Algebra: Groups. Group Theory a. Examples of Groups. groups. The inverse of a is simply a, which exists.

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1 Group Theory a Let G be a set and be a binary operation on G. (G, ) is called a group if it satisfies the following. 1. For all a, b G, a b G (closure). 2. For all a, b, c G, a (b c) = (a b) c (associativity). Algebra: Groups 3. There exists e G with a e = e a = a for all a G (identity). 4. For each a G, there is an element b G such that a b = b a = e (inverse). G is commutative or abelian if a b = b a for all a, b G. a Niels Henrik Abel ( ) and Evariste Galois ( ). This formal definition is by Arthur Cayley in c 2004 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 627 c 2004 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 629 Examples of Groups Under ordinary +, (Z, +), (Q, +), (R, +), (C, +) are groups. The inverse of a is simply a, which exists. The pursuit of mathematics is a divine madness of the human spirit. Alfred North Whitehead ( ), Science and the Modern World Under ordinary, none of (Z, ), (Q, ), (R, ), (C, ) are groups. The number 0 has no inverses. Under ordinary, (Q, ), (R, ), (C, ) are groups. A denotes the nonzero elements of A. Under ordinary, (Z, ), (Q, ), (R, ) are not groups. The associative axiom fails: a (b c) (a b) c. c 2004 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 628 c 2004 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 630

2 Properties of Groups a The identity of G is unique. If e 1, e 2 are both identities, then e 1 = e 1 e 2 = e 2 by the identity condition. The inverse of each element of G is unique (it is a 1 under and a under +, e.g.). Suppose b, c are both inverses of a G. Then b = b e = b (a c) = (b a) c = e c = c. a Properties must be proved using only the four axioms or their logical corollaries. (a 1 ) 1 = a. Inverses Both are inverses of a 1 and inverses are unique. (a b) 1 = b 1 a 1. (b 1 a 1 ) (a b) = b 1 (a 1 a) b = b 1 b = e. (G, ) is abelian if and only if (a b) 1 = a 1 b 1. If (G, ) is abelian, then (a b) 1 = (b a) 1 = a 1 b 1. If (a b) 1 = a 1 b 1, then a b = ((a b) 1 ) 1 = (a 1 b 1 ) 1 = (b 1 ) 1 (a 1 ) 1 = b a. c 2004 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 631 c 2004 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 633 Powers The Cancellation Properties The left-cancellation property: If a, b, c G and a b = a c, then b = c. b = (a 1 a) b = a 1 (a b) = a 1 (a c) = (a 1 a) c = c. The right-cancellation property: If a, b, c G and b a = c a, then b = c. The associative property implies that a 1 a 2 a n is well-defined. For n > 0, define For n < 0, define a n = a n = n {}}{ a a a. n {}}{ a 1 a 1 a 1. a n = (a 1 ) n because (a b) 1 = b 1 a 1. Define a 0 = e. c 2004 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 632 c 2004 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 634

3 Operations on Powers Lemma 90 a n a m = a n+m for n, m Z. For n, m 0, n m n+m a n a m {}}{{}}{{}}{ = a a a a = a a = a n+m. For n 0, m < 0, m n a n a m {}}{{}}{ n+m = a a a 1 a 1 {}}{ = a a = a n+m. The other two cases are similar. Criteria for Being a Subgroup Only two axioms need to be checked. Theorem 91 Let H be a nonempty subset of a group (G, ). Then H is a subgroup of G if and only if (1) for all a, b H, a b H, and (2) for all a H, a 1 H. Proof ( ): Assume that H is a subgroup of G. Then H is a group. So H satisfies, among other things, the closure property (1) and the inverse property (2). c 2004 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 635 c 2004 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 637 Let (G, ) be a group. Let H G. Subgroups If H is a group under, we call it a subgroup of G. For example, the set of even integers is a subgroup of (Z, +). H inherits from G in that it is the same operation, producing the same result in both G and H wherever applicable. Proof ( ): The Proof (concluded) Let H satisfy (1) and (2). We need to verify the associative property and the existence of identity. Associativity: For all a, b, c H, (a b) c = a (b c) G, hence in H by (1). Identity: For any arbitrary a H, a 1 a H by (2) and is an identity. c 2004 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 636 c 2004 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 638

