Group Theory Notes, Study notes of Law

Download Group Theory Notes and more Study notes Law in PDF only on Docsity! Group Theory Notes 0 1 2 3 4 5 6 7 Donald L. Kreher March 18, 2020 ii Contents Ackowledgements iii 1 Introduction 1 1.1 What is a group? . . . . . . . . . . . . . . . . . . . . . . . . 1 1.1.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . 3 1.2 Some properties are unique. . . . . . . . . . . . . . . . . . . 4 1.2.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . 6 1.3 When are two groups the same? . . . . . . . . . . . . . . . . 8 1.3.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . 10 1.4 The automorphism group of a graph . . . . . . . . . . . . . 11 1.4.1 One more example. . . . . . . . . . . . . . . . . . . . 12 1.4.2 Exercises . . . . . . . . . . . . . . . . . . . . . . . . 13 2 The isomorphism theorems 15 2.1 Subgroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 2.1.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . 17 2.2 Cosets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 2.2.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . 20 2.3 Cyclic groups . . . . . . . . . . . . . . . . . . . . . . . . . . 21 2.3.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . 22 2.4 How many generators? . . . . . . . . . . . . . . . . . . . . . 24 2.4.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . 27 2.5 Normal subgroups . . . . . . . . . . . . . . . . . . . . . . . 28 2.6 Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 2.6.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . 32 2.7 Conjugation . . . . . . . . . . . . . . . . . . . . . . . . . . . 34 2.7.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . 36 3 Permutations 37 3.1 Even and odd . . . . . . . . . . . . . . . . . . . . . . . . . . 37 3.1.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . 40 v vi CONTENTS 3.2 Group actions . . . . . . . . . . . . . . . . . . . . . . . . . . 41 3.2.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . 51 3.3 Cayley’s theorem . . . . . . . . . . . . . . . . . . . . . . . . 53 3.3.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . 54 3.4 The Sylow theorems . . . . . . . . . . . . . . . . . . . . . . 55 3.4.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . 57 3.5 Some applications of the Sylow theorems . . . . . . . . . . . 58 3.5.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . 62 3.6 Simplicity of the alternating group . . . . . . . . . . . . . . 64 4 Finitely generated abelian groups 65 4.1 The Basis Theorem . . . . . . . . . . . . . . . . . . . . . . . 65 4.1.1 How many finite abelian groups are there? . . . . . . 69 4.1.2 Exercises . . . . . . . . . . . . . . . . . . . . . . . . 71 4.2 Generators and relations . . . . . . . . . . . . . . . . . . . . 72 4.2.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . 74 4.3 Smith normal form . . . . . . . . . . . . . . . . . . . . . . . 75 4.4 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . 79 4.4.1 The fundamental theorem of finitely generated abelian groups . . . . . . . . . . . . . . . . . . . . . 79 4.4.2 Systems of Diophantine Equations . . . . . . . . . . 81 4.4.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . 83 5 Fields 85 5.1 A glossary of algebraic systems . . . . . . . . . . . . . . . . 85 5.2 Ideals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86 5.3 The prime field . . . . . . . . . . . . . . . . . . . . . . . . . 88 5.3.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . 91 5.4 Algebraic extensions . . . . . . . . . . . . . . . . . . . . . . 92 5.5 Splitting fields . . . . . . . . . . . . . . . . . . . . . . . . . 94 5.6 Galois fields . . . . . . . . . . . . . . . . . . . . . . . . . . . 95 5.7 Constructing a finite field . . . . . . . . . . . . . . . . . . . 95 5.7.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . 97 6 Linear groups 99 6.1 The linear fractional group and PSL(2, q) . . . . . . . . . . 99 6.1.1 Transitivity . . . . . . . . . . . . . . . . . . . . . . . 104 6.1.2 The conjugacy classes . . . . . . . . . . . . . . . . . 108 6.1.3 The permutation character . . . . . . . . . . . . . . 116 6.1.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . 118 7 Automorphism groups 119 7.1 Inner and outer automorphisms . . . . . . . . . . . . . . . . 119 CONTENTS vii 7.1.1 Semidirect products . . . . . . . . . . . . . . . . . . 120 7.1.2 Automorphism of Sn . . . . . . . . . . . . . . . . . . 121 7.1.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . 125 1.1. WHAT IS A GROUP? 3 5. The set of complex numbers G = {1, i,−1,−i} under multiplication. The multiplication table for this group is: ∗ 1 i −1 −i 1 1 i −1 −i i i −1 −i 1 −1 −1 −i 1 i −i −i 1 i −1 6. The set Sym (X) of one to one and onto functions on the n-element set X, with multiplication defined to be composition of functions. (The elements of Sym (X) are called permutations and Sym (X) is called the symmetric group on X. This group will be discussed in more detail later. If α ∈ Sym (X), then we define the image of x under α to be xα. If α, β ∈ Sym (X), then the image of x under the composition αβ is xαβ = (xα)β .) 1.1.1 Exercises 1. For each fixed integer n > 0, prove that Zn, the set of integers modulo n is a group under +, where one defines a+ b = a+ b. (The elements of Zn are the congruence classes a, a ∈ Z.. The congruence class ā is {x ∈ Z : x ≡ a (modn)} = {a+ kn : k ∈ Z}. Be sure to show that this addition is well defined. Conclude that for every integer n > 0 there is a group with n elements. 2. Given integer n > 0 let G be the subset of complex numbers of the form e 2kπ n i, k ∈ Z. Show that G is a group under multiplication. How many elements does G have? 4 CHAPTER 1. INTRODUCTION 1.2 Some properties are unique. Lemma 1.2.1. If (G, ∗) is a group and a ∈ G, then a∗a = a implies a = e. Proof. Suppose a ∈ G satisfies a∗a = a and let b ∈ G be such that b∗a = e. Then b ∗ (a ∗ a) = b ∗ a and thus a = e ∗ a = (b ∗ a) ∗ a = b ∗ (a ∗ a) = b ∗ a = e Lemma 1.2.2. In a group (G, ∗) (i) if b ∗ a = e, then a ∗ b = e and (ii) a ∗ e = a for all a ∈ G Furthermore, there is only one element e ∈ G satisfying (ii) and for all a ∈ G, there is only one b ∈ G satisfying b ∗ a = e. Proof. Suppose b ∗ a = e, then (a ∗ b) ∗ (a ∗ b) = a ∗ (b ∗ a) ∗ b = a ∗ e ∗ b = a ∗ b. Therefore by Lemma 1.2.1 a ∗ b = e. Suppose a ∈ G and let b ∈ G be such that b ∗ a = e. Then by (i) a ∗ e = a ∗ (b ∗ a) = (a ∗ b) ∗ a = e ∗ a = a Now we show uniqueness. Suppose that a ∗ e = a and a ∗ f = a for all a ∈ G. Then (e ∗ f) ∗ (e ∗ f) = e ∗ (f ∗ e) ∗ f = e ∗ f ∗ e = e ∗ f Therefore by Lemma 1.2.1 e ∗ f = e. Consequently f ∗ f = (f ∗ e) ∗ (f ∗ e) = f ∗ (e ∗ f) ∗ e = f ∗ e ∗ e = f ∗ e = f and therefore by Lemma 1.2.1 f = e. Finally suppose b1 ∗ a = e and b2 ∗ a = e. Then by (i) and (ii) b1 = b1 ∗ e = b1 ∗ (a ∗ b2) = (b1 ∗ a) ∗ b2 = e ∗ b2 = b2 1.2. SOME PROPERTIES ARE UNIQUE. 5 Definition 1.4: Let (G, ∗) be a group. The unique element e ∈ G satisfying e ∗ a = a for all a ∈ G is called the identity for the group (G, ∗). If a ∈ G, the unique element b ∈ G such that b ∗ a = e is called the inverse of a and we denote it by b = a−1. If n > 0 is an integer, we abbreviate a ∗ a ∗ a ∗ · · · ∗ a︸ ︷︷ ︸ n times by an. Thus a−n = (a−1)n = a−1 ∗ a−1 ∗ a−1 ∗ · · · ∗ a−1︸ ︷︷ ︸ n times Let (G, ∗) be a group where G = {g1, g2, . . . , gn}. Consider the multiplica- tion table of (G, ∗). gj gi gi ∗ gj Let [x1 x2 x3 · · · xn] be the row labeled by gi in the multiplication table. I.e. xj = gi ∗ gj . If xj1 = xj2 , then gi ∗ gj1 = gi ∗ gj2 . Now multiplying by g−1 i on the left we see that gj1 = gj2 . Consequently j1 = j2. Therefore every row of the multiplication table contains every element of G exactly once a similar argument shows that every column of the multiplication table contains every ele- ment of G exactly once A table satisfying these two properties is called a Latin Square. Definition 1.5: A latin square of side n is an n by n array in which each cell contains a single element form an n-element set S = {s1, s2, . . . , sn}, such that each element occurs in each row exactly once. It is in standard form with respect to the sequence s1, s2, . . . , sn if the elements in the first row and first column are occur in the order of this sequence. 