MATH3071 Groups and Symmetry PDF

MATH3071 Groups and Symmetry © Kevin Houston University of Leeds November 17, 2024 Contents 0 Module information and Resources 1 0.1 Module details............................ 1 0.2...

MATH3071 Groups and Symmetry © Kevin Houston University of Leeds November 17, 2024 Contents 0 Module information and Resources 1 0.1 Module details............................ 1 0.2 Books and Online notes....................... 1 0.3 Videos................................ 2 0.4 Do I care what you think?..................... 2 1 Basic notions and revision 3 1.1 Definition of a group........................ 3 1.2 Symmetry and Rotation Groups.................. 5 1.3 Symmetric and Alternating Groups, Sn and An.......... 6 1.4 Subgroups.............................. 11 1.5 Cyclic Groups............................ 15 1.6 Equivalence Relations........................ 16 1.7 An equivalence relation partitions the set............. 18 2 Quotient Groups 21 2.1 Right cosets............................. 21 2.2 Left cosets.............................. 26 2.3 Conjugation of subgroups...................... 26 2.4 Normal groups............................ 28 2.5 Quotient groups........................... 30 3 The Three Isomorphism Theorems 35 3.1 Homomorphisms........................... 35 3.2 Isomorphisms............................ 36 3.3 The First Isomorphism Theorem.................. 37 3.4 The preimage of a subgroup under a homomorphism is a subgroup 38 3.5 The Second Isomorphism Theorem................. 39 3.6 The Third Isomorphism Theorem.................. 41 4 Group actions 43 4.1 Group actions............................ 43 4.2 Orbits................................ 48  5 Orbit-Stabiliser Theorem 52 5.1 Stabilisers.............................. 52 5.2 The Orbit-Stabiliser Theorem................... 56 5.3 Fixed Point Sets........................... 58 5.4 Burnside’s Lemma.......................... 60 5.5 Applications of Burnside’s Lemma................. 60 6 Transitive, Faithful, and Free Actions 66 6.1 Transitive group actions....................... 66 6.2 The underlying homomorphism of an action............ 67 6.3 Faithful group actions........................ 69 6.4 Cayley’s Theorem.......................... 71 6.5 Free group actions......................... 71 Today 7 Conjugacy 7.1 Conjugacy of elements....................... 72 72 7.2 The conjugate group action and conjugacy classes........ 73 7.3 Centralisers............................. 75 7.4 Centre of a group, Z(G)...................... 76 7.5 The Class Equation......................... 80 Today 8 Sylow Theorems 8.1 p-subgroups and Sylow p-subgroups................ 83 83 8.2 Sylow Theorems........................... 84 8.3 Proofs of the Sylow Theorems................... 88 8.4 Applications of the Sylow Theorems................ 90 9 Simple Groups 93 9.1 Definition of simple groups and finite classification........ 93 9.2 Abelian groups of non-prime order................. 94 9.3 Groups of order p r......................... 94 9.4 The groups Sn and An....................... 95 9.5 A5 is simple............................. 95 9.6 Counting and kernel methods.................... 97 9.7 Concluding remark......................... 99 Index 100 Chapter 0 Module information and Resources This is a quick summary. Please see the Module Handbook on Minerva for full details. 0.1 Module details Module leader and first person to contact : Prof Kevin Houston [email protected] What to call me: Kevin or Prof Houston (the Hou part of my name is pronounced as Who.) Office Hours: In Mathematics Building 8:20 o Monday 11-12 o Wednesday 11-12 Lectures: o Monday 9-10 RSLT02 o Tuesday 12-1 Chemistry LT D (G.35) o Thursday 12-1 RSLT11 0.2 Books and Online notes The reading list for the module can be found at MATH3071 Reading List In addition there is book with online notes freely available: Thomas Judson, Abstract Algebra Theory and Applications. o Pdf version of book. http://abstract.ups.edu/index.html o Online version: http://abstract.ups.edu/sage-aata.html We’ll mainly be looking at the topics up to Chapter 15. The online version has lots of questions to work through. 1 2 CHAPTER 0. MODULE INFORMATION AND RESOURCES 0.3 Videos Look in comments as mistakes are often caught there. o Richard Borcherds: Group Theory Playlist A series of videos on Group Theory by a Fields Medal winner. We’ll cover up to about video 15. o Professor Macauley Visual Group Theory Playlist We’ll cover up to Lecture 5.7. 0.4 Do I care what you think? Of course I do! Student feedback is very important. For example, following student attainment in the module in previous years I have rewritten the notes from start to finish. (No doubt, there will still be some typos. Let me know if you find any! [email protected]. I will change the colour of text which has been changed to make it easier to find changes.) I will run a mid-module survey to collect feedback but feel free to inform me of problems at any time in the module. There will also be the standard end-of-module questionnaire in Week 10 or Week 11. Chapter 1 Basic notions and revision At the start of the module we will revise various notions from the groups part of MATH2022 Groups and Vector Spaces. In this chapter we’ll look at the basic notions and in the next we will look again at quotient spaces. Mostly, the viewpoint and explanations will be di↵erent to those in MATH2022 but will include some new material which I hope will help you consolidate what you already know. 1.1 Definition of a group Recall the definition of a group from MATH2022. Definition 1.1 A group, denoted (G, ⇤), is a set G and operation ⇤ on that set that combines two elements such that they satisfy the following: AX1 Closure: For all x, y 2 G, we have x ⇤ y 2 G. In this case we say ⇤ is a binary operation. AX2 Associativity: For all x, y , z 2 G, we have x ⇤ (y ⇤ z) = (x ⇤ y ) ⇤ z. AX3 Identity: There exists an element e 2 G such that for all x 2 G we have e ⇤ x = x ⇤ e = x. Such an element is called the identity of G. AX4 Inverses: For each x 2 G there exists a y 2 G such that x ⇤ y = y ⇤ x = e (where e is the element in AX3.) Such a y is called the inverse of x. The inverse is denoted x 1. We usually drop the reference to the binary operation ⇤ if no confusion will result and hence write xy rather than x ⇤ y. The identity and the inverse for each element are unique and in the equations xz = y z and z x = z y we can cancel to get x = y. Also, (xy ) 1 = y 1 x 1 for all x, y 2 G. Group structures appear almost everywhere in mathematics and so are very important to study. If G is a group and we have xy = y x for all x, y 2 G, then we say G is an Abelian group. The number of elements in G is called the order of the group and is denoted |G|. If this number of elements if finite, then we say G is a finite group and if not, then we say G is an infinite group. 3 4 CHAPTER 1. BASIC NOTIONS AND REVISION Examples 1.2 (i) The integers under addition from a group, denoted (Z, +). This group is Abelian and |Z| = 1. (ii) The non-zero real numbers under multiplication (R⇤ , ⇥) form a group.1 This is an infinite Abelian group. (iii) The integers modulo n under addition, denoted (Zn , +) where Zn = {, , ,... , [n 1]}. forms a group under the binary operation +n by [x] +n [y ] := [x + y ]. Often we will drop the reference to n in +n and just write +. As we have |Zn | = n, this is a finite group. (iv) The general linear group, denoted GLn (R), the set of invertible n ⇥ n matrices, is a group. The identity is In , the identity matrix, and the inverse of the matrix X is just the matrix inverse X 1 of X. (v) Vectors spaces over a field are Abelian groups with a bit of extra structure – scalar multiplication is defined and the distributive law holds. Vectors spaces can be finite or infinite. This usually depends on whether or not the field has a finite or infinite number of elements. We can use groups to define other structures. For examples, we can define fields in a rather concise way. (Recall that fields are just objects in algebra where we have some notion of addition, subtraction, multiplication and division (for non- zero elements of course).) Definition 1.3 A field is a set F with at least two distinct elements 0 and 1 and with two binary operations + and ⇥ such that the following hold. (i) (F, +) is an Abelian group. The binary operation + is called addition and the identity element is 0. (ii) (F\{0}, ⇥) is an Abelian group. The binary operation ⇥ is called multipli- cation and the identity element is 1. (iii) For all a, b, c 2 F the distributive law holds: (a + b)c = ab + bc. Examples are Q, R, C with the usual addition and multiplication. The group Zn is a field when n is prime with the multiplication being multiplication mod n. The set Z is not a field under the usual addition and multiplication since most numbers do not have a multiplicative inverse, e.g., there is no integer x such that x ⇥ 2 = 1. For a group with a finite number of elements we can draw up a table of the products. This is called a Cayley Table for the group. Example 1.4 The Klein Four-Group V (also called the Klein Vierergruppe) has four elements such that every element x has x 2 = e. Let V = {e, a, b, c}. Then we can describe V by writing out its Cayley table as seen in Table 1.1. 1 We usually denote a set X with the zero removed as X ⇤. E.g., C⇤ = C\{0}. 