4 Cyclic Groups Simpler Criterion for Being a Subgroup Theorem 92 Let H be a nonempty subset of a group (G, ). Then H is a subgroup of G if and only if a b 1 H for all a, b H. First, a a 1 H for any a H. So e = a a 1 H. A group G is called cyclic if there is an element x G such that for each a G, a = x n for some n Z. In other words, G = {x k : k Z}. G is said to be generated by x, denoted by G = x. x is called a generator, primitive root, or primitive element. a a Paolo Ruffini ( ). c 2004 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 639 c 2004 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 641 The Proof (concluded) By Theorem 91 (p. 637), we only need to prove that if a b 1 H for all a, b H, then the closure and inverse properties hold. Inverse: For any b H, b 1 = e b 1 H. Closure: For any arbitrary a, b H, a b = a (b 1 ) 1 H. Orders a of Groups and Group Elements For every group G, the number of elements in G is called the order of G, denoted by G. The order of a G, written o(a), is the least positive integer n such that a n = e. If a finite n does not exist, a has infinite order. a Paolo Ruffini. c 2004 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 640 c 2004 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 642

5 Criterion for Being a Subgroup: The Finite Case Orders of Groups and Group Elements (concluded) Lemma 93 If n is a s order and a k = e, then n k. Assume otherwise. So e = a k = a qn+r = a r, where 0 < r < n. This is a contradiction because a s order is now at most r < n. Corollary 95 Let H be a nonempty subset of a finite group (G, ). Then H is a subgroup of G if and only if for all a, b H, a b H (the closure property). By Theorem 91 (p. 637), we only need to prove that if a b H for all a, b H, then a 1 H for all a H. Let a H. Then a m = e for some m Z by Lemma 94 (p. 644). Hence a 1 = a m 1 H. c 2004 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 643 c 2004 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 645 Finiteness of Orders of Groups and Group Elements a Lemma 94 If G is a finite group, then the order of every element a G must be finite. Consider the chain a 1, a 2, a 3,.... Because G is finite, the chain must eventually repeat itself. So there must be distinct i < j such that a i = a j. By the cancellation property, a j i = e. a Contributed by Mr. Bao (B ) on December 23, Finite Cyclic Groups Lemma 96 Suppose G is a finite group and a G. (1) a = {a k : k Z + }. (2) a = o(a). The set a {a k : k Z} contains a, a 2, a 3,...,a o(a) = e. But no two of them are identical. Otherwise, a i = a j for 1 i < j o(a), and a j i = e, a contradiction because j i < o(a). The set s other o(a) elements, a 1, a 2,...,a o(a) = e are not new because a m = a o(a) m. As a m = a m mod o(a), there are no other elements. c 2004 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 644 c 2004 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 646

6 Cosets as Partitions Cyclic Subgroups Lemma 97 Let (G, ) be a group and a G. Then ({a k : k Z}, ) is a subgroup of G. For a i, a j G, we have a i a j = a i j by Lemma 90 (p. 635). Theorem 92 (p. 639) then implies the lemma. Corollary 98 Every finite group can be decomposed into disjoint cyclic subgroups. Let G be a finite group. For a, b G, either ah = bh or ah bh =. Assume ah bh. Let c = a h 1 = b h 2 for some h 1, h 2 H. If x ah, then x = a h for some h H and x = (b h 2 h 1 1 ) h = b (h 2 h 1 1 h) bh, which implies ah bh. Similarly, we can prove that bh ah. As a ah for any a G, G can be partitioned by cosets. c 2004 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 647 c 2004 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 649 Cosets a If H is a subgroup of G, the set ah = {a h : h H} is called a (left) coset of H in G. ah = H when G is finite. ah H by definition. If ah < H, then a h i = a h j for some distinct h i, h j H, which implies h i = h j by the left-cancellation property, a contradiction. Similarly, we can also define a right coset of H in G, denoted by Ha. a Augustin Louis Cauchy ( ), who published more than 800 papers. 1: print H; 2: G := G H; 3: while G do 4: Pick a G; 5: print ah; 6: G := G ah; 7: end while Constructing a Coset Partition e a b H ah bh c 2004 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 648 c 2004 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 650

