sylow p-subgroup is of the form ∩ ′ ,QG for some conjugate Q′ of Q in .K
Let us now look at some uses of these theorems for analysing the structure of finite groups.
3.4 APPLICATIONS OF SYLOW’S THEOREMS
Let us now use Sylow’s three theorems to analyse the structure of some groups of small orders. The following proposition will be used crucially.
Proposition 2: Let H and N be subgroups of a finite group G. Suppose that H normalises ,N that is, −1 = NhNh for all h in .H Then: i) The set = ∈ ∈ }Nn,Hh|hn{HN is a subgroup of G. In particular, if N ,G then ≤ .GH N ⋅ |N||H| ii) |HN| = . ∩ |NH| iii) If N also normalises ,H and ∩ NH is trivial, then NH is isomorphic 70
to the direct product × .NH This happens, in particular, when H and N The Sylow Theorems iv) are both normal subgroups of ,G and their intersection is trivial. v) If both H and N are normal in ,G then so is .NH
The proof of this proposition is left to you. Let us now see how to use this, and the Sylow theorems.
Example 11: Going beyond Example 7, how many Sylow 2-subgroups and
Sylow 3-subgroups does S4 have? Solution: Applying Sylow’s third theorem, you know that the number of Sylow 2-subgroups is )2(mod1 and divides .3 Hence it is 1 or ,3 each of order .8 The number of Sylow 3-subgroups is )3(mod1 and divides .8 Thus, it is 1 or .4 From Example 7, you know that one is ( ) .321 You can see that ( ) ( 431,421 ) and ( 432 ) are also Sylow 3-subgroups. So, the number is .4 Call them T,T,T 321 , and 4 .T
From Example 7, you also know at least one Sylow 2-subgroup, call it 1 .H Is 4 it unique? Let’s see. Now, UTi has 9 elements, of which one is e and the =1i other 8 are all of order .3 Also H1 has 7 elements of order 2 or .4 So, we have covered 16 elements of 4 .S There are still 8 others left, of order a power of .2 Thus, there are two other Sylow 2-subgroups, H 2 and 3 .H ***
Example 12: Construct the Sylow 2-subgroups of 5 .S
3 3 Solution: Since 5 == ,5.3.2! 5)S(o each Sylow 2-subgroup has order 2 = 8. By Sylow’s third theorem, their number is 1(mod 2) and divides 15. Thus, the number can be 1, 3, 5 or 15. In fact, you will see that it is 15. Consider the subgroup consisting of those permutations that leave one of the digits fixed (say, 5). This subgroup is isomorphic to 4 ,S and thus, it has three Sylow 2-subgroups (see Example 11). These subgroups are of order 8, and hence are Sylow 2-subgroups of S5 as well. There are 5 different subgroups of 5 ,S each isomorphic to S4 (by the choice of the digit to be fixed). An inspection shows that no two of these subgroups share a Sylow 2-subgroup. We have, thus, produced 15 distinct Sylow 2-subgroups. ***
Example 13: Apply the Sylow theorems to analyse the structure of groups of order 15. Solution: Let G be a group of order = × .531 5 Let H be a Sylow 3-subgroup and N a Sylow 5-subgroup. They are both unique by Sylow’s third theorem, and hence normal. Since the order of H and N are coprime to each other, it follows that ∩ NH is trivial. So, G is isomorphic to × .NH It follows that G is cyclic, as both H and N are cyclic (being of prime order) and their orders are coprime. *** 71
Group Theory This leads us to an important point to note.
Remark 5: The argument in Example 13 applies more generally to groups of order pq, , where p and q are primes such that p < q and ≡/ 1q (mod p). For instance, to groups of order 51, 77 and = × 317391 .2 The conclusion is that there is only one group (up to isomorphism) of order pq , where p and q are prime, < ≡/ 1qq,p p),(mod namely, the cyclic one.
Try some exercises now.
a a 21 E20) Let G be a group of order 1 2 ,pp where p1 and p2 are two distinct N primes and 21 ∈ .a,a Let P1 and 2 ,P respectively, be a Sylow
p1-subgroup and a Sylow p2-subgroup of .G Suppose that both P1 and P2
are normal. Show that ×= 21 .PPG Generalise the result to the case when G has arbitrary order and for all p, its Sylow p-subgroups are normal.