1.2. SYMMETRY AND ROTATION GROUPS 5 ⇤ e a b c e e a b c a a e c b b b c e a c c b a e Table 1.1: Cayley table for Klein Four-Group. Here x ⇤ y has x from the left side column and y from the top row. (Down the stairs and out the house!) Definition 1.5 Let G be a group. The set generated by x, denoted by hxi, is the set hxi = {x n | n 2 Z} = {... , x 2 , x 1 , e, x, x 2 ,... }. Note that, by convention, x 0 = e, the identity of the group. The element x is a called a generator of hxi. Definition 1.6 The order of an element x 2 G is the number of elements of hxi. That is, Order(x) = |hxi|. We will also write |x| when it is clear that the order of an element is meant. Examples 1.7 (i) The order of e is equal to 1 and this is the only element of order 1. (ii) The order of each of the non-identity elements in the Klein Four-Group is 2. Proposition 1.8 Let x be an element of finite order in a group. Then, |hxi| = min{n 2 N | x n = e}. Example 1.9 Every element in Z5 = {0, 1, 2, 3, 4} has order 5 except 0. For Z6 = {0, 1, 2, 3, 4, 5} we have |0| = 1, |1| = 6, |2| = 3, |3| = 2, |4| = 3, and |5| = 3. 1.2 Symmetry and Rotation Groups A good source of groups is the group of symmetries of some object. At this point we should say that Group Theory is often said to be the study of symmetry. But what does that mean? What is symmetry? What is a symmetry? The idea of symmetry is that if we do something to an object, then something of interest remains unchanged. That is of course very vague so let us look at some examples. Probably the first time we meet symmetry is to do with reflectional symmetry in some axis, i.e., for a picture in the plane there is some line along which we can reflect so that the picture looks like the same but is, well, ‘reflected’. It is a bit easier to define rotational symmetry. If we have a square and rotate it about its centre by 90 degrees, then it looks the same as before. If someone had 6 CHAPTER 1. BASIC NOTIONS AND REVISION looked away at the right moment, they may not have realised that it had been rotated. This is of course still very vague and so we use groups to define what we mean by symmetry and a symmetry is just an element of a group. (Some people say that Group Theory is the study of symmetry.) This perspective will become clearer through the module, particularly when we introduce group actions. First let us consider symmetry groups. Suppose we have an object in Rn for some n, for example a rectangle in R2 or a tetrahedron in R3. What transforma- tions of Rn can we make so that the object looks the same after the transforma- tion. For example, we can rotate the rectangle by 180 degrees or reflect in one of two axes going through the midpoints of opposite edges. For the tetrahedron we can rotate it about an axis going through a vertex and through the centre of an opposite face. We can also rotate by 180 degrees about an axis through the midpoints of opposite edges. By composing these reflections and rotations we can create a group called the symmetry group of the object. If we restrict to only rotations, then we get its rotation group. Consider a cube and the four axes that go through opposite vertices (each axis goes through a pair of vertices and as there are 8 vertices in a cube, there are 4 axes). When we reflect or rotate so that the cube looks unchanged, then we will exchange the axes. So if we label the axes 1 to 4 we can describe the symmetries by what happens to these numbers. In other words, a reflection or a rotation can be described by a permutation of the numbers 1 to 4. Similarly, if we label the four corners of a tetrahedron we can describe all rotations using where the corners go. Again we can describe the rotations using the permutations of the numbers 1, 2, 3, 4. If we look at a triangle, the we can label the vertices with the numbers 1 to 3 and describe the symmetries with permutations of 1, 2 and 3. This looks like a good time revisit permutations on the set {1, 2,... , n}. 1.3 Symmetric and Alternating Groups, Sn and An Recall the definition of a bijection. Definition 1.10 A map f from the set X to the set Y is called a bijection if f has an inverse map. We also say the f is bijective. Recall that f is bijective if and only if f is injective (f (x) = f (y ) =) x = y for all x, y ) and f is surjective (for all y 2 Y there exists x 2 X such that f (x) = y ). Note that I avoid the use of the name one-to-one for injective as this name is very misleading. Let X be a non-empty set. The symmetric group for X, denoted Sym(X), is defined by Sym(X) = {the set of bijections on X} = {f : X ! X | f is a bijection} with the binary operation being composition of maps. With this operation the set Sym(X) really is a group as its name suggests. (Composition of two bijections 1.3. SYMMETRIC AND ALTERNATING GROUPS, SN AND AN 7 is a bijection, composition of maps is associative, the identity map is a bijection and serves as the identity of the group, the inverse of the element f is just the inverse map f 1 which is of course a bijection.) The group Sn = Sym({1, 2, 3,... , n}) is a special case of this and is called the symmetric group of degree n. This is a core definition in Group Theory. Our definition is that an element of Sn is a bijective map but like any core definition, there are many ways of describing a permutation of the numbers 1 to n. For example, we can see Sn as describing how we can permute the numbers from 1 to n. This allows us to count the number of bijections, i.e., calculate |Sn |. When we permute 1, 2,... , n, we are just picking an order for the numbers. We have n ways of picking the first number, then n 1 for the second as we can’t choose the first number again. For the third number we have n 2 choices, and so on. Hence, |Sn | = n(n 1)(n 2) · · · 2 · 1 = n!. Another popular and useful way for representing permutations is via two row notation. For example, the 2 S5 written as ✓ ◆ 1 2 3 4 5 = 4 2 1 5 3 permutes 1 to 4; 2 to 2; 3 to 1; 4 to 5 and 5 to 3. Another way of thinking of elements of Sn is to think of 2 Sn as a function : {1, 2,... , n} ! {1, 2,... , n} where (i ) is the number that permutes i to. So for the two-rotation example above we have (1) = 4, (2) = 2, (3) = 1, (4) = 5, (5) = 3. We will mostly use cycle notation. In the above we have 1 goes to 4, 4 goes to 5, 5 goes 3 and 3 goes to 1. (It goes round in a cycle – hence the name!) We write this as (1453). The cycle ‘2 goes to 2’ can be written as (2) or omitted when written as part of the full permutation. That is, above is written as = (1453)(2) or = (1453). Cycle notation is not unique in the sense that (1453) = (5314), (123) = (231), etc. But note that (123) and (321) are not the same. The set Sn can be given a group structure by defining multiplication as com- position when describing elements as a function. However, we will mostly use cycle notation and so being able to multiply cycles without thinking of them as functions is important. For example, let , ⌧ 2 S6 be given by = (13)(265) and ⌧ (146)(253). We multiply and ⌧ by writing ⌧ and starting from the right work out what number permutes to which number. (This makes sense if you think of the elements as functions.) Thus, ⌧ = (13)(265)(146)(253) = (145)(2)(36) = (145)(36) 8 CHAPTER 1. BASIC NOTIONS AND REVISION 1 (12) (13) (23) (123) (132) 1 1 (12) (13) (23) (123) (132) (12) (12) 1 (132) (123) (23) (13) (13) (13) (123) 1 (132) (12) (23) (23) (23) (132) (123) 1 (13) (12) (123) (123) (13) (23) (12) (132) 1 (132) (132) (23) (12) (13) 1 (123) Table 1.2: Cayley Table for S3. The identity is given as 1 to make it easier to find inverses. and ⌧ = (146)(253)(13)(265) = (12)(246)(5) = (12)(346). Note that Sn is not Abelian as this example shows that ⌧ 6= ⌧. We call a k-cycle if it consists of k distinct numbers in a cycle, i.e., = (a1 a2 · · · ak ) where ai 6= ai for i 6= j. For example, (132) is a 3-cycle, (3741) is a 4-cycle and (12)(34) is not a k-cycle for any k. The identity of Sn is just the identity permutation. We write this as Id or (1). Inverses are easy to handle. If = (abcd), then the inverse is (dcba). For example, (14253) 1 = (35241) as is easily checked. For a composition of cycles, then we just use the (xy ) 1 = y 1 x 1 rule. For example, 1 1 1 ((13)(462)) = (462) (13) = (264)(31). Examples 1.11 (i) For S3 we have S3 = {(1)(2)(3), (12)(3), (13)(2), (1)(23), (123), (132)} = {(1), (12), (13), (23), (123), (132)}. The Cayley table for S3 is given in Table 1.2. (ii) Elements of S4 are given in the first column of Table 1.3. Two cycles 1 and 2 are called disjoint cycles if they do not contain any numbers the same. For example, (132) and (74) are disjoint but (132) and (375) are not disjoint as they both contain 3. Suppose that can be written as a product of cycles 1 2 3... r where the i are pairwise disjoint and the length of i  j for all i  j. We define the cycle type of a permutation to be the r -tuple (k1 , k2 , k3 ,... kr ) where ki is the length of the cycle i. For example, (12)(34) 2 S5 is of type (1, 2, 2). Where this does not cause confusion we shall drop the 1’s so that (12)(34) 2 S5 is of type (2, 2). The  Elements of S4 Order cycle type Even/odd In A4 Cube rotation Tet rotation e = (1)(2)(3)(4) 1 (1, 1, 1, 1) even X Identity Identity (12) (23) (34) (13) (24) 2 (1, 1, 2) odd 180 through opposite edges midpoints (14) (12)(34) (13)(24) (14)(23) 2 (2,2) even X 180 through opposite faces midpoints Edge midpoints axis (123) (124) (134) (234) 3 (1,3) even X 120 through opposite corners Vertex to Face centre axis (132) (142) (143) (243) (1234) (1243) (1324) 4 (4) odd 90 through opposite faces midpoints (1432) (1342) (1423) Table 1.3: Elements of S4 and A4. The order 2 elements are self-inverses. The inverses for the cycle types (1, 3) and (4) are positioned vertically, e.g., the inverse of (123), i.e., (132), is below it in the table. 