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1 [section] [section] 2 Lecture Notes in Abstract Algebra I AUREA Z. ROSAL Polytechnic University of the Philippines Sta. Mesa, Manila Contents 1 Basic Necessities 1 1.1 The n−ary Operations . . . . . . . . . . . . . . . . . . . . . . 1 1.2 Algebraic Systems . . . . . . . . . . . . . . . . . . . . . . . . . 3 2 Groups 6 2.1 Deﬁnitions and Examples . . . . . . . . . . . . . . . . . . . . . 6 2.2 Cayley Table . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 2.3 Some Group Properties . . . . . . . . . . . . . . . . . . . . . . 9 2.4 More Properties . . . . . . . . . . . . . . . . . . . . . . . . . . 15 3 Subgroups 21 4 Cyclic Groups 26 4.1 Division Algorithm and Integers modulo n . . . . . . . . . . . . 26 4.2 Order of an Element and Cyclic Groups . . . . . . . . . . . . . . 30 4.3 Classiﬁcation of Cyclic Groups . . . . . . . . . . . . . . . . . . . 37 4.3.1 The Fundamental Theorem of Finite Cyclic Groups . . . . 40 5 Permutation Groups 44 5.1 Functions, Injection, Surjection, Bijection . . . . . . . . . . . . . 44 5.2 Permutations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47 5.3 Symmetries of a Regular Polygon . . . . . . . . . . . . . . . . . 54 5.4 Cycles and Cyclic Notations . . . . . . . . . . . . . . . . . . . . 56 5.5 Even and Odd Permutations . . . . . . . . . . . . . . . . . . . . 60 5.6 The Alternating Groups . . . . . . . . . . . . . . . . . . . . . . 62 ii Chapter 1 Basic Necessities The study of abstract algebra is concerned primarily in the general properties satisﬁed by algebraic systems such as groups, rings, ﬁelds among others. Precisely what is meant by an algebraic system and their types shall be discussed in this chapter. 1.1 The n−ary Operations Deﬁnition 1.1.1. Given A, B = ∅. A function γ from A to B is a rule that assigns to each element in A a unique element in B. Another name for a function is a mapping . In symbols, γ : A −→ B. We call A as the domain and B as the codomain of γ. If x ∈ A and y ∈ B, such that γ assigns x to y, then we write γ : x −→ y or γ(x) = y. The ele- ment y is known as the image of x with respect to γ while x is called the pre-image of y under γ. The set Imγ := {y ∈ B|y = γ(x) for some x ∈ A} is called the image set or range of γ. Another symbol for range of γ is γ[A]. The function γ may also be expressed as a set of ordered pairs γ = {(x, y)|y = γ(x)}. Notice that γ ⊆ A × B. 1 2 CHAPTER 1. BASIC NECESSITIES Hb 1. 1. Let ρ : A −→ B, where A and B are non empty subsets of R. If ρ(x) = x2 − 6x + 5, then ρ is a function from A to B. What are A and B equal to? 2. Let C = {1, 2, 3, 4, 5} and D = {a, b, c, d} and α = {(1, b), (2, d), (3, a), (4, c), (5, b)}. Then α is a function from C to D. 3. Using the same sets as in (2), let δ = {(1, a), (1, b), (2, c), (3, d), (4, a), (5, d)}. Is δ a function from C to D? A function f whose domain and codomain are equal is said to be a func- tion deﬁned on the set. If A = ∅ we deﬁne An to be the set of n−tuples An := {(a1 , a2 , . . . , an )|ai ∈ A for all i = 1, 2, . . . , n} Deﬁnition 1.1.2. An n−ary operation on a set A = ∅ is a function φ from An to A. In other words, if φ is an n − ary operation on A, then for each ordered n − tuple (a1 , a2 , . . . , an ) a unique element φ(a1 , a2 , . . . , an ) = a ∈ A. If n = 1, we call the operation as unary. If n = 2, then we call the operation as binary. If n = 3, we call the operation ternary or 3 − ary. In this course, we shall concentrate basically on unary and binary operations. Note that a function from a set into itself is unary. Thus a function on set can be considered as an operation on the set. If ∗ is a binary operation on a non empty set A, then we say that A is closed with respect to (or under) ∗. In other words, if x, y ∈ A, then A is closed under ∗ iﬀ ∗(x, y) is a unique element in A. It is customary to write x ∗ y in lieu of ∗(x, y). Hb 2. 1. Let f be a mapping on R. Deﬁne f such that it assigns to each element x ∈ R its additive inverse −x. f is a unary operation on R. 2. The operation ∗ in N such that m ∗ n = mn , is a binary operation on N. Assume N = {1, 2, . . .}. 1.2. ALGEBRAIC SYSTEMS 3 dp 3. Let R[x] be the set of all polynomials over R. If g : p(x) → . What dx type of operation is g on R[x]? 4. Let µ be a rule that assigns (p1 , p2 , p3 ) ∈ R3 to the real number p1 p2 −p3 . Then µ is a ternary operation on R. 1.2 Algebraic Systems Deﬁnition 1.2.1. An algebraic system or algebraic structure consists of a non empty set S, together with: 1. set O of operations on S 2. set of relations R on S 3. set of properties P satisﬁed by the operations on S. We denote the algebraic system by S; O, R, P . However, if no confusion will arise, we simply denote it by its domain S. Algebraic systems are classiﬁed according to the properties they satisfy. Deﬁnition 1.2.2. Let S = ∅. If S is closed under a binary operation ∗, then we call S a groupoid or a magma. Since every non empty set S with a binary operation ∗ is closed under this operation, it follows that every set with a binary operation is a groupoid or a magma. Thus, the basic property required for a set to be a groupoid is simply the closure property . Deﬁnition 1.2.3. Let S, ∗ be a magma, then 1. S is abelian iﬀ for any x, y ∈ S, x ∗ y = y ∗ x. In this case, we say that ∗ satisﬁes the commutative property or is commutative. 2. S is a semigroup iﬀ for any x, y and z in S, (x ∗ y) ∗ z = x ∗ (x ∗ y). In this case, we say that ∗ is associative or satisﬁes the associative property. 3. S has an identity element if there exists an element e ∈ S such that for all x ∈ S, x ∗ e = e ∗ x = x. e is the identity element with respect to ∗. 4. S is invertible if it has an identity element, say e, and for each x ∈ S, there exists an element y ∈ S such that x∗y = y∗x = e. It is customary to denote y by x−1 and call it the inverse of x with respect to ∗. 4 CHAPTER 1. BASIC NECESSITIES 5. S is a monoid iﬀ it is a semigroup with identity element. 6. S is a quasigroup iﬀ for all x, y ∈ S, there exists unique elements z, w ∈ S such that (a) x ∗ z = y (b) w ∗ x = y 7. S is a loop if it is a quasigroup with identity element. 8. S is a pseudogroup iﬀ it is ﬁnite and non associative invertible quasi- group. Hb 3. 1. Let G = {a, b, c} and let × be deﬁned using the following table. × a b c a a b c b b c a c c a b What properties are satisﬁed by G; × ? G is clearly an abelian in- vertible quasigroup. It is quite tedious, but easy to show that it is a semigroup. Thus, we say that G is an abelian invertible loop. 2. R is an abelian invertible loop under + but it is not invertible with respect to multiplication. Why is this so? 3. N is an abelian semigroup with respect to addition, but is not monoid, since it has no identity element. 4. However, N0 is an abelian monoid under addition since it has an identity element 0. 5. The set of odd integers is not a groupoid with respect to addition, but it is with respect to multiplication. 6. Let Z = {(a, b)|a, b ∈ Z}. We deﬁne “=” by (a, b) = (c, d) if and only if a = c and b = d. Deﬁne ⊕ by (a, b) ⊕ (x, y) = (a + x, b + y). Prove that (a) Z is a magma with respect to ⊕ (b) ⊕ is commutative 1.2. ALGEBRAIC SYSTEMS 5 (c) ⊕ is associative (d) Z, ⊕ has an identity element (e) Z, ⊕ is invertible. 7. What kind of an algebraic system is Z, ⊕ ? Chapter 2 Groups 2.1 Deﬁnitions and Examples Deﬁnition 2.1.1. A group G; ∗ consists of a non empty set G together with a binary operation ∗ deﬁned on its elements such that the following properties are satisﬁed: (G1 ) ∗ is associative, i.e. if a, b, c are in G, then a ∗ (b ∗ c) = (a ∗ b) ∗ c (G2 ) There exists an element e ∈ G such that for any element a ∈ G, a ∗ e = e ∗ a = a. e is called the identity element in G with respect to ∗. (G3 ) For any a ∈ G, there exists an element b ∈ G such that a ∗ b = b ∗ a = e. The element b is called the inverse of a with respect to ∗. Clearly, a group may also be deﬁned as an invertible monoid. Unless confusion may arise, a group G; ∗ , like any algebraic system with one binary operation shall be denoted simply by its domain G. Remark 1. A word of caution! A group operation need not be commutative. Should the binary operation of the group is commutative, we say then that the group is abelian. Deﬁnition 2.1.2. The order of a group G; ∗ is equal to the order |G| of its domain. The group is ﬁnite if |G| = n for some nonnegative integer n. Otherwise, we say that the group is inﬁnite. Hb 4. 1. One famous group is R; + . 6 2.1. DEFINITIONS AND EXAMPLES 7 2. On the other hand, R; · , where · is the multiplicative operation, is not a group. This is so since for any r ∈ R, r · 0 = 0 · r = 0. Hence, we can not ﬁnd an inverse of 0 with respect to ·. 3. Let V4 = {e, a, b, c} such that e a b c e e a b c a a e c b b b c e a c c b a e Cayley Table of Klein 4 Group It can be veriﬁed, though tediously, that the system is a semigroup. Note that e is an identity element and for each x ∈ V4 , the inverse of x with respect to is itself. In cases such as this, we say that the element is a self - inverse element. Furthermore, is commutative. Thus, V4 is an abelian group wrt . Group V4 ; is known as the Klein - 4 group or Viergruppe. 4. Let Z4 = {0, 1, 2, 3} such that +4 0 1 2 3 0 0 1 2 3 1 1 2 3 0 2 2 3 0 1 3 3 0 1 2 Assuming associativity of +4 it can be easily veriﬁed that Z4 ; +4 is a an abelian group. This group is called the cyclic group of order 4. 1 V4 and Z4 are the only two groups of order 4 up to isomorphism. 1 Isomorphism will be discussed later. 8 CHAPTER 2. GROUPS Since we do not want to be burdened with so many symbols that may yet get to be too confusing, we shall use the following standard notations for groups. Notation 2.1.1. In general, group G; ∗ shall be identiﬁed by its domain. Through out the text, unless speciﬁed, a group shall be denoted by G. 1. Multiplicative Notation. This system of notation is generic, mean- ing, it may be used for any type of group. (a) ab := a ∗ b. ab shall be read as “The product ab or a times b.” (b) 1 shall be used to denote the identity element of the group. (c) x−1 shall denote the inverse of x. 2. Additive Notation. This system of notation is used solely for abelian groups. (a) a + b := a ∗ b. a + b shall be read as a plus b and a + b is called the sum of a and b. (b) 0 shall denote the identity element of an abelian group under the additive notation. (c) −x shall denote the inverse of x under the additive notation. Remarks 1. 