8 CHAPTER 1. INTRODUCTION 1.3 When are two groups the same? When ever one studies a mathematical object it is important to know when two representations of that object are the same or are different. For example consider the following two groups of order 8. G =  g1= [ 1 0 0 1 ] , g2= [ 0 −1 1 0 ] , g3= [ −1 0 0 −1 ] , g4= [ 0 1 −1 0 ] , g5= [ 1 0 0 −1 ] , g6= [ −1 0 0 1 ] , g7= [ 0 1 1 0 ] , g8= [ 0 −1 −1 0 ]  (1.1) (G, ·) is a group of 2 by 2 matrices under matrix multiplication. H = { h1 : x 7→ x, h2 : x 7→ ix, h3 : x 7→ −x, h4 : x 7→ −ix, h5 : x 7→ x̄, h6 : x 7→ −x̄, h7 : x 7→ ix̄, h8 : x 7→ −ix̄ } (1.2) (H, ◦) is a group complex functions under function composition. Here i =√ −1 and a+ bi = a− bi. The multiplication tables for G and H respectively are: g1 g2 g3 g4 g5 g6 g7 g8 g1 g1 g2 g3 g4 g5 g6 g7 g8 g2 g2 g3 g4 g1 g7 g8 g6 g5 g3 g3 g4 g1 g2 g6 g5 g8 g7 g4 g4 g1 g2 g3 g8 g7 g5 g6 g5 g5 g8 g6 g7 g1 g3 g4 g2 g6 g6 g7 g5 g8 g3 g1 g2 g4 g7 g7 g5 g8 g6 g2 g4 g1 g3 g8 g8 g6 g7 g5 g4 g2 g3 g1 h1 h2 h3 h4 h5 h6 h7 h8 h1 h1 h2 h3 h4 h5 h6 h7 h8 h2 h2 h3 h4 h1 h7 h8 h6 h5 h3 h3 h4 h1 h2 h6 h5 h8 h7 h4 h4 h1 h2 h3 h8 h7 h5 h6 h5 h5 h8 h6 h7 h1 h3 h4 h2 h6 h6 h7 h5 h8 h3 h1 h2 h4 h7 h7 h5 h8 h6 h2 h4 h1 h3 h8 h8 h6 h7 h5 h4 h2 h3 h1 Observe that these two tables are the same except that different names were chosen. That is the one to one correspondence given by: x g1 g2 g3 g4 g5 g6 g7 g8 θ(x) h1 h2 h3 h4 h5 h6 h7 h8 1.3. WHEN ARE TWO GROUPS THE SAME? 9 carries the entries in the table for G to the entries in the table for H. More precisely we have the following definition. Definition 1.7: Two groups (G, ∗) and (H, ◦) are said to be isomor- phic if there is a one to one correspondence θ : H → G such that θ(g1 ∗ g2) = θ(g1) ◦ θ(g2) for all g1, g2 ∈ G. The mapping θ is called an isomorphism and we say that G is isomorphic to H. This last statement is abbreviated by G ∼= H. If θ satisfies the above property but is not a one to one correspondence, we say θ is homomorphism. These will be discussed later. A geometric description of these two groups may also be given. Consider the square drawn in the [ x y ] –plane with vertices the vectors in the set: V = {[ 1 0 ] , [ 0 1 ] , [ −1 0 ] , [ 0 −1 ]} . y x [ 0 1 ] [ 1 0 ] [ −1 0 ] [ 0 −1 ] The set of 2 by 2 matrices that preserve this set of vertices is the the group (G, ·) specified in 1.1. Thus (G, ·) is the group of symmetries of the square. Now consider the square drawn in the complex–plane with vertices the complex numbers in the set: V = {1, i,−1,−i}. The set of complex functions that preserve this set of vertices is the the group (H, ◦) as specified in 1.2. Thus (H, ◦) is also the group of sym- metries of the square. Consequently it is easy to see that these two groups are isomorphic. = < i 1 −1 −i 10 CHAPTER 1. INTRODUCTION 1.3.1 Exercises 1. The groups given in example 1.1.3 and 1.1.5 are nonisomorphic. 2. The groups given in example 1.1.5 and Z4 are isomorphic. 3. Symmetries of the hexagon (a) Determine the group of symmetries of the hexagon as a group G of two by two matrices. Write out multiplication table of G. (b) Determine the group of symmetries of the hexagon as a group H of complex functions. Write out the multiplication table of H. (c) Show explicitly that there is an isomorphism θ : G→ H. 1.4. THE AUTOMORPHISM GROUP OF A GRAPH 13 The image of x under θ is written in the bottom row. below x in the top row. Although this is simple an even simpler notation is cycle notation. The cycle notation for θ is θ = (1, 11, 3, 4)(2)(5, 6)(7, 8, 9)(10) To see how this notation works we draw the diagram for the graph with edges: {x, xθ} for each x. But instead of drawing a line from x to xθ we draw a directed arc: x→ θ(x). 1 4 3 11 2 5 6 9 7 8 10 The resulting graph is a union of directed cycles. A sequence of vertices enclosed between parentheses in the cycle notation for the permutation θ is called a cycle of θ. In the above example the cycles are: (1, 11, 3, 4), (2), (5, 6), (7, 8, 9), (10). If the number of vertices is understood the convention is to not write the cycles of length one. (Cycles of length one are called fixed points. In our example 2 and 10 are fixed points.) Thus we write for θ θ = (1, 11, 3, 4)(5, 6)(7, 8, 9) Now we are in good shape to give the example. The automorphism group of Γ1 in figure 1.1 is Aut (Γ1) = { e, (1, 2), (5, 6), (1, 2)(5, 6), (1, 5)(2, 6)(3, 4), (1, 6)(2, 5)(3, 4), (1, 5, 2, 6)(3, 4), (1, 6, 2, 5)(3, 4) } e is used above to denote the identity permutation. The product of two permuations α and β is function composition read from left to right. Thus xαβ = (xα)β For example: (1, 2)(3, 6, 5, 4) (1)(2, 6, 5, 4, 3) = (1, 6, 4, 2)(3, 5) as illustrated in Figure 1.2. 1.4.2 Exercises 1. Write the permutation that results from the product( 1 2 3 4 5 6 7 8 9 10 11 11 2 4 1 6 5 8 9 7 10 3 )( 1 2 3 4 5 6 7 8 9 10 11 3 6 4 11 9 7 8 10 5 2 1 ) 14 CHAPTER 1. INTRODUCTION 6 5 4 3 2 1 6 5 4 3 2 1 6 5 4 3 2 1 α β αβ Figure 1.2: The product of permutations α and β. in cycle notation. 2. Show that Aut (Γ1) is isomorphic to the group of symmetries of the square given in Section 1.3. 3. What is the automorphism group of the graph Γ = (V, E) for which V = {1, 2, 3, 4, 5, 6}; and E = {{1, 2}, {2, 3}, {1, 3}, {4, 5}, {4, 6}, {5, 6}, {1, 4}, {2, 5}, {3, 6}} Chapter 2 The isomorphism theorems Through out the remainder of the text we will use ab to denote the product of group elements a and b and we will denote the identity by 1. Also we will denote a group (G, ·) by G the name given to its element set. 2.1 Subgroups Definition 2.1: A nonempty subset S of the group G is a subgroup of G if S a group under binary operation of G. We use the notation S ≤ G to indicate that S is a subgroup of G. If S is a subgroup we see from Lemma 1.2.1 that 1 the identity for G is also the identity for S. Consequently the following theorem is obvious: Theorem 2.1.1. A subset S of the group G is a subgroup of G if and only if (i) 1 ∈ S; (ii) a ∈ S ⇒ a−1 ∈ S; (iii) a, b ∈ S ⇒ ab ∈ S. 15 18 CHAPTER 2. THE ISOMORPHISM THEOREMS 2.2 Cosets Definition 2.3: If S is a subgroup of G and a ∈ G, then Sa = {xa : x ∈ S} is a right coset of S. If S is a subgroup of G and a, b ∈ G, then it is easy to see that Sa = Sb whenever b ∈ Sa. An element b ∈ Sa is said to be a coset representative of the coset Sa. Lemma 2.2.1. Let S be a subgroup of the group Gand let a, b ∈ G. Then Sa = Sb if and only if ab−1 ∈ S. Proof. Suppose Sa = Sb. Then a ∈ Sa and so a ∈ Sb. Thus a = xb for some x ∈ S and we see that ab−1 = x ∈ S. Conversely, suppose ab−1 ∈ S. Then ab−1 = x, for some x ∈ S. Thus a = xb and consequently Sa = Sxb. Observe that Sx = S because x ∈ S. Therefore Sa = Sb. Lemma 2.2.2. Cosets are either identical or disjoint. Proof. Let S be a subgroup of the group G and let a, b ∈ G. Suppose that Sa ∩ Sb 6= ∅. Then there is a z ∈ Sa ∩ Sb. Hence we may write z = xa for some x ∈ S and z = yb for some y ∈ S. Thus, xa = yb. But then ab−1 = x−1y ∈ S, because x, y ∈ S and S is a subgroup. Definition 2.4: The number of elements in the finite group G is called the order of G and is denoted by |G|. If S is a subgroup of the finite group G it is easy to see that |Sa| = |S| for any coset Sa. Also because cosets are identical or disjoint we can choose coset representatives a1, a2, . . . , ar so that G = Sa1∪̇Sa2∪̇Sa3∪̇ · · · ∪̇Sar. Thus G can be written as the disjoint union of cosets and these cosets each have size |S|. The number r of right cosets of S in G is denoted by |G : S| 2.2. COSETS 19 and is called the index of S in G. This discussion establishes the following important result of Lagrange (1736-1813). Theorem 2.2.3. (Lagrange) If S is a subgroup of the finite group G, then |G : S| = |G| |S| Thus the order of S divides the order of G. Definition 2.5: If x ∈ G and G is finite, the order of x is |x| = | 〈x〉 |. Corollary 2.2.4. If x ∈ G and G is finite, then |x| divides |G|. Proof. This is a direct consequence of Theorem 2.2.3. Corollary 2.2.5. If |G| = p a prime, then G is cyclic. Proof. Let x ∈ G, x 6= 1. Then |x| = p, because p is a prime. Hence 〈x〉 = G and therefore G is cyclic. A useful formula is provided in the next theorem. If X and Y are subgroups of a group G, then we define XY = {xy : x ∈ X and y ∈ Y }. Lemma 2.2.6. (Product formula) If X and Y are subgroups of G, then |X Y ||X ∩ Y | = |X||Y | Proof. We count pairs [(x, y), z] (2.1) such that xy = z, x ∈ X, y ∈ Y in two ways. First there are |X| choices for x and |Y | choices for y this determines z to be xy, and so there are |X||Y | pairs 2.1. Secondly there are |XY | choices for z. But given z ∈ XY there may be many ways to write z as z = xy, where x ∈ X and y ∈Y Let z ∈ XY be given and write z = x2y2. If x ∈ X and y ∈ Y satisfy xy = z, then x−1x2 = yy−1 2 ∈ X ∩ Y. 20 CHAPTER 2. THE ISOMORPHISM THEOREMS Conversely if a ∈ X ∩ Y , then because X ∩ Y is a subgroup of both X and Y , we see that x2a ∈ X and a−1y2 ∈ Y thus the ordered pair (x2a, a −1y2) ∈ X × Y is such that (x2a)(a−1y2) = x2y2. Thus given z ∈ XY the number of pairs (x, y) such that x ∈ X, y ∈ Y and xy = z is |X ∩ Y |. Thus there are |X ∩ Y ||XY | pairs 2.1. 2.2.1 Exercises 1. Let G = Sym ({1, 2, 3, 4}) and let H = 〈(1, 2, 3, 4), (2, 4)〉. Write out all the cosets of H in G. 2. Let |G| = 15. If G has only one subgroup of order 3 and only one subgroup of order 5, then G is cyclic. 3. Use Corollary 2.2.5 to show that the Latin square given in Exer- cise 1.2.1.4 cannot be the multiplication table of a group. 4. Recall that the determinant map δ : GLn(R) → R is a homomor- phism. Let S = ker δ. Describe the cosets of S in GLn(R). 2.3. CYCLIC GROUPS 23 (b) Draw the subgroup lattice for a cyclic group of order p2q; where p and q are distinct primes. 24 CHAPTER 2. THE ISOMORPHISM THEOREMS 2.4 How many generators? Let G be a cyclic group of order 12 generated by a. Then G = {1, a1, a2, a3, a4, a5, a6, a7, a8, a9, a10, a11} Observe that〈 a5 〉 = {1, a5, a10, a3, a8, a, a6, a11, a4, a9, a2, a7} = G Thus a5 also generates G. Also, a7, a11 and a generate G. But, the other elements do not. Indeed: 〈1〉 = {1}〈 a6 〉 = {1, a6}〈 a4 〉 = 〈 a8 〉 = {1, a4, a8}〈 a3 〉 = 〈 a9 〉 = {1, a3, a6, a9}〈 a2 〉 = 〈 a10 〉 = {1, a2, a4, a6, a8, a10} Definition 2.6: The Euler phi function or Euler totient is φ(n) = |{x : 1 ≤ x ≤ n and gcd (x, n) = 1}| the number of positive integers x ≤ n that have no common divisors with n. For example when n=12 we have: {x : 1 ≤ x ≤ n and gcd (x, n) = 1} = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} \ {2, 3, 4, 6, 8, 9, 10, 12} = {1,