1.3. SYMMETRIC AND ALTERNATING GROUPS, SN AND AN 9 10 CHAPTER 1. BASIC NOTIONS AND REVISION 1 (12)(34) (13)(24) (14)(23) (123) (243) (142) (134) (132) (143) (234) (124) 1 1 (12)(34) (13)(24) (14)(23) (123) (243) (142) (134) (132) (143) (234) (124) (12)(34) (12)(34) 1 (14)(23) (13)(24) (243) (123) (134) (142) (143) (132) (124) (234) (13)(24) (13)(24) (14)(23) 1 (12)(34) (142) (134) (123) (243) (234) (124) (132) (143) (14)(23) (14)(23) (13)(24) (12)(34) 1 (134) (142) (243) (123) (124) (234) (143) (132) (123) (123) (134) (243) (142) (132) (124) (143) (234) 1 (14)(23) (12)(34) (13)(24) (243) (243) (142) (123) (134) (143) (234) (132) (124) (12)(34) (13)(24) 1 (14)(23) (142) (142) (243) (134) (123) (234) (143) (124) (132) (13)(24) (12)(34) (14)(23) 1 (134) (134) (123) (142) (243) (124) (132) (234) (143) (14)(23) 1 (13)(24) (12)(34) (132) (132) (234) (124) (143) 1 (13)(24) (14)(23) (12)(34) (123) (142) (134) (243) (143) (143) (124) (234) (132) (12)(34) (14)(23) (13)(24) 1 (243) (134) (142) (123) (234) (234) (132) (143) (124) (13)(24) 1 (12)(34) (14)(23) (142) (123) (243) (134) (124) (124) (143) (132) (234) (14)(23) (12)(34) 1 (13)(24) (134) (243) (123) (142) Table 1.4: Cayley Table for A4 identity in S4 is of type (1, 1, 1, 1). The cycles types for the elements of S4 are given in Table 1.3. Cycles of the form (ab) for some a and b are called transpositions. We can write any cycle as a product of transpositions. For example, in S5 , (1354) = (14)(15)(13) and (234) = (24)(23). Every k-cycle can be written as a product of transpositions like so: (a1 a2 a3... ak ) = (a1 ak )... (a1 a3 ) (a1 a2 ). We can then write any product of cycles as a product of transpositions, e.g., (1354)(234) = (14)(15)(13)(24)(23). Of course, as you can see this can be simplified to get a smaller number of transpositions. However, the parity of the number of transpositions remains the same and so we introduce even and odd transpositions. If can be written as an even number of transpositions, then we say is an even permutation. If can be written as an odd number of transpositions, then we say is an odd permutation. So for example (1354) is odd and (234) is even. Every permutation is either odd or even, and not both. The sign of a per- mutation is 1 if the permutation is even and 1 if the permutation is odd. In particular, the sign of a permutation is always in { 1, 1}. As k-cycle has sign( ) = ( 1)k 1. The sign of a product can then be calculated from the product of signs. Definition 1.12 The alternating group of degree n, denoted An , is the set of all even permutations. Exercise 1.13 Prove that this is a group. We have |An | = n!/2. The Cayley table for A4 is given in Table 1.4. We can describe the symmetry groups of the cube and the tetrahedron using S4 and A4 respectively. 1.4. SUBGROUPS 11 For the cube imagine four axes going through the four pairs of opposite cor- ners. Label these 1 to 4. Then every rotation can be described by how it permutes the numbers 1 to 4. The descriptions are listed in a column of Table 1.3. Con- versely, every permutation in S4 gives a rotation of the cube. Now take the tetrahedron and this time label the vertices 1 to 4. In this case the rotations can be described by elements of S4. In contrast to the cube, not every permutation in S4 gives a rotation. In this case only the elements of A4 give rise to rotations of the tetrahedron. See the final column of Table 1.3. 1.4 Subgroups The set An is a subset of Sn which is a group in its own right. This leads us to the definition of a subgroup. Definition 1.14 Let G be a group. A subset H of G is called a subgroup of G if H is a group using the same binary operation as G. Examples 1.15 (i) {e} and G are always subgroups of G. (ii) An is subgroup of Sn. (iii) (R+ , ⇥) is a subgroup of (R⇤ , ⇥), where R⇤ = R\{0} and R+ is the set of positive real numbers. (iv) The dihedral group is a very important groups and can be interpreted as a subgroup of a symmetric group. (Cayley’s Theorem from MATH2022 tells us that this holds for all groups but this is particularly simple to see.) Recall that the dihedral group of order 2n is generated by two elements a rotation r and flip s associated to a n-regular polygon. The group is D2n = {e, r, r 2 ,... , r n 1 , s, r s, r 2 s,... , r n 1 s} and we have r n = s 2 = (sr )2 = 1. (As s 2 = 1 and (sr )2 = 1 we also get the useful relation sr s 1 = r 1.) We can represent D8 using symmetries of a square. This is illustrated in Figure 1.1. Let r = (1234) and s = (13) (or equivalently, s = (13)(2)(4)). We calculate s 2 = (13)(13) = (1)(3) = Id r 2 = (1234)(1234) = (13)(24) r 3 = (13)(24)(1234) = (1432) r 4 = (1432)(1234) = (1)(2)(3)(4) = Id r s = (1234)(13) = (14)(23) r 2 s = (1234)(14)(23) = (1)(24)(3) = (24) r 3 s = (1234)(24) = (12)(34). 12 CHAPTER 1. BASIC NOTIONS AND REVISION Figure 1.1: Symmetries of a square and the dihedral group D8 Thus, D8 = {Id, (13), (24), (13)(24), (14)(23), (12)(43), (1234), (1432)}. Therefore, we can view D8 as a subgroup of S4. An important result is Lagrange’s Theorem which states that H is a subgroup of the finite group G, then |H| divides |G|. In fact, we will do a little better than this when we state the theorem precisely (Theorem 2.9). Example 1.16 Let us list the subgroups of A4. This has 12 elements and these are listed in the description of elements of S4 in Table 1.3. The Cayley Table is given in Table 1.4. By Lagrange’s Theorem we know that the order of any subgroup divides 12 and hence the only possibilities are subgroups of order 1, 2, 3, 4, 6, and 12. We shall show that all possibilities occur with the exception of 6. In other words, there is no (straightforward) converse to Lagrange’s Theorem, i.e., given n divides |G| for a group, there does not necessarily exist a subgroup of order n. First, {e} = {Id} = {(1)(2)(3)(4)} and A4 are subgroups. Next, every in- dividual element will generate a subgroup given by powers of that element. (In MATH2022 we said that the group was cyclic). The orders of this group are given by the orders of the elements. These can be found in Table 1.3. Let H be a subgroup of G and let us consider what happens when we have di↵erent types of cycles in the group. From Table 1.3 we can see that we can have cycle types (1, 1, 1, 1) (i.e., the identity), (2, 2), and (1, 3) (i.e., the 3-cycles). First, suppose that H is a subgroup of G such that H contains two 3-cycles, e.g., (123) and (134). Then their inverses (132) and (143) must also be in H if H is a subgroup. Since (123)(134) = (234) we also get this element which is not equal to any of the other 3-cycles or their inverses. So now we have 3 non-trivial elements and their inverses plus the identity in H. This means that |H| > 7. 1.4. SUBGROUPS 13 ⇤ (1) (12)(34) (13)(24) (14)(23) (1) (1) (12)(34) (13)(24) (14)(23) (12)(34) (12)(34) (1) (14)(23) (13)(24) (13)(24) (13)(24) (14)(23) (1) (12)(34) (14)(23) (14)(23) (13)(24) (12)(34) (1) Table 1.5: Cayley Table for V , the Klein Four-Group Lagrange’s Theorem tells us that |H| divides G and so we must have |H| = |G|, i.e., H = G. This reasoning holds for any two distinct 3-cycles. Now suppose we have two (2, 2)-cycles for example (12)(34) and (13)(42). Then, (12)(34)(13)(24) = (14)(23) and we can see that this means that all three (2, 2) cycles are in H. This holds for any pair of (2, 2)-cycles. Hence, from the Cayley table for A4 , Table 1.4, we can see see that V = {Id, (12)(34), (13)(24), (14)(23)} generates a subgroup. The Cayley table for this is given in Table 1.5. As we can see this is the same table as the one for the Klein Four-Group.. Hence, this subgroup is the same as the Klein Four-Group. [Aside: Note that this description also gives us that the Four-Group V can be viewed as a subgroup of D8 (which is not a subgroup of A4 – the order doesn’t divide 12 – but of S4 ). So if r = (1234) and s = (13) we have V = {Id, (12)(34), (13)(24), (14)(23)} = {Id, r 3 s, r 2 , r s}. 