1. The multiplicative notation may be used also for abelian groups. 1 2. x−1 = . This notation is used solely by the group of non zero numbers x under multiplication. Recall that if a matrix M is nonsingular, then 1 its inverse is also denoted by M −1 . However, does not make any M sense at all. 3. 1 and 0 are mere symbols, they do not necessarily denote the real numbers 1 and 0 respectively. Notice that V4 is an abelian group, but + was not used for its binary operation in order to avoid confusion with the binary operation for Z4 . 2.2. CAYLEY TABLE 9 2.2 Cayley Table As we have seen in the previous section, a group may be represented by a table known as the multiplication table or Cayley table. This can be done only with ﬁnite groups. If |G| = n, then we may write G as {a1 , a2 , . . . , an . The rows and columns of the Cayley table may be labelled accordingly as a1 , a2 , . . . , an . The entry in the ith row and j th column is the entry aij := ai aj . ∗ a1 a2 . . . an a1 a11 a12 . . . a1n a2 a21 a22 . . . a2n . . . . .. . . . . .. an an1 an2 . . . ann It shall be proven later that each row and each column of a Cayley table of a group G must contain the elements of G exactly once. If the group is abelian, then ai aj = aj ai . Consequently, aij = aji . Hence, a ﬁnite group is abelian if its multiplication table is symmetric. 2.3 Some Group Properties Recall in College Algebra that in R, if 3 + x = 10, then equivalently, 3 + x = 3 + 7. Solving for x will give us x = 7. However, a “shorter” way to solve the ﬁrst equation is simply by cancelling 3 from both sides. Cancellation property of nonzero real numbers under multiplication is very famous among mathematics students. However, cancellation laws apply to all groups. Theorem 2.3.1. Given a group G and a, b, c ∈ G then (Left Cancellation Law) ab = ac implies b = c. (Right Cancellation Law) ba = ca implies b = c. Note 1. Why is there a need for two cancellation laws? The proof to be presented is not complete and the reader is expected to complete it. 10 CHAPTER 2. GROUPS Proof. Suppose ab = ac. Since a ∈ G, then a−1 ∈ G. Then a−1 (ab) = a−1 (ac). By associativity, (a−1 a)b = (a−1 a)c. Consequently, eb = ec or b = c, where e is the identity element in G. This proves the Left Cancellation Law or (LCL). (Proof of Right Cancellation Law (RCL) is left as an exercise.) -TN- Hb 5. The set of all 2 × 2 matrices is closed under matrix multiplication. If 1 2 2 1 −2 7 A= B= C= , 2 4 3 2 5 −1 8 5 2 then, AB = AC = but B = C. Why is this so? 16 10 We introduce another important theorem. Again as a motivation, let us consider an example in college algebra. Suppose 3x = 6, can we ﬁnd x ∈ R such that equation is true? By experience, we know it is true. Furthermore, we know that the solution is 1 unique. This solution is x = 2. Now, 2 = 3 6. We were also taught that 3−1 = 1 . If we let x = 3−1 6, then 3x = 3 3−1 · 6 = 3 3 · 6 = 3(2) = 6. 3 1 Theorem 2.3.2. Quasigroup Property or Unique Solution to Linear Equa- tion Property If G is a group and a, b ∈ G, then we can ﬁnd unique elements x and y in G such that ax = b and ya = b. Proof. Let a, b ∈ G. then a−1 is in G by G3 . By the closure property, a−1 b is again in G. If we let x = a−1 b, then x ∈ G, and the product ax ∈ G. But ax = a(a−1 b) = (aa−1 )b = e(b) = b, where e is the identity element in G. Hence, we have shown that there exists an element x ∈ G such that ax = b. We next show uniqueness. Suppose x ∈ G such that ax = b. Then it follows that ax = a(a−1 b) = b. By the LCL, it follows that x = a−1 b. Hence, the solution to the linear equation ax = b is unique. 2 Kolman B. and David Hill, Introductory Linear Algebra with Applications, Pearson Education Asia Pte Ltd, 2002 2.3. SOME GROUP PROPERTIES 11 (The proof to the remaining part of the theorem shall be left as an exercise.) -TN- We shall use the above theorem to prove a very important consequence, i.e., the identity element is unique. Corollary 2.3.3. If G is a group, then the identity element is unique with respect to its binary operation. Proof. For all a ∈ G, there exists a unique elements x, y ∈ G such that ax = a and ya = a. Let e be the identity in G, then ae = ea = a. Thus, e is the unique solutions we are looking for. Hence, the identity element is unique. -TN- An alternative proof: Proof. Let e be the identity in G. Suppose there is an identity element e ∈ G, i.e, ae = a = e a for all a in G. Now, ee = e and e e = e , since e is the identity in G. However, since e is also an identity element, then, ee = e and e e = e. Consequently, e = e . This contradicts our assumption. Thus, the identity is unique. -TN- Can you present another proof? (Use the cancellation laws.) Another consequence of the quasigroup property is the uniqueness of the inverse elements. Corollary 2.3.4. If G is a group a ∈ G, then the inverse of a with respect to the binary operation is unique. Proof. Let a ∈ G, then there exists an element a−1 ∈ G such that aa−1 = −1 a a = e, where e is the identity element. Thus, a−1 is the unique solution to both equations ax = e and ya = e. Hence, the inverse of a is unique. -TN- Construct two alternative proofs for this corollary. Deﬁnition 2.3.1. An element i of a magma G with binary operation ∗ is said to be idempotent if and only if i ∗ i = i. 12 CHAPTER 2. GROUPS Consider the set of all propositions P under operation ∨. If p ∈ P, then p ∨ p ⇔ p. Thus, every proposition is an idempotent element with respect to ∨. Exercises 2.3.1. I. True or False. Justify answers for false statements. 1. Every group has at least one identity element. 2. Every group has at most one identity element. 3. R, · is a group. 4. If G is a group and a ∈ G, such that a−1 = a, then a is the identity element in G. 5. Every invertible magma is a group. 6. Every group is a monoid. 7. The identity element of a group is unique. 8. If a, b ∈ R, then ax = b has a unique solution. 9. Every invertible quasigroup is a semigroup. 10. There exists a group which is a quasigroup. II. Mechanical 1. The set N is not a group under addition. Justify why not. 2. Consider the following Cayley table, ∗ ∗ ∗ ∗ ∗ Does the above table represent a group or not? Why or why not? Is it a quasigroup? 3. Construct a 5 × 5 Cayley table representing an invertible quasi- group that is not a group. (Hint: make the elements self-inverse and show that the resulting table is not associative) 2.3. SOME GROUP PROPERTIES 13 4. Let Q = {±1, ±i, ±j, ±k} be a set with multiplication determined by the following: i2 = j 2 = k 2 = −1, ij = k = −ji, jk = i = −kj, ki = j = −ik and 1and −1 multiply as in real number multiplication. To help remember these products, follow the diagram .......... . ......... ..... . ............ . i ............ .......... ......... ........ ....... ..... ..... ... .... ... ... ... ... . ...... . . . .. . . . . . . k ............. .. j ......... . . ....... ......... ........ ............ ......... ......................................... The product of any two adjacent i, j, k taken clockwise is the next one, the product counterclockwise is the negative of the next one. Construct a Cayley table for the above algebraic system. Assum- ing that the product is associative, show that Q is a group. This group is called the quaternion group 3 and has much application in natural and physical sciences. 1 a b 5. Show that the set G = 0 1 c a, b, c ∈ R is a group under 0 0 1 matrix multiplication. What kind of matrices are the elements in G? 6. Let 2Z = {2k | k ∈ Z}. List at least 10 elements of the set. Show that 2Z is a group under real number addition. 7. Fill the Cayley table below, given that G = {1, a, b, c, d} is a group. • 1 a b c d 1 1 a b c d a a 1 b b b c c d d 3 Nicholson, Keith W., Introduction to Abstract Algebra 2nd ed., John Wiley & Sons, 1999 14 CHAPTER 2. GROUPS 8. A non empty subset H of a group G is a subgroup of the G if H also a group under the operation in G. In symbols, we write H ≤ G. Let Z8 = {0, 1, 2, 3, 4, 5, 6, 7} be a magma under +8 , where a +8 b = r such that r is the remainder when a + b is divided by 8. Show that the subset {0, 2, 4, 6} ≤ Z8 . Construct the Cayley table for this subgroup. III. Proving 1. Every group has a unique idempotent element. 2. Let G be a group with identity e, such that a2 = e for all a ∈ G. Show that G is abelian. 3. In any group G, (a−1 )−1 = a. 4. Prove that in any group H if a2 = aa = 1, then a−1 = a, where a ∈ H. 5. In a group G, if a and b are in G, then (ab)−1 = b−1 a−1 6. If K is a group, show that the equation a−1 xb = e has a unique solution, where a, b ∈ K and e is the identity. 7. Show that if H and K are subgroups of a group G, then H ∩ K ≤ G. 8. Let GLn (R) be the set of all n × n nonsingular matrices. This is called the general linear group. Prove speciﬁcally that GL2 (R) is a group under matrix multiplication. Arthur Cayley was a mathematical genius who studied mathematics at Cambridge at the age of 17. He pub- lished nearly 300 papers in his lifetime. He was a hiker and mountaineer and a literati. At the age of 25, since no one would hire him as a mathematician, he studied law and became a lawyer 3 years later. In 1863 he even- tually became a professor of mathematics in Cambridge and remained there for the rest of his life. Although Cayley introduced the concept of abstract group, he to- gether with his friend J. J. Sylvester founded the theory of invariants. He was one of the ﬁrst to consider geom- etry of more than three dimensions. 2.4. MORE PROPERTIES 15 2.4 More Properties In college algebra, students are very familiar with symbols, x4 , y −5 , 30 and the likes. The operation involved in all of these is the multiplication of real numbers. Precisely what do these symbols mean? Deﬁnition 2.4.1. Let G be a group with identity element e, a ∈ G, n ∈ N0 , we deﬁne the following notation: a0 = e a1 = a a2 = aa a3 = (a2 )a a4 = (a3 )a . . . an = (an−1 )a for all n ≥ 1 If n ∈ Z− , then n = −m or −n = m for some m ∈ Z+ . We deﬁne a−m := (a−1 )m Hb 6. 