7 Lagrange s a Theorem Theorem 99 If G is a finite group with subgroup H, then H divides G. G can be partitioned by cosets of H Each coset of H has the same order, H. Hence H divides G. a Joseph Louis Lagrange ( ). First Corollary of Lagrange s Theorem a Corollary 100 If G is a finite group and a G, then o(a) divides G. The set generated by a, {a k : k Z}, has size o(a) by Lemma 96 (p. 646). Set {a k : k Z} is a subgroup of G by Lemma 97 (p. 647). Lagrange s theorem thus implies our claim. a See also Lemma 93 (p. 643). c 2004 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 651 c 2004 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 653 The Fermat a -Euler Theorem Applications of Lagrange s a Theorem Suppose G = 16, then the orders of its subgroups must be 1, 2, 4, 8, or 16 by Lagrange s Theorem (p. 651). Suppose G = 18, then the orders of its subgroups must be 1, 2, 3, 6, 9, or 18 by Lagrange s Theorem (p. 651). a Joseph Louis Lagrange ( ). Theorem 101 If G is a finite group, then every a G satisfies a G = e. By Corollary 100 (p. 653), o(a) divides G. Let G = o(a) k, where k Z +. Now, a G = a o(a) k = (a o(a) ) k = e k = e. a Pierre de Fermat ( ). c 2004 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 652 c 2004 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 654

8 Number of Generators in Finite Cyclic Groups Lemma 102 Let G be a finite cyclic group with order m and g be a generator of G. Then the generators are where 1 i < m and gcd(i, m) = 1. Hence the number of generators is φ(m), Euler s phi function (p. 342). g i, Suppose 1 i < m is relatively prime to m. Let j = o(g i ). So e = g ij = g ij mod m by Corollary 100 (p. 653). As g is a generator, ij mod m = 0. Permutations a Let function f : {1, 2,...,n} {1, 2,...,n} be one-to-one and onto. f must be a permutation on {1, 2,...,n}. Write f as I = 1 2 n f(1) f(2) f(n) 1 2 n 1 2 n, the identity permutation. a Lagrange (1770); Ruffini (1799); Cauchy (1815). c 2004 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 655 c 2004 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 657 The Proof (concluded) This implies that m divides ij. As m cannot divide i by assumption, m divides j. As j > 0, we must have j = m and g i is a generator. Conversely, assume 1 i < m but gcd(i, m) = d > 1. Define j = m/d. But g i cannot be a generator. Indeed, 0 < j < m and (g i ) j = g ij = g im/d = g (i/d)m = (g m ) i/d = e. Permutation Groups Let f and g be two permutations on {1, 2,...,n}. Then f g is defined as 1 2 n. (79) g(f(1)) g(f(2)) g(f(n)) Note that f is applied first. This is unlike other books or function composition on p c 2004 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 656 c 2004 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 658

9 For example, Permutation Groups (concluded) = When a set of permutations forms a group under, we have a permutation group.. Cycle Decomposition of Permutations A permutation like represented as There are 3 cycles above (1 3)(2 4)(5). can be 5 is a fixed point; it is invariant under the permutation. The cycle decomposition can be calculated efficiently. c 2004 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 659 c 2004 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 661 Permutation Group as a Multiplication Table g 1 g 2 g 3 g 4 g 5 g 6 Another Cycle Decomposition = (1 2 3)(4 5)(6) There are 3 cycles above. Equivalent cycle decompositions: (3 1 2)(5 4)(6), (4 5)(1 2 3)(6),. The cycle decomposition is essentially unique. c 2004 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 660 c 2004 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 662