2 Note that each non-trivial element has order 2. (Use s 2 = Id and r 2 = r 4 = Id.) End of aside.] Now suppose H contains a 3-cycle and a (2, 2)-cycle, for example (123) and (12)(34). Then, (123)(12)(34) = (134). So, H contains two 3-cycles and, by the reasoning above, we have H = G. This exhausts all possibilities for subgroups of A4. Note, as we claimed earlier, that A4 has no subgroup of order 6. It can be tedious to check the axioms for a subset to be a subgroup. The following proposition gives some simpler conditions instead. These are very useful. Part (ii) should be your go-to method for showing a subset is a subgroup. Proposition 1.17 Let G be a group and H be a subset of G. Then, the following are equivalent. (i) H is a subgroup of G. (ii) The following are true: (a) H 6= ;, (b) xy 1 2 H for all x, y 2 H. 14 CHAPTER 1. BASIC NOTIONS AND REVISION (iii) The following are true: (a) e 2 H, (b) xy 2 H for all x, y 2 H, (c) x 1 2 H for all x 2 H. Proof. (i) =) (ii) is true by the properties of a group. (ii) =) (iii) If (ii)(a) is true, then there exists x 2 H. By (ii)(b) we get e = xx 1 2 H, i.e., (iii)(a) holds. As we now know that e 2 H, from (ii)(b) we get y 1 = ey 1 2 H for all y 2 H. So (iii)(c) holds. As y 1 2 H when y is, then xy = x(y 1 ) 1 2 H by (ii)(b), so (iii)(b) holds. (iii) =) (i): The subset inherits the property of associativity from G. The operation is closed by (iii)(b). The inverses are in H by (iii)(c). Since e 2 H, there exists an identity in H as he = e = eh for all h 2 H as G is a group. ⇤ Examples 1.18 (i) Let K be a field. The special linear group of degree n, denoted SLn (K), is the set of n ⇥ n matrices with entries in the field K with determinant equal to 1. We shall show that SLn (K) is a subgroup of the general linear group of degree n, GLn (K). The field contains elements 0 and 1 which we can use to construct an identity matrix, i.e., we have 1s down the diagonal and 0s elsewhere. This matrix has determinant 1 so SLn (K) 6= ;. If A, B 2 SLn (K), then 1 1 1 det(AB ) = det(A) det(B) =1·1 = 1. Hence, AB 1 2 SLn (K) and so SLn (K) is a subgroup of GLn (K). (ii) Elements of O(n) are real orthogonal matrices and so det(X) = ±1 for all X 2 O(n). We define the special orthogonal group of degree n, denoted SO(n), to be the subset of O(n) defined by SO(n) = {X 2 O(n) | det(X) = 1}. This is a subgroup of O(n). We have det (In ) = 1 so In 2 SO(n). If X, Y 2 SO(n), then det(X) = det(Y ) = 1. We have 1 1 det XY = det(X)(det(Y )) =1 so XY 1 2 SO(n). Hence, SO(n) is a subgroup of O(n) (and hence a subgroup of GLn (R)). (iii) The Heisenberg group, denoted by H3 (R), is represented by 3 ⇥ 3 upper triangular matrices of the form 0 1 1 x z @0 1 y A 0 0 1 1.5. CYCLIC GROUPS 15 where x, y , z 2 R. This is a subgroup of GL3 (R). Clearly, if we take x = y = z = 0 (or any real numbers), then we see that H3 (R) 6= ;. 0 1 0 1 1 x z 1 a c For matrices @0 1 y A and @0 1 b A in H3 (R) we have 0 0 1 0 0 1 0 10 1 1 0 10 1 1 x z 1 a c 1 x z 1 a ab c @0 1 y A @0 1 b A = @0 1 y A @0 1 b A 0 0 1 0 0 1 0 0 1 0 0 1 0 1 1 x a ab bx c + z = @ 0 1 y b A 0 0 1 2 H3 (R). Therefore, H3 (R) is a subgroup of GL3 (R). Exercises 1.19 (i) For n 2 N, let Rn = {z 2 (C⇤ , ⇥) | z n = 1}. This is the set of the roots of unity. Show that Rn is a finite subgroup of C⇤. (ii) Let SLn (Z) be the n ⇥ n matrices with integer entries such that det is equal to 1. (Note that this is not an example of SLn (K) as Z is not a field.) Show that SLn (Z) is a subgroup of GLn (R). We call this the special linear group over Z. (iii) Let H and K be subgroups of a group G. Show that H \ K is a subgroup of G. Generalise this to: If H , 2 ⇤, is a (possibly infinite) family of subgroups of G, then show \ 2⇤ H is a a subgroup of G. 1.5 Cyclic Groups Cyclic groups are groups which can be generated by a single element. Recall that hxi = {... , x 3 , x 2 , x 1 , e, x, x 2 , x 3 ,... } and that we say x generates hxi. Definition 1.20 Suppose that G = hxi for some x 2 G. Then, we say that G is a cyclic group. We denote by C(n) the cyclic group of order n. From Proposition 1.8 we know that if x is an element of finite order in a group, then, |hxi| = min{n 2 N | x n = e}. In MATH2022 we saw that if C is non-trivial and cyclic, then there exists x 2 C such that x n = e and C = hxi and n is the smallest such natural number. 16 CHAPTER 1. BASIC NOTIONS AND REVISION We also saw that every subgroup of a cyclic group is cyclic. By Lagrange’s Theorem the order of the element generating the subgroup must divide the order of the element generating the group. In Example 1.16 we saw that there is no subgroup of order 6 for A4 and so the converse to Lagrange’s Theorem is false. However, it is true for cyclic groups! Theorem 1.21 Let C be a finite cyclic group. If k is a positive integer that divides |C|, then C has exactly one subgroup of order k. Proof. If C is trivial or k = 1, then the statement is obvious. Therefore, we may assume that C is non-trivial and k > 1. Suppose C = hxi for some e 6= x 2 C and |C| = n. Given any k we have n = kr for some positive integer r. As n is the smallest positive integer such that x n = e it follows that k is the smallest positive integer such that (x r )k = e. Therefore, hx r i is a subgroup of C of order k. Suppose H is a subgroup of C with |H| = k. Let s be the minimal positive integer such that x s 2 H. As H is cyclic of order k we must have (x s )k = e. Thus, n|sk. n | sk =) sk = nj, for some integer j, =) sk = (kr )j =) s = r j. So, x s = (x r )j 2 hx r i. Hence, H ✓ hx s i ✓ hx r i. But |H| = k = |hx r i| so H = hx r i, i.e., there is only one subgroup of order k. ⇤ 1.6 Equivalence Relations We will have need of equivalence relations throughout this module (we care a lot about their equivalence classes and the fact that they partition sets) and so now is a good time to review the relevant notions related to relations. Definition 1.22 An equivalence relation, denoted ⇠, on a set X, which we always assume to be non-empty, satisfies the following properties. o Reflexivity: For all x 2 X, x ⇠ x. o Symmetry: For all x, y 2 X, if x ⇠ y , then y ⇠ x. o Transitivity: For all x, y , z 2 X, if x ⇠ y and y ⇠ z, then x ⇠ z. For each element x 2 X, the equivalence class of x, denoted [x], is defined to be [x] = {y 2 X | y ⇠ x}. That is, the equivalence class of x is the set of all points equivalent to x. The set of equivalence classes, denoted X/ ⇠, is called the quotient set. 1.6. EQUIVALENCE RELATIONS 17 Examples 1.23 (i) Equivalence relations can help define the rational numbers. If we define rational numbers by saying they are quotients of integers m and n with n 6= 0 in the form m/n, then, even if one can express what is meant by a quotient, that still doesn’t help with the idea that 4/8 should be the same as 1/2. Equivalence relations allow us to overcome these problems. Let X be the set of ordered pairs of integers (m, n) with non-zero n. (So, X = Z ⇥ (Z\{0}).) Define ⇠ on X by (m, n) ⇠ (p, q) if and only if mq = np. (Note that this doesn’t have to introduce anything about mysterious ‘quotients’ – it just uses multiplication of integers.) This is an equivalence relation: o Reflexive: (m, n) ⇠ (m, n) as mn = nm. o Symmetry: Suppose (m, n) ⇠ (p, q). Then, mq = np implies qm = pn so (p, q) ⇠ (m, n). o Transitivity: Suppose (m, n) ⇠ (p, q) and (p, q) ⇠ (r, s). Then, mq = np =) mqr = npr =) mps = npr =) ms = nr, for p 6= 0 so (m, n) ⇠ (r, s). The case p = 0 is straightforward. Define the rational numbers Q to be the equivalence classes of ⇠ and denote the equivalence class of (m, n) with the notation m/n instead of the usual [(m, n)]. In this case (4, 8) ⇠ (1, 2) as 4 · 2 = 8 · 1 so both represent 1/2 as we would hope! That is, [(4, 8)] = [(1, 2)] = 1/2. We can identify the integer m 2 Z with the equivalence classes of (m, 1) and so can view Z as a subset of Q. Thus, we have rigorously defined Q.2 We can then define addition and multiplication on the equivalence classes so that Q is a field. (ii) Let n be a natural number and X = Z. Define ⇠ by x ⇠ y if and only if x ⌘ y mod n. The set of equivalence classes is denoted Zn and is given by Zn = Z/ ⇠= {, , ,... , [n 1]}. We’ve already seen that this is a group. (iii) Let G be a group and H be a subgroup of G. Define the relation ⇠R on G by x ⇠R y () y = hx for some h 2 H. This is an equivalence relation: o Reflexive: As x = ex and e 2 H then x ⇠R x. 2 We can then define the reals R by Dedekind cuts or something similar. From the reals we can define C as the field R[x]/hx 2 + 1i and so on. 18 CHAPTER 1. BASIC NOTIONS AND REVISION o Symmetric: Assume x ⇠R y. Then y = hx for some h 2 H. Then, x = h 1 y and so y ⇠R x as h 1 2 H. o Transitive: Assume x ⇠R y and y ⇠R z. Then y = h1 x and z = h2 y for some h1 , h2 2 H. We have z = h2 y = h2 (h1 x) = (h2 h1 ) x as h2 h1 2 H and so x ⇠R z. Therefore, ⇠R is an equivalence relation. The equivalence classes are called right cosets and [x] is traditionally de- noted Hx as its elements are of the form hx where h runs through all of H. 1.7 An equivalence relation partitions the set The importance of equivalence relations is that they allows us to gather together elements of the set that are equivalent. That is, the equivalence classes are sets of elements that are all equivalent. We can now do our maths on those classes, i.e., do our maths on the quotient set. Hopefully, there are fewer elements in the quotient set and grouping equivalent objects makes things clearer. The technical way of describing this is that an equivalence relation divides the set into subsets that have certain properties. The following will be used repeatedly in the module so take note. You don’t need to know the proof (although that is always helpful!) though some of you will have seen it in MATH1025 Number Systems. You can also find a proof (and an analysis of the proof!) in my book How to Think Like a Mathematician. First let’s give a name to a technical concept. Definition 1.24 Let X be a (non-empty) set and ⇠ and equivalence relation on X. A transversal for ⇠ is the choice of a representative for each equivalence class in X/ ⇠. We can specify a transversal using a subset T ✓ X so that T contains the choices. Examples 1.25 (i) We’ve actually already done this for Zn when we choose 0, 1, 2,... , n 1 to represent the equivalence classes. For example, a transver- sal for Z6 is T = {0, 1, 2, 3, 4, 5} ✓ Z. Also, T = {6, 1, 2, 123, 10, 1} will do. (ii) Let X = R and define x ⇠ y () x = y + 3n, for some n 2 Z. (Note that this is not the same as Z3 as here X = R not X = Z.) Let T = {3 n + 12 | n 2 Z}. This is a transversal. (iii) Consider the equivalence relation we used to define the rationals. Here a transversal would be an infinite set as the rationals are infinite in number and could be given by, for example, each m/n = [(m, n)] is written in its lowest terms, i.e., m and n are coprime. 