1. In any group G, a ∈ G, a3 = a2 a = (aa)a = a(aa) = a(a2 ) by associa- tivity. 2. a4 = a3 (a) = (a2 a)a = a2 (aa) = a2 a2 . Also, a2 a2 = (aa)(a2 ) = a(aa2 ) = a(a3 ). 3. In general, using associativity, an = a(an−1 ). 4. Because of associativity, the grouping symbol is usually dropped and express an = aaa . . . a n factors The following are what we usually call the Laws of Exponents. 16 CHAPTER 2. GROUPS Theorem 2.4.1. Let a and b be elements of a group G with identity e, m and n be in N0 . Then, 1. am an = am+n 2. (am )n = amn 3. If ab = ba, then (ab)n = an bn . Proof. 1. For n ≥ 0 proof is by induction on n. Basis of Induction. Let n = 0, then am a0 = am e = am = am+0 . Thus theorem is true for n = 0. Hypothesis of Induction. Suppose theorem is true for some n = k, that is am ak = am+k . We need to show that theorem is true for n = k + 1. Induction Step. am ak+1 = am (ak a) = (am ak )a = am+k a = a(m+k)+1 = am+(k+1) . Thus, we have shown that theorem is true for all n ∈ N0 . 2. Exercise 3. Exercise -TN- Theorem 2.4.2. If G is a group, n a positive integer, then (an )−1 = a−n . A direct consequence to this theorem is the following. Corollary 2.4.3. (a−1 )−1 = a Deﬁnition 2.4.2. If G is an abelian group, a ∈ G and n ∈ N0 , then we use the additive notation and instead of power, we shall talk of multiplicity. We now deﬁne the following notations: 0(a) = 0 1(a) = a 2a = a + a 3a = a + 2a 2.4. MORE PROPERTIES 17 . . . na = (n − 1)a + a, for all n ≥ 1. (−n)a = n(−a) Just as in the multiplicative notation above, because of associativity, we may write na as na = a + a + . . . + a n terms A counterpart of Theorem 5. is the following, which holds true whenever the group is abelian. Theorem 2.4.4. If G is an abelian group, a, b ∈ G and m, n ∈ N0 , then 1. (m + n)a = ma + na 2. (mn)a = m(na) 3. m(a + b) = ma + mb 4. −(ma) = m(−a) A direct consequence of the above is that (−1)a = −a. 18 CHAPTER 2. GROUPS Hb 7. 1. Consider the cyclic group of order 6, Z6 = {0, 1, 2, 3, 4, 5} with the given Cayley table. +6 0 1 2 3 4 5 0 0 1 2 3 4 5 1 1 2 3 4 5 0 2 2 3 4 5 0 1 3 3 4 5 0 1 2 4 4 5 0 1 2 3 5 5 0 1 2 3 4 (a) 5(4) = 4 +6 4 +6 4 +6 4 +6 4 = 2 (b) 3(−5) = (c) 7(3) +6 (−5)(2) = 5 2. If G is the set of all m × n matrices, and M = [mij ] is in G, then kM is deﬁned as kM = [k(mij )]. Remark 2. Note that ka = ak as ak may not make any sense at all. Considering the last example above, take a = M , then M k is not deﬁned. Hb 8. 1. In matrices, if A = [aij ] is n × m then k ∈ Z then kA = [kaij ] for all i = 1, 2, . . . n and j = 1, 2, . . . m. 2. In V4 = {e, a, b, c}, nx for x ∈ V4 is not deﬁned since although V4 is abelian, the multiplicative notation is used. Instead, we use the power notation, xn . 3. If G = R under addition, then na is the same as taking the product of n and a. 2.4. MORE PROPERTIES 19 Exercises 2.4.1. I. True or False. Justify answers for false statements. 1. If G is a group and a, b ∈ G, (ab)n = bn an . 2. The set S = {0, 1, −1} is a group under real number addition. 3. If a and b are in a group H, then a(ba−1 ) = b. 4. It is possible for a group to have more than one identity element. 5. A group is abelian if for some elements a and b, ab = ba. 6. In a group K, if (ab)−1 = a−1 b−1 , then K is abelian. 7. In Z9 = {0, 1, 2, 3, 4, 5, 6, 7, 8}, under +9 , then −6+9 3+9 5+8 = 10. 8. In GL2 (R), (A + B)−1 = B −1 + A−1 . 9. The set P (R) of all polynomials over R is a group under addition of functions. 10. The set P (R) of all polynomials over R is a group under multipli- cation of functions. II. Mechanical. Do as instructed. 1. In Z9 , ﬁnd the following sums. We shall write +9 as simply +. (a) (−5)(3 + (−8) + 2 + 1) (b) (−2) + 7 + (−4) + 5 + −1 (c) 12(−2) + −10(3) (d) (7 + 8 + 3 + −4) + −7(−7) −1 √ 0 −2 2. In GL2 (R), ﬁnd 5 3 3. Let f (x) = x + 1. Find g(x) such that (f ◦ g)(x) = (g ◦ f )(x) = x. 20 CHAPTER 2. GROUPS III. Proving. 1. If a ∈ R, (−1)a = −a. a − bi 2. If a, b ∈ R, then (a + bi)−1 = . a2 + b 2 3. If G is a group and a1 , a2 , . . . , an are in G, then (a1 a2 . . . an )−1 = (an )−1 (an−1 )−1 . . . (a2 )−1 (a1 )−1 . Chapter 3 Subgroups R is a group under +. Notice that so are Z and Q. Notice too that Z and Q are non empty subsets of R. These subsets are also known as subgroups of R; + . Formally, we deﬁne a subgroup as follows. Deﬁnition 3.0.3. If G; ∗ is a group and φ = H ⊆ G is also a group under ∗, then we say that H is a subgroup of G. In symbols, we write H ≤ G. Hb 9. 1. In Z8 , the set {0, 2, 4, 6} is a subgroup under +8 2. 2Z ≤ Z 3. Q∗ is a subgroup of R∗ with respect to multiplication. 4. Z is not a subgroup of R∗ ; · 5. Is {1, −1, i, −i} a subgroup of C under multiplication? Why or why not? Given a group G and a non empty subset H, how do we determine whether H is a subgroup or not? This question shall be answered by the following theorem. Theorem 3.0.5. Subgroup Test. A subset H of a group G is a subgroup if and only if the following are satisﬁed. 1. The identity element e ∈ G is in H. 2. If h1 and h2 are in H, so is h1 h2 . 3. If h ∈ H, then so is h−1 . 21 22 CHAPTER 3. SUBGROUPS Proof. Let G be a group under ∗. To show H is a subgroup with respect to ∗, we need to show that H; ∗ is also a group. This requires 4 conditions: closure, associativity, existence of identity element and existence of inverse element for each element in H. Clearly, all conditions are satisﬁed by con- ditions (1), (2) and (3). Associativity follows since H is a subset of G and thus, if a, b, c ∈ H, then these are also in G. Hence, a ∗ (b ∗ c) = (a ∗ b) ∗ c. Conversely, if H is a subgroup of G, then H is a group with respect to the operation ∗ of G. Clearly, conditions (2) and (3) are satisﬁed. For condition (1), since H is a group, then there exists an element e ∈ H such that for all h ∈ H, e ∗ h = h ∗ e = h. Also e , h ∈ G. Thus, e ∗ h = h. Since h = h, then e ∗ h = e ∗ h. By RCL, cancelling h, it follows that e = e . Thus, e ∈ H. -TN- Hb 10. 1. Is S = {0, 1, 3, 4} a subgroup of Z8 ? 0 ∈ S. If s ∈ S, then so is −s. But, 1 +8 4 = 5 is not in S. Thus, condition (2) is not satisﬁed. Therefore S is not a group with respect to +8 2. Let n ∈ Z. We deﬁne nZ := {nk | k ∈ Z} = {0, ±n, ±n2, ±n3, . . .}. If n = 2, then 2Z is the set of even integers, both positive and negative. Show that nZ is a subgroup of Z under +. 3. If G is group, then it is a subgroup of itself. G is known as the im- proper subgroup of itself. 4. If e is the identity in a group G, then {e} is a subgroup of G, known as the trivial subgroup. A more compact way of performing a subgroup test is using the following: Theorem 3.0.6. Compact Criterion for Subgroup Test. Let G be a group and H be a nonempty subset of G. Then H ≤ G if and only if, whenever h, g ∈ H, then hg −1 ∈ H. Proof. Exercise. -TN- The Compact Criterion or CC makes testing for subgroup a lot easier. 23 Hb 11. In one of the exercises, you were asked to prove H ∩ K is a subgroup of a group G whenever H and K are. Using CC, this can be done easily. If h, g ∈ H ∩ K, then g, h ∈ H and g, h ∈ K. By CC, hg −1 ∈ H and −1 hg ∈ K. Thus, hg −1 ∈ H ∩ K. Hence, H ∩ K ≤ G. p p q Hb 12. Let x = ∈ Q∗ ⊆ R∗ . Then pq = 0. Since · = 1, then q q p by the uniqueness of the inverse of an element of a group, it follows that p −1 q x−1 = = . q p We now show that Q∗ ≤ R∗ under multiplication. m p Let y = and x = be in Q∗ . It follows that mn = 0 and pq = 0. n q m q mq Now, yx−1 = = (Recall College Algebra). Also mq and np n p np are non zero integers since none of m, n, p or q is zero. mq Thus, yx−1 = ∈ Q∗ . Hence, by CC, Q∗ ≤ R∗ under multiplication. np Theorem 3.0.7. Let H be a subgroup of a group G. If g ∈ G, then gHg −1 = {ghg −1 | h ∈ H} is a subgroup of G. These subgroups are called the conjugates of H. Proof. Using CC, we need to show that if x, y ∈ gHg −1 , then so is xy −1 . Now, there exist h1 and h2 in H such that x = gh1 g −1 and y = gh2 g −1 . Then, y −1 = (gh2 g −1 )−1 = g(h2 )−1 g −1 (Proof is left to the reader). Hence, xy −1 = (gh1 g −1 )(gh2 −1 g −1 ). By associativity, (gh1 g −1 )(gh2 −1 g −1 ) = (gh1 )(g −1 g)(h2 −1 g −1 ) = (gh1 )(e)(h2 −1 g −1 ) = g(h1 h2 −1 )g −1 24 CHAPTER 3. SUBGROUPS Since H ≤ G, it follows that h1 h2 −1 ∈ H. Thus, xy −1 = g(h1 h2 −1 )g −1 is in gHg −1 . By CC, gHg −1 is a subgroup of G. -TN- Remark 3. If G is abelian, then gHg −1 , then gHg −1 = H(gg −1 ) = He = H. Thus, for abelian groups, the conjugate of a subgroup of a group is the subgroup itself. Exercises 3.0.2. I. True or False. 1. Every group has at least two subgroups. 2. A subset H of a group G is a subgroup provided it contains the identity element in G and for every x ∈ H, then x−1 ∈ H. 3. 2Z ≤ 4Z. 4. 4Z ≤ 2Z. 5. If L and M are subgroups of a group J , then so is L ∪ M. 6. Every subgroup of a group is closed with respect to the operation in the group. 7. If P is a group and T is a subgroup of P, deﬁne jT := {jt | t ∈ J }, where j ∈ P. Then jT ≤ P. 8. A subgroup H of a group G is a conjugate of itself. 9. GL2 (R) ≤ M2 (R), where M2 (R) is the set of all 2 × 2 matrices under matrix multiplication. 10. The subgroups of an inﬁnite group are always inﬁnite. II. Mechanical.In each of the following, determine if the given subset is a subgroup of the given group. Show solution. 1. Z; + , H = N0 2. C∗ ; · , H = R∗ . 1 0 −1 0 0 −1 0 1 3. GL2 (R), H = , , , 0 1 0 −1 1 0 −1 0 p(x) 4. G = f (x) = p(x) and q(x) are polynomials over R and q(x) p(x)q(x) = 0 under composition of functions. Let H = {f1 , f2 , f3 , f4 , f5 , f6 }1 such that 1 Snaith, Victor P., Groups, Rings and Galois Theory, 2nd ed., World Scientiﬁc Pub- lishing Singapore, 1998 25 f1 = x f2 = x−1 x f3 = 1 1−x 1 x f4 = x f5 = 1 − x f6 = x−1 Form the Cayley table for H. 5. G = R∗ under multiplication, H = {5n | n ∈ Z} III. Proving. 1. Prove compact criterion for subgroup test. 2. Show, if G is a group and a, b, c ∈ G, then (abc)−1 = c−1 b−1 a−1 . 