A famous theorem by Burnside tells us that if G E21) How many Sylow 5- and Sylow 3-subgroups does S5 have, and why? is a finite non-abelian simple group, then at least E22) Prove that no group of order pq is simple, where p and q are distinct three distinct primes divide )G(o . primes. What about a group of order 2 qp ?
E23) Prove Proposition 2.
Now, we shall use the Sylow theorems to understand the structure of groups of order 34, or 38, or 62, for example.
Theorem 8 (Groups of order 2p for a prime p ): There are precisely two groups of order ,p2up to isomorphism, where p is a prime.
Proof: If = ,2p then you know from Example 5 that the group can either be the cyclic group or the Klein 4-group. Now, let p be odd and G be a group of order .p2 The number of Sylow p-subgroups is of the form 1(mod p) and divides 2. Hence, the Sylow p-subgroup of G is unique, and hence normal. Denote it by .N Since ( ) = N,pNo is cyclic of order p , say = ,rN where r(o = .p) Let = { s,eH }, where 2 = ,es be a Sylow 2-subgroup of .G Since ∩ NH is trivial (by considerations of order), Proposition 2 (ii) applies. Hence,= ,NHG so that G is generated by s and r. Now define =ϕ→ϕ −1 .srs)r(:NN: Since N is normal, ϕ is an automorphism of .N This automorphism is determined by what it does to r, since r generates .N − m1 In Unit 6 you will study Choose m to be an integer such that = .rsrs We have, on the one hand, 2 about generators of −1m − === mmmm1 .r)r ()srs(ssr And on the other hand, −1m = −− 11 = ,rs)srs(sssr groups, and the relations 2 they satisfy, in detail. since 2 = .es Thus, m = rr . So, 2 − ),1m(|)r(o that is, 2 − ⇒ |p)1m(|p − )1m( or |p + ).1m( 72
If − ),1m(|p then −1 = ,rsrs which means that sr = rs, that is, elements of H The Sylow Theorems commute with those of .N In this case, = × ,NHG and G is cyclic. If + ),1m(|p then we have = −− 11 .rsrs This means that G is generated by elements r and s subject to the relations 2 2 == er,es and rs = −− 11 .rs Thus, G is isomorphic to the dihedral group of order p2 .
We shall now state a result, without proof, regarding the types of groups of order 12. This, as well as many similar results, will be proved in Unit 6, while studying generators and relations.
Theorem 9: There are five isomorphism classes of groups of order 12. These are represented by: i) Z Z × Z Z;34 ii) Z Z Z Z×× Z Z;322 iii) A4 ; iv) D12 ; v) the group generated by x, y such that 4 3 === 2 .xyxy,yex
From this theorem, you can see, for example, that two types of groups of order 12 are abelian and 3 types are not.
Try this exercise now, looking at all the results you have studied in this unit.
E24) What are the possible structures of groups of order n, where = K ≠ 8n,10,,1n ? [You will study groups of order 8 in Unit 6.]
Let us now end our discussion on the analysis of the structure of some finite groups, with a summary of what you have studied in this unit.
3.5 SUMMARY
In this unit, we have discussed the following points.
1. The definition, and examples, of external direct products of groups.
2. The definition, and examples, of internal direct products of normal subgroups.
Z Z Z 3. If = 1)n,m( , then × ~ mnnm .
4. × = ).K(o)H(o)KH(o
5. The converse of Lagrange’s theorem is not true in general.
6. Cauchy’s theorem: If G is a finite group and p is a prime such that | )G(op , then G has an element of order p.
73
Group Theory 7. Sylow’s theorems: If G is a group such that o = r ,mp)G( where p is a prime, r ≥ 1 and = ,1)m,p( then
i) G has p-subgroups of orders p 2 K r .p,,p, A subgroup of order pr is called a Sylow p-subgroup. ii) Any two Sylow p-subgroups are conjugates in .G iii) The number of distinct Sylow p-subgroups of G is mod1 )p( and divides m.