1.7. AN EQUIVALENCE RELATION PARTITIONS THE SET 19 Figure 1.2: A transversal (iv) Let X = R and define a relation by x ⇠ y if x y 2 Q. It’s easy to show it is an equivalence relation. But how do we choose elements for the equivalence classes. Here we need to use the Axiom of Choice3. We will not be requiring such deep dives into mathematical logic so don’t need to worry about this. One may ask where the name transverse comes from. The word ‘transverse’ in plain English usually means ‘cutting across’ and is often used to describe some- thing cutting across at right angles. If our set X has equivalence classes consisting of lots of curves, then a line cutting across them creates a transversal. See Fig- ure 1.2. This picture can be misleading as not all transversals are constructed by cutting across in this manner. See the example of Z6 above. We now come to the crucial result. An equivalence relation divides up the set, we say it partitions the set, into disjoint pieces. The pieces are made up of equivalent elements. Theorem 1.26 (Partition of set by equivalence classes) Let X be a non-empty set and ⇠ be an equivalence relation on X. Then, for all x, y 2 X, the following are true. [ (i) X = [x]. x2X (ii) Either [x] = [y ] or [x] \ [y ] = ; but not both. (iii) [x] = [y ] () x ⇠ y. (iv) [x] \ [y ] = ; () x 6⇠ y. (v) [x] = [y ] () y 2 [x]. 3 The Axiom of Choice is the seemingly obvious but actually quite deep statement ‘Given any collection of non-empty sets, including infinite collections, we can construct a new set by choosing one element from each original set.’. It’s important but we’ll leave it to the logicians to worry about. 20 CHAPTER 1. BASIC NOTIONS AND REVISION (vi) If X is finite, then there exists a transversal {x1 ,... , xq } ✓ X such that q G X= [xi ]. i=1 Consequently, o X/ ⇠= {[x1 ], [x2 ],... , [xq ]} , q X o |X| = [xi ] , and i=1 o every transversal has the same number of elements. Example 1.27 For the example of Z6 we partition Z into the following equivalence classes (a representative from each is highlighted) {... , 18, 12, 6, 0, 6, 12, 18,... } {... , 17, 11, 5, 1, 7, 13, 19,... } {... , 16, 10, 4, 2, 8, 14, 20,... } {... , 15, 9, 3, 3, 9, 15, 21,... } {... , 14, 8, 2, 4, 10, 16, 22,... } {... , 13, 7, 1, 5, 11, 17, 23,... }. Chapter 2 Quotient Groups Quotient groups are fundamental objects in Group Theory. We met them in MATH2022 but will go over them again in a slightly di↵erent way. This will prepare us for the notion of orbit when we do group actions. The basic idea behind a quotient group is that we take a special type of subgroup H of a group G and create a new group where the group structure is inherited from G and where the subgroup H is the identity element. This has the e↵ect of allowing us to study G with the e↵ect of H removed. This will rely heavily on the partition of G by right cosets. 2.1 Right cosets Let H be a subgroup of the group G. Recall the relation ⇠R : x ⇠R y () y = hx for some h 2 H. This is an equivalence relation (Examples 1.23), the equivalence class [x] is de- noted Hx, (i.e., [x] = Hx), and the quotient set is G/ ⇠R. (This will be the quotient group G/H under special conditions.) Examples 2.1 (i) Consider C12 , the cyclic group with generator x of order 12, the cyclic subgroup H = e, x 4 , x 8 generated by x 4 has four di↵erent right cosets He = 1, x 4 , x 8 , Hx = x 1 , x 5 , x 9 , Hx 2 = x 2 , x 6 , x 10 , Hx 3 = x 3 , x 7 , x 11. We can then calculate that Hx 4 = Hx 8 = He, Hx 5 = Hx 9 = Hx, Hx 6 = Hx 10 = Hx 2 , Hx 7 = Hx 11 = Hx 3. 21 22 CHAPTER 2. QUOTIENT GROUPS That we have distinct sets and all those equalities follows from ⇠R being an equivalence relation and hence gives a partition. This will be made explicit in the next theorem. (ii) We always have He = H: For all h 2 H, he = h so h 2 He. Conversely, if x 2 He, then x = he for some h 2 H. But as he = h, this means x = h, i.e., x 2 H. Exercise 2.2 Consider S3 and its subgroup H = { Id, (12)}. Show that there are three distinct cosets H = H Id = H(12), H(23) = H(123), H(13) = H(132). Theorem 2.3 Let H be a subgroup of the group G. The set of right cosets, X/ ⇠R partitions G. That is, for all x, y 2 G, the following are true. [ (i) G = Hx. x2G (ii) Either Hx = Hy or Hx \ Hy = ; but not both. (iii) Hx = Hy () y = hx for some h 2 H. (iv) Hx \ Hy = ; () y 6= hx for all h 2 H. (v) Hx = Hy () y 2 Hx. (vi) If G is finite, then there exists a transversal {x1 ,... , xq } ✓ G such that q G G= Hxi. i=1 Consequently, o G/ ⇠R = {Hx1 , Hx2 ,... , Hxq } , q X o |G| = Hxi , and i=1 o every transversal has the same number of elements. Proof. This is just an application of Theorem 1.26 where X = G and [x] = Hx. The partition is pictured in Figure 2.1. ⇤ This may seem like a lot to learn but all you really need to learn is Theo- rem 1.26. The statements follow simply from the statements there! However, the theorem also gives you a helpful summary of the elementary properties of right cosets that you can refer to if you ever find yourself confused as to why something about a right coset is true in a proof. In addition to these properties arising from the fact that ⇠R is an equivalence relation we also have some extra properties arising from the nature of the relation. 2.1. RIGHT COSETS 23 Figure 2.1: Right cosets give a partition of G and |Hx| = |H| Corollary 2.4 We have (i) Hx = Hy () xy 1 2 H, (ii) Hx = H () x 2 H. Proof. (i) This is because, by Theorem 2.3(iii), Hx = Hy () y = hx for some h 2 H 1 1 () xy =h for some h 2 H 1 () xy 2 H. (ii) As He = H we just take y = e in (i) of this corollary. ⇤ Definition 2.5 The cardinality of the set G/ ⇠R is called the index of H in G and is denoted [G : H]. If this is finite, then we say that H has finite index in G. Examples 2.6 (i) Let G = O(n) be the orthogonal group and H = SO(n) the special orthogonal group defined in Examples 1.18. For X 2 O(n) we have det(X) = ±1 and (by definition) have X 2 H () det(X) = 1. We have 1 HX = HY () XY 2H 1 () det XY =1 1 () det(X) det(Y ) =1 () det(X) = det(Y ) Thus as det(x) = ±1 for X 2 O(n), we have two cosets, G/ ⇠R = {SO(n), O(n)\SO(n)}. Thus, [G : H] = [O(n) : SO(n)] = 2. 24 CHAPTER 2. QUOTIENT GROUPS (ii) Let’s do an example of cosets where the group is using the addition notation. In this case we write H + x for the right coset Hx. Consider G = (Z, +) and H = (2Z, +). Let’s consider a representative of of each class. Since 0 is the identity in Z we have 2Z + 0 = 2Z as one right coset. In other words, one coset is the even numbers. Consider now 2Z + 1. Then, 2Z + y = 2Z + 1 if and only if y = h + 1 for some h 2 2Z. In other words, y 2 2Z + 1 if and only if y is odd. We’ve got the even numbers and the odd numbers so we’ve got all members of Z. Since the right cosets partition Z we must have all right cosets. Therefore, G/ ⇠R = {2Z + 0, 2Z + 1} = {{Even numbers}, {Odd numbers}} = {2Z, Z\2Z}. Therefore, [G : H] = [Z : 2Z] = 2. Note that the set {0, 1} is a transversal. Transversals are not unique and we can pick any even and odd pair as a transversal. That is {35 167, 5} would do just as well as a transversal. More generally, [Z : nZ] = n for n 2 N. (This is a straightforward exercise. Don’t overcomplicate it!) (iii) Let G = (C⇤ , ⇥) and H be the nth roots of unity for n 2, so H = 2 {1, !, ! ,... , ! n 1 ⇤ } for some ! 2 C such that ! = 1. This is a subgroup n of G. Let x, y 2 C⇤. Then, Hx = Hy if and only if y = hx for some h 2 H. For any h 2 H we have |h| = 1 and so |y | = |hx| = |h||x| = |x|. In other words, x and y are the same distance from the origin – they lie on the same circle centred at the origin. Looking at the arguments1 of the complex numbers we see that arg(y ) = arg(hx) = arg(x) + arg(h). The argument of any h 2 H is a multiple of 2⇡/n. Thus, x and y are equivalent if they lie on the same circle (centred at the origin) and are some multiple of 2⇡/n apart. In particular, every right coset (i.e., an equivalence class) is a finite set of points, all equally spaced on the same circle. Therefore, there are an infinite number of right cosets and so [G : H] = [C⇤ : H] = 1. Recall that the argument of z 2 C is just the angle between the ray from 0 to z with the 1 positive x-axis. If z = 0, then we will let arg(z) = 0. 2.1. RIGHT COSETS 25 Exercise 2.7 Let G be a group and H is a subgroup. Show the following. (i) If H = {e}, then G/ ⇠R = { {g} | g 2 G} and so [G : H] = |G|. (ii) If H = G, then G/ ⇠R = {G} and so [G : H] = 1. Lemma 2.8 For all x 2 G, we have |Hx| = |H|. Proof. This was proved in MATH2022. You may like to try to prove it yourself before you look up the proof there or online. ⇤ We can deduce a familiar result from the preceding! Theorem 2.9 (Lagrange’s Theorem) Let H be a subgroup of the finite group G. Then, |G| = [G : H] · |H|. In particular, |H| divides |G| and [G : H] (i.e., the index of H in G) divides |G| too. Proof. By Theorem 2.3 there is a transversal for the quotient set [G : H] given by {x1 ,... xq } for some xi 2 G where q = [G : H]. We have q X |G| = Hxi i=1 Xq = |H|, by Lemma 2.8, i=1 = q|H| = [G : H] · |H|, as q = [G : H], by definition. ⇤ Remark 2.10 Lagrange’s Theorem helps explains the use of colon in [G : H]. From the theorem we have [G : H] = |G|/|H| so [G : H] is in some sense a quotient number. Now, consider the notation for ratio. The notation A : B representing a ratio for the numbers A and B is equivalent to the quotient A/B. E.g., a ratio of 3 : 2 is equivalent to the number 3/2. Examples 2.11 (i) Groups of prime order are cyclic groups. Suppose that G is of prime order p, then consider an element g that is not the identity. This generates a subgroup that is not the trivial group and its order must divide p. Hence, the order of the subgroup is p. In other words, G has an element of order p that generates G. Hence, G is cyclic. (ii) The alternating group An is a subgroup (of order n!