A2,
A3,
A4, 5,
A6, 7,
A8,
A9,Z10, 11,Z12} = {1, 5, 7, 11} and so φ(12) = 4. When n is a prime then gcd (x, n) = 1 unless n divides x. Hence φ(n) = n−1 when n is a prime. 2.4. HOW MANY GENERATORS? 25 Theorem 2.4.1. Let G be a cyclic group of order n generated by a. Then G has φ(n) generators. Proof. Let 1 ≤ x < n and let m = |ax|. Then m is the smallest positive integer such that amx = 1. Moreover amx = 1 also implies n divides mx. Thus ax has order n if and only if x and n have no common divisors. Thus gcd (x, n) = 1 and the theorem now follows. Corollary 2.4.2. Let G be a cyclic group of order n. If d divides n, the number of elements of order d in G is φ(d). It is 0 otherwise. Proof. If G has an element of order d, then by Lagrange’s theorem (Theo- rem 2.2.3) d divides n. We now apply Theorem 2.3.3 to see that G has a unique subgroup H of order d. Hence every element of order d belongs to H. Therefore by Theorem 2.4.1 H has exactly φ(d) generators and so G has exactly φ(d)elements of order d. Theorem 2.4.1 won’t do us any good unless we can efficiently compute φ(n). Fortunately this is easy as Lemma 2.4.3 will show. Lemma 2.4.3. (i) φ(1) = 1; (ii) if p is a prime, then φ(pa) = pa − pa−1; and (iii) if gcd (m,n) = 1, then φ(mn) = φ(m)φ(n). Proof. (i) It is obvious that φ(1) = 1. (ii) Observe that gcd (x, pa) 6= 1 if an only if p divides x. Thus crossing out every entry divisible by p from the pa−1 by p array 1 2 3 . . . p− 1 p p+ 1 p+ 2 p+ 3 . . . 2p− 1 2p 2p+ 1 2p+ 2 2p+ 3 . . . 3p− 1 3p ... ... (pa−1 − 1)p+ 1 (pa−1 − 1)p+ 2 (pa−1 − 1)p+ 3 . . . pa − 1 pa delete the last column leaving an array of size pa−1 by p− 1. Thus φ(n) = pa−1(p− 1) = pa − pa−1. 28 CHAPTER 2. THE ISOMORPHISM THEOREMS 2.5 Normal subgroups Definition 2.7: A subgroup N of the group G is a normal subgroup if g−1Ng = N for all g ∈ G. We indicate that N is a normal subgroup of G with the notation N E G. Example 2.3: Some normal subgroups 1. Every subgroup of an abelian group is a normal subgroup. 2. The subset of matrices of GL2(R) that have determinant 1 is a normal subgroup of GL2(R). Lemma 2.5.1. The subgroup N of G is a normal subgroup of G if and only if g−1Ng ⊆ N for all g ∈ G. Proof. Suppose N is subgroup of G satisfying g−1Ng ⊆ N for all g ∈ G. Then for all n ∈ N and all g ∈ G, we have gng−1 = (g−1)−1n(g−1) = n′ ∈ N for some n′, because (g−1) ∈ G. Solving for n we find n = g−1n′g ∈ g−1Ng. Hence N ⊆ g−1Ng and so, N = g−1Ng. Therefore N is a normal subgroup of G. The converse is obvious. The multiplication of two subsets A and B of the group G is defined by AB = {ab : a ∈ A and b ∈ B} This multiplication is associative because the multiplication in G is asso- ciative. Thus, if a collection of subsets of G are carefully chosen, then it may be possible that they could form a group under this multiplication. Theorem 2.5.2. If N is a normal subgroup of G, then the cosets of N form a group. If G is finite, this group has order |G : N |. Proof. Let x, y ∈ G. Then NxNy = NxNx−1xy = NNxy = Nxy 2.6. LAWS 29 because N is normal in G. Thus the product of two cosets is a coset. It is easy to see N is the identity and Nx−1 is (Nx)−1 for this multiplication. Thus the cosets form a group as claimed. Furthermore when G is finite Theorem 2.2.3 applies and the number of cosets is |G : N |. Definition 2.8: The group of cosets of a normal subgroup N of the group G is called the quotient group or the factor group of G by N . This group is denoted by G/N which is read “G modulo N” or “G mod N”. Notice how this definition closely follows what we already know as modular arithmetic. Indeed Zn (the integers modulo n) is precisely the factor group Z/nZ. 2.6 Laws The most important elementary theorem of group theory is: Theorem 2.6.1. (First law) Let θ : G → H be a homomorphism. Then N = kernel (θ) is a normal subgroup of G and G/N ∼= image (θ) . Proof. In Example 2.1.3 we have already seen that N is a subgroup of G. To see that N is a normal subgroup, let g ∈ G and n ∈ N . Then θ(g−1ng) = θ(g−1)θ(n)θ(g) = θ(g−1)θ(g) = θ(g−1g) = θ(1) = 1. Thus g−1ng ∈ N and hence by Lemma 2.5.1 N is normal in G. Now define Ψ : G/N → image (θ) by Ψ(Ng) = θ(g). To see that Ψ well defined suppose Nx = Ny. Then, xy−1 ∈ N . So, 1 = θ(xy−1) = θ(x)θ(y)−1. Therefore θ(x) = θ(y) and hence Ψ(Nx) = Ψ(Ny). Also, Ψ is a homomorphism, for Ψ(NxNy) = Ψ(Nxy) = θ(xy) = θ(x)θ(y) = Ψ(Nx)Ψ(Ny). Moreover Ψ is one to one since Ψ(Nx) = Ψ(Ny) implies θ(x) = θ(y). So, xy−1 ∈ kernel (θ) = N . But then, Nx = Ny. Clearly image (Ψ) = image (θ). Therefore Ψ is an isomorphism between G/N and image (θ). 30 CHAPTER 2. THE ISOMORPHISM THEOREMS Suppose K E G, and consider the mapping π : G → G/K defined by π(x) = Kx. Observe that π(xy) = Kxy = Kxky and π(x) = K ⇔ Kx = K ⇔ x ∈ K. Thus π is a homomorphism with kernel K. The mapping π is called the natural map. Theorem 2.6.2. If H ≤ G and N E G, then HN = NH is a subgroup of G. Proof. Let S = 〈H,N〉 be that smallest subgroup of G that contains H and N . (I.e. S is the intersection over all subgroups of G, that contain H and also N .) Certainly H,N ⊆ NH and HN,NH ⊆ S. Hence it suffices to show that HN and NH are subgroups of G. If h1n1, h2n2 ∈ HN , then (h1n1)(h2n2)−1 = h1(n1n −1 2 h−1 2 ) = h1(h−1 2 n3) ∈ HN for some n3 ∈ N , because N E G. Therefore by Theorem 2.1.2 HN is a subgroup. A similar argument will show that NH is also a subgroup. Remark: It follows from Theorem 2.6.2 and the product formula (Theo- rem 2.2.6) that if H ≤ G and N E G, then |NH|/|N | = |H|/|H ∩N |. This suggests the second isomorphism law. Theorem 2.6.3. (Second law) Let H and N be subgroups of G with N normal. Then H ∩N is normal in H and H/(H ∩N) ∼= NH/N . Proof. Let π : G→ G/N be the natural map and let π↓H be the restriction of π to H. Because π↓H is a homomorphism with kernel H ∩N we see by Theorem 2.6.1, that H ∩N E H and that H/(H ∩N) ∼= image (π↓H). But by the above remark we know that the image of π↓H is just the collection of cosets of N with representatives in H. These are the cosets of of N in HN/N . Theorem 2.6.4. (Third law) Let M ⊂ N be normal subgroups of G. Then N/M is a normal subgroup of G/M and (G/M)/(N/M) ∼= G/N . Proof. Define f : G/M → G/N by f(Mx) = Nx. Check that f is a well- defined homomorphism with kernel N/M and image G/N . Apply The First law. 2.6. LAWS 33 Show that R/Z ∼= T , where R is the additve group of real numbers. (If z = a+ bi, then ||z|| = √ a2 + b2.) 34 CHAPTER 2. THE ISOMORPHISM THEOREMS 2.7 Conjugation Definition 2.10: Let x and y be elements of the group G. If there is a g ∈ G such that g−1xg = y, then we say that x is conjugate to y. The relation “x is conjugate to y” is an equivalence relation and the equivalence classes are called conjugacy classes. We denote the conjugacy class of x by K(x). Thus, K(x) = {g−1xg : g ∈ G} If x is an element of the group G, then it is easy to see that K(x) = {x} if and only if x commutes with every element of G. So, in particular, conjugacy classes of abelian groups are not interesting. Definition 2.11: The center of G, is Z(G) = {x ∈ G : xg = gx, for all g ∈ G}. It is the set of all elements of G that commute with every element of G. Observe that for x ∈ G, |K(x)| = 1 if and only if x ∈ Z(G). Consequently if the group G is finite we can write G = Z(G)∪̇K(x1)∪̇K(x2)∪̇ · · · ∪̇K(xr) where x1, x2, . . . , xr are representatives one each from the distinct conju- gacy classes with |K(xi)| > 1. Thus |G| = |Z(G)|+ r∑ i=1 |K(xi)|. (2.2) This is called the class equation. We will use it later. Definition 2.12: If x is an element of the group G, then the cen- tralizer of x in G is the subgroup CG(x) = {g ∈ G : gx = xg} the set of all elements of G that commute with x. 2.7. CONJUGATION 35 Theorem 2.7.1. Let x be an element of the finite group G. The number of conjugates of x is the index of CG(x) in G. That is, |K(x)| = |G : CG(x)|. Proof. Exercise 2.7.3 shows that CG(x) is a subgroup of G. Observe that for two elements g1, g2 ∈ G: g−1 1 xg1 = g−1 2 xg2 ⇔ g1g −1 2 x = xg1g −1 2 ⇔ g1g −1 2 ∈ CG(x) ⇔ CG(x)g1 ∈ CG(x)g2a (See Lemma 2.2.1.) Thus the mapping F : g−1xg 7→ CG(x)g is a one to one correspondence from K(x) to the right cosets of CG(x). Thus |K(x)| is the number of cosets of CG(x) in G and this is |G : CG(x)| by Lagrange’s theorem (Theorem 2.2.3). Theorem 2.7.2. If G is a group of order pn for some prime p, then |Z(G)| > 1. Proof. Write the class equation for G and apply Theorem 2.7.1: |G| = |Z(G)|+ r∑ i=1 |K(xi)| = |Z(G)|+ r∑ i=1 |G|/|CG(xi)| By Lagrange we know that |K(xi)| = pj for some j > 0. (Note j > 0, because |K(xi)| 6= 1.) Thus the sum is divisible by p, |G| is divisible by p and therefore |Z(G)| is divisible by p. Consequently |Z(G)| > 1. Lemma 2.7.3. If G is a finite abelian group whose order is divisible by a prime p, then G contains an element of order p. Proof. Let x ∈ G have order t > 1. If p t, say t = mp, then xm has order p. So suppose p - t. Then because G is abelian, 〈x〉 is a normal subgroup of G and G/ 〈x〉 is an abelian group of order |G|/t. Now |G|/t < |G|, so by induction there is an element y ∈ G/ 〈x〉 of order p. Then y = y 〈x〉 for some y ∈ G and |y| = p, says yp = xi for some i. Hence ypt = 1. Thus |yt| divides p. But |yt| 6= 1, because |y| = p, and p - t. Thus |yt| = p. Theorem 2.7.4. (Cauchy) If G is a finite group whose order is divisible by a prime p, then G contains an element of order p. 38 