10 Cycles and Transpositions Call a cycle (i 1 i 2 i n ) an n-cycle, where i 1, i 2,...,i n are distinct. The order of an n-cycle π is n as π n = I. A 2-cycle is called a transposition. (1 2 3) = (1 2)(1 3) (from left to right always). In general, (i 1 i 2 i n ) = (i 1 i 2 )(i 1 i 3 ) (i 1 i n ). So every permutation is a product of transpositions. Expected Number of Cycles Theorem 103 The expected number of cycles of a random permutation of {1, 2,...,n} is n i=1 (1/i) lnn. Define the indicator function: 1, if i is contained in a cycle of length k I i = 0, otherwise I 1 + I I n is the number of points in {1, 2,...,n} contained in a cycle of length k. (I 1 + I I n )/k is the number of cycles of length k.. c 2004 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 663 c 2004 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 665 Cycle Length Pick a permutation of {1, 2,...,n} at random. We claim that the probability that the cycle containing 1 has length k is 1/n. There are ( n 1 k 1) ways to choose the elements of the cycle containing 1. There are (k 1)! ways to order them. There are (n k)! ways to permute the rest. Hence the desired probability equals ( n 1 k 1) (k 1)! (n k)! = 1 n! n. The Proof (concluded) The expected number of cycles of length k is, by Adam s theorem, [ ] I1 + I I n E = E[ I 1 ] + E[ I 2 ] + + E[ I n ]. k k From p. 664, we know that E[ I i ] = 1/n. Hence the expected number of cycles of length k is 1/k. Finally, the expected number of cycles is n k=1 1 k lnn. c 2004 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 664 c 2004 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 666

11 S n : The Symmetric Group of Degree n There are n! permutations on {1, 2,...,n}. These permutations form a group denoted by S n (verify it). This group is called the symmetric group of degree n. S n = n!. A key result by Cayley: Every finite group is isomorphic to a permutation group (i.e., a subgroup of S n ). Stabilizer of 111 g 1 g 2 g 3 g 4 g 5 g 6 Orbit of 111 Permutation character of g 1 c 2004 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 667 c 2004 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 669 Of Orbits, Stabilizers, and Characters Let G be a permutation group on a finite set X. Let x X. O x = { g(x) : g G } is called the orbit of x with respect to G. Note that x O x. G x = { g G : g(x) = x } is called the stabilizer of x in G. F(g) = { z X : g(z) = z } is called the permutation character of g in X. Orbits as Partitions Lemma 104 If G be a permutation group on set X, then G s orbits partition X. x O x = X because x O x for all x X. If O x O y, then O x O y. For any a O x, a = g (x) for some g G. Suppose z O x O y. Then z = g(x) = g (y) for some g, g G. Hence a = g (g 1 (z)) = g (g 1 (g (y))) O y. The other direction O y O x is symmetric. c 2004 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 668 c 2004 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 670

12 Orbithood as an Equivalence Relation Lemma 105 Suppose G be a permutation group on set X. Two i, j X are in the same orbit if and only if there is a g G such that g(i) = j. Suppose i, j X are in the same orbit O x. i = g 1 (x) and j = g 2 (x) for some g 1, g 2 G. Hence j = g 2 (x) = g 2 (g 1 1 (i)). Suppose there is a g G such that g(i) = j. Then j O i and i O i. Stabilizers of Elements of an Orbit Lemma 107 Suppose G be a permutation group. If x, y are in the same orbit, then G x = G y. By Lemma 105 (p. 671), y = h(x) for some h G. G(x, y) { g G : g(x) = y } is then nonempty as it contains h. If g G(x, y), then g(x) = h(x) or h 1 (g(x)) = x. We conclude that g h 1 G x. Then g G x h. Recall that G x h = { g h : g G, g(x) = x }. c 2004 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 671 c 2004 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 673 Grouphood of Stabilizers Lemma 106 A stabilizer is a subgroup. Let G be a permutation group on set X. Consider a stabilizer G x = { g G : g(x) = x } for x X. For all g 1, g 2 G x, g 1 g 2 G x because g 1 g 2 fixes x. For all g G x, g 1 G x because g 1 fixes x. The lemma follows by Theorem 91 (p. 637). The Proof (concluded) As a result G(x, y) G x h. Similarly, we can prove that G x h G(x, y). Hence G(x, y) = G x h, a right coset of G x. By the coset partition theorem (p. 649), G(x, y) = G x. By the same argument (why?), G(x, y) = hg y, a left coset of G y. Hence G(x, y) = G y. c 2004 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 672 c 2004 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 674