/2) in the group Sn (which of order n!). Therefore, |Sn | n! [Sn : An ] = = = 2. |An | n!/2 That is, the index of An in Sn is 2. 26 CHAPTER 2. QUOTIENT GROUPS (iii) Note that if k divides |G|, then it does not mean there is a subgroup of G of order k. We saw that there is a subgroup in the case of a cyclic group (Theorem 1.21) but this is exceptional. For example, 6 divides 12 but we see from Example 1.16 that A4 has no subgroup of order 6. Investigating when subgroups exist will be a central concern of the latter part of the module. (iv) If H and K are subgroups of G such that |H| and |K| are coprime, then H \ K = {e}. This follows from the fact that H \ K is a subgroup of both H and K so by Lagrange’s Theorem |H \ K| must divide both |H| and |K|. Since those numbers are coprime, then |H \ K| = 1, i.e., H \ K = {e}. 2.2 Left cosets We can do all the preceding but for left cosets rather than right cosets. (Try writing out the equivalent of Theorem 2.3.) That is, for a subgroup H of a group G and using the equivalence relation x ⇠L y if and only if y = xh for some h 2 H, the sets xH = {xh | h 2 H} form a partition of G. We denote the set of left cosets with G/ ⇠L. We can prove Lagrange’s Theorem (again!) using left cosets. Since the proofs are nearly identical we shall not write them down. One consequence of the theorem is that the number of left cosets is equal to the number of right cosets, i.e., equal to [G : H]. However, an important point to note is that, in general, xH 6= Hx and so G/ ⇠R and G/ ⇠L are generally di↵erent. Subgroups where we have equality of left and right cosets for all x 2 G are called normal and will be investigated shortly. 2.3 Conjugation of subgroups Before we formally define normal subgroups we will define the conjugation of a subgroup. This is not the same as conjugation of elements. Definition 2.12 Let H be a subgroup of the group G. For any g 2 G we can define a subset of G, denoted gHg 1 , by gHg 1 = ghg 1 | h 2 H. This is the conjugation of H by g. Example 2.13 Recall that the dihedral group D8 can be viewed as a subgroup of S4 : D8 = {Id, (13), (24), (13)(24), (14)(23), (12)(43), (1234), (1432)} We have D2n = {e, r, r 2 ,... , r n 1 , s, r s, r 2 s,... , r n 1 s} and so D8 = {e, r, r 2 , r 3 , s, r s, r 2 s, r 3 s}. 2.3. CONJUGATION OF SUBGROUPS 27 Let H = hsi and g = r. Then, 1 1 1 1 r Hr = r {e, s}r = {r er , r sr }. It is easy to see that r er 1 = e. What about r sr 1? We can calculate that 1 r sr = (1234)(13)(1432) = (1)(24)(3) = r 2 s. Hence, 1 r Hr = {e, r 2 s} = hr 2 si = 6 H. Now consider consider the Klein Four-Group. Let H = V = {Id, r 3 s, r 2 , r s} again let’s take g = r. This gives 1 rV r = r e, r 3 s, r 2 , r s r 1 1 = r er , r r 3 sr 1 , r r 2r 1 , r r sr 1 1 = e, sr , r 2 , r 2 sr 1. But, 1 sr = (13)(1432) = (14)(23) = r s and so r 2 sr 1 = r 3 s. Thus, 1 rV r = e, r s, r 2 , r 3 s = V. Hence, we have examples where gHg 1 = H and gHg 1 6= H. Theorem 2.14 Let H be a subgroup of the group G. Then, gHg 1 is a subgroup of G. Proof. As e 2 H we have geg 1 2 gHg 1 but geg 1 = e so e 2 gHg 1. Suppose x, y 2 gHg 1. Then, x = gh1 g 1 and y = gh2 g 1 for some h1 , h2 2 H. Then, 1 1 1 1 xy = gh1 g gh2 g 1 1 1 = gh1 g (g ) h2 1 g 1 1 = gh1 g gh2 1 g 1 = gh1 h2 1 g 1 1 2 gHg as H is a subgroup of G. Thus, gHg 1 is a subgroup of G. ⇤ Definition 2.15 We say that subgroups H and K of G are conjugate if there exists g 2 G such that K = gHg 1. Exercise 2.16 Show that the relation on subgroups given by H ⇠ K if H and K are conjugate subgroups is an equivalence relation. 28 CHAPTER 2. QUOTIENT GROUPS 2.4 Normal groups Recall the following definition of normal group. It is a crucial ingredient in defining quotient groups. Definition 2.17 Let H be a subgroup of the group G. We say that H is a normal subgroup in G if the left and right cosets of g coincide for every g in G. That is, gH = Hg for all g 2 G. Theorem 2.18 Let H be a subgroup of the group G. The following are equivalent. (i) H is normal. (ii) gH = Hg for all g 2 G. (iii) ghg 1 2 H for all h 2 H and for all g 2 G. (iv) gHg 1 = H for all g 2 G. Proof. The equivalence of (i), (ii) and (iii) was given in MATH2022. We can show the equivalence of (ii) and (iv): 1 1 1 gH = Hg () gHg = Hgg () gHg = He = H. ⇤ Examples 2.19 (i) If G is Abelian, then all subgroups of G are normal. To see this we just use that for all h 2 H and g 2 G we have gh = hg as G is Abelian, so ghg 1 = hgg 1 = h 2 H. (ii) We found the subgroups of A4 in Example 1.16. The group V = {(Id), (12)(34), (13)(24), (14)(23)} is a normal subgroup of the alternating group A4. We obviously have (Id)V = {(1), (12)(34), (13)(24), (14)(23)} = V (Id) and we can (quickly ) calculate that (123)V = {(123), (134), (243), (142)} = V (123) and (132)V = {(132), (234), (124), (143)} = V (132). We can see that this accounts for all 12 elements of A4 (because we have a partition) and hence all possible left and right cosets. Since the left and right cosets are equal we deduce that V is normal in A4. In fact, it is the only proper non-trivial normal subgroup in A4 (i.e., it isn’t {Id} or V itself). 2.4. NORMAL GROUPS 29 (iii) The alternating group A4 has a number of cyclic subgroups: three arising from (2, 2)-cycles, h(12)(34)i, h(13)(24)i, and h(14)(23)i plus four arising from h(abc)i for distinct a, b, c 2 {1, 2, 3, 4}. For the latter we can see that 1 (abd)(abc)(abd) = (abd)(abc)(adb) = (a)(bdc) 2 / h(abc)i as powers of habci only include a, b, c in their representations while (bdc) obviously contains a d. Thus, these cyclic subgroups of A4 are not normal in A4. Similarly, 1 (123)h(12)(34)i(123) = h(14)(23)i = 6 h(12)(34)i 1 (123)h(13)(24)i(123) = h(12)(34)i = 6 h(13)(24)i 1 (123)h(14)(23)i(123) = h(13)(24)i = 6 h(14)(23)i Hence, the other three cyclic subgroups are not normal either. (iv) The relation H in normal is G is not transitive. As a counterexample to the relation being transitive consider o h(12)(34)i is normal in V , o V is normal in A4 , o but h(12)(34)i is not normal in A4. The first fact follows from V being Abelian and so any subgroup is normal in it. The second two facts were shown in the previous examples. This non-transitivity means we have to be careful about specifying which group H is normal in. Exercise 2.20 Show that {id, (123), (132)} is normal in S3 but {id, (ab)}, for a, b distinct, are not normal in S3. Theorem 2.21 If H is a subgroup of the group G and [G : H] = 2, then H is a normal subgroup of G. Proof. Since [G : H] = 2, G can be partitioned into two distinct left cosets of H. One of these must be H and, consequently, the other is G\H. Let g 2 G. We have g 2 H if and only if Hg = He = H by Corollary 2.4(ii) so ⇢ H, if g 2 H, Hg = G\H, if g 2 G\H. A similar argument gives ⇢ H, if g 2 H, gH = G\H, if g 2 G\H. Thus, gH = Hg for all g 2 G and so H is normal in G. ⇤ 30 CHAPTER 2. QUOTIENT GROUPS Examples 2.22 (i) The alternating group An is normal in the symmetric group Sn as [Sn : An ] = 2 by Examples 2.11. (ii) Consider the special orthogonal group SO(n) in the orthogonal group O(n). Examples 2.6 tells us that the quotient set is {SO(n), O(n)\SO(n)}. That is, [O(n) : SO(n)] = 2 and so SO(n) is normal in O(n). Proposition 2.23 Let H and K be normal subgroups of G. Then, H \K is normal in G. Proof. Suppose that x 2 H \ K. Then, by Theorem 2.18, we want to show that gxg 1 2 H \ K for all g 2 G. Let g 2 G. As x 2 H \ K, then x 2 H and so we have 1 gxg 2 H as H is normal in G, and since x 2 K we have 1 gxg 2 K as K is normal in G. Thus, gxg 1 2 H \ K for all g 2 G. Therefore, H \ K is normal in G. ⇤ 2.5 Quotient groups We are now in a position to define quotient groups. Theorem 2.24 Suppose that H is a normal subgroup of the group G. Denote the quotient G/ ⇠R by G/H. We can define a group structure on G/H using the binary operation Hx ⇤ Hy = H(xy ). In this case we call G/H a quotient group and say G/H is the quotient of G by H. The identity in G/H is the right coset He and the inverse of Hx is Hx 1. Proof. First we need to show that the map is well-defined. It is defined for the right H-cosets by taking a representative from the coset and so maybe the mapping we have defined depends on the representative chosen. We have to show that this is not the case. ⇤ is well-defined: We need to show that if Hu = Hx and Hv = Hy , then H(uv ) = H(xy ). Well, these assumptions imply, by Theorem 2.3(iii), that there are h1 , h2 2 H such that x = h1 u and y = h2 v. Then, xy = h1 uh2 v = h1 h3 uv , for some h3 2 H as uH = Hu since H is normal. Thus, H(xy ) = H(uv ) and so we conclude that ⇤ is well-defined. 