8. If = q|G| ,p where p and q are primes such that p < q and q ≡/ ,)p(mod1 then G is cyclic.
9. If G| = ,p2| where p is an odd prime, then G is cyclic or G~ p2 .D
10. If = 2|G| ,1 then G has 5 possible structures.
3.6 SOLUTIONS/ANSWERS
E1) ∗ is associative: Let 332211 ∈ .G)b,a(),b,a(),b,a(
Use the fact that ∗1 and ∗2 are associative to show that
2211 1133 ∗∗=∗∗ 3322 )).b,a()b,a(()b,a()b,a())b,a()b,a((
The identity element of G is e( 21 ),e, where e1 and e2 are the identities
in G1 and 2 ,G respectively. The inverse of ∈ G)y,x( is −− 11 ),y,x( where −1 −1 1 21 =∗=∗ 2.eyy,exx
E2) Define 21 ×→× 12 = b()b,a(f:GGGG:f )a, . Then check that f is well-defined, 1-1, surjective and a homomorphism, that is, f is an isomorphism.
GG ~ ××∴ GG 1221 .
E3) We need to show that any element of ×GG 21 is of the form k ,h where
h ∈ H and ∈ .Kk For this, consider any )y,x( in ×GG 21 .
= 12 )y,e()e,x()y,x( , with 2 ∈ 1 y,e(,H)e,x( ∈ .K)
21 =×∴ .HKGG Now, let us look at H ∩ K . Let ∈ ∩ .KH)y,x(
Since =∈ 2.ey,H)y,x( Since y,x( =∈ 1.ex,K)
=∴ 21 )e,e()y,x( . =∩∴ e{(KH 21 )}.e,
E4) Now, ×∈ 21 )GG(Z)y,x(
⇔ = ×∈∀ GG) b,a()y,x()b,a()b,a()y,x( 21
⇔ y,ax()yb,xa( 1 ∈∈∀= Gb,Ga)b 2
∈∀=⇔ xxa Gaa 1 and ∈∀= Gbb yyb 2
∈⇔ 1 )G(Zx and ∈ 2 )G(Zy
1 ×∈⇔ 2 ).G(Z)G(Z)y,x( ×=×∴ ).G(Z)G(Z)GG(Z 74 21 1 2
E5) Let = xA and = ,yB where = y(o,m)x(o = .n) The Sylow Theorems Z ~ Z Then A ~ m and B n . Z Z ~ Z If we prove that × mnnm , then we will have proved that Z × BA ~ mn , that is, A × B is cyclic of order mn. Z ~ Z Z So, let us prove that if = 1)n,m( , then mn × nm . Z Z Z Z Z Define :f nm ++=×→ ).nr,mr()r(f: Now, you can check that f is well-defined and f is a homomorphism. = ∈Z ∈ Z ∩ Z}nmr|r{fKer = ∈Z ∈ Z }mnr|r{ , since = 1)n,m( = Zn.m Finally, to show that f is surjective, take any element Z Z Z Z Z ( )nv,mu ×∈++ nm . Since = ∃ t,s,1)n,m( ∈ such that + =tms 1n . So, − + − )nt1(v)ms1(u ∈Z such that − + − = + Z + Z).nv,mu())nt1(v)ms1(u(f Thus, f is surjective. Now, we apply the Fundamental Theorem of Homomorphism to find Z Z Z Z Z Z ~ Z Z that fKer ~ fIm , that is, mn ~ × nm , that is, mn × nm . ∴A × B is cyclic of order mn.
E6) Since H1 and H2 are non-empty, so is × 21 .HH
Also, for 21 )h,h( ( and 43 )h,h in −1 −1 −1 × 432121 = ()h,h()h,h(,HH 31 42 ×∈ 21 .HH)hh,hh So,
×≤× 2121 .GGHH
Next, for ×∈ GG) y,x( 21 (and ×∈ 2121 ,HH)h,h −1 21 x()h,h()y,x( )y, −1 −1 = 1 2 ×∈ 21 .HH)yyh,xxh(
Thus, if H1 G1 , H2 G2 , then × 21 )HH( × 21 ).GG(
Z Z Z Z E7) Take 21 == .2GG Note that 2 has two subgroups: the trivial
one and the whole group. So the subgroups of type × HH 21 are four in
number. But ×GG 21 is the Klein 4-group (non-cyclic of order 4) which has five subgroups: the trivial group, the whole group, and three of order 2.