CHAPTER 3. PERMUTATIONS be a factorization of the identity I where β1, β2, . . . , βn are transpositions. Now n 6= 1, because β1 6= 1. If n = 2, then I has been factored in to 2 transpositions, and this is an even number. Suppose n > 2 and observe that (w, x)(w, x) = I (w, x)(y, z) = (y, z)(w, x) (w, x)(x, y) = (x, y)(w, y) (w, x)(w, y) = (x, y)(w, x) Let w be one of the two symbols moved by β1. Then we can “push” w to the right until two transpositions cancel and we reduce to a factorization into n−2 transpositions. Consequently by induction n−2 is even and therefore n is even. There must be such a cancellation, because the identity fixes w. The following algorithm makes this process clear: Let w be one of the two symbols moved by β1. i← 1 x← wβi ; (Thus βi = (w, x).) while i < n do  if βi+1 = βi then  comment: { π = β1β2 · · ·βi−1βi+2, . . . , βn, and so, by induction n− 2 is even. return (n is even) if βi+1 = (y, z), where y, z /∈ {w, x} then replace βiβi+1 with (y, z)(w, x) if βi+1 = (x, y), where y /∈ {w, x} then replace βiβi+1 with (x, y)(w, y) if βi+1 = (w, y), where y /∈ {w, x} then replace βiβi+1 with (x, y)(w, x) i← i+ 1 Theorem 3.1.3. Let π = β1β2 · · ·βn = γ1γ2 · · · γm be two factorizations of the permutation π where the βis and the γjs are transpositions. Then either n and m are both even or they are both odd. 3.1. EVEN AND ODD 39 Proof. Observe that because γ−1 j = γj we have: I = ππ−1 = β1β2 · · ·βn(γ1γ2 · · · γm)−1 = β1β2 · · ·βnγm · · · γ2γ1 Therefore by Lemma 3.1.2, m+ n is even and the result follows. Now that we have Theorem 3.1.3 the following definition makes sense. Definition 3.2: A permutation is an even permutation if it can be written as the product of an even number of transpositions; otherwise it is an odd permutation. If X is a finite set, then Alt (X) is the set of all even permutations in Sym (X) and is called the alternating group. Theorem 3.1.4. Let X be a set, |X| = n. Then Alt (X) is a subgroup of Sym (X) of order n! 2 . Proof. Clearly the product of two even permutations is an even permutation and Lemma 3.1.2 shows that the identity is even. Consequently by Theo- rem 2.1.3 Alt (X) is a subgroup of Sym (X). Let Θ : Sym (X)→ {1,−1} be defined by Θ(π) = { 1 if π is even; −1 if π is odd. Then it is easy to see that Θ is a homomorphism on to the multiplicative group {1,−1} with kernel (Θ) = Alt (X). Thus by the First law (Theo- rem 2.6.1) Sym (X) /Alt (X) ∼= {1,−1}. Hence, |Sym (X) /Alt (X) | = 2 and so Alt (X) = n! 2 as claimed. 40 CHAPTER 3. PERMUTATIONS 3.1.1 Exercises 1. Write the following permutation as a product of transpositions and determine if it is even or odd.( 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 11 10 9 3 2 1 15 4 12 5 16 13 14 6 8 7 ) 2. LetH be a subgroup of Sym (X). Show that either all the permuations in H are even or that exactly half of them are. 3. Let X be a finite set. A matrix M : X ×X → {0, 1} satisfying for each x ∈ X there is exactly one y ∈ X such that M [x, y] = 1 and for each y ∈ X there is exactly one x ∈ X such that M [x, y] = 1 is called a permutation matrix on X. If P (X) is the set of permutation matrices on X, prove that P (X) is a multiplicative group and that θ : Sym (X)→ P (X) defined by θ(α)[x, y] = { 1 if y = xα 0 otherwise is an isommorphism. Prove that α is even (or odd) if and only if det(θ(α)) is 1 (or −1). 4. An r-cycle is even if and only if r is odd. 5. If |X| > 2, then Alt (X) is generated by the 3-cyles on X. 3.2. GROUP ACTIONS 43 that xgi = xi. Suppose that Gxgi = Gxgj . Then gig −1 j ∈ Gx, and hence xgig −1 j = x. Thus xi = xgi = xgjxj , and consequently xi = xj . Therefore the cosets Gxgi, 1 ≤ i ≤ m are pairwise disjoint. Furthermore, if g ∈ G, then xg = xi for some i, 1 ≤ i ≤ m. Hence xgg −1 i = x. Thus gg−1 i ∈ Gx, and so g ∈ Gxgi. Consequently G = Gxg1∪̇Gxg2∪̇Gxg3∪̇ · · · ∪̇Gxgm Therefore by Lagrange’s Theorem (Theorem 2.2.3)m = |G : Gx| = |G| |Gx| If G acts on Ω, then the orbits under G partition the the objects in Ω. Counting the number of orbits is very useful. For example the number of orbits of subgraphs under the action of S4 is the number of non-isomorphic subgraphs of K4. We will use Lemma 3.2.1 to establish the beautiful and useful theorem of Frobenius, Cauchy and Burnside that counts the number of orbits. First observe that G acts on the subsets of Ω in a natural way. Also, if g ∈ G, let χk(g) denote the number of k-element subsets fixed by g. χk(g) = |{S ⊆ Ω : |S| = k and Sg = S}| If S ⊆ Ω, then Sg = {xg : x ∈ S}. Theorem 3.2.2. Let G be a group acting on the set Ω. Then the number of orbits of ( Ω k ) (the k–element subsets of Ω) under G is ∣∣∣∣(Ω k ) /G ∣∣∣∣ = 1 |G| ∑ g∈G χk(g) Proof. Let Nk = ∣∣∣∣(Ω k ) /G ∣∣∣∣. Define an array whose rows are labeled by the elements of G and whose columns are labeled by the k element subsets of Ω. The (g, S)–entry of the array is a 1 if Sg = S and is 0 otherwise. Thus the sum of the entries of row g is precisely χk(g) and the sum of the entries in column S is |GS |. Hence∑ g∈G χk(g) = ∑ S⊆Ω,|S|=k |GS | (3.1) Now partition the k-element subsets into the Nk orbits O1,O2, . . . ,ONk under G. Choose a fixed representative Si ∈ Oi for each i = 1, 2, . . . , Nk. Then for all S ∈ Oi, |GS | = |GSi | and the right hand side of Equation 3.1 44 CHAPTER 3. PERMUTATIONS may be rewritten and Lemma 3.2.1 can be applied. ∑ g∈G χk(g) = Nk∑ i=1 |GSi | · |SGi | = Nk∑ i=1 |G| = Nk|G| This establishes the result. Example 3.3: Number of non-isomorphic graphs To count the number of graphs on 4 vertices Theorem 3.2.2 can be used as follows. Let G = Sym ({1, 2, 3, 4}) and label the edges of K4 as in Figure 3.1. 3 0 1 2 a c b e d f Figure 3.1: Edge labeling of K4 Each permutation can be mapped to a permutation of the edges. For example g = (1, 2, 3) 7→ (a, b, c)(d, e, f). Thus for instance χ2(g) = 0 and χ3(g) = 2. That is g fixes no subgraphs with 2 edges and 2 subgraphs with 3 edges. We tabulate this information in Table 3.1 for all elements of G. The last row of Table 3.1 Gives Nk the number of non-isomorphic subgraphs of K4 with k–edges, k = 0, 1, 2, . . . , 6. 3.2. GROUP ACTIONS 45 Table 3.1: Numbers of non isomorphic subgraphs in K4 g χ0 χ1 χ2 χ3 χ4 χ5 χ6 (0)(1)(2)(3) 7→ (a)(b)(c)(d)(e)(f) 1 6 15 20 15 6 1 (0)(1)(2, 3) 7→ (a)(b, c)(d, e)(f) 1 2 3 4 3 2 1 (0)(1, 2)(3) 7→ (a, b)(c)(d)(e, f) 1 2 3 4 3 2 1 (0)(1, 2, 3) 7→ (a, b, c)(d, f, e) 1 0 0 2 0 0 1 (0)(1, 3, 2) 7→ (a, c, b)(d, e, f) 1 0 0 2 0 0 1 (0)(1, 3)(2) 7→ (a, c)(b)(d, f)(e) 1 2 3 4 3 2 1 (0, 1)(2)(3) 7→ (a)(b, d)(c, e)(f) 1 2 3 4 3 2 1 (0, 1)(2, 3) 7→ (a)(b, e)(c, d)(f) 1 2 3 4 3 2 1 (0, 1, 2)(3) 7→ (a, d, b)(c, e, f) 1 0 0 2 0 0 1 (0, 1, 2, 3) 7→ (a, d, f, c)(b, e) 1 0 1 0 1 0 1 (0, 1, 3, 2) 7→ (a, e, f, b)(c, d) 1 0 1 0 1 0 1 (0, 1, 3)(2) 7→ (a, e, c)(b, d, f) 1 0 0 2 0 0 1 (0, 2, 1)(3) 7→ (a, b, d)(c, f, e) 1 0 0 2 0 0 1 (0, 2, 3, 1) 7→ (a, b, f, e)(c, d) 1 0 1 0 1 0 1 (0, 2)(1)(3) 7→ (a, d)(b)(c, f)(e) 1 2 3 4 3 2 1 (0, 2, 3)(1) 7→ (a, d, e)(b, f, c) 1 0 0 2 0 0 1 (0, 2)(1, 3) 7→ (a, f)(b)(c, d)(e) 1 2 3 4 3 2 1 (0, 2, 1, 3) 7→ (a, f)(b, d, e, c) 1 0 1 0 1 0 1 (0, 3, 2, 1) 7→ (a, c, f, d)(b, e) 1 0 1 0 1 0 1 (0, 3, 1)(2) 7→ (a, c, e)(b, f, d) 1 0 0 2 0 0 1 (0, 3, 2)(1) 7→ (a, e, d)(b, c, f) 1 0 0 2 0 0 1 (0, 3)(1)(2) 7→ (a, e)(b, f)(c)(d) 1 2 3 4 3 2 1 (0, 3, 1, 2) 7→ (a, f)(b, c, e, d) 1 0 1 0 1 0 1 (0, 3)(1, 2) 7→ (a, f)(b, e)(c)(d) 1 2 3 4 3 2 1 Sum 24 24 48 72 48 24 24 Sum/|G| 1 1 2 3 2 1 1 48 CHAPTER 3. PERMUTATIONS Example 3.5: Quick conjugation in Sn To find an element g ∈ S12 to conjugate α = (1, 2, 3, 4)(5, 6, 7, 8)(9, 10)(11)(12) onto β = (5)(3, 1, 2, 6)(10, 4, 7, 11)(9, 8)(12). We first arrange, anyway we like, the cycles of β under the cycles of α so that k-cycles are under k-cycles k = 1, 2, 3, . . . , n. Remember there are k different ways to write the same k-cycle. α = ( 1 , 2 , 3 , 4 )( 5 , 6 , 7 , 8 )( 9 , 10 )( 11 )( 12 ) β = ( 4 , 7 , 11 , 10 )( 2 , 6 , 3 , 1 )( 9 , 8 )( 12 )( 5 ) Now define g ∈ S12 by g : x 7→ y if x in α appears directly above y in β. In our example we get g = ( 1 2 3 4 5 6 7 8 9 10 11 12 4 7 11 10 2 6 3 1 9 8 12 5 ) = (1, 4, 10, 8)(2, 7, 3, 11, 12, 5)(6)(9) Then g−1αg = β. Indeed g is precisely the permutation defined in Theo- rem 3.2.3. The computation in Example 3.5 also tells us how to compute the centralizer CSn(α) of α in Sn. For after all g ∈ CSn(α) if and only if g conjugates α onto itself. Thus we let α play also the role of β in the above computation. Example 3.6: Computing the centralizer in Sn To compute the central- izer of α = (1, 2, 3)(4, 5, 6)(7, 8)(9). in S9 we use the technique shown in Example 3.2. 3.5. Thus we arrange α under itself in all possible ways and write down the mapping g from one arrangement to the other. The set of all these gs is the centralizer CSn(α). The computations are done in Table 3.3 The centralizer of α are the 36 permutations that appear in the last column of the table. Notice that the number of permutations in Sn that centralize a permutation α ∈ Sn is just the number of ways to arrange the cycles of α under itself so that k-cycles are below k-cycles. If α has tj j-cycles, there are jtj tj ! ways to arrange them, since each can be put in anyone of tj positions and each j-cycle has j equivalent descriptions. Thus we have the following theorem. 