2.5. QUOTIENT GROUPS 31 To show ⇤ is associative we use the following: Hx⇤(Hy ⇤ Hz) = Hx⇤H(y z) = Hx(y z) = H(xy )z = (H(xy ))⇤Hz = (Hx ⇤ Hy )⇤Hz. The identity and inverse are easy to show: He ⇤ Hx = H(ex) = Hx = H(xe) = Hx ⇤ He and 1 1 1 1 Hx ⇤ Hx = Hxx = He = Hx x = Hx ⇤ Hx. ⇤ Examples 2.25 (i) Let’s have another look at Zn. Let G = Z and H = nZ for some n 2 N. Then, (as sets) Z/nZ = {, , ,... , [n 1]}. This inherits a group structure via [x] ⇤ [y ] = [x + y ]. We call the resulting group the integers modulo n and denote it Zn. In this case we denote ⇤ with +, i.e., we have (the rather trivial looking but isn’t) [x] + [y ] = [x + y ]. Some authors use +n to denote this binary operation. (ii) Consider again the special orthogonal group SO(n) in the orthogonal group O(n). For this we have that O(n)/SO(n) = {SO(n), O(n)\SO(n)}. This has two elements and so O(n)/SO(n) is a cyclic group of order 2 generated by the non-trivial element O(n)\SO(n). (iii) Consider the group S3 and the subgroup H = {Id, (123), (132)}. This is a normal subgroup by Exercise 2.20 and we can take the quotient. There are two right cosets H Id = {Id, (123), (132)} and H(12) = {(12), (13), (23)} and so the quotient has two elements. In Table 2.1 elements in the first coset are coloured white and those in the second coset are coloured grey. Hopefully it can be seen that this colouring is essentially giving the Cayley table for Z2. (iv) Attempting to quotient S3 by the subgroup H = {Id, (12)} does not lead to a group as the operation Hx ⇤ Hy = H(xy ) is not well-defined. We can see this in Table 2.2. There are three right cosets {Id, (12)}, {(13), (132)} and {(123), (23)}. The elements for these in the Cayley table for S3 are coloured the same. As can be seen in the table H(13) = H(132) but H(13) ⇤ H(13) = H(Id), H(13) ⇤ H(132) = H(23) 6= H(Id). 32 CHAPTER 2. QUOTIENT GROUPS 1 (123) (132) (12) (13) (23) 1 1 (123) (132) (12) (13) (23) (123) (123) (132) 1 (13) (23) (12) (132) (132) 1 (123) (23) (12) (13) (12) (12) (23) (13) 1 (132) (123) (13) (13) (12) (23) (123) 1 (132) (23) (23) (13) (12) (132) (123) 1 Table 2.1: The quotient of S3 by the normal (and cyclic) subgroup H = {Id, (123), (132)}. The binary operation ⇤ is well-defined when the subgroup is normal. The Cayley Table for resulting group S3 / ⇠R = S3 /H can be seen in the colouring of entries. Entries in the same coset are coloured the same. 1 (12) (13) (132) (123) (23) 1 1 (12) (13) (132) (123) (23) (12) (12) 1 (132) (13) (23) (123) (13) (13) (123) 1 (23) (12) (132) (132) (132) (23) (12) (123) 1 (13) (123) (123) (13) (23) 1 (132) (12) (23) (23) (132) (123) (12) (13) 1 Table 2.2: The binary operation is not well-defined when the subgroup is not normal. Here S3 has the subgroup H = {Id, (12)} but the right cosets of the quotient S3 / ⇠R do not form a group as the map is not well-defined. This is because H is not normal by Exercise 2.20. 2.5. QUOTIENT GROUPS 33 1 (12)(34) (13)(24) (14)(23) (123) (243) (142) (134) (132) (143) (234) (124) 1 1 (12)(34) (13)(24) (14)(23) (123) (243) (142) (134) (132) (143) (234) (124) (12)(34) (12)(34) 1 (14)(23) (13)(24) (243) (123) (134) (142) (143) (132) (124) (234) (13)(24) (13)(24) (14)(23) 1 (12)(34) (142) (134) (123) (243) (234) (124) (132) (143) (14)(23) (14)(23) (13)(24) (12)(34) 1 (134) (142) (243) (123) (124) (234) (143) (132) (123) (123) (134) (243) (142) (132) (124) (143) (234) 1 (14)(23) (12)(34) (13)(24) (243) (243) (142) (123) (134) (143) (234) (132) (124) (12)(34) (13)(24) 1 (14)(23) (142) (142) (243) (134) (123) (234) (143) (124) (132) (13)(24) (12)(34) (14)(23) 1 (134) (134) (123) (142) (243) (124) (132) (234) (143) (14)(23) 1 (13)(24) (12)(34) (132) (132) (234) (124) (143) 1 (13)(24) (14)(23) (12)(34) (123) (142) (134) (243) (143) (143) (124) (234) (132) (12)(34) (14)(23) (13)(24) 1 (243) (134) (142) (123) (234) (234) (132) (143) (124) (13)(24) 1 (12)(34) (14)(23) (142) (123) (243) (134) (124) (124) (143) (132) (234) (14)(23) (12)(34) 1 (13)(24) (134) (243) (123) (142) Table 2.3: The binary operation is defined when the subgroup is normal. The Klein Four-Group V is a subgroup of A4. There are three cosets in the quotient A4 /V. This can be seen in the colouring. (v) We can do a similar example for A4. This has the Klein Four-Group V = {(Id), (12)(34), (13)(24), (14)(23)} as a normal subgroup. The Cayley Table for this situation is shown in Table 2.3. There are three right cosets (again we colour their elements in the table) H = H(Id) = {(Id), (12)(34), (13)(24), (14)(23)}, H(123) = {(123), (243), (142), (134)}, H(132) = {(132), (143), (234), (124)}. Hence, the quotient group A4 /V has three elements. The Cayley Table for this quotient can be seen in the colouring in Table 2.3. The di↵erent coloured blocks correspond to di↵erent elements. Thus, A4 /V = {e, x, x 2 } where x = H(123) = {(123), (243), (142), (134)}. The Cayley table for this is e x x2 e e x x2 x x x2 e x2 x2 e x This looks the same as the colouring in Table 2.3. (vi) We have a table for an attempt at the quotient of A4 by the subgroup h(123)i. This subgroup is not normal and so the set of four cosets do not form a group as the binary operation is not well-defined. This is pictured in Table 2.4. Proposition 2.26 If H is a normal subgroup of the finite group G, then |G| = |G/H| · |H|. 34 CHAPTER 2. QUOTIENT GROUPS 1 (123) (132) (12)(34) (134) (234) (13)(24) (243) (124) (14)(23) (142) (143) 1 1 (123) (132) (12)(34) (134) (234) (13)(24) (243) (124) (14)(23) (142) (143) (123) (123) (132) 1 (134) (234) (12)(34) (243) (124) (13)(24) (142) (143) (14)(23) (132) (132) 1 (123) (234) (12)(34) (134) (124) (13)(24) (243) (143) (14)(23) (142) (12)(34) (12)(34) (243) (143) 1 (142) (124) (14)(23) (123) (234) (13)(24) (134) (132) (134) (134) (124) (14)(23) (123) (143) (13)(24) (142) (132) (12)(34) (243) (234) 1 (234) (234) (13)(24) (142) (132) (14)(23) (243) (143) 1 (134) (124) (12)(34) (123) (13)(24) (13)(24) (142) (234) (14)(23) (243) (132) 1 (134) (143) (12)(34) (123) (124) (243) (243) (143) (12)(34) (142) (124) 1 (123) (234) (14)(23) (134) (132) (13)(24) (124) (124) (14)(23) (134) (143) (13)(24) (123) (132) (12)(34) (142) (234) 1 (243) (14)(23) (14)(23) (134) (124) (13)(24) (123) (143) (12)(34) (142) (132) 1 (243) (234) (142) (142) (234) (13)(24) (243) (132) (14)(23) (134) (143) 1 (123) (124) (12)(34) (143) (143) (12)(34) (243) (124) 1 (142) (234) (14)(23) (123) (132) (13)(24) (134) Table 2.4: The binary operation is not defined when the subgroup is not normal. This shows the operation on elements for the subgroup h(123)i which is not normal in A4. Proof. This follows from [G : H] = |G/ ⇠R | = |G/H| and Lagrange’s Theorem. ⇤ Examples 2.27 We can calculate this for the examples in Examples 2.25. (i) For G = S3 and H = h(123)i we have |S3 | 6 |G/H| = |S3 /h(123)i| = = = 2. |h(123)i| 3 (ii) For G = A4 and H = V = {(Id), (12)(34), (13)(24), (14)(23)} we have |A4 | 12 |G/H| = |A4 /V | = = = 3. |V | 4 Hence, A4 /V has prime order and hence is cyclic. Chapter 3 The Three Isomorphism Theorems In this chapter we will revise homomorphisms and the First Isomorphism Theorem. We shall introduce the Second and Third Isomorphisms which we will deduce from the First Isomorphism. 3.1 Homomorphisms Recall the following. Definition 3.1 Let (G, ⇤) and (H, ) be two groups. A map ' : G ! H is called a homomorphism if we have '(g1 ⇤ g2 ) = '(g1 ) '(g2 ) for all g1 , g2 2 G. Examples 3.2 (i) The determinant map det : (GLn (R), ⇥) ! (R⇤ , ⇥) is a homomorphism as det(AB) = det(A) det(B) for all matrices A and B. (ii) Let G = Sn. Every element 2 Sn can be written as a product of transpo- sitions. Define the sign of to be sign( ) = ( 1)k where k is the number of transpositions. This does not depend on the representation of chosen. Then, sign : Sn ! ({ 1, 1}, ⇥) is a homomorphism: Suppose , ⌧ 2 Sn are written as a product of j and k transpositions respectively. So, ⌧ can be written as j + k transpositions. Thus, sign( ⌧ ) = ( 1)j+k = ( 1)j ( 1)k = sign( ) sign(⌧ ). (iii) Suppose that H is a normal subgroup of G. Then, natural map ' : G ! G/H given by '(g) = Hg is a homomorphism: '(g1 g2 ) = H(g1 g2 ) = Hg1 Hg2 = '(g1 )'(g2 ). 35 36 CHAPTER 3. THE THREE ISOMORPHISM THEOREMS Definition 3.3 Let ' : G ! H be a homomorphism. The kernel of ', denoted Ker('), is the set Ker(') = {g 2 G | '(g) = eH }. The image of ', denoted Im('), is the set Im(') = {'(g) | g 2 G}. Proposition 3.4 Let ' : G ! H be a homomorphism. The following are true. (i) '(eG ) = eH. 1) 1 (ii) '(g = ('(g)) for all g 2 G. (iii) The kernel Ker(') is a normal subgroup of G. (iv) The image Im(') is a subgroup of H. (v) ' is injective if and only if Ker(') = {eG }. Example 3.5 The kernel of sign : Sn ! { 1, 1} is An and hence An is normal in Sn. 3.2 Isomorphisms Definition 3.6 We say that ' : G ! H is an isomorphism if (i) ' is a homomorphism, and (ii) ' is a bijection. If there is an isomorphism from G and H, then we say that G and H are isomorphic and write G ⇠ = H. Exercise 3.7 Show that the relation given by G is isomorphic to H is an equiva- lence relation. Examples 3.8 (i) We can see Z2 = {0, 1} ⇠ = { 1, 1} via the map x 7! ( 1)x. (ii) If H and K are conjugate subgroups, then they are isomorphic. That is, K = gHg 1 for some g 2 G =) H ⇠ = K. This isomorphism comes from the map ' : H ! K defined by '(x) = gxg 1 for all x 2 G. We see this as follows. ' is a homomorphism: 1 1 1 '(xy ) = g(xy )g = gxg gy g = '(x)'(y ). ' is bijective: Consider the map : K ! H given by (k) = g 1 kg. This is an inverse to ': 1 1 1 (' )(k) = '( (k)) = g ( (k)) g =g g kg g = k, 3.3. THE FIRST ISOMORPHISM THEOREM 37 and 1 1 1 ( ')(h) = ('(h)) = g '(h)g = g ghg = h. Thus, conjugate subgroups H and K are isomorphic. This result has implications for the order of the subgroups. That is, if H and K are conjugate, then they have the same order, and so |gHg 1 | = |K| 3.3 The First Isomorphism Theorem For any homomorphism ' : G ! H the kernel is normal and hence we can always take the quotient G/ Ker('). The First Isomorphism says that this quotient is isomorphic to the image of '. Theorem 3.9 Let ' : G ! H be a group homomorphism with kernel Ker('). Then, the quotient group G/ Ker(') is isomorphic to Im('), the image of '. In symbols, G/ Ker(') ⇠= Im(') and this isomorphism is provided by the map ' : G/ Ker(') ! Im(') defined by '(Ker(')g) = '(g). Proof. For simplicity let us denote the group Ker(') by K. ' is well-defined: We need to show that ' does not depend on the choice of representative for the coset. Suppose Kg1 = Kg2 , i.e., g1 and g2 are two representatives for the same coset. Then, g1 = kg2 for some k 2 K by Theorem 2.3(iii). Hence, g1 g2 1 2 K, i.e., ' g1 g2 1 = eH. This implies ' (g1 ) = ' (g2 ) and so ' (Kg1 ) = ' (g1 ) = ' (g2 ) = ' (Kg2 ). Thus, ' is well-defined. ' is a homomorphism: ' (Kg1 Kg2 ) = ' (Kg1 g2 ) , by definition of multiplication in G/K, = ' (g1 g2 ) , by definition of ', = ' (g1 ) ' (g2 ) , as ' is a homomorphism, = ' (Kg1 ) ' (Kg2 ) , by definition of '. Thus, ' is a homomorphism. ' is injective: '(Kg) = eH =) '(g) = eH =) g 2 K =) Kg = K = Ke, by Corollary 2.4, =) Ker(') = {Ke}. 