3. Show that subgroup relation is transitive, i.e. if G is a group such that H ≤ G and K ≤ H, then K ≤ G. Chapter 4 Cyclic Groups 4.1 Division Algorithm and Integers modulo n In elementary years, we were taught mixed numbers, where an improper fraction is expressed as a sum of a whole number and a proper fraction. For example, 47 5 5 = 7 + or simply 7 . 6 6 6 Notice that 47 = 7(6) + 5. We have learned then that 7 is the quotient and 5 is the remainder when 47 is divided by 6. In general, Theorem 4.1.1. Division Algorithm. Given any integers n and d ≥ 1, there exist unique integers q and r such that n = qd + r. Furthermore, 0 ≤ r < d. Proof. Let X = { n − td | t ∈ Z, n − td ≥ 0 }. If n ≥ 0, then for t = 0, n − 0d = n ≥ 0 and hence, n − 0d ∈ X . If n < 0, then n(1 − d) ≥ 0 since The well-ordering d ≥ 1. Consequently, n(1−d) = n−nd ∈ X . In any case, X = φ. Notice that principle states that X ⊆ N. By the well-ordering principle, there exists a least element r in X . every non empty sub- Hence, r = n−qd for some q ∈ Z and that r ≥ 0. We have established the ex- set A of N, contains istence of q and r. It remains to show that r < d and that q and r are unique. a least element. Suppose d ≤ r, then 0 ≤ r − d = n − qd − d = n − (q + 1)d. Hence, r − d ∈ X . Since d ≥ 1, it follows that r − d < r. This contradicts our choice of r, being the least element in X . Therefore r ≤ d. 26 4.1. DIVISION ALGORITHM AND INTEGERS MODULO N 27 For uniqueness, suppose for some integers r and q, n = qd + r. Then either r ≤ r or r ≤ r. Without loss of generality, we shall prove theorem using only the case r ≤ r. If r ≤ r, then r − r = −qd + qd = (q − q)d ≥ 0. Hence, r − r is a non negative multiple of d but is less than d. This can only happen if and only if r − r = 0 or r = r. Consequently, q = q. Thus, uniqueness is proved and the proof is complete. -TN- In the above theorem, d is known as the divisor , q is the quotient and r is the remainder . Hb 13. In each of the following, ﬁnd q and r 1. n = 45, d = 7, then n = 6(7) + 3. So q = 6 and r = 3. 2. If n = 6, d = 12, then q = 0 and r = 6. 3. If n = −28, d = 5, then q = −6 and r = 2. If n = qd, then it follows that the remainder is 0. In this case we say that d divides n, symbolized as d n. Deﬁnition 4.1.1. Let m and n be any integers, then we say that they are congruent modulo d if and only if m − n = qd for some integer q.We write, m ≡ n( mod d). Equivalently, we say m ≡ n( mod d) if and only if d (m − n). Hb 14. 1. 5 ≡ 9( mod 2) since 5 − 9 = −4 = −2(2). 2. 6 ≡ (−9)( mod 5) since 6 − (−9) = 15 = 3(5). 3. For any integer n, n ≡ n( mod d) for any non zero integer d. Theorem 4.1.2. Congruence modulo d, is an equivalence relation on Z, i.e., if d is a positive integer and m, n and p are any integers, then Reﬂexivity. n ≡ n( mod d) Symmetry. If m ≡ n( mod d) then n ≡ m( mod d). 28 CHAPTER 4. CYCLIC GROUPS Transitivity. If m ≡ n( mod d) and n ≡ p( mod d), then m ≡ p( mod d). Proof. Reﬂexivity. Since n − n = 0 = 0 · d, then n ≡ n( mod d). Symmetry. If m ≡ n( mod d), then for some q, m−n = q·d. Consequently, n − m = (−q) · d. Hence, by deﬁnition, n ≡ m( mod d). Transitivity. If m ≡ n( mod d), then there m − n = q · d for some integer q. Hence, m = n + q · d. Also if n ≡ p( mod d), then n − p = t · d for some integer t. Thus, n = p + t · d. Substituting this in the above equation, m = (p + t · d) + q · d. Simplifying, m = p + (t + q) · d. Thus, by deﬁnition m ≡ p( mod d). Hence, congruence modulo d is an equivalence relation on Z. -TN- Since congruence modulo d is an equivalence relation on Z, it induces a partition of Z through its equivalence classes. We shall call the congruence class [a] as residue class modulo d. We also use the notation a for this. Formally, a = [a] = { x ∈ Z | x ≡ a( mod d) } Hb 15. Let d = 5, then [4] = {4, −1, 9, −6, 14, −11, 19, −16, . . .} [9] = {9, 4, 14, −1, 19, −6, 24, −11, . . .} [−1] = {−1, 4, −6, 9, −11, 14, −16, . . .} What do you notice about these residue classes modulo 5? Theorem 4.1.3. For every integer d ≥ 2, [a] = [b] if and only if a ≡ b( mod d) Proof. Exercise -TN- Remark 4. [a] = [b] does not mean that a = b. From division algorithm, if a ∈ Z and n is a positive integer, then there exists unique integers q and r such that a = qn + r. Furthermore, 0 ≤ r ≤ (n − 1). Hence, a ≡ r( mod n). This is true for all r = 0, 1, 2, . . . , n − 1. 4.1. DIVISION ALGORITHM AND INTEGERS MODULO N 29 Theorem 4.1.4. If n ≥ 2 is an integer, then 1. If a ∈ Z, then [a] = [r] for some r, 0 ≤ r ≤ n − 1 2. The residue classes [0], [1], [2],. . . [n − 1] are distinct. Proof. The proof of (1.) follows from the discussion before the theorem. For (2.), suppose there exists t and s integers such that 0 ≤ t, s ≤ n − 1 with [t] = [s]. Then t ≡ s( mod n). Without loss of generality, assume t ≥ s. It follows that 0 ≤ t − s = q · n for some nonnegative integer q. But t and s are nonnegative integers less than n, thus, t − s is a nonnegative multiple of n that is less than n. This can happen only if t = s. Hence, theorem is proved. -TN- Deﬁnition 4.1.2. The set of all residue classes modulo n is denoted by Zn = {[0], [1], [2], . . . [n − 1]} It is called the set of all integers modulo n. We now induce an operation on Zn , known as addition modulo n, +n deﬁned as follows: If [a] and [b] are in Zn , then [a] +n [b] = [a + b] We now introduce a very important theorem. Theorem 4.1.5. Zn is a group under +n for all integer n ≥ 1. Proof. Let Zn = {[0], [1], [2], . . . , [n − 1]}, Closure. Let [t], [s] ∈ Zn . then [t] +n [s] = [t + s]. We need to show that [t+s] ∈ Zn . Since s and t are also in Z, then by division algorithm, there exist integers q and r such that t + s = qn + r, where 0 ≤ r ≤ n − 1. Hence, t + s ≡ r( mod n). Therefore, [t + s] = [r] ∈ Zn . Thus, [t] +n [s] ∈ Zn . Associativity. (Left as exercise) Existence of Identity Element. [0] ∈ Zn and if [a] ∈ Zn , then [a]+n [0] = [a + o] = [a] = [0 + a]. Hence. [0] is an identity element with respect to +n . 30 CHAPTER 4. CYCLIC GROUPS Existence of Inverse Element. Let [a] ∈ Zn . Then [n − a] ∈ Zn . This is so since 0 ≤ a ≤ n − 1. Claim: [n − a] is the inverse of [a]. By deﬁnition, [n − a] +n [a] = [(n − a) + a] = [n]. Now, n ≡ 0( mod n), hence, [n] = [0]. Therefore [n − a] +n [a] = [0]. Likewise, [a] + [n − a] = [0]. Thus, [n − a] is the inverse of [a]. From the above, we have proven that Zn is a group under +n . -TN- Since mathematicians have the tendency to be slack, somewhere in time, the set Zn has been reduced to simply, {0, 1, 2, . . . n − 1}. and +n becomes just +. This slackened notations will be used later in our discussions. Remark 5. It can be shown that addition modulo n is equivalent to: a +n b = r if and only if a − b ≡ r( mod n), 0 ≤ r ≤ n − 1, as shown in the proof of the previous theorem. Exercises 4.1.1. 1. Use division algorithm to ﬁnd the quotient and remainder in the fol- lowing. (a) n = 56, d = 12 (b) n = −32, d = 7 (c) n = 124, d = 38 (d) n = −205, d = 17 (e) n = 89, d = 102 2. Construct the Cayley table for Z8 under +8 . Identify the inverse of each element. Compare this with the quaternions. 3. Prove that addition modulo n is associative. 4.2 Order of an Element and Cyclic Groups Let us consider the following group table, S3 = {ι, ρ1 , ρ2 , µ1 , µ2 , µ3 } 4.2. ORDER OF AN ELEMENT AND CYCLIC GROUPS 31 · ι ρ1 ρ2 µ1 µ2 µ3 ι ι ρ1 ρ2 µ1 µ2 µ3 ρ1 ρ1 ρ2 ι µ3 µ1 µ2 ρ2 ρ2 ι ρ1 µ2 µ3 µ1 µ1 µ1 µ2 µ3 ι ρ1 ρ2 µ2 µ2 µ3 µ1 ρ2 ι ρ1 µ3 µ3 µ1 µ2 ρ1 ρ2 ι The above table represents the structure of S3 , the symmetric group on 3 elements. This is a group of order 6. There are only two groups of order 6 up to isomorphism, and this is one of them. Note that this is our ﬁrst non abelian ﬁnite group. Consider the following products: (ρ1 )1 = ρ1 , (ρ1 )2 = ρ2 , (ρ1 )3 = ι, (ρ1 )4 = ρ1 , (ρ1 )5 = ρ2 , (ρ1 )6 = ι, . . . Notice that there is a pattern to the powers of ρ1 . For any n ∈ N, n (ρ1 ) = ρ1 , ρ2 or ι. If n is negative, then n = −k, where k ∈ N. Hence, (ρ1 )n = (ρ1 )−k = ((ρ1 )−1 )k . Since (ρ1 )−1 = ρ2 (As seen in the table), then for n ∈ Z− , (ρ1 )n is just a positive power of ρ2 . Taking the powers of ρ3 , (ρ2 )1 = ρ2 , (ρ2 )2 = ρ1 , (ρ2 )3 = ι, (ρ2 )4 = ρ2 , (ρ2 )5 = ρ1 , (ρ2 )6 = ι, . . . Hence, just like in the above, for n ∈ N, (ρ2 )n = ρ2 , ρ1 or ι. Also, by deﬁnition, (ρ1 )0 = ι. Thus, combining this with the results above, for all n ∈ Z, (ρ1 )n = ρ1 , ρ2 or ι. Let us deﬁne ρ1 as ρ1 = {ρ1 n | n ∈ Z } = {ι, ρ1 , ρ2 }. It can be easily veriﬁed that ρ1 is a subgroup of S3 . Deﬁnition 4.2.1. If G is a group and a ∈ G, the order of the element a is the smallest positive integer m such that am = e. If no such number exists, then we say that the order of a is inﬁnity. We write o(a) or |a| for the order of a. In S3 , the orders of ρ1 and ρ2 are 3. The orders of µ1 , µ2 and µ3 are 2. The order of ι is 1. In general, the order of the identity element of any group is 1. Using additive notation, in an abelian group G, the order of an element a ∈ G is n, if it is the smallest positive integer such that na = 0. In the Z4 = {0, 1, 2, 3}: 32 CHAPTER 4. CYCLIC GROUPS Order of 0 is 1. Order of 1 is 4, since 1(1)=1, 2(1)=2, 3(1)=3 and 4(1)=0. Order of 2 is 2, since 1(2)=2 and 2(2)=0. Order of 3 is 4, since 1(3)=3, 2(3)=2, 3(3)=1 and 4(3)=0. Some more examples: Hb 16. In R∗ , only 1 and -1 have ﬁnite orders. o(1) = 1 since (1)1 = 1. Also, (−1)1 = −1, and (−1)2 = 1, hence, o(−1) = 2. For any x ∈ R∗ , xn = 1 for all n ∈ Z+ . Thus, o(x) is inﬁnite. Hb 17. Consider Z and deﬁne d = dZ by d = dZ = { nd | n ∈ Z}. So, 3Z = {0, 3, −3, 6, −6, 9, −9, . . .}, −4Z = {0, −4, 4, −8, 8, −12, 12, . . .}. For any x ∈ dZ, x = 0, o(x) is inﬁnity. Only 0 has a ﬁnite order of 1. Hb 18. Let H = {2n | n ∈ Z}. 