3.2. GROUP ACTIONS 49 Table 3.3: The centralizer of α = (1, 2, 3)(4, 5, 6)(7, 8)(9) in S9. Arrangement of α Centralizing element 1. (1, 2, 3)(4, 5, 6)(7, 8)(9) (1)(2)(3)(4)(5)(6)(7)(8)(9) 2. (2, 3, 1)(4, 5, 6)(7, 8)(9) (1, 2, 3) 3. (3, 1, 2)(4, 5, 6)(7, 8)(9) (1, 3, 2) 4. (1, 2, 3)(5, 6, 4)(7, 8)(9) (4, 5, 6) 5. (2, 3, 1)(5, 6, 4)(7, 8)(9) (1, 2, 3)(4, 5, 6) 6. (3, 1, 2)(5, 6, 4)(7, 8)(9) (1, 3, 2)(4, 5, 6) 7. (1, 2, 3)(6, 4, 5)(7, 8)(9) (4, 6, 5) 8. (2, 3, 1)(6, 4, 5)(7, 8)(9) (1, 2, 3)(4, 6, 5) 9. (3, 1, 2)(6, 4, 5)(7, 8)(9) (1, 3, 2)(4, 6, 5) 10. (1, 2, 3)(4, 5, 6)(8, 7)(9) (7, 8) 11. (2, 3, 1)(4, 5, 6)(8, 7)(9) (1, 2, 3)(7, 8) 12. (3, 1, 2)(4, 5, 6)(8, 7)(9) (1, 3, 2)(7, 8) 13. (1, 2, 3)(5, 6, 4)(8, 7)(9) (4, 5, 6)(7, 8) 14. (2, 3, 1)(5, 6, 4)(8, 7)(9) (1, 2, 3)(4, 5, 6)(7, 8) 15. (3, 1, 2)(5, 6, 4)(8, 7)(9) (1, 3, 2)(4, 5, 6)(7, 8) 16. (1, 2, 3)(6, 4, 5)(8, 7)(9) (4, 6, 5)(7, 8) 17. (2, 3, 1)(6, 4, 5)(8, 7)(9) (1, 2, 3)(4, 6, 5)(7, 8) 18. (3, 1, 2)(6, 4, 5)(8, 7)(9) (1, 3, 2)(4, 6, 5)(7, 8) 19. (4, 5, 6)(1, 2, 3)(7, 8)(9) (1, 4)(2, 5)(3, 6) 20. (4, 5, 6)(2, 3, 1)(7, 8)(9) (1, 4, 2, 5, 3, 6) 21. (4, 5, 6)(3, 1, 2)(7, 8)(9) (1, 4, 3, 6, 2, 5) 22. (5, 6, 4)(1, 2, 3)(7, 8)(9) (1, 5, 2, 6, 3, 4) 23. (5, 6, 4)(2, 3, 1)(7, 8)(9) (1, 5, 3, 4, 2, 6) 24. (5, 6, 4)(3, 1, 2)(7, 8)(9) (1, 5)(2, 6)(3, 4) 25. (6, 4, 5)(1, 2, 3)(7, 8)(9) (1, 6, 3, 5, 2, 4) 26. (6, 4, 5)(2, 3, 1)(7, 8)(9) (1, 6)(2, 4)(3, 5) 27. (6, 4, 5)(3, 1, 2)(7, 8)(9) (1, 6, 2, 4, 3, 5) 28. (4, 5, 6)(1, 2, 3)(8, 7)(9) (1, 4)(2, 5)(3, 6) 29. (4, 5, 6)(2, 3, 1)(8, 7)(9) (1, 4, 2, 5, 3, 6) 20. (4, 5, 6)(3, 1, 2)(8, 7)(9) (1, 4, 3, 6, 2, 5) 31. (5, 6, 4)(1, 2, 3)(8, 7)(9) (1, 5, 2, 6, 3, 4) 32. (5, 6, 4)(2, 3, 1)(8, 7)(9) (1, 5, 3, 4, 2, 6) 33. (5, 6, 4)(3, 1, 2)(8, 7)(9) (1, 5)(2, 6)(3, 4) 34. (6, 4, 5)(1, 2, 3)(8, 7)(9) (1, 6, 3, 5, 2, 4) 35. (6, 4, 5)(2, 3, 1)(8, 7)(9) (1, 6)(2, 4)(3, 5) 36. (6, 4, 5)(3, 1, 2)(8, 7)(9) (1, 6, 2, 4, 3, 5) 50 CHAPTER 3. PERMUTATIONS Table 3.4: Numbers of non isomorphic subgraphs in K4 type(g) |K(g)|vert.edges|K(g)|χ0|K(g)|χ1|K(g)|χ2|K(g)|χ3|K(g)|χ4|K(g)|χ5|K(g)|χ6 1 14 16 1 · 1 1 · 2 1 · 3 1 · 4 1 · 3 1 · 2 1 · 1 6 12 21 12 22 6 · 1 6 · 2 6 · 3 6 · 4 6 · 3 6 · 2 6 · 1 8 11 31 32 8 · 1 8 · 0 8 · 0 8 · 2 8 · 0 8 · 0 8 · 1 3 22 12 22 3 · 1 3 · 2 3 · 3 3 · 4 3 · 3 3 · 2 3 · 1 6 41 21 41 6 · 1 6 · 0 6 · 1 6 · 0 6 · 1 6 · 0 6 · 1 Sum 24 24 48 72 48 24 24 Sum |G| 1 1 2 3 2 1 1 Theorem 3.2.4. If g ∈ Sn has type(g) = 1t1 2t2 · · · ntn , then the order2 of the centralizer of g is |CSn(g)| = n∏ j=1 jtj tj !. Putting this all together we obtain the following very useful corollary. Corollary 3.2.5. The number of element of type 1t1 2t2 · · · ntn in Sn is n!/ ∏n j=1 j tj tj !. Proof. If g ∈ Sn has type(g) = 1t1 2t2 · · · ntn , then by Theorem 3.2.3 all elements of this type are conjugate to g. The number of these is thus the size |KSn(g)| of the conjugacy class of g. By Theorem 2.7.1 we have |KSn(g)| = (n!)/|CSn(g)|. Now apply Theorem 3.2.4 to obtain the desired result. Using this concept of type the computations in Table 3.1 can be simplified. The new calculations are presented in simple Table 3.1. We close this section with an application of Corollary 3.2.5. There is im- portant discussion in the proof of the following theorem and the reader is encouraged to study it. Theorem 3.2.6. The alternating group A4 is a group of order 12 with no subgroup of order 6. Proof. If H is a subgroup of order 6 in A4, then the |A4 : H| = 2, and thus H is normal in A4. Consequently, H is a union of conjugacy classes in A4. 2This is not formal notation but the actual product of the numbers involved in the notation for the type. 3.3. CAYLEY’S THEOREM 53 3.3 Cayley’s theorem In 1854 Author Cayley gave a one-to-one correspondence between an ar- bitrary finite group G and a subgroup of the symmetric group degree |G|. Burnside attributes the first proof that correspondence was a homomor- phism to Jordan, but the first published proof is by Walther Dyck in 1882. Nevertheless it has become known as Cayley’s theorem. If G is a finite group, then G acts on the the elements of G by right mul- tiplication: g 7→ xg. The kernel of the action is K = {x ∈ G : xg = x}. But if xg = x, then g = 1 and hence K = 1. Furthermore xg = yg if and only if x = y and so the right multiplication map g 7→ xg is a one-to-one homomorphism and we have Cayley’s theorem. Theorem 3.3.1. Every finite group G is isomorphic to a subgroup of Sym (G). This representation of G as a group of permutations of degree |G| is called the right regular representation of G. In Exercise 3.1.1.3 it is shown that there is a isomorphism between Sym (X) and P(X) the set of permutation matrices index by X. Because the entries of a permutation matrix are only 0 and 1, we may regard them as living in an arbitrary field F. Thus we have the following corollary to Cayley’s theorem. Theorem 3.3.2. If G is a finite group of order n and F is a field, then G is isomorphic to a subgroup of GLn(F) the multiplicative group of invertible n by n matrices with entires in F . In general the group GLd(F) of invertible d by d matrices is called the general linear group of degree d over the filed F and the determinant map det : GLd(F)→ F has kernel SLd(F) the special linear group of matrices with determinant 1. Note that GLd(F)/SLd(F) ∼= F? the multiplicative group of non-zero elements of the filed F. If ∆ : G 7→ GLd(R), for some degree d then ∆ is said to be a representation of G of degree d. The degree d need not be the order |G|. Now we consider the action on the right cosets of a subgroup. Let Sn denote the symmetric group of degree n. Theorem 3.3.3. Let H be a subgroup of index n in a group G There is a homomorphism of G into Sn whose kernel is CoreG (H) = ⋂ x∈G x−1Hx. 54 CHAPTER 3. PERMUTATIONS Proof. Let Ω = {Hg1, Hg2, . . . ,Hgn} the n cosets of H in G and let G act on Ω by right multiplication inducing the homomorphism into Sym (Ω) ∼= Sn given by z 7→ ( Hg1 Hg2 · · · Hgi · · · Hgn Hg1z Hg2z · · · Hgiz · · · Hgnz ) The z ∈ G is in the kernel of this homomorphism if and only if Hgiz = Hgi for all i = 1, 2, . . . , n Hence z ∈ g−1 i Hgi for all i = 1, 2, . . . , n . Thus z ∈ x−1Hx for all x and so z ∈ ∩x∈Gx−1Hx as was to be shown. 3.3.1 Exercises 1. Prove that every finite group can be embedded into a group that can be generated by 2 elements. 2. Let H be a subgroup of G and prove that CoreG (H) is the laregest normal subgroup of G contained in H. 3. Let H be a subgroup of G and suppose CoreG (H) = {1}. Prove that G has a faithful representation of degree |G : H| over any field F. 4. A simple group G containing a proper subgroup of index n can be embedded in Sn. 5. Let H be a subgroup of index p in the finite group G, where p is the smallest prime divisor of |G|. Prove that H is a normal subgroupof G. 3.4. THE SYLOW THEOREMS 55 3.4 The Sylow theorems Definition 3.6: A finite group G is a p-group if |G| = px, for some prime p and positive integer x. A maximal p-subgroup of a finite group G is called a Sylowp-subgroup subgroup-of G. If P is a Sylow p-subgroup of G and H is a p-subgroup of G such that P ⊆ H, then H = P . Definition 3.7: Let H be a subgroup of a group G. A subgroup S of G is conjugate to H if and only if S = g−1Hg for some g ∈ G. Notice that conjugate subgroups are isomorphic. Definition 3.8: Let H be a subgroup of G. The normalizer of H in G is NG(H) = {g ∈ G : g−1Hg = H} The normalizer NG(H) of H in G is the largest subgroup of G in which H is normal. We establish two technical lemmas. Lemma 3.4.1. Let P be a Sylow p-subgroup of G. Then NG(P )/P has no element whose order is a power of p except for the identity. Proof. Suppose g ∈ NG(P )/P has order a power of p. Let S = 〈g〉 a subgroup of NG(P )/P . Then there is a subgroup S of G containing P such that S = S/P , (See Theorem 2.6.5.) Because S and P are both p-groups, it follows that S is a p-group. But the maximality of P implies P = S. Therefore S = {1} and g = 1. Lemma 3.4.2. Let P be Sylow p-subgroup of G and let g ∈ G have order a power of p. If g−1Pg = P , then g ∈ P . Proof. Because g ∈ NG(P ), then gP ∈ NG(P )/P . Furthermore g has order a power of p, so therefore gP has order a power of p But by Lemma 3.4.1 gP is P the identity of NG(P )/P . Consequently g ∈ P . 58 CHAPTER 3. PERMUTATIONS 3.5 Some applications of the Sylow theorems Definition 3.9: Let H and K be groups the direct product of H and K is the group H ×K H ×K = {(h, k) : h ∈ H and k ∈ K} with multiplication (h1, k1)(h2, k2) = (h1h2, k1k2). Theorem 3.5.1. Let H and K be subgroups of the group G. If (1) G = HK, (2) H and K are both normal subgroups of G, and (3) H ∩K = {1}, then G ∼= H ×K Proof. First of all we have from (1) that every element g ∈ G can be written as a product g = hk where h ∈ H and k ∈ K. Property (3) shows that the choice of h and k is unique, for if h1k1 = h2k2, then h−1 2 h1 = k2k −1 1 ∈ H∩K. And so h1 = h2 and k1 = k2. This says that the map θ : G → H × K given by hk 7→ (h, k) is well defined. It is obviously onto. To see that it is a homomorphism first consider arbitrary elements h ∈ H and k ∈ K. Then h−1k−1hk = h−1(k−1hk) ∈ H because H is normal h−1k−1hk = (h−1k−1h)k ∈ K because K is normal Thus by (3) h−1k−1hk = 1 and so hk = kh for all h ∈ H, k ∈ K. Now let g1 = h1k1 g2 = h2k2 be elements of G, h1, h2 ∈ H and k1, k2 ∈ K. Then θ(g1g2) = θ(h1k1h2k2) = θ(h1h2k1k2) = (h1h2, k1k2) = (h1, k1)(h2, k2) Therefore θ is a homomorphism. Furthermore g = hk ∈ kernel (θ) if and only if θ(hk) = (h, k) = (1, 1). Thus kernel (θ) = {1}, and therefore θ : G→ H ×K is an isomorphism. Corollary 3.5.2. If gcd (m,n) = 1, then Zmn ∼= Zm × Zn. Proof. We know by Theorem 2.3.3 that Zmn has a subgroup H ∼= Zm and a subgroup K ∼= Zn. These are normal subgroups because Zmn is abelian. Furthermore, gcd (m,n) = 1 so H ∩ K = {1}. Therefore Theorem 3.5.1 gives the result. 