E8) We know that each ∈Gx can be expressed as hk , where h ∈H and k ∈K . ∴ = .H KG We need to show that ∩ = }.e{KH For this, take ∈ ∩ .KHx Then x ∈H and ∈ .Kx ∴ ∈HKx e and ∈ .H Kex So, x has two representations, xe and ex , as a product of an element of H and an element of .K But we have assumed that each element must have only one such representation. So the two representations xe and ex must coincide, that is x = e . ∴ ∩ = }.e{KH ∴ = × .KHG
E9) Let )n(P be the statement that if 1 K××= n ,HHG then n = ∏ i .)H(o)G(o =1i 75
Group Theory )1(P is trivially true, as you can see. Now, assume that )k(P is true for some ∈N.k
Consider 1 K××= HHG +1k now. k Since H ,G G ~ K×× .HH Thus, oG = ,)H(o so +1k H 1 k H ∏ i +1k +1k =1i +1k that = ∏ i o)G(o )H( , i.e., + )1k(P is true. =1i Hence )n(P is true ∀ ≥ .1n
3 E10) 5 ××= .235|S| Now, }e{ and S5 are subgroups of 5 ,S of orders 1 and
120, respectively. By Sylow’s first theorem, S5 must have subgroups of Z Z ~Z order 4,2,3,5 and .8 Further, since × mnnm for = S, 1)n,m( 5 must also have subgroups of order × × ×23,25,35 , i.e., 10,15 and 6.
We also know that A,S 44 and A5 are subgroups of 5 ,S of orders 12,24 and ,60 respectively. Thus, the required values of n are .120,60,24,15,12,10,8,6,5,4,3,2,1
E11) No. For example, consider S3 and 4 .S The order of a Sylow 2-subgroup
in S3 is 2, and in S4 is 8. So they cannot be the same.
e1 e1 E12) Let 1 = mp|G| 1 and 2 = 2 ,mp|G| with m1 and m2 coprime to p.
+ee 21 Since G,|G|p has a Sylow p-subgroup say ,P where = .p|P| Let π1
denote the projection of G onto 1 ,G that is, →×π GGG: 1211 given by
a 121 .g)g,g( Similarly, let π 2 denote the projection of G onto 2 ,G
that is, 212 →×π GGG: 2 given by ( 21 ) a 2.gg,g The projections π1
and π2 , being homomorphisms, their images π1(P) and π 2 )P( are
d1 p-subgroups of G1 and 2 ,G respectively. Thus, 2 =π p| )P(| and
d 2 2 =π ,p|)P(| with ≤ ed 11 and ≤ 22 .ed But, since P is contained in the
+ee 21 +dd 21 subgroup 1( ) π×π ( PP 2 ) , 1 2 =π⋅π≤= .p|)P(||)P(||P|p
Hence == 2211 ,ed,ed π 1 )P( and π2 )P( are Sylow p-subgroups of P1
and P2 , respectively, and 1 P(P π×π= 2 .)P()
Conversely, if P1 and P2 are Sylow p-subgroups of G1 and 2 ,G
+ee 21 respectively, then 21 =× ,p|PP| so that × PP 21 is a Sylow p-subgroup of .G
−1 E13) Let ∈ G ).P(Nx Then = .PxPx −1 Now, for any ∈ G ),P(Ny ∈ G )P(Nxyx . −1 So, G = G )P(Nx)P(xN , that is, ∈ GG .))P(N(Nx
To prove the other inclusion, let g be an element in GG .))P(N(N Then −1 −1 −1 G = G P(Ng)P(gN ) and ⊆ G .)P(NP So, ⊆ G )P(NgPg . So gPg is
a Sylow p-subgroup of G .)P(N
But ,Pbeing normal in G ),P(N is the only Sylow p-subgroup of G .)P(N 76
−1 The Sylow Theorems So = PgPg , that is, ∈ G )P(Ng .