20 = 1 ∈ H. Hence, H = φ. If x ∈ H, then x = 2k for some k ∈ Z. Thus, x−1 = (2k )−1 = 2−k . Since k ∈ Z, then so does −k. Consequently, 2−k ∈ H. Thus, x−1 ∈ H. Suppose x, y ∈ H. Then there exist k and t in Z such that x = 2k and y = 2t . xy −1 = 2k 2−t = 2k+−t ( this is true since x, y ∈ G and by a law of exponent). Since Z is a group with respect to addition, then k + −t is again in Z. Thus, xy −1 = 2k+−t ∈ G. Therefore by the compact criterion for subgroup test, H ≤ G. In general, we have, Theorem 4.2.1. Let G be a group and let H = {an | a ∈ G and n ∈ Z } Then H ≤ G. We denote H = a = {an | a ∈ G and n ∈ Z } call it the cyclic subgroup of G generated by a. The element a is called the generator of H. Proof. Let G be a group, then H = {an | a ∈ G and n ∈ Z } is a non empty subset of G since a0 = e is in H. If x, y ∈ H, then there exist k and t in Z such that x = ak and y = at . xy −1 = ak a−t = ak+−t (this is true since 4.2. ORDER OF AN ELEMENT AND CYCLIC GROUPS 33 x, y ∈ G and by a law of exponent). Since Z is a group with respect to addition, then k + −t is again in Z. Hence, xy −1 = ak+−t ∈ H. Therefore by CC, H ≤ G. -TN- Remark 6. H is the smallest subgroup of G that contains a. Why? If G is abelian, we write H = {na| n ∈ Z}. If G = a = {an | n ∈ Z} for some a ∈ G, then we say G is cyclic. Hb 19. 1. In R∗ , the subgroup H = {2n | n ∈ Z} is cyclic with generator 2. Now, 1 1 2−1 = . It is easy to show that also generates the group. 2 2 Proof. Let x ∈ H, then there exists k ∈ Z such that x = 2k . Since −(−k) = k, 1 k −1 1 −k then 2k = 2−(−k) = (2−k )−1 = ((2−1 )k )−1 = = . 2 2 1 −k Thus, x = . Since x is arbitrary, it follows that every element in 2 1 H can be expressed as an integral power of . 2 1 Therefore, generates H. 2 -TN- 2. Z is cyclic generated by 1 and -1. Also, for all d ∈ Z, d = −d = dZ ≤ Z. 3. i = −i = {1, −1, i, −i} ≤ C 4. R is not cyclic under addition, neither is R∗ under multiplication. 5. Z6 is cyclic, generated by 1 and 5. Notice that every cyclic group given above are abelian. This is not a coincidence, but rather the rule in cyclic groups. We state a theorem to this eﬀect. Theorem 4.2.2. Every cyclic group is abelian. 34 CHAPTER 4. CYCLIC GROUPS Proof. Let G = a be a cyclic group generated by a. Let x and y be in G. Then for some integers n and m, x = an and y = am . Thus, xy = an am = an+m = am+n = am an = yx. Hence, G is abelian. -TN- However, the converse is not true as we have seen in the two examples above. Hb 20. Consider Z6 = {0, 1, 2, 3, 4, 5}. Z6 is a group with respect to +6 . +6 0 1 2 3 4 5 0 0 1 2 3 4 5 1 1 2 3 4 5 0 2 2 3 4 5 0 1 3 3 4 5 0 1 2 4 4 5 0 1 2 3 5 5 0 1 2 3 4 Cayley Table of Z6 Consider 1. 1 = {1, 2, 3, 4, 5, 0} = Z6 2. 2 = {2, 4, 0} 3. 3 = {3, 0} 4. 4 = {4, 2, 0} 5. 5 = {5, 4, 3, 2, 1, 0} = Z6 6. 0 = {0} We could see that Z6 is cyclic generated by 1 and 5. From Theorem 17., all the other sets above are subgroups of Z6 . Can we ﬁnd other subgroups of Z6 ? Theorem 4.2.3. Every subgroup of a cyclic group is cyclic. Proof. Let G = a be a cyclic group generated by a. Let H ≤ G. We need to show that there exists an element c ∈ H such that c = H. If H = {e}, then H = e and hence, cyclic. Suppose H = {e}. If x ∈ H, then x ∈ G and there exists n ∈ Z such that x = an . Let m be the least positive integer such that am ∈ H. 4.2. ORDER OF AN ELEMENT AND CYCLIC GROUPS 35 Claim: Let c = am , then c = H. To show this, we need to show that if b ∈ H, then b = cr for some integer r. Since b ∈ H, then b ∈ G. Hence, there exists an n ∈ Z such that b = an . By division algorithm, there exist unique integers q and r, 0 ≤ r ≤ m − 1, such that n = qm + r. Thus, an = aqm+r = aqm ar = amq ar = (am )q ar . Since am ∈ H, and H is a subgroup of G, then by closure property, (am )q ∈ H. Consequently, ((am )q )−1 an = ar ∈ H. But r < m, thus, r = 0 by the choice of m. Therefore, an = (am )q = cq . Hence, H = c . -TN- Remark 7. If G is a group, then for each element a ∈ G, a generates a cyclic subgroup of G, speciﬁcally a . However, it does not follow that these are the only subgroups of G. We shall have more of this later. Corollary 4.2.4. The subgroups of Z under addition are precisely the sets nZ. Theorem 4.2.5. If G is a nontrivial group generated by a, then G = a−1 . Proof. Exercise -TN- 36 CHAPTER 4. CYCLIC GROUPS Exercises 4.2.1. I. True or False. 1. Every abelian group is cyclic. 2. If every proper subgroup of a group is cyclic, then the group itself is cyclic. 3. If [a] = [b] ∈ Zn , then a = b. 4. In Z15 , 8 = 7 . 5. In R, | − 1| = 1 6. Every group has a cyclic subgroup. 7. Q∗ is cyclic with respect to multiplication. 8. S3 is cyclic. 9. Z5 has 4 generators. 10. The group of quaternions has 6 cyclic subgroups. II. Mechanical. 1. Determine which of the following groups is cyclic. Justify your answers. For cyclic groups, identify all generators. (a) R under addition. (b) {( 1 )n | n ∈ Z} 2 (c) Q∗ under multiplication. (d) {−5k | k ∈ Z} (e) The quaternions. 2. Find the subgroups of the given groups generated by the given elements. List at least 10 elements. (a) G = R under addition, H = 2 3 (b) G = Z12 under modulo addition, H = 10 (c) G = group of quaternions, H = i (d) G = Z under addition, H = −3 (e) G = S3 , H = ρ1 III. Proving 1. If G is a group and a ∈ G, then a is the smallest subgroup of G containing a. 4.3. CLASSIFICATION OF CYCLIC GROUPS 37 2. If G = a , then G = a−1 . 3. Let G be a cyclic group of order n. Then for all g ∈ G, g n = 1. 4. If n ∈ N, then Zn = r whenever r and n are relatively prime. 4.3 Classiﬁcation of Cyclic Groups Theorem 4.3.1. Let G be a group and g ∈ G. We consider two cases. Case 1. The cyclic subgroup g is ﬁnite. Then there exists a smallest pos- itive integer n such that g n = 1, i.e., |g| = n and we have (a) g k = 1 if and only if n|k. (b) g k = g m if and only if k ≡ m( mod n). (c) g = {1, g, g 2 , . . . , g n−1 } and these elements are distinct. Case 2. The cyclic subgroup a is inﬁnite. Then (d) g k = 1 if and only if k = 0. (e) g k = g m if and only if k = m. (f ) g = {. . . , g −3 , g −2 , g −1 , 1, g, g 2 , g 3 , . . .} and these powers of g are distinct. Proof. Case 1. Since g is ﬁnite, then the powers of g, g, g 2 , g 3 , . . . , are not all distinct. Then there exist distinct integers l and m such that g l = g m . Without loss of generality, assume l > m. Consequently, g l g −m = g l−m = 1. Since l > m, then l − m > 0. Let {t ∈ Z|g t = 1}. By the well ordering principle, there exists a smallest positive integer n such that g n = 1. (a) Let g k = 1. From division algorithm, there exist q and r, where 0 ≤ r < n such that k = nq + r. Now, g k = g nq+r = g nq g r . Since g nq = (g n )q = 1q = 1. Hence, g nq g r = g r , or g k = g r = 1. By choice of n and since r < n, then r = 0. Therefore, k = nq, that is, n|k. Conversely, suppose n|k. Then there exists an integer q such that k = nq. Then g k = g nq = (g n )q = 1q = 1. Therefore, g k = 1. 38 CHAPTER 4. CYCLIC GROUPS (b) Let g k = g m . Then, as in above, g k−m = 1. Therefore, n|(k − m) from (a). Therefore, k ≡ m( mod n). On the other hand, suppose k ≡ m( mod n), then n|(k − m). Hence, there exists an integer t such that k − m = nt. It follows that g k−m = g nt . But g nt = 1. Therefore, g k−m = 1. Conse- quently, g k = g m . (c) Since g is group, then {1, g, g 2 , . . . , g n−1 } ⊆ g . It suﬃces to show that g ⊆ {1, g, g 2 , . . . , g n−1 }. If g t ∈ g for some integer t, then there exist q and r integers such that 0 ≤ r < n and t = nq +r. Thus, as in above, g t = g r . Since 0 ≤ r < n, it follows that g t = g r ∈ {1, g, g 2 , . . . , g n−1 }. Hence, g ⊆ {1, g, g 2 , . . . , g n−1 }. Therefore, g = {1, g, g 2 , . . . , g n−1 }. Assume there exist two elements of {1, g, g 2 , . . . , g n−1 } that are equal, say, g l = g t . Then g l−t = 1. WLOG, assume l ≥ t. Then, 0 ≤ l − t < n. By choice of n, l − t = 0 or l = t. Hence, all the elements of {1, g, g 2 , . . . , g n−1 } are distinct. Case 2. Suppose g is inﬁnite, then (d) Suppose g k = 1. If k = 0, then we are done. Otherwise, if k = 0, then as in above, there exists a least element n such that g n = 1. Consequently, |g| = n and | g | = n. This contradicts the fact that | g | is inﬁnite. Hence, k = 0. Conversely, if k = 0, clearly, g k = 1. (e) If k = m, then clearly, g k = g m . If g k = g m , then g k−m = 1. Thus, from (d), k − m = 0 or k = m. (f ) Suppose, there exist distinct k and m such that g k = g m . Then from Case 1., g will be ﬁnite. Since this is not the case, then it must follow that the powers of g are all distinct. -TN- Remark 8. Notice that from the above theorem that we can deduce that |g| = |g | In Z8 , The reader can verify that 1 = 3 = 7 = 5 . The question now is, what are the other generators of a any given cyclic group? Theorem 4.3.2. Let G = g and |g| = n. Then G = g k if and only if gcd(n, k) = 1. 4.3. CLASSIFICATION OF CYCLIC GROUPS 39 Proof. Number Theory: Suppose G = g = g k . This implies that g ∈ g k . Hence, there is an If a = bq +r, then integer t such that g = (g k )t = g kt . Consequently, g 1−kt = 1. Then, from d(a, b) = gcd(b, r).Theorem 22, n|(1 − kt). It follows that 1 − kt = nq = 1 + k(−t) for some If gcd(mt, k) = 1, integer q. Thus, gcd(n, k) = 1 n gcd(m, k) = 1. If gcd(a, b) = d,, Conversely, if gcd(n, k) = 1, then there exist q and p integers such that n d = aq + pb for 1 = nq + kp. Thus, g 1 = g nq+kp = g nq g kp = 1(g kp ) = g kp . Hence, g = g kp = me integers q and k p k k k (g ) . Therefore, g ∈ g . Consequently, G = g ⊆ g . Since g ⊆ G, therefore, G = g k . -TN- Hb 21. 1. Find all the generators of Z12 . Since Z12 is of order 12 and is generated by 1, then its other generators are elements of the form k(1) where, gcd(12, k) = 1. Hence, solving for k, we have, k = 1, 5, 7, 11. Thus, all the generators of Z12 are: 1(1), 5(1), 7(1) and 11(1). Equivalently, 1,5,7 and 11. 2. In general, if The number of k integers less than n= pi qi , where pi s are distinct primes and qi s are positive integers, and relatively prime i=1 to a given inte- then the number of generators of Zn is ger n is ϕ(n) = k 1 k n 1− . 1 pi i=1 ϕ(n) = n 1− This equation is i=1 pi called Euler’s ϕ (phi) function. 3. Let G be a group and g ∈ G. Suppose |g| = n and d is an integer such n that d|n. Then |g d | = . d Proof. k Let n = k. Then, n = dk and g n = g d = 1. We need to show d that |g d | = k. Suppose |(g d )| = r. By deﬁnition of order, r ≤ k. By Theorem 22, it follows that for some q, dr = nq = dkq. Simplifying, r = kq. Thus, k|r or k ≤ r. This implies that r = k by assumption on r. Therefore, k = n = |g d |. d -TN- 40 CHAPTER 4. CYCLIC GROUPS 4.3.1 The Fundamental Theorem of Finite Cyclic Groups Theorem 4.3.3. Let G be a ﬁnite cyclic group of order n generated by g. 1. If H ≤ G, then H = g d for some d|n. 2. If H ≤ G, |H| = k, then k|n. 3. If k|n, then H = g n/k is the unique subgroup of G of order k. Proof. 