3.5. SOME APPLICATIONS OF THE SYLOW THEOREMS 59 Definition 3.10: The dihedral group Dn, n ≥ 2 is a group of order 2n generated by two elements a and b satisfying the relations an = 1, b2 = 1, and bab = a−1 The relations for the dihedral group show that ba = a−1b and hence any product written in the generators a and b is equal to a product of the form aibj where 0 ≤ i < n and 0 ≤ j < 2. Thus Dn will have 2n elements should it exist. But of course it exists. It is the symmetry group of an n-gon is a dihedral group Dn. In fact we may take a and b to be the functions on Zn defined by a : x 7→ x+ 1 (modn) and and b : x 7→ −x (modn) Then an(x) = b2(x) = x for all x ∈ Zn. Hence an and b2 are the identity function. Also, xbab = (xb)ab = (−x)ab = ((−x)a)b = (−x+ 1)b = (x− 1) = xa −1 Theorem 3.5.3. Every group of order 2p is either cyclic or dihedral, when p is an odd prime. Proof. Let G be a group of order 2p. Then by Cauchy’s Theorem (Theo- rem 2.7.4), G contains an element a of order p and an element b of order 2. Let H = 〈a〉, then |G : H| = 2 and so H is normal in G. Therefore bab = ai for some i, because b−1 = b. Now a = b2ab2 = b(bab)b = b(ai)b = (bab)i = (ai)i = ai 2 Thus ai 2−1 = 1, and so p (i2 − 1). Consequently, p (i− 1) or p (i+ 1). If p (i−1), then ai−1 = 1, so ai = a, hence bab = a. Therefore G is abelian. So 〈b〉 is normal in G and therefore applying Theorem 3.5.1 we have that G is isomorphic to the direct product 〈a〉 × 〈b〉 ∼= Zp × Z2 ∼= Z2p, because gcd (2, p) = 1. Therefore G is cyclic. If p (i + 1), then ai+1 = 1, so ai = a−1, hence bab = a−1. Therefore G is dihedral. Theorem 3.5.4. Let G be a group of order |G| = pq, where p > q are primes. If q does not divide p− 1, then G is cyclic. Proof. Let Np be the number of Sylow p-subgroups, then Np ≡ 1 mod p and Np divides q. Hence Np = 1 because p > q. Therefore the Sylow p-subgroup H is normal in G. 60 CHAPTER 3. PERMUTATIONS Let Nq be the number of Sylow q-subgroups, then Nq ≡ 1 mod q and Nq divides p. Hence Nq = 1 or p because p is a prime. If Nq = p, we have p ≡ 1 mod q and so q divides p − 1 contrary to the hypothesis. Therefore Nq = 1 and the Sylow q-subgroup K is normal in G. Obviously H ∩K = {1} so Theorem 3.5.1 applies and we see that G ∼= H ×K ∼= Zp × Zq ∼= Zpq. Consequently, G is cyclic as claimed. Theorem 3.5.5. Let G is a group of order |G| = pq, where p > q are primes. If q divides p− 1, then either G is cyclic or G is generated by two elements a and b satisfying ap = 1, bq = 1, and b−1ab = ar, where r 6≡ 1 (mod p), but rq ≡ 1 (mod p). Proof. By Cauchy’s Theorem (Theorem 2.7.4) there exists an elements a, b ∈ G of order p and q respectively. The proof of Theorem 3.5.4 shows that 〈a〉 E G and if 〈b〉 E G, then G is cyclic. Furthermore if q - p − 1, then necessarily 〈b〉 E G. So we suppose q (p−1) and 〈b〉 is not a normal a subgroup. In particular G is not abelian. Then because 〈a〉 E G, we have bab−1 = ar for some r. Hence r 6≡ 1 (mod p), because G is not abelian. Furthermore for all j bjab−j = bj−1(bab−1)b−(j−1) = bj−1arb−(j−1) = bj−2(barb−1)b−(j−2) = bj−2ar 2 b−(j−2) = bj−3(bar 2 b−1)b−(j−3) = bj−3ar 3 b−(j−3) ... = ar j In particular if j = q, we obtain a = ar q and therefore rq ≡ 1 (mod p). Definition 3.11: The quaterians Q, is a group of order 8 generated by two elements a and b satisfying the relations a4 = 1, b2 = a2, and b−1ab = a−1 3.5. SOME APPLICATIONS OF THE SYLOW THEOREMS 63 Order Number of distinct groups Groups 1 1 {1} 2 1 Z2 3 1 Z3 4 2 Z4,Z2 × Z2 5 1 Z5 6 2 Z6,S3 ∼= D3 7 2 Z7 8 5 Z8, Z2 × Z4, Z2 × Z2 × Z2,D4,Q 9 2 Z9, Z3 × Z3 10 2 Z10, D5 11 2 Z11 12 5 Z12,Z2 × Z6,Z2 × S3,A4, G12 13 2 Z13 14 2 Z14, D7 15 2 Z15 64 CHAPTER 3. PERMUTATIONS 3.6 Simplicity of the alternating group Theorem 3.6.1. If |X| ≥ 5, then An ∼= Alt (X) is a simple group. Proof. We prove this by induction on n; the case n = 5 is already done. (See Exercise 3.2.1 3b.) Let n ≥ 6, and G = Alt (X), where X = {1, 2, . . . , n}. For each x ∈ X the stabilizer Gx is isomorphic to An−1 and by induction is simple. Given x, y ∈ X, x 6= y, choose z ∈ X \ {x, y} and set σ = (x, y, z), Then σ ∈ G and Gy = σ−1Gxσ. Thus the stabilizers Gx, x ∈ X are all conjugate in G . Suppose G is not simple. Then there a subgroup N E G and thus for all x ∈ X, N ∩Gx = {I} or N ∩Gx = Gx. If N ∩Gx = Gx for some x, then in fact N ∩Gy = Gy for all y ∈ X, because Gy is conjugate to Gx for each y ∈ X. So because N E G, if N contains Gx for some x, it contains Ny for all y ∈ X. But, because n ≥ 6, any product of two transpositions is in Gx for some x, and any element of G = An is a product of such permutations. So, N = An. On the other hand, if N ∩ Gx = {I} for each x ∈ X, then no element of N has a fixed point, except I. Consider some σ ∈ N and write σ as a product of disjoint cycles. If σ has an r-cycle (i1, i2, . . . , ir), where r ≥, let ρ = (i3, j, k),,where j, k 6∈ {i1, i2, i3}. This is possible, because n ≥ 6. Let τ = ρ−1σρ. Then τ ∈ N , and σ 6= τ , because σ : i2 7→ i3 and τ : i2 7→ j. But στ−1 fixes i1, a contradiction. Consequently any σ ∈ N is eithr I or a product of transpositions. Suppose σ = (i, j)(k, `) · · · ∈ N . Let ρ = (`, p, q) with p, q 6∈ {i, j, k, `}; again, this is possible, because n ≥ 6. Then τ = ρ−1σρ ∈ N , and στ−1 fixes i. But σ : k 7→ `, while τ : k 7→ p, so σ 6= τ . This is a contradiction and thus in case we must have N = {I}. Therefor, we have proven that either N = G or N = {I}. That is, An ∼= G is simple. Chapter 4 Finitely generated abelian groups 4.1 The Basis Theorem The usual custom for abelian groups is to adopt additive notation instead of multiplicative. The following table provides the translation: Multiplication ←→ Addition ab ←→ a+ b 1 ←→ 0 a−1 ←→ −a an ←→ na ab−1 ←→ a− b aH ←→ a+H direct product ←→ direct sum H ×K ←→ H ⊕K m∏ i=1 Hi ←→ m∑ i=1 Hi Let A be an abelian group. 1. If a, b ∈ A and n ∈ Z, then n(a+ b) = na+ nb. 2. If A = 〈a1, a2, . . . , ak〉, then A = {n1a1 + n2a2 + · · · + nnak : n1, n2, . . . , nk ∈ Z} the set of all linear combinations of the elements a1, a2, . . . , ak with integer coefficients. 65 68 CHAPTER 4. FINITELY GENERATED ABELIAN GROUPS Proof. (Induction on |A|.) If A is cyclic we’re done, so suppose A is not cyclic. Then by Lemma 4.1.4 A has at least two cyclic subgroups of order p while C can have only one. Hence there exist a cyclic subgroup K ≤ A of order p, that is not contained in C. Consequently by Theorem 2.6.3 (Second law) (C +K)/K ∼= C/(C ∩K) = C Given any g ∈ A, by the Law of correspondence (Theorem 2.6.5) the order of g+K in A/K divides |g| which is at most |C|, because C is a cyclic subgroup of maximal order. Thus the subgroup (C + K)/K is a cyclic subgroup of maximal order in A/K and we can apply the inductive hypothesis to obtain A/K = (C +K)/K ⊕H/K where K ≤ H ≤ A. But this means for all a ∈ A there are c ∈ C, k ∈ K and h ∈ H such that a+K = ((c+k)+K)+(h+K) = (c+h+k)+K = (c+h)+k+K = (c+h)+K. Thus A ⊆ (C +H) +K = C + (H +K) = C +H, and so A = C +H. Now because (C +K)/K intersects H/K trivially we have H ∩ (C +K) = K, and thus H ∩C = {0}, and hence the sum C +H is a direct sum, i.e. A = C ⊕H. Recursively applying Lemma 4.1.5 yields corollary 4.1.6. Corollary 4.1.6. Every finite abelian p-group is the direct sum of cyclic groups. Theorem 4.1.7. (Basis theorem) Every finite abelian group is the direct sum of cyclic p-groups. Proof. Let A be a finite abelian group. Use Theorem 4.1.3 to write A = P1 ⊕ P2 ⊕+ · · ·+ Pk, where Pi is the Sylow pi-subgroup of A, i = 1, 2, . . . , k. Use Corollary 4.1.6 to replace each Pi with a direct sum of cyclic groups. Corollary 4.1.8. (Canonical decomposition) Every finite abelian group A is isomorphic to a direct sum of cyclic groups satisfying A ∼= ZD1 ⊕ ZD2 ⊕ · · · ⊕ ZDs , (4.2) where Di Di+1, i = 1, 2, . . . , s − 1. Furthermore the positive integers Di, i = 1, 2, . . . , s are unique. 4.1. THE BASIS THEOREM 69 Proof. Use the Basis theorem to decompose A into cyclic p-group sum- mands. For each prime divisor pi of |A|, i = 1, 2, . . . , k let Zdi be be a cyclic summand of largest order with pi di. Then because direct sums are commutative and associative, we have A ∼= (Zd1 ⊕ Zd2 ⊕ · · · ⊕ Zdk)⊕A1, where A1 is the direct sum of the remaining cyclic summands. By Theo- rem 3.5.2 A ∼= Zδ1 ⊕A1 where δ1 = ∏k i=1 di. Repeating this process on A1, we may write A1 ∼= Zδ2 ⊕ A2. Moreover δ2 δ1, because we always choose from what remains the cyclic p-group summands of largest order. Because A is finite, this process will end in a finite number of say s steps. Setting Di = δs−i+1, i = 1, 2, . . . , s, we obtain (4.2). To see that the Di are unique. Observe that |A| = ∏s i=1Di and that DsA = {0}, because Ds = δ1 is the order of the largest cyclic subgroup. Let m be the smallest positive integer such that mA = {0}. Then m ≥ Ds. Write by the division algorithm Ds = qm+ r, where 0 ≤ r < m. Then {0} = DsA = (qm+ r)A = qmA+ rA = {0}+ rA = rA Consequently, r = 0, so m Ds and thus m = Ds. Therefore if A ∼= ZE1 ⊕ ZE2 ⊕ · · · ⊕ ZEt , where Ei Ei+1, i = 1, 2, . . . , t − 1 is some other decomposition of A into cyclic groups, then Et = Ds. Proceeding inductively we see that t = s and Ei = Di for i = 1, 2, . . . , s. The decomposition of A given in Theorem 4.1.8 is called the canonical decomposition. 