E14) No. Take, for instance, = = ,1d,2p and G to be the Klein 4-group. There are three subgroups of order 2 of .G Since G is abelian, each of these is normal, so no two of them are conjugate.
E15) Let 2 ×== .7228|G| By Sylow’s third theorem, the number of Sylow 7-subgroups is )7(mod1 and divides 4. Thus, it is 1. Call this unique subgroup .H Since H is normal in ,G G is not simple.
Now, any element of order 7 generates a subgroup of order 7, which must be .H Also every element of ,H apart from e, is of order 7. Thus, G has 6 elements of order 7.
2 E16) D12 has order 2 ×= .321 The number of Sylow 3-subgroups is 1 or 4.
If it is 4, then D12 has 8 elements of order 3. However, by considering its elements, you know that it has only two elements of order 3, namely, the rotations by π 3/2 and π .3/4 So there is a unique Sylow 3-subgroup. As to the Sylow 2-subgroups, there are either 1 or 3 of them. The
number cannot be 1 since D12 is not abelian. So the number is 3. The rotation by an angle π belongs to the centre of the group. It is contained in all 3 Sylow 2-subgroups. Each of these subgroups consists of e, the rotation by angle π, and a pair of reflections. To describe such a pair, consider the line through a pair of opposite vertices. The reflection in this line, along with that in the line perpendicular to it, forms a pair of reflections. Note that the perpendicular line passes through the midpoints of the two edges that are not incident on the chosen pair of opposite vertices. There being three such pairs (of opposite vertices and therefore of reflections), the Sylow 2-subgroups are all described.
E17) This group has order 2 ×=× .3262 This has Sylow 2- and Sylow 3-subgroups of orders 4 and 3, respectively. Along the lines of E12, you
can show that the Sylow 3-subgroups are of the form × P} e{ 1 where P1 Z is a Sylow 3-subgroup of 3 .S Since P1 is unique, ×S32 has a unique Sylow 3-subgroup. Using E12, there are three Sylow 2-subgroups, each of which is isomorphic to the Klein 4-group.
E18) i) Note that if a term in the numerator is divisible by p, it is of the form − sr ,npmp where = .1)n,p( Then in the denominator there is a corresponding term − sr ).npp( So ps divides both terms and p +1s does not divide either. So the powers of p in the numerator and denominator are the same and cancel out. Hence the result.
ii) G ×℘(G)→℘(G)a gS)S,g(: is an action. If =⊆= r ,}p|S|GS{X then × ⊆ .XXG
So X is a G-invariant subset of ℘(G). 77
Group Theory rmp iii) The number of distinct subsets in X is . r p Now /| ,|X|p by (i) above. Also X is a union of its G-orbits. So there is at least one G-orbit in ,X say ,Y such that /| .|Y|p
r iv) = G )S(StabH for ∈ ,YS where /| |Y|p and = .p|S| Now, by E6 (ii) of Unit 2, as S is H-invariant, it is the union of right cosets of .H So = r .p|S||H|
v) Y~ HG as G-sets. Since /| r Hp ,,|Y|p .||
vi) From (iv) and (v), = r ,p|H| so that H is a Sylow p-subgroup of .G
E19) i) Let X denote the set of cosets ,QK with the natural action of K on it. Restrict to G the action of K on ,X and consider the G-orbits in .X Since Q is a Sylow p-subgroup of ,K it follows that |X| is coprime to p. So there exists a G-orbit Y of X such that |Y| is coprime to p. Let y be a point of Y and consider the
stabiliser, G ).y(Stab Call it H. Now, if y represents the coset kQ of Q in ,Kthen the stabiliser of y in K is −1 ,kQk so ∩= kQkGH −1 is a p-group (being a subgroup of the p-group kQk−1 ). And, on the order hand, Y −~ ,HG as G-sets. So the index of H in G (being equal to |Y| ) is coprime to p. So H is a Sylow p-subgroup of .G