1. Since G is cyclic and H ≤ G, then H is cyclic by Theorem 19. From the proof of that theorem, we see that H = g d such that n = dq for some integer q. Hence, d|n. 2. Suppose H ≤ G of order k. Then H = g t for some integer t and |g t | = k = n by Hb 18. Hence, k|n. t 3. Let H = g d be a subgroup of G. Then d|n from (1). If |H| = k, then n k = n or d = n . Hence, H = g k . If F = g d ≤ G of order k, using d k the same argument above, d = n . Therefore, d = d and H = F . k -TN- Remark 9. Property number (2) of Theorem 22 above is true for all ﬁnite groups. This property is known as Lagrange’s Theorem Hb 22. 1. Find all distinct subgroups of Z12 . From above, the subgroups of Z12 = 1 are of orders k, where k di- vides 12. Thus, k = 1, 2, 3, 4, 6, 12. The subgroup of order 1 is {0} and the subgroup of order 12 is Z12 itself. For each k, let Hk be the corresponding subgroup. From above, these subgroups are unique. We now have the following: (a) H2 = ( 12 )(1) = 6 = {0, 6} 2 (b) H3 = ( 12 )(1) = 4 = {0, 4, 8} 3 (c) H4 = ( 12 )(1) = 3 = {0, 3, 6, 9} 4 (d) H6 = ( 12 )(1) = 2 = {0, 2, 4, 6, 8, 10} 6 4.3. CLASSIFICATION OF CYCLIC GROUPS 41 A lattice diagram is a line diagram showing the relationship between a group and its subgroups such that a line connects two subgroups H and K iﬀ H ≤ K (or vice versa) and no other subgroup lies between them. The diagram is topped by the group itself and has bottom entry the trivial subgroup. The lattice diagram of Z12 is as follows: Z12 ....... ....... ....... ....... ....... .... ....... .... ............ .... .... .... .... H 6 ..... .... .... .... .... .... . . . .... . . .... . . . . . . .... .... .... .... H4 .... . .... . . . H 3 .... .... .... .... . . . . . . .... .... .... .... . .... .... .... H2 .... .. .... ....... .... ...... .... ...... .... ....... .... ............ .... ........ ... {0} 2. Since Z12 is also generated by 7, then 3. H2 = ( 12 )(7) = 6 2 4. H3 = ( 12 )(7) = 4 3 5. H4 = ( 12 )(7) = 9 4 6. H6 = ( 12 )(7) = 2 6 Observe that H4 = 3 = 9 . Can you make other representations based on the generators 5 and 11? 7. Can the group of quaternions contain a subgroup of order 6? Why or why not? 8. If p is a prime integer, how many generators will Zp have? Construct the lattice diagram. 9. Consider Z80 . Now, Theorem 4.3.4. Let G be a cyclic group of order n, generated by a. Let b = as be an element of G. Then b generates a cyclic subgroup H of G with n d elements and d = gcd(n, s). 42 CHAPTER 4. CYCLIC GROUPS Proof. Relax -TN- A direct consequence of the theorem is the following corollary. Corollary 4.3.5. If G is a ﬁnite cyclic group of order n, then G is generated by g k where gcd(n, k) = 1. I. True or False 1. Q, + is cyclic. 2. If G = a and |G| = n, then H = an/d is a subgroup of G of order n d 3. For every positive integer n, there exists a cyclic group of order n. 4. Z93 has 7 generators. 5. If G is a ﬁnite group of order n, and a ∈ G, then an = 1. 6. nZ = d for some d ∈ Z and gcd(n, d) = 1. 7. If K = a is a cyclic group of order t and suppose p is a prime, then ap generates K. 8. Z is generated by 1 and -1 only. 9. All proper subgroups of R is cyclic. 10. There exists a non abelian cyclic group. II. Mechanical 1. Let G = a be of order 30. Determine the following subgroups. (a) a6 (b) a7 (c) a24 (d) Construct the lattice diagram of G 2. Find all elements of order 6 of G in number 1. 3. Find the number of elements of order 3 of G in number 1. 4. Give an example of an inﬁnite group which contains a nontrivial ﬁnite cyclic subgroup. 5. In each of the following, ﬁnd all subgroups of G = a and draw the lattice diagram. 4.3. CLASSIFICATION OF CYCLIC GROUPS 43 (a) |a| = 8 (b) |a| = p3 where p is prime. (c) |a| = pq and p and q are distinct primes. (d) |g| = 18 III. Proving 1. Let G be a group. If G has at most two nontrivial subgroups, then G is cyclic. 2. Let G be a ﬁnite group. Show that if G has exactly one nontrivial subgroup, then |G| = p2 for some prime p. 3. Show that an inﬁnite cyclic group has exactly two generators. 4. Prove Theorem 25. 5. Let G be a cyclic group of order n. If for all, g m = 1 where gcd(m, n) = 1, then g = 1. Augustin -Louis Cauchy (1789 - 1857) was born on August 21, 1789 in Paris, France. He was ﬁrst trained by his father. They were neighbors with Laplace and Berthollet. Cauchy became acquainted with mathemati- cians and scientists at a young age. Lagrange is said to have warned Augustin’s father not to show him any mathematics book until he reaches the age of 17. His mathematical career started when he solved a problem on convex polygon sent to him by Lagrange in 1811. In algebra, the notion of the order of an element, a subgroup and conjugates are found in his papers. He proved the famous Cauchy’s Theorem: If the order of ﬁnite group G is n, and p is a prime that divides n, then G has a subgroup of order p. Cauchy enjoyed teaching and has published more than 800 papers and has written 8 books. He died on May 22, 1857. Chapter 5 Permutation Groups Algebra is the intellectual instrument which has been created for rendering clear the quantitative aspect of the world. -Alfred North Whitehead. Mathematics takes us still further from what is human into the region of absolute necessity, to which not only the actual world, but every possible world must conﬁrm. - Bertrand Russell. In the past, we have been dealing mainly with abelian groups and cyclic groups. In this chapter, we are going to deal with ﬁnite non abelian groups known as permutation groups. Historically, group theory originated in the study of permutation groups. Before we begin our discussion on permutations, we ﬁrst review some concepts on functions. 5.1 Functions, Injection, Surjection, Bijection Deﬁnition 5.1.1. A function φ from a non empty set A into a non empty set B is a rule that assigns to each element in A a unique element in B. Another name for a function is a map or mapping . 44 5.1. FUNCTIONS, INJECTION, SURJECTION, BIJECTION 45 We denote a function φ from A to B by the following notations. maps −→ (i.) φ : A −→ B or f : A − − B A is called the domain (domφ) and B is the codomain with respect to φ. (ii.) If a ∈ A and b ∈ B such that φ assigns a to b, then we write φ(a) = b. or φ : a → b. a is called the pre-image of b and b is the image of a with respect to φ. (iii.) Given the above notation, φ may be written as φ = {(a, b)|b = φ(a)}. maps −→ Deﬁnition 5.1.2. If f : A − − B, then the range or image set of f is the set im(f ) = {y ∈ B|y = f (x), for some x ∈ A}. Hb 23. 1. Let A = {a, b, c, d, e, f } and B = {1, 2, 3, 4, 5, 6, 7, 8.9.10}. Then f = {(a, 7), (b, 1), (c, 10), (d, 4), (e, 5), (f, 3)} is a mapping from A to B. 2. g = {(a, 3), (b, 4), (c, 2), (c, 1), (d, 8), (e, 6), (f, 6)} is not a mapping from A to B. Why so? maps 3. Let h : R − − R such that h(x) = x2 − 2x − 4. The domain of h is R −→ and the range is the interval [−5, ∞). 4. The set j = {(x, y)|y = x3 } is again a function from R to itself. In Hb 20.3., h(−1) = −1 = h(3), but −1 = 3. In Hb 20.4, if j(a) = j(b), then it follows that a3 = b3 . Consequently, a3 − b3 = (a − b)(a2 + ab + b2 ) = 0. But from algebra, we know that a2 + ab + b2 = 0. Thus, it follows that a − b = 0 or a = b. How do we call functions having similar characteristic as of j? Deﬁnition 5.1.3. A function φ : A −→ B is said to be one to one (1 − 1) or an injection, if and only if, whenever φ(x1 ) = φ(x2 ) for all x1 and x2 in A, then x1 = x2 . Thus, f and j are one to one functions while h is not in Hb 20 above. Equivalently, we say that a function φ : A −→ B is one to one if and only if x1 = x2 implies that φ(x1 ) = φ(x2 ). Consider again Hb 20.2, note that 6 has no pre image under f . Also in Hb 20.3 if y < −5 then there is no x ∈ R such that y = h(x). 46 CHAPTER 5. PERMUTATION GROUPS √ √ However, for Hb 20.4. if y ∈ R, and x = 3 y, then j(x) = ( 3 y)3 = y. This is so since the cube root of any real number exists. Thus, im(j) = R, which is also the codomain of j. In this case, we say that j is onto. Deﬁnition 5.1.4. A function φ : A −→ B is said to be onto or a surjection, if and only if im(φ) = B, i.e., for all b ∈ B, there exists an a ∈ A such that φ(a) = b. A function which is both an injection and a surjection is called a bijec- tion. Hb 24. 1. Thus, j is a bijection, h is neither one to one nor onto, and therefore not a bijection, while f is one to one but not onto and there- fore not a bijection also. 2. f : R −→ R such that f (x) = ax + b for a a non zero real number and b any real number is a bijection on R. 3. g : R −→ Z such that g(x) = x is not one to one since g(3.5) = g(3.7) = 3, but 3.5 = 3.7. It is onto since if n ∈ Z, then for all x ∈ R such that x ∈ [n, n + 1) is a pre image of n. Nonetheless, g is not a bijection. Functions may be combined using an operation called composition. Formally we deﬁne it as, maps maps −→ −→ Deﬁnition 5.1.5. If f : A − − B and g : B − − C, then the composition g ◦ f is a function deﬁned as maps −→ g ◦ f : A − − C such that (g ◦ f )(x) = g(f (x)). We may illustrate composition using the following diagram: A B C ...................... ...................... ...................... ......... ..... ......... ..... ......... ..... ..... .... ..... .... ..... .... ... .... ... .... ... .... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... .. ... .. ... .. ... .. .. .. ... .. .. .. ... .. .. .. .. . . . . . . .. . . . . . . . . .. . . .. . . . . . . . . . . . . .. . . . f . . . . . . . . .. . . . . g . . . . . . . . .. . . . . . . . ............ . . ........ ..................... .. . ............ . . ........ ..................... .. . . . .............. . .............. . . . ................... . ..................... . ................... . ..................... . . . . . . . . . a ...... . . . . . . . . . . .. . . .. b . . . . . . . . . . . . . . .. .. c . . . . . . . . . . . ....... ....... . . . . . . . . . . . . . . .......... . .... . . . . . . ....... ........ .. . . . . . . . . . . . ........... . . ... . . . . . . . .... . . ........ .. . . .. .. . .. .. . ........ . . . .. ........ . . .. .. .. . . ........... .. .. . ........ .... . ... ... ... . .. .......... ............ .... ................. g◦f .. . .. .............. ................ ... . ... ... ... . .. ... ... ... ........................................................... .. ... ... ... ... ... ... ... ... ... .... ... .... ... .. .... ... .... .... .... ... .... ... ... ....... .... .... .... ..... ..... .............................. ........ .......................... ......... ...................... f g − −→ a −→ b − c and f (a) = b, g(b) = c, (g ◦ f )(a) = c We now go to the main topic of our chapter, which is permutation groups. To do so, let us now deﬁne what a permutation is and its properties. 5.2. PERMUTATIONS 47 5.2 Permutations Deﬁnition 5.2.1. A permutation of a non empty set A is a bijective mapping from A onto itself. 1−1 − We sometimes denote a permutation ρ of A by ρ : A − → A. onto Hb 25. maps −→ 1. If A is a non empty set and ι : A − − A, such that ι(a) = a, then it is a permutation on A known as the identity permutation or identity function. 2. Let C = {a, b, c}. The pairing: ρ a −→ c b −→ b c −→ a a b c is a permutation of C which may be denoted by ρ = . c b a 3. Why is f = { (x, y) | y = 3x + 4} a permutation on R? Deﬁnition 5.2.2. The function idA : A −→ A such that idA (x) = x for all x ∈ A is known as the identity function on A. Deﬁnition 5.2.3. A function f : A −→ B is said to be invertible if there exists a function g : B −→ A such that g ◦ f = idA and f ◦ g = idB . We denote g by f −1 and call it the inverse of f or f -inverse. Hb 26. 1. In R, the function y = x is the identity function. What is the graph of this function? √ 2. In R, if f (x) = x3 , then f −1 = 3 x. 3. If h(x) = x2 for all x ∈ R, does h−1 exists? The set of all permutations of a non empty set will be proven to be a group under composition of functions. To do this, we have to prove the following. 48 CHAPTER 5. PERMUTATION GROUPS Theorem 5.2.1. If f : A −→ B is onto, then, im(f ) = B bijection − −→ Lemma 5.2.2. f is invertible if and only if f : A − − − B, . Proof. The proof consists of two parts, i.e., ﬁrst we show that if f is invertible, then it is one to one and onto. Next, we show if f is one to one and onto, then it is invertible Suppose f is invertible, Show one to one Let f (x1 ) = f (x2 ) ∈ B. Since f is invertible, f −1 exists and f −1 (f (x1 )) = f −1 (f (x2 )) = idA (x1 ) = idA (x2 ). Thus, x1 = x2 . Therefore, f is one to one. Show ontoness Let y ∈ B. We need to show that there exists an x ∈ A such that maps f (x) = y. Since f is invertible, f −1 exists, where f −1 : B − − A. −→ −1 Thus, for each y ∈ B, f (y) is a unique element in A. Let f −1 (y) = x, then, (f ◦ f −1 )(y) = f (x). But f ◦ f −1 = idB . Hence f ◦ f −1 (y) = y. Therefore, we have found an x such that y = f (x). Hence, f is a bijection. Suppose f is a bijection, show f is invertible. Suppose f is a bijection, then it is one to one and onto. Deﬁne g : B −→ A such that g(y) = x if and only if f (x) = y. We need to show that 1. g is well deﬁned, i. e., g is a function. 2. g ◦ f = idA and f ◦ g = idB . To show g is well deﬁned, we need to show that for each y ∈ B, then g assigns a unique x in A. Since f is onto, then for each y ∈ B, there is an x ∈ A such that f (x) = y. Thus, for each y ∈ B, g assigns an element x in A. Suppose there exists a w in A such that g assigns y to w also. We need to show x = w. Since g assigns y to w, by deﬁnition of g, f (w) = y. Thus, f (x) = f (w). Since f is one to one, x = w. Hence, g assigns to each y in B a unique element x inA. To show (2): Exercise. -TN- 5.2. PERMUTATIONS 49 Corollary 5.2.3. If f : A −→ B is invertible, then f −1 is a bijection. Since a permutation ρ on a nonempty set A is a bijection, it follows that −1 ρ exists and it is a bijective function from A onto itself. Consequently, ρ−1 is again a permutation of A. We now go to our very important theorem. Theorem 5.2.4. Let A be a non empty set and G be the set of all permutations of A. Then G is a group under composition of permutations, also known as permutation multiplication. Proof. To show that G is a group, we need to show: 1. Closure Let α and β be permutations of A. We need to show that α ◦ β again permutes A. Let x and w be in A. Suppose (α ◦β)(x) = (α ◦β)(w). We need to show x = w. Since α and β are bijections, then α−1 and β −1 exist and are both bijective functions on A. Thus, both are permutation of A. Furthermore, (β −1 ◦ α−1 ) ◦ (α ◦ β) = idA (Why?). Consequently, (β −1 ◦ α−1 ) ◦ (α ◦ β)(x) = (β −1 ◦ α−1 ) ◦ (α ◦ β)(w) implies x = w. Thus, α ◦ β is one to one. 2. Existence of Identity Element Let idA be the identity function on A, clearly, it is a permutation of A. It is easy to see that idA ◦ α = α ◦ idA = α for all permutation α of A. Thus, the identity permutation exists. 3. Existence of Inverse Elements From above, we see that if α is a permutation of A, then α−1 is again a permutation of A, and by deﬁnition of invertible functions, α ◦ α−1 = α−1 ◦ α = idA . 4. Associativity (Exercise) Thus, theorem is proved. -TN- Deﬁnition 5.2.4. If A is a non empty set, we call denote the group of all permutations of A by SA . If A is a ﬁnite set of order n, then we write SA as Sn and call it the symmetric group on n letters ( n elements). Hb 27. On A = {a, b, c}, the permutations are: 50 CHAPTER 5. PERMUTATION GROUPS a b c 1. ι = a b c a b c 2. ρ1 = b c a a b c 3. ρ2 = c a b a b c 4. µ1 = a c b a b c 5. µ2 = c b a a b c 6. µ3 = b a c Since |A| = 3, then S3 is the group of all permutations of A such that S3 = {ι, ρ1 , ρ2 , µ1 , µ2 , µ3 }. What do you notice about this group and the S3 that you have learned in the previous chapters. Consider ρ1 and µ1 above, what will be ρ1 ◦ µ1 equal to? (ρ1 ◦ µ1 )(a) = ρ1 (µ1 (a)) = ρ1 (a) = b (ρ1 ◦ µ1 )(b) = ρ1 (µ1 (b)) = ρ1 (c) = a (ρ1 ◦ µ1 )(c) = ρ1 (µ1 (c)) = ρ1 (b) = c Thus, ρ1 µ1 = µ3 . Using the standard notation a b c a b c a b c = b c a a c b b a c For facility, we shall denote the product α◦β by αβ for any permutations α and β on the same set. Using the standard notation, multiplication of permutations on a ﬁnite set will be done ”right to left”. Suppose A = {a1 , a2 , . . . , an } and α and β are permutations of A given by a1 a2 ... an a1 a2 ... an α= and β = α(a1 ) α(a2 ) . . . α(an ) β(a1 ) β(a2 ) . . . β(an ) then α ◦ β is 5.2. PERMUTATIONS 51 a1 a2 ... an a1 a2 ... an α(a1 ) α(a2 ) . . . α(an ) β(a1 ) β(a2 ) . . . β(an ) a1 a2 ... an = α(β(a1 )) α(β(a2 )) . . . α(β(an )) 1 2 3 4 1 2 3 4 Hb 28. On B = {1, 2, 3, 4}, if γ = and τ = , 2 4 1 3 4 1 3 2 1 2 3 4 1 2 3 4 1 2 3 4 then γτ = = 2 4 1 3 4 1 3 2 3 2 1 4 Using the above method, construct the table of SA where A = {a, b, c}. Since a permutation ρ of a set A is invertible, then ρ−1 exists and if A is ﬁnite, ρ−1 may be written in standard form. Hb 29. 1 2 3 4 5 6 In A = {1, 2, 3, 4, 5, 6}, let α = . Then 3 1 4 2 6 5 1 2 3 4 5 6 α−1 = 2 4 1 3 6 5 Notice that the rows of α were interchanged to form α−1 , and then the elements on the top row were arranged in the standard form. Hb 30. In A = {1, 2, 3, 4, 5, 6, 7, 8}, let 1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8 α= β= 3 5 8 1 7 2 4 6 8 7 6 2 4 1 5 3 1 2 3 4 5 6 7 8 δ= 3 6 7 8 5 2 1 4 1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8 1. αβ = 3 5 8 1 7 2 4 6 8 7 6 2 4 1 5 3 1 2 3 4 5 6 7 8 = 6 4 2 5 1 3 7 8 1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8 2. δ −1 α2 = 7 6 1 8 5 2 3 4 8 7 6 3 4 5 1 2 1 2 3 4 5 6 7 8 = 4 3 2 1 8 5 7 6 3. β −2 δα3 Exercise 4. (αβ 2 )−3 Exercise 52 CHAPTER 5. PERMUTATION GROUPS 5. (δ 2 α−1 )2 Exercise 1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8 6. β = , , 8 7 6 2 4 1 5 3 3 5 1 7 2 8 4 6 1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8 , 6 4 8 5 7 3 2 1 1 2 3 4 5 6 7 8 7. What is α ? 5.2. PERMUTATIONS 53 Exercises 5.2.1. I. True or False. 1. Let f assigns to each element x ∈ R an element y ∈ Z such that x ≤ y. Then f is a function from R to Z. 2. Every permutation of a set J is one to one. 3. Every one to one function on a set T is a permutation of T . 4. Every permutation on a nonempty set may be written in standard form. 5. Every onto mapping is one to one. 6. If A and B are ﬁnite sets of the same order, and if one to one −−− g : A − − − → B, then g is onto. 7. f (x) = x3 − 2x2 + 4 is a permutation of R. 8. Every bijection is a permutation. 9. Every permutation is a bijection. maps maps 10. If f : M − − N , then f −1 : N − − M −→ −→ II. Mechanical 1. Do (a) Hb 30.3 (b) Hb 30.4 (c) Hb 30.5 a b c d e f g h 2. Let ρ = Find h e g b d a c f (a) ρ−2 (b) |ρ| (c) ρ 1 2 3 4 5 6 1 2 3 4 5 6 3. Let γ = and δ = . 6 4 2 5 3 1 4 6 1 3 2 5 Find (a) (γ 2 δ)−3 (b) |γ| (c) |δ| (d) |γδ| 54 CHAPTER 5. PERMUTATION GROUPS III. Proving 1. Prove that f (x) = 2x − 5 is a permutation of R 2. Prove that permutation multiplication is associative. 3. Show why ex is not a permutation of R 4. If A is a ﬁnite set of order n ≥ 1, then |SA | = n! 5.3 Symmetries of a Regular Polygon The next example is quite interesting, as this will show us another way of forming S3 . Consider the following equilateral triangle. . 2 .. ... . . ... ..... ... ... ... ... ... ... ... ... ... ... ... .. ........................................................ 1 3 An equilateral triangle is symmetric with respect to four axes, namely, the three medians and the line through its center. Thus, if we imagine the above triangle as a puzzle piece which may be placed in a hole of similar shape, we may position the piece into 6 ways. Assume that the numbers 1, 2 and 3 are vertex labels of the triangular hole and the colored bullets are the vertices of the triangular puzzle piece. Consider the above drawing as the puzzle in its original position. 1. Original position . 2 . . . ... ... ... .... ... ... ... ... ... ... ... ... ... ... .. .. ........................................................ 1 3 2. A 120◦ rotation clockwise about the center from original position . 2 .. ... . . ... .... ... ... ... ... ... ... ... ... ... ... ... ... ........................................................ 