4.1.1 How many finite abelian groups are there? Consider an abelian group A of order pn, pprime. The canonical decom- position of A is of the form A ∼= Zpn1 × Zpn2 × · · · × Zpns , where n1, n2, . . . , ns are such that n1 + n2 + . . .+ ns = n, n1 ≤ n2 ≤ . . . ≤ ns } (4.3) 70 CHAPTER 4. FINITELY GENERATED ABELIAN GROUPS Solutions to Equation 4.3 are called integer partitions. The notation P(m) is used to denote the number of partitions of m; P(m) is called a partition number . The first few partition numbers are P(1) = 1, P(2) = 2, P(3) = 3, P(4) = 5, P(5) = 7 and P(6) = 11. As an example, we list the 11 different partitions of the integer 6: 6, 1 + 5, 2 + 4, 1 + 1 + 4, 3 + 3, 1 + 2 + 3, 1 + 1 + 1 + 3, 2 + 2 + 2, 1 + 1 + 2 + 2, 1 + 1 + 1 + 1 + 2, 1 + 1 + 1 + 1 + 1 + 1 This means there are 11 abelian groups of order p6, namely: Zp6 Zp ⊕ Zp5 Zp2 ⊕ Zp4 Zp ⊕ Zp ⊕ Zp4 Zp3 ⊕ Zp3 Zp ⊕ Zp2 ⊕ Zp3 Zp ⊕ Zp ⊕ Zp ⊕ Zp3 Zp2 ⊕ Zp2 ⊕ Zp2 Zp ⊕ Zp ⊕ Zp2 ⊕ Zp2 Zp ⊕ Zp ⊕ Zp ⊕ Zp ⊕ Zp2 Zp ⊕ Zp ⊕ Zp ⊕ Zp ⊕ Zp ⊕ Zp Although partitions have been studied by mathematicians for hundreds of years and many interesting results are known, there is no known formula for the values P(m). The growth rate of P(m) is known however; it can be shown that P(m) ∼ 1 4m √ 3 eπ √ 2m/3 as m→∞ For a discussion on computing integer partitions see Section 3.1 of Com- binatorial algorithms: generation, enumeration and search by D.L. Kreher and D.R. Stinson. The following Theorem should be apparent. Theorem 4.1.9. The number of abelian groups order n = pe11 p e2 2 · · · p ek k is P(e1)P(e2) · · ·P(ek). Example 4.1: Abelian groups of order 72. Writing 72 = 2332 we see that the Sylow 2-subgroup is an abelian group of order 8 and there are P(3) = 3 such groups, namely Z8,Z2 ⊕ Z4 and Z2 ⊕ Z2 ⊕ Z2 Also the Sylow 3-subgroup is an abelian group of order 9 and there are P(2) = 2 such groups, namely Z9,Z3 ⊕ Z3. Combining these decompositions and using the proof of Corollary 4.1.8 we find the 6 groups of order 54. They are displayed below. 4.2. GENERATORS AND RELATIONS 73 for some ni ∈ Z, i ∈ I and finitely many non-zero. Then by Theorem ??, mi = pni for all i ∈ I. Thusmi ≡ 0 (mod p) for all i ∈ I. This contradiction shows independence. Hence {xi : i ∈ I} is a basis for F/pF as a vector space over Zp. Thus |I| = Dim (F/pF ). Consequently if F ∼= G, then |I| = Dim (F/pF ) = |J |. Definition 4.3: If F is free abelian on {xi : i ∈ I} and G is free abelian on {yj : j ∈ J} and I and J have the same cardinality, then we say F and G have the same rank. If I is finite and |I| = n, then we say that F has rank n. The set {xi : i ∈ I} is called a basis. Theorem 4.2.3. Le F be free abelian with basis X = {xi : i ∈ I}, G an arbitrary abelian group and f : X → G any function. Then there is a unique homomorphism θ : F → G such that θ(xi) = f(xi), for all i ∈ I Proof. If Zi = 〈xi〉, define fi : Zi → G by fi(mxi) = mf(xi). It is easy to see that fi is a homomorphism. To define θ let x ∈ F . Then then there are uniquely determined integer coefficients such that x = ∑ i∈I cixi. We define θ by θ(x) = ∑ i∈I fi(cixi) = ∑ i∈I cif(xi). Because each fi is a homomorphism it follows that θ is a homomorphism. If ψ : F → G is another homomorphism such that ψ(xi) = f(xi), for all i ∈ I, then ψ(x) = ∑ i ψ(cixi) = ∑ i ciψ(xi) = ∑ i cif(xi) = θ(x). Theorem 4.2.4. Every abelian group A is a quotient of a free abelian group. Proof. Let Za be the infinite cyclic group with generator a and set F =∑ a∈A Za, the free group with basis A. Let f : A → A be the identity function, i.e., f(a) = a for all a ∈ A. Clearly f is onto. By Theorem 4.2.3 f extends to a homomorphism θ onto A. Therefore by Theorem 2.6.1 A ∼= F/ ker θ. Hence A is a quotient of the free abelian group F . 74 CHAPTER 4. FINITELY GENERATED ABELIAN GROUPS Definition 4.4: An abelian group A has generators X = {x1, x2, . . . , xn} and relations k∑ j=1 aijxn = 0, i = 1, 2, . . . ,m in case A ∼= F/R, where F is a free abelian on X and R is the subgroup generated by { ∑n j=1 aijxn : i = 1, 2, . . . ,m} Example 4.2: 1. A = Z15 has generator x and relation 15x = 0. 2. A = Z15 has generators x, y and relations 3x = 0, 5x = 0. 3. A = Zp∞ has generators {x1, x2, x3, . . .} and relations px1 = 0, px2 = x1, px3 = x2, . . . 4.2.1 Exercises 1. Prove that a direct summand of a finitely generated abelian group is also finitely generated. 2. Show that every subgroup H of a finitely generated abelain group A is itself finitely generated, and furthermore if A can be generated by r elements, then H can be generated by r elements. 3. Show that the multiplicative group Q? of positive rationals is a free abelian group of (countably) infinite rank. 4.3. SMITH NORMAL FORM 75 4.3 Smith normal form In 1858 Heger formulated conditions for the solvability of the Diophantine equations M~x = ~b in the case where M has full rank over Z. In 1861, the problem was solved in full generality by H.J.S. Smith. Theorem 4.3.1 appeared in a form close to the one above in an 1868 treatise by Frobenius who generalized Heger’s theorem and emphasized the unimodularity of the transformations. Theorem 4.3.1. (Smith normal form) Suppose M is an m by n integer matrix (m ≤ n). Then there exist integer matrices P and Q that have determinant ±1 such that PMQ = D where D = [∆, Z], ∆ is a diagonal matrix with entries d1, d2, . . . , ds, 0, . . . , 0, satisfying d1 ≥ 1 and d1 | d2 | . . . | ds and where Z is the m by n − m matrix of zeros. This is called the Smith normal form of M . Proof. The matrix P will be a product of matrices that define elementary row operations and Q will be a product corresponding to elementary column operations. The elementary operations are: 1. Add an integer multiple of one row to another (or a multiple of one column to another). 2. Interchange two rows (or two columns). 3. Multiply a row (or column) by −1. Each of these operations is given by left or right multiplying by an invertible matrix E with integer entries, where E is the result of applying the given operation to the identity matrix, and E is invertible because each operation can be reversed using another row or column operation over the integers. It also easy to see that the determinant of E is ±1. To see that the proposition must be true, assume M 6= 0 and perform the following steps. 1. By permuting rows and columns, move a nonzero entry of M with smallest absolute value to the upper left corner of M . Now attempt to make all other entries in the first row and column 0 by adding multiples of row or column 1 to other rows (see step 2 below). If an operation produces a nonzero entry in the matrix with absolute value smaller than |m11|, start the process over by permuting rows and columns to move that entry to the upper left corner of M . Because 78 CHAPTER 4. FINITELY GENERATED ABELIAN GROUPS Transpose (To perform column operations.) 1 0 0 0 0 2 0 0 0 0 18 36 0 0 −6 −12 0 0 −6 −12 1 0 0 0 0 1 0 0 −1 0 8 0 1 −5 0 −3 1 0 2 0 0 0 0 −1 1 ← QT 1 0 0 0 0 2 0 0 0 0 −6 −12 0 0 18 36 0 0 −6 −12 1 0 0 0 0 1 0 0 −1 0 −3 1 0 2 0 8 0 1 −5 0 0 0 0 −1 1 ← QT 3 1 −1 1 −1 1 0 0 0 0 2 0 0 0 0 6 12 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 −1 0 3 −1 0 −2 0 −1 3 1 1 0 3 −1 0 −3 1 ← QT Transpose (To perform row operations.) 1 0 0 0 0 0 2 0 0 0 0 0 6 0 0 0 0 12 0 0 0 0 1 0 0 1 2 0 1 −2 −2 0 0 −3 −3 1 ← P −2 1 1 0 0 0 0 0 2 0 0 0 0 0 6 0 0 0 0 0 0 0 0 0 1 0 0 1 2 0 1 −2 −2 0 −2 1 1 1 ← P 4.4. APPLICATIONS 79 Then P =  0 0 1 0 0 1 2 0 1 −2 −2 0 −2 1 1 1  Q =  1 1 3 −1 3 0 0 −1 3 −1 0 0 0 1 0 0 −1 −2 1 −3 0 0 0 0 1  PMQ =  1 0 0 0 0 0 2 0 0 0 0 0 6 0 0 0 0 0 0 0  4.4 Applications 4.4.1 The fundamental theorem of finitely generated abelian groups Corollary 4.4.1. ( Fundamental theorem of finitely generated abelian groups ) If A is a finitely generated abelian group with generators x1, x2,. . ., xn. and relations 0 = n∑ j=1 mi,jxj for i = 1, 2, 3, . . . ,m, mi,j ∈ Z, then A ∼= Zd1 ⊕ Zd2 ⊕ · · · ⊕ Zdt ⊕ Z⊕ Z⊕ · · ·Z︸ ︷︷ ︸ m−t times , where d1|d2| · · · |dt and are as obtained in Theorem 4.3.1, and t = rank([mi,j ]). Proof. Let M = [mi,j ], 1 ≤ i ≤ m, 1 ≤ j ≤ n and use Smith Normal Form (Theorem 4.3.1) to find unimodular matrices P and Q such that PMQ = [∆, Z] = D, where ∆ is a diagonal matrix with entries d1, d2, . . . , dt, 0, . . . , 0, 80 CHAPTER 4. FINITELY GENERATED ABELIAN GROUPS satisfying d1 ≥ 1 and d1 | d2 | . . . | dt and where Z is the m by n − m matrix of zeros. Let F be the free abelian group with basis {x1, x2, . . . , xn}. Taking advan- tage (or abuse) of linear algebra notation. Let ~x = (x1, x2, . . . , xn), then F = {~z · ~x : ~z ∈ Zn} and N = {~w · M~x : ~w ∈ Zm} is a subgroup of F such that A ∼= F/N . Let ~y = (y1, y2, . . . , yn) = Q−1~x and let F ′ be the free abelian group with basis {y1, y2, . . . , yn}. Define θ : F → F ′, by θ(~z · ~x) = ~zQ · ~y. Then because Q is invertible, θ is an isomorphism. Claim θ(N) = N ′ = {~w′ ·D~y : ~w′ ∈ Zm}. To see this consider ~w ·M~x an arbitrary element of N . Then θ(~w ·M~x) = θ(~wM · ~x)~wMQ · ~y) = ~wP−1PMQ · ~y = (~wP−1)D · ~y = (~wP−1) ·D~y = ~w′ ·D~y ∈ N ′ where ~w′ = ~wP−1, and thus the claim is true. Hence A = F/N ∼= F ′/N ′. To finish the proof we observe that