ii) Let P be a p-subgroup of G and let X be as in (i) above. Restrict to P the action of K on .X Since P is a p-group, we may invoke E17 of Unit 2 to conclude that P ≠ .0|X| Letting kQ be the coset corresponding to a fixed point of P in ,X we obtain ⊆ −1.kQkP
iii) Let P be a Sylow p-subgroup of .G By (ii) above, there exists a conjugate Q′ of Q in K such that ⊆ ′ ,QP so that ⊆ ∩ ′ .QGP But ∩ QG ′ is a p-subgroup of G (since Q′ is a p-group), so its cardinality cannot exceed that of .P Hence, = ∩ ′ .QGP
E20) We apply Proposition 2, with 1 = HP and 2 = .NP For reasons of order,
∩ PP 21 is trivial. So = 21 ,|PP||G| so that = 21 .PPG Thus, G is
isomorphic to × 21 .PP
For the general case, let P, ...,P,Pk21 be Sylow subgroups of ,G corresponding one each to the various prime divisors of .|G| Suppose that they are all normal. The argument of the previous paragraph shows that PP is × .PP . It follows, from Proposition 2(iv), that PP is 78 21 21 21
normal in .G We now apply Proposition 2 again, this time with The Sylow Theorems
= PPH 21 and = 3 ,PN to conclude that PPP 321 is normal and isomorphic ~ to PPP ××−× 321321 .PPP Proceeding thus, we conclude that G is
isomorphic to the direct product 1 ×⋅⋅⋅× k .PP
3 E21) The order of 5S5 is ⋅⋅= .532! The number of Sylow 5-subgroups can be 1 or 6. The number of Sylow 3-subgroups can be 4,1 or 10. Now, the order of a Sylow 5-subgroup is 5. Every subgroup of order 5 consists of four 5-cycles in addition to the identity. No two distinct subgroups of order 5 intersect non-trivially – the identity is the only
element common to them both. The number of 5-cycles in S5 being 24,
we conclude that there are 6 Sylow 5-subgroups in 5 .S
The analysis for Sylow 3-subgroups is similar. The 3-cycles in S5 are × × 345 = 20 in number. They are distributed over 10 Sylow 3 3-subgroups.
E22) Let G be a group of order pq. Assume p < q. (The same argument works if p > q. ) Then G has a Sylow q-subgroup and the number of such subgroups is )q(mod1 and divides p. Hence the number is 1. So the Sylow q-subgroup has order q and is normal in .G Hence G is not simple.
Now consider a group G of order 2 .qp If p > q , then, as above, G will not be simple. Now assume q > p. The number of Sylow q-subgroups can be p,1 or 2.p If the number is 1, then G is not simple. Since q > p, the number cannot be p. If the number of Sylow q-subgroups is p2 , then 2 − ,)1p(q that is +− .)1p()1p(q Since /| +− .)1p(q),1p(q This is only possible if = .2p Then = ,3q so that = 12|G| and G has
4 Sylow 3-subgroups. Let the Sylow 3-subgroups be 4321 .T,T,T,T 4
Then UTi consists of the identity and 8 elements of order 3. Any =1i Sylow 2-subgroup of G intersects this union trivially because any such
group intersects each Ti trivially. So a Sylow 2-subgroup must contain the 3 remaining elements in UT\G i and the identity. Thus, the i Sylow 2-subgroup is normal, and G is not simple.
Hence, in all the cases, G is not simple.
E23) i) Firstly, check that ≠ « .HN Next, use the fact that H normalises N to show that if −1 2211 ∈ ,HNnh,nh then 2211 ∈ .HN)nh()nh( Hence ≤ GHN . 79
Group Theory ii) You have seen this in Unit 1.
iii) Since ∩ = HN},e{NH is a direct product.
iv) From (i) you know that ≤ .GHN Further, for any ∈ ∈ ∈ ,Nn,Hh,Gg ghng −1 = −− 11 ∈ .HNgngghg Hence HN .G
E24) If n = 1, then = }.e{G • If = ,7,5,3,2n then G ~ Z Z.n • If = ,4n then G is cyclic or the Klein 4-group (see Example 5). • If = 6n or 10, then by Theorem 8, G is either cyclic or isomorphic to the dihedral group. • If = ,9n then G is abelian. If G has an element of order 9, it is cyclic. If not, consider its subgroup = ,xH where 3 = .ex As HG ~ Z 3Z , G ~ Z Z × Z Z.33
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