1 3 3. A 240◦ rotation clockwise about the center from the original position 5.3. SYMMETRIES OF A REGULAR POLYGON 55 . 2 . ... . . ... .... ... ... ... ... ... ... ... ... ... ... ... ... ........................................................ 1 . 3 4. A 180◦ ﬂip about the median through vertex (1) from original position . 2 .. ... . . ... ..... ... ... ... ... ... ... ... ... ... ... ... .. ........................................................ 1 3 5. A 180◦ ﬂip about the median through vertex (2) from original position . 2 . . . ... ... ... ..... ... ... ... ... ... ... ... ... ... ... .. . ........................................................ 1 3 6. A 180◦ ﬂip about the median through vertex (3) from original position . 2 .. . . ... .... ... .... ... ... ... ... ... ... ... ... ... ... . ........................................................ 1 .. 3 Above are the only possible ways of positioning the triangular puzzle piece into similarly shaped puzzle hole. Would the same apply to a triangular puzzle piece which is not equilat- eral? The above ”movements” of the equilateral triangular puzzle piece are what we call the symmetries of an equilateral triangle. These symmetries form a group of order 6, known as the dihedral group on 3 elements, denoted by D3 . It is isomorphic to S3 Can you construct D4 , the group representing the symmetries of a square? Generalizing the above example, consider a puzzle piece in the shape of regular n − gon. The number of ways that this piece can be placed in a hole of similar shape forms a group, a subgroup of Sn , known as the nth dihedral group Dn . This group is of order 2n. 56 CHAPTER 5. PERMUTATION GROUPS 5.4 Cycles and Cyclic Notations When I am working on a problem, I never think about beauty. But when I have ﬁnished, if the solution is not beautiful, I know it is wrong.- Buckminster Fuller We now show another way of representing a permutation on a ﬁnite set. It will be shown also that this notation is not only a convenient way of representation, but will also help us classify a ﬁnite permutation into two possible manners. Deﬁnition 5.4.1. Let ∅ = A and let σA ∈ SA . Let a be a ﬁxed element in A. The orbit of a under σ is the set, Oa,σ = {σ n (a)|n ∈ Z} 1 2 3 4 5 6 7 Hb 31. Let σ = 3 7 4 1 2 6 5 The distinct orbits under σ are: O1,σ = {1, 3, 4} = O3,σ = O4,σ O2,σ = {2, 7, 5, } = O7,σ = O5,σ O6,σ = {6} In the above example, notice that given any two orbits, they are either equal or are disjoint. Furthermore, Oa,σ = Ob,σ whenever b ∈ Oa,σ . This characteristic is not unique in our example but is generally the case in any orbit of a given permutation. Let us generalize the observation through the following lemma and the- orem. Lemma 5.4.1. Let ∅ = A and σ ∈ SA . (i) If b ∈ Oaσ , then Oa,σ = Ob,σ . (ii) If Oa,σ ∩ Ob,σ = ∅, then Oa,σ = Ob,σ . Proof. 5.4. CYCLES AND CYCLIC NOTATIONS 57 (i) Let b ∈ Oaσ , then for some integer n, b = σ n (a). Since σ ∈ SA , then so is σ n . Consequently, (σ n )−1 = σ −n is in SA . Hence, σ −n (b) = σ −n (σ n (a)) = (σ −n σ n )(a) = idA (a) = a. Thus, σ −n (b) = a. Let x ∈ Oa,σ . For some integer m, x = σ m (a) = σ m (σ −n (b)) = (σ m σ −n )(b) = (σ m−n )(b). Since Z is a group under addition, then m − n ∈ Z. It follows that x ∈ Ob,σ . Therefore, Oa,σ ⊆ Obσ . Proof of Obσ ⊆ Oaσ , is an Exercise. Therefore, Oa,σ = Ob,σ . (ii) Suppose Oa,σ ∩ Ob,σ = φ, then there exists x ∈ A such that x ∈ Oa,σ ∩ Ob,σ . Consequently, x = σ m (a) and x = σ n (b) for some integers m and n. Thus, σ m (a) = σ n (b). Hence, a = σ n−m (b). Therefore, a ∈ Ob,σ . Therefore from (i), Oa,σ = Ob,σ . -TN- Theorem 5.4.2. If ∅ = A, and σ ∈ SA , then Π = { Ox,σ | x ∈ A} partitions A. Proof. To show Π partitions A, we need to show the following, (i) If Oa,σ and Ob,σ are elements of Π, then either Oa,σ ∩ Ob,σ = φ or Oaσ = Ob,σ (ii) Oa,σ = A. a∈A (i) follows directly from the above lemma. For (ii), clearly, for each a ∈ A, Oa,σ ⊆ A. Thus, a∈A Oa,σ ⊆ A. It remains to be shown that A ⊆ a∈A Oa,σ = A. Now, let x ∈ A, then x ∈ Ox,σ . This is so since x = σ 0 (x). Thus, x ∈ a,σ Oa,σ . Therefore A ⊆ a,σ Oa,σ Consequently, Oa,σ = A. a∈A -TN- The concept of orbits shall be used in the following topic. 58 CHAPTER 5. PERMUTATION GROUPS Deﬁnition 5.4.2. Let A = ∅ be a ﬁnite set. A permutation of A is said to be a cycle if it has at most one orbit containing more than one element. Equivalently, we say a permutation on a ﬁnite set is not a cycle if it has two or more orbits with more than one element. Hb 32. Let A = {1, 2, 3, 4, 5, 6, 7} 1 2 3 4 5 6 7 1. Let α = 5 1 7 2 6 3 4 O1,α = {1, 5, 6, 3, 7, 4, 2} = A. From Theorem 22, it follows that this is the only distinct orbit of α. By deﬁnition, α has only one orbit of more than one element. Thus, α is a cycle. 1 2 3 4 5 6 7 2. Let β = . The distinct orbits of β are: 7 2 3 1 5 4 6 O1,β = {1, 7, 6, 4} O2,β = {2} O3,β = {3} O5,β = {5} Since β has only one orbit of more than one element, then β is a cycle. 1 2 3 4 5 6 7 3. Let δ = . The distinct orbits of β are: 3 7 5 4 1 2 6 O1,δ = {1, 3, 5} O2,δ = {2, 7, 6} O4,δ = {4} Since δ has more than one orbit with more than one element, then δ is not a cycle. 1 2 3 4 5 6 7 4. Consider the identity permutation ι = . The 1 2 3 4 5 6 7 distinct orbits of ι are O1,ι = {1}, O2,ι = {2}, O3,ι = {3}, O4,ι = {4}, O5,ι = {5}, O6,ι = {6}, O7,ι = {7} Notice that every orbit of ι is of length 1, thus, it has no orbit containing more than one element. Hence, by deﬁnition, ι is a cycle. 5.4. CYCLES AND CYCLIC NOTATIONS 59 The length of a cycle is equal to the order its longest orbit, ie. the orbit with the most number of elements. A cycle ρ of length n can be compactly denoted by ρ = (a1 a2 a3 . . . an ) where ai+1 = ρi (a1 ), for all i = 1, 2, . . . , n − 1 and ρ(an ) = a1 . Hb 33. In Hb 29.1: α = (1 5 6 3 7 4 2) β = (1 7 6 4) ι = (1) = (2) = (3) = (4) = (5) = (6) = (7) Two cycles are disjoint if and only if they do not move a common element. Formally, Deﬁnition 5.4.3. Let ∅ = A be a ﬁnite set and α and β be cycles in SA . Then α and β are disjoint, if and only if for all x ∈ A, whenever α(x) = x then β(x) = x. Hb 34. In S7 α = (1 4 6 2) and δ = (5 7) and β = (3) are disjoint cycles. In Hb 29.3, δ is product of three disjoint cycles, δ = (1 3 5)(2 7 6)(4). But since (4) is just the identity permutation, we may write δ as a product of just two disjoint cycles. So, δ = (1 3 5)(2 7 6) In general, Theorem 5.4.3. Every non identity permutation σ of a ﬁnite set can be ex- pressed uniquely (up to the number of factors) as a product of disjoint cycles of length at least 2. Proof. Let ∅ = A be a ﬁnite set and let σ ∈ SA . Let O1 , O2 , . . . , Ok be the distinct orbits of σ. Let us deﬁne σi such that σ(x) , if x ∈ Oi σi (x) = x , if otherwise Clearly, σi is a cycle for each i = 1, 2 . . . , k. Furthermore, these k cycles are disjoint; since, suppose there exists an x ∈ A such that σi (x) = σj (x), for i = j, then it will imply that Oi ∩ Oj = ∅. This contradicts our assumption. 60 CHAPTER 5. PERMUTATION GROUPS Now, let the product σk σk−1 . . . σ1 = β. Then β ∈ SA and for each x ∈ A, β(x) = y ∈ A. Thus, there exists a unique i such that y ∈ Oi . Consequently, β(x) = σi (x) = σ(x) and σj (x) = x for all i = j. Therefore, β = σ. Hence, σ is expressed as a product of disjoint cycles. Since the number of orbits of any permutation on a ﬁnite set is ﬁxed, then the number of corresponding disjoint cycles is also ﬁxed. -TN- How do we multiply cycles? Just like in ordinary permutation multipli- cation, process is done from right to left. Hb 35. In A = {1, 2, 3, 4, 5, 6, 7, 8} 1 2 3 4 5 6 7 8 1. ζ = (1 4 8 5)(3 1 6 7 4 2) = = (1 6 7 8 5)(2 3 4) 6 3 4 2 1 7 8 5 1 2 3 4 5 6 7 8 2. µ = (1 5 6 8)(3 7 4)(2) = 5 2 7 3 6 8 4 1 Perfection is achieved, not when there is nothing more to add, but when there is nothing left to take away. - Antoine de Saint Exupery 5.5 Even and Odd Permutations Deﬁnition 5.5.1. A cycle of length two is called a transposition. If α = (a1 a2 ) is a transposition, then α = (a2 a1 ). Thus, α2 = (a1 a2 )(a2 a1 ) = (a1 ) = ι. It follows that α = α−1 . Let δ = (a1 a2 a3 . . . an ) be a cycle of length n, then we may write δ as (a1 a2 a3 . . . an ) = (a1 an ) . . . (a1 a3 )(a1 a2 ) Thus, a cycle is a product of transpositions (not necessarily disjoint). Corollary 5.5.1. Every permutation of a ﬁnite set can be expressed as a prod- uct of transpositions. 5.5. EVEN AND ODD PERMUTATIONS 61 Proof. From Theorem 33, if α ∈ Sn for some n ∈ N, then α can be expressed as a product of disjoint cycles, From above, each of these αj may be expressed a a product of transposi- tions. Thus, every permutation of a ﬁnite set can be expressed as a product of transpositions. -TN- Hb 36. In S7 1. α = (4 6 1 2 7 5) = (4 5)(4 7)(4 2)(4 1)(4 6) 2. ϕ = (7 8 1 3 2 4 6 5) = (7 5)(7 6)(7 4)(7 2)(7 3)(7 1)(7 8) Theorem 5.5.2. If α is a permutation of a ﬁnite set such that α = α1 α2 . . . αk , ( where αi is a transposition for each i = 1, 2 . . . k), then k is either always even or odd. Above theorem may be reworded as: Every permutation of a ﬁnite set may be expressed uniquely as a product of either odd or even number of transpositions. Deﬁnition 5.5.2. A permutation of a ﬁnite set is even or odd according to whether is can be expressed as a product of even or odd number of transpo- sitions respectively. In Hb. 33, α and ϕ are both odd. What about ι? Note that if α = (a1 a2 ) is a transposition of a ﬁnite set, then, α2 = ι. Thus, ι is even. Deﬁnition 5.5.3. Let α and β be permutations in Sn , then α and β are said to be conjugates if there exists a permutation δ in Sn such that δσδ −1 = β 1 2 3 4 5 6 1 2 3 4 5 6 Hb 37. α = and β = are con- 5 3 1 2 4 6 4 6 1 2 5 3 jugates, since 1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 6 2 1 4 3 6 5 5 3 1 2 4 6 2 1 4 3 6 5 1 2 3 4 5 6 = 4 6 1 2 5 3 62 CHAPTER 5. PERMUTATION GROUPS 5.6 The Alternating Groups Let Sn be the symmetric group on n letters, we claim that for n ≥ 2, the number of even permutations is the same as the number of odd permutations in Sn .

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