13 Burnside s Lemma a Theorem 108 G is a permutation group on {1, 2,...,n}. The average number of fixed points of permutations in G equals the number of orbits. Let O 1, O 2,...,O k be the distinct orbits with O i = o i. They partition X {1, 2,...,n} by Lemma 104 (p. 670). Stabilizer G x is the set of permutations fixing x. G x is a subgroup of G (Lemma 106 on p. 672). Consider x X (note that x O x ). a William Burnside ( ) in The theorem is actually due to Cauchy and later Ferdinand Frobenius ( ) in 1896! The Proof (concluded) Let k(π) denote permutation π s number of fixed points. Then the average number of fixed points is 1 k(π) = 1 G G π G = 1 G = 1 G = k. x {1,2,...,n} k i=1 k i=1 x O i G x o i G o i G x (80) c 2004 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 675 c 2004 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 677 The Proof (continued) Each right coset of G x consists of those permutations in G that map x to a given element of O x. Elements of O x are mapped only to elements of O x by G (Lemma 105 on p. 671). Consider the right coset G x g for g G. Every permutation in G x g maps x to the same g(x) O x (p. 650). Hence G x has o x right cosets. By the coset partition theorem (p. 649), G x = G o x. An Alternative Form of Burnside s Lemma (p. 675) Theorem 109 (Burnside s lemma) Let k is the number of distinct orbits of permutation group G. Then g G F(g) = k. G Corollary 110 Let G be a permutation group on X. Then x X G x = g G F(g) Let k be the number of distinct orbits of G. By Eq. (80) on p. 677, x X G x = k G. The corollary then follows from Theorem 109. c 2004 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 676 c 2004 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 678

14 Circular Seating (concluded) How To Use Burnside s Lemma By Lemma 105 (p. 671), i, j in the same orbit are considered identical (under G). Hence the number of distinct orbits is the number of distinct configurations such as colorings, seatings, etc. (under G). The identity permutation g has F(g) = n! because every seating is fixed by g. All other permutations g have F(g) = 0 because no seatings are fixed by g. As the number of permutations is n, Burnside s lemma says that the number of distinct seatings is n 1 {}}{ n! = (n 1)!. n This agrees with the result on p. 20. c 2004 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 679 c 2004 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 681 Circular Seating (p. 20) We want to seat n people around a circle. Two seatings are equivalent if one is the result of rotation of the other (i.e., in the same orbit). How many distinct seatings (i.e., orbits) are there? The permutation group consists of n clockwise rotations. Note that a rotation moves a seating into another So the permutation group acts on the n! possible seatings, not the n people. Bracelet Coloring How many ways are there to color a bracelet of n beads with k colors, where n is an odd prime? The bracelet can be rotated but not flipped over. The permutation group consists of n clockwise rotations. The identity permutation g has F(g) = k n because there are k n colorings. c 2004 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 680 c 2004 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 682

15 Bracelet Coloring (continued) All nonidentity permutations g have F(g) = k. Suppose coloring C does not paint beads with the same color, say those at positions a and b. There exists a i N such that g i moves the bead at location a to location b. Solve di (b a) mod n for i if g rotates the bracelet by d > 0 positions. If coloring C is a fixed point under g, then it is also a fixed point under g i. But this is impossible. c 2004 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 683 c 2004 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 685 Bracelet Coloring (continued) All nonidentity permutations g have F(g) k. All colorings with beads painted with the same color are in the same equivalence class (orbit). There are k of them. Bracelet Coloring (concluded) The number of distinct colorings is n 1 {}}{ k n + k + k + + k n = kn n + n 1 k. (81) n c 2004 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 684 c 2004 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 686

16 Bracelet Coloring with n = 5 and k = 2 c 2004 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 687 Finis c 2004 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 688

Cyclic Groups. AgroupG is called cyclic if there is an element x G such that for each a G, a = x n for some n Z. G = { x k : k Z }.

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