38 CHAPTER 3. THE THREE ISOMORPHISM THEOREMS Therefore, ' is injective. ' is surjective: For any h 2 Im('), there exists g 2 G such that '(g) = h. Then '(Kg) = '(g) = h, showing that ' is surjective. Since ' is a well-defined homomorphism that is both injective and surjective, it is an isomorphism and so G/ Ker(') ⇠= Im('). ⇤ Examples 3.10 (i) As sign : Sn ! { 1, 1} is a surjective homomorphism for n 2 and An = Ker(sign) we have Sn /An ⇠ = { 1, 1} ⇠ = Z2. (ii) Let det : GLn (R) ! R⇤ be the determinant map. Then, Ker(det) is the set of matrices of determinant 1. That is, Ker(det) = SLn (R), the special linear group. The map det is surjective since if y 2 R⇤ , then the identity matrix with one of the 1s replaced with y will have determinant y. Thus, GLn (R)/ SLn (R) ⇠ = R⇤ and the map is SLn (R)X 7! det(X) for any SLn (R)X 2 GLn (R)/ SLn (R). (iii) Consider the group S 1 = {z 2 C | |z| = 1}, under multiplication. Let n be a positive integer and define ' : S 1 ! S 1 by '(x) = z n. This is easily checked to be a surjective homomorphism. We have Ker(') = Rn the nth -roots of unity and the First Isomorphism Theorem gives S 1 /Rn ⇠ = S1. 3.4 The preimage of a subgroup under a homomorphism is a subgroup The image of a group under a homomorphism is a group. We are now in a position to show that the pre-image of a group is too. Definition 3.11 Let f : X ! Y be a map from the set X to the set Y. Let A be subset of Y. The preimage of A under f , denoted f 1 (A), is defined to be 1 f (A) = {g 2 G : f (g) 2 A}. Remark 3.12 Note that this is di↵erent to the inverse of f – it is a set not a map. Note also that f may not be a bijection so the inverse f 1 may not even exist. Mathematics notation can be confusing sometimes! Theorem 3.13 Let f : G ! H be a homomorphism and K be a subgroup of H. Then, (i) The set f 1 (K) is a subgroup of G, 3.5. THE SECOND ISOMORPHISM THEOREM 39 (ii) Ker(f ) ✓ f 1 (K). Proof. (i) f 1 (K) is a subgroup of G: Since f is a homomorphism, it preserves the identity, thus f (eG ) = eH where eG and eH are the identity elements of G and H, respectively. Since K is a subgroup of H, eH 2 K. Therefore, eG 2 f 1 (K), so f 1 (K) 6= ;. Consider any x, y 2 f 1 (K). Then, there exists a, b 2 K such that f (x) = a and f (y ) = b. We have 1 1 1 1 f (xy ) = f (x)f (y ) = f (x) (f (y )) = ab 2K since K is a group. Thus, xy 1 2 f 1 (K). Therefore, f 1 (K) is a subgroup of G. (ii) The kernel of f is a subset of f 1 (K): Suppose x 2 Ker(f ). Then, 1 x 2 Ker(f ) =) f (x) = eH 2 K =) x 2 f (K) so Ker(f ) ✓ f 1 (K). ⇤ 3.5 The Second Isomorphism Theorem Definition 3.14 Let H and K be subgroups of the group G. We define the set HK to be HK = {hk | h 2 H and K 2 K}. Exercise 3.15 Prove that HK is a subgroup of G if, and only if, HK = KH. Remarks 3.16 (i) Note that HK = KH does not say that elements of H and K commute. We could have hk = kh0 where h 6= h0. (ii) Note also that the groups HN and H ⇥ N are not the same. The former is a subset of G while H ⇥ N is the set-theoretic product of H and K which is not a subset of G. Lemma 3.17 Suppose G is a group. If H and N are subgroups of G such that N is normal in G, then (i) HN is a subgroup of G, and (ii) N is a normal subgroup of HN. Proof. (i) HN is a subgroup G: The identity element e 2 G is in HN as e 2 H and e 2 N and so e = ee 2 HN. Thus, HN 6= ;. We now show that if x, y 2 HN, then xy 1 2 HN: If x, y 2 HN, then x = h1 n1 and y = h2 n2 for some h1 , h2 2 H and n1 , n2 2 N. So, 1 1 xy = h1 n1 (h2 n2 ) = h1 n1 n2 1 h2 1 = h1 h2 1 h2 n1 n2 1 h2 1 2 HN 40 CHAPTER 3. THE THREE ISOMORPHISM THEOREMS where h2 n1 n2 1 h2 1 2 N as N is normal. Therefore, HN is a group. (ii) N is a normal subgroup of HN: Clearly, N is a subset of HN as for any n 2 N we have n = en 2 HN. Hence, as N is a group (it is a subgroup of G) it is a subgroup of HN. Normality is automatic: Since N is a normal subgroup of G we have 1 gNg = N for all g 2 G. So, obviously this will hold if we restrict g to any subset, such as HN. ⇤ This lemma allows us to show that the quotients in the next theorem really are quotient groups, i.e., the quotients make sense. Theorem 3.18 (The Second Isomorphism Theorem) Suppose G is a group. If H and N are subgroups of G such that N is normal, then H \ N is a normal subgroup of H and H ⇠ HN =. H\N N Proof. By Lemma 3.17, we have N is a normal subgroup of HN and so can define a map HN ':H! N defined by '(h) = Nhe = Nh. This is a homomorphism: ' (h1 h2 ) = Nh1 h2 e = Nh1 eh2 e = Nh1 eNh2 e = ' (h1 ) ' (h2 ). This is surjective since if N (hn) 2 HN/N, then, as N is normal in HN, N(hn) = (hn)N = hN = Nh so '(h) = Nh = Nhn. Now for the kernel of ': Ker(') = {h 2 H | '(h) = Ne} = {h 2 H | Nh = Ne} = {h 2 H | h 2 N} = H \ N. Hence, H \ N is normal in H as kernels are normal. Finally, by the First Isomorphism Theorem we have H ⇠ HN = Im(') =. H\N N ⇤ 3.6. THE THIRD ISOMORPHISM THEOREM 41 Example 3.19 Consider the group (Z, +). For any integer x we have the sub- group xZ = hxi = {y 2 Z | y = mx for some m 2 Z}. For aZ and bZ we have aZ + bZ = gcd(a, b)Z and aZ \ bZ = lcm(a, b)Z. From the Second Isomorphism Theorem we get aZ aZ ⇠ aZ + bZ gcd(a, b)Z = = = lcm(a, b)Z aZ \ bZ bZ bZ Calculating the orders we see b lcm(a, b) = gcd(a, b) a which can be arranged to give the more familiar gcd(a, b) lcm(a, b) = ab. In other words we can see the Second Isomorphism Theorem as generalising this result. 3.6 The Third Isomorphism Theorem Theorem 3.20 Let G be a group and let N and K be normal subgroups of G such that K ✓ N. Then, (G/K)/(N/K) ⇠ = G/N. Proof. First we define the map ' : G/K ! G/N by '(Kg) = Ng for all g 2 G. ' is well-defined: Suppose Kg1 = Kg2. This means g1 = kg2 and since K ✓ N we then have Ng1 = Ng2. So the map is well-defined. ' is a homomorphism: For any g1 , g2 2 G we have '((Kg1 )(Kg2 )) = '(Kg1 g2 ) = N(g1 g2 ) = (Ng1 )(Ng2 ) = '(Kg1 )'(Kg2 ). Thus, ' is a homomorphism. Kernel of ': The kernel is given by Ker(') = {Kg 2 G/K | '(Kg) = Ne} = {Kg 2 G/K | Ng = N}. This means g 2 N, so Ker(') = {Kg | g 2 N} = N/K. 42 CHAPTER 3. THE THREE ISOMORPHISM THEOREMS ' is surjective: Since ' maps cosets Kg to cosets Ng, the image of ' is all of G/N and so ' is surjective. Apply the First Isomorphism Theorem: By the First Isomorphism Theorem, since ' is a homomorphism from G/K onto G/N with kernel N/K, we have (G/K)/(N/K) = (G/K)/ Ker(') ⇠ = Im(') = G/N. ⇤  Chapter 4 Group actions Every chapter after this will rely heavily on group actions and so this essential material. Actions allow groups to interact with other objects. These objects could be some set like a set of vertices or faces of a polyhedron, a set of maps, a vector space, or the group itself. There are plenty of possibilities. In particular the object could be something geometric (such as a polyhedron) and so we have a link between algebra and geometry. Furthermore, a group is itself a set and so can interact with itself or its subgroups. This will be particularly useful in the last two chapters of the module. 4.1 Group actions Definition 4.1 Let G be a group and X be a non-empty set. We say that G acts on X (on the left), if there exists a map ↵ : G ⇥ X ! X, where ↵(g, x) is usually denoted g · x, such that (ACT1) (gh) · x = g · (h · x), for all g, h 2 G, and all x 2 X, (ACT2) e · x = x, for all x 2 X. In this case we say ↵ is a group action, or more simply, an action. Usually we will say just G acts on X and not specify any particular notation for the action other than a dot and drop the ↵. The following set of examples shows how a number of familiar situations can all be united under the same umbrella of being examples of an action. Examples 4.2 (i) Let G be a group and X any non-empty set. The mapping ↵ : G ⇥ X ! X defined by ↵(g, x) = x, also denoted g · x = x, is called the trivial action on X. This example is quite boring but shows that G can interact with any non-empty set. (ii) Consider the general linear group of degree n over the real numbers, GLn (R). This is just the set of invertible n ⇥ n real matrices with the matrix multi- plication structure. Then, GLn (R) acts on the vector space Rn by ↵ : GLn (R) ⇥ Rn ! Rn by ↵(A, v) = Av. 43 44 CHAPTER 4. GROUP ACTIONS That is, matrix-vector multiplication can be viewed as an action. For sim- plicity we can write this as A · v = Av. Let’s show that this is an action: (ACT1) For all A, B 2 GLn (R) and v 2 Rn we have A · (B · v) = A · (Bv) = A (Bv) = (AB) v = (AB) · v. (ACT2) The identity of GLn (R) is the n ⇥ n identity matrix In and for all