the epimorphism ϕ : F ′ → Zd1 ⊕ Zd2 ⊕ · · · ⊕ Zdt ⊕ Z⊕ Z⊕ · · ·Z︸ ︷︷ ︸ m−t times Given by ϕ(z1y1 + z2y2 + · · ·+ ztyt + zt+1yt+1 + · · ·+ znyn) = (z1, z2, . . . , zn) has kernel N ′. Example 4.4: Identifying an abelian group Consider the abelian group A with generators X = {a, b, c, d, e} and rela- tions −2a +4c−6d−12e = 0 −2a+2b −4c−4d −4e = 0 1a+1b −3c+1d +1e = 0 −3a−3b+15c−9d−21e = 0 Thus the generators satisfy the matrix equation M · (a, b, c, d, e) = 0, where M is as given in Example 4.3. Computing the Smith canonical form of M see Example ?? we find that PMQ = D = [diag (1, 2, 6, 0) ,~0] (See Example ??). Therefore M ∼= Z1 ⊕ Z2 ⊕ Z6 ⊕ Z0 ∼= Z2 ⊕ Z6 ⊕ Z 4.4. APPLICATIONS 83 Corollary 4.4.3. The m by n system of Diophantine equations M~x = ~b has a solution ~x ∈ Zn if and only if gcd (~yM) divides ~y ·~b for every vector ~y ∈ Zm. We provide two proofs. First proof. Suppose M~x = ~b has integral solution ~x and that ~b ∈ Zm. If g divides each entry of ~yM , then g divides the integral linear com- bination (~yM) · ~x = ~y · (M~x) = ~y ·~b. Conversely suppose gcd (~yM) divides ~y · ~b for every vector ~y ∈ Zm and let ~u ∈ Qm be such that ~uM ∈ Zn. Let 0 < δ ∈ Z be such that ~y = δ~u ∈ Zm. Then because ~uM ∈ Zm, we see that δ gcd (~uM) = gcd (δ~uM) = gcd ((δ~u)M) divides δ~u ·~b. That is the integer δ~u ·~b = Rδ gcd (~uM) for some integer R. Hence ~u ·~b = R gcd (~uM) ∈ Z. Thus Statement (3) of Corollary 4.4.2 holds and therefore so does Statement (4). Second proof. According to Statements (1) and (2) of Corollary 4.4.2, M~x = ~b has has an integer solution ~x if and only if D~y = ~c has an integer solution ~y. The latter is uncoupled and has solution ~y = (y1, y2, . . . , yn) if and only if diyi = ci, for i = 1, 2, . . . , n. For a fixed i and a fixed prime p, this equation has a solution modulo p if and only if gcd (di, p) divides ci. Hence if this equation has a solution modulo p for all primes p, then si must divide ci, or equivalently D~y = ~b has an integer solution ~y. The converse is easier. You can use the integer solution to give you a solution For every prime p. Remarks: The second proof was received by e-mail communication from Bryan Shader. 4.4.3 Exercises 1. Show that every integer matrix M has a unique Smith normal form. 84 CHAPTER 4. FINITELY GENERATED ABELIAN GROUPS 2. Compute the Smith normal form of the matrix M =  2 4 −4 −2 −4 −12 12 8 2 −4 4 6  3. Let A be the abelian group with presentation A = 〈 a, b, c, d : 2a+ 4b− 4c− 2d = 0, −4a− 12b+ 12c+ 8d = 0, 2a− 4b+ 4c+ 6d = 0 〉 . What is the canonical decomposition for A. 4. Let Γ be a graph, set g = gcd (deg(u) : u ∈ V (Γ)) and choose positive integer n satisfying ( n 2 ) ≡ 0 (mod |E(Γ)|) (4.4) n− 1 ≡ 0 (mod g) . (4.5) Let B be the set of subgraphs of the complete graph Kn that are isomorphic to Γ and define the matrix M : E(Kn)× B → {0, 1} by M [e, Y ] = { 1 if e ∈ E(Y ); 0 if e /∈ E(Y ) . (a) Show that there is an integer valued vector ~x such that M~x = J , where J [X] = 1, for all X ∈ B. (b) Show that M has constant row sum c for some positive integer c. (c) Show that there is a constant λ and a positive integer valued vector ~y such that M~y = λJ . (d) Conclude that the complete multi-graph λKn can be edge de- composed into subgraphs isomorphic to X. (The multiigraph λKn has each edge repeated λ times.) Chapter 5 Fields 5.1 A glossary of algebraic systems Semigroup: A semigroup is a set with an associative binary operation. Ring: A ring is a set with two binary operations, addition and multiplica- tion, linked by the distributive laws: a(b+ c) = ab+ ac (b+ c)a = ba+ ca Rings are abelian groups under addition and are semigroups under multiplication. We will assume our rings have the multiplicative iden- tity 1 6= 0. Commutative ring: A commutative ring is a ring in which the multipli- cation is commutative. Domain: A domain (or integral domain) is a ring with no zero divisors, that is ab = 0⇒ a = 0 or b = 0 for all a, b in the domain . Field: A field is a commutative ring in which every nonzero element has a multiplicative inverse. Skew field: A skew field (or division ring) is a ring (not necessarily com- mutative) in which the nonzero elements have a multiplicative inverse. 85 88 CHAPTER 5. FIELDS Proof. Let R be a Euclidean Domain and let deg be the degree function for R. Suppose I C R, choose a ∈ I, a 6= 0 to have the smallest degree. If b ∈ I write b = aq + r where q, r ∈ R and r = 0 or deg(r) < deg(a). But then r = b − aq ∈ I so deg(r) < deg(a) is impossible and thus r = 0 and b = aq ∈ (a). Hence I ⊆ (a) and therefore I = (a). Conversely not every PID is an ED. Some examples of PIDs are Z and F[x]. For example in Z, the ideal (a, b) = (g) where g = gcd (a, b). Thus it is common to drop the prefix and just write (a, b) for the greatest common divisor of a and b. Definition 5.4: An ideal P is a prime ideal , if whenever ab ∈ P , then either a ∈ P or b ∈ P . For example the prime ideals of Z are (p) = pZ = {xp : x ∈ Z}, where p is prime integer. Theorem 5.2.3. 0 6= P C R is a prime ideal if and only if R/P is a domain. Proof. Suppose R = R/P is a domain. Denote α by α = α + P ∈ R. If αβ = 0, then α = 0 or β = 0 but αβ = αβ by definition of multiplication so translating to R we see that αβ ∈ P ⇒ α ∈ P or β ∈ P so P is a prime ideal. Conversely, if P prime ideal, then αβ ∈ P ⇒ α ∈ P or β ∈ P . Thus αβ = 0 ⇒ α = 0 or β = 0. Hence R can have no zero divisors and thus R is a domain. 5.3 The prime field Definition 5.5: A prime field is a field with no proper subfields. Theorem 5.3.1. Every prime field Π is isomorphic to Zp or Q. Proof. Let 1 be the multiplicative identity of Π and set R = {n · 1 : n ∈ Z} ⊂ Π. It is easy to see that Π is a domain. The map θ : Z→ R given by n 7→ n · 1 is a homomorphism of rings and is onto. Let P = kernel (θ), then 5.3. THE PRIME FIELD 89 R = Zp because R is a domain by the previous theorem P is a prime ideal of Z. Thus P = {0} or P = (p), p a prime. If P = {0} then Z ∼= R and therefore Π ⊇ S where S ∼= Q. But it has no proper subfields so Π = S ∼= Q. If P = (p) then R ∼= Zp and therefore Π ∼= Zp. Theorem 5.3.2. Every field F contains a unique prime field Π. Proof. Let Π be the intersection of all subfields of F, 1 ∈ F so Π 6= {0}. This is a subfield having no proper subfields. Therefore Π is a prime field and is clearly unique. Definition 5.6: If the prime field of F is Π ∼= Zp we say F has characteristic p otherwise we say F has characteristic 0. Theorem 5.3.3. Every finite field F has pn elements for some prime p and natural number n. Proof. F is a finite dimensional vector space over its prime field Π. Then |F| = |Π|dimΠF = pn where n = dimΠF. Theorem 5.3.4. The commutative ring R is a field if and only if R con- tains no proper ideals. Proof. Suppose R is a field and let a ∈ R, a 6= 0, then 1 = a−1a ∈ (a) ⇒ R ⊆ (a)⇒ R = (a). Thus R has no proper ideals. Conversely suppose R contains no proper ideal. Then for all a 6= 0, (a) = R. Hence 1 ∈ (a) ⇒ 1 = ra for some r ∈ R and so a has an inverse and therefore R is a field. Definition 5.7: An ideal M of R is a maximal ideal if M 6= R and there is no proper ideal of R that contains M Theorem 5.3.5. M is a maximal ideal of the commutative ring R if and only if R/M is a field. Proof. M C R is a maximal ideal if and only if by the law of correspondence R/M has no ideals if and only if by Theorem 5.3.4 R/M is a field. Corollary 5.3.6. Maximal ideals of commutative rings are prime ideals. 90 CHAPTER 5. FIELDS Proof. If M C R is a maximal ideal, then R/M is a field but fields are domains so therefore M is prime ideal. Theorem 5.3.7. Every prime ideal of a PID is a maximal ideal. Proof. Let R be a PID and suppose P is a prime ideal of R. Let I C R such that P C 6= I C 6=R. Because R is a PID we may choose p, a ∈ R such that P = (p) and I = (a). Then p ∈ (a), so p = ra for some r ∈ R. Thus because P is a prime ideal either r ∈ (p) or a ∈ (p). But if a ∈ (p) then I = (a) ⊆ (p) = P which is a contradiction. Thus r ∈ (p) so r = sp for some s ∈ R and therefore p = spa so p(1 − sa) = 0. Therefore sa = 1 because p 6= 0 and R is a domain. Therefore 1 ∈ (a)⇒ I = (a) = R which is a contradiction. Hence P is maximal. (There are no proper ideals of R that properly contain P .) An element u ∈ R is called a unit if it has an inverse. The units of Z are ±1. In a field F every non-zero element is a unit. In F[x] the constant polynomials are the units. Definition 5.8: An element p ∈ R is an irreducible if and only if in every factorization p = ab either a or b is a unit. If p = uq where u is a unit then p and q are said to be associates. Theorem 5.3.8. If R is a PID, then the non-zero prime ideals of R are the ideals (p), where p is irreducible. Proof. Suppose (p) 6= 0 is a prime ideal of R and let p = ab be a factoriza- tion in R then either a ∈ (p) or b ∈ (p). Say a ∈ (p), then a = sp for some s. Hence p = spb ⇒ p(1 − sb) = 0 ⇒ sb = 1, because p 6= 0, and R is a domain. Therefore b is a unit and hence p is irreducible. Corollary 5.3.9. If R is PID and p ∈ R is irreducible, then R/(p) is a field. Proof. p irreducible⇒ by Theorem 5.3.8 that (p) is a prime ideal. Therefore by Theorem 5.3.7 (p) is a maximal ideal. Consequently by Theorem 5.3.5 R/(p) is a field. Lemma 5.3.10. If F is a field and f(x) is irreducible in F[x] then F[x]/(f(x)) is a field containing a root of f(x) and a subfield isomorphic to F.