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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1 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. 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, orprimitive element. a a Paolo Ruffini ( ). c 2017 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 825

2 Cyclic Groups Are Abelian Lemma 116 Every cyclic group is abelian. Let x, y G, a cyclic group g. Let x = g m and y = g n for some m, n Z. Now, x y = g m g n = g m+n = g n+m = g n g m = y x. c 2017 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 826

3 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 Ruffini (1799). c 2017 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 827

4 The Group (Z n, +), n>1 Under ordinary + modulo n, (Z n, +) is an additive abelian group for n>1 (p. 809). Its order is n. Its generator is 1. Every i Z n can be expressed as i {}}{ c 2017 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 828

5 Orders of Group Elements Lemma 117 If x s order is n and x k = e, thenn k. Assume otherwise and k = qn + r, where0<r<n. So e = x k = x qn+r = x qn x r = x r. So x s order is at most r<n, a contradiction. c 2017 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 829

6 Finiteness of Orders of Groups and Group Elements a Lemma 118 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<jsuch that a i = a j. Then e = a j i by Lemma 112 (p. 818). As a result, a s order is at most j i, which is finite. a Contributed by Mr. Bao (B ) on December 23, c 2017 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 830

7 Criterion for Being a Subgroup: The Finite Case Corollary 119 Let H be a nonempty subset of a finite group (G, ). ThenH is a subgroup of G if and only if for all a, b H, a b H (closure). By Theorem 113 (p. 820), we only need to prove that if the closure property holds, then a 1 H for all a H (inverse). Let a H. Then a m = e for some m Z by Lemma 118 (p. 830). Hence a 1 = a m 1 H. This is because a a m 1 = a m = e. c 2017 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 831

8 Remarks Corollary 119 may not hold for infinite groups. For example, (Z, +) is a group (p. 808). Its subset (N, +) is closed under +. But (N, +) is not a subgroup of (Z, +) (p. 808)! c 2017 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 832

9 Cyclic Subsets of Finite Groups Lemma 120 Suppose G is a finite group and a G. (1) { a k : k Z } = { a k : k Z + }.(2) {a k : k Z } = o(a). The set a Δ = { a k : k Z } contains at least They are all distinct. a, a 2,a 3,...,a o(a) = e. Otherwise, a i = a j for 1 i<j o(a), and e = a j i, a contradiction because j i<o(a). It is easy to see that a m = a m mod o(a) for all m 0. Similarly, a m = a ( m) modo(a) for all m 0. Hence there are no other elements in { a k : k Z }. c 2017 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 833

10 Remarks Lemma 120(1) (p. 833) does not (yet) say { a k : k Z + } must be a subgroup for every a G. It only asserts that the cyclic set { a k : k Z } is identical to { a k : k Z + }. Recall our definition on p c 2017 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 834

11 Cyclic Subgroups Lemma 121 Let (G, ) be a group and a G. Then H =({ a k : k Z }, ) is a subgroup of G. For a i,a j H, wehave by Lemma 112 (p. 818). a i a j = a i j H Theorem 114 (p. 822) then implies the lemma. c 2017 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 835

12 Remarks Lemma 121 (p. 835) does imply { a k : k Z } (hence { a k : k Z + } as well) is a subgroup of G for every a G. Corollary 122 Let (G, ) be a finite group and a G. Then H =({ a k : k Z + }, ) is a subgroup of G. c 2017 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 836

13 Cyclic Structures Must Form a Group? a Must a cyclic structure ({ a k : k Z }, ) be a group without restrictions on and entity a? Note that Lemma 121 (p. 835) does impose some restrictions: (G, ) must be a group. Consider algebraic structure ({ 2 k : k Z }, mod 12). Note that 2 cannot have an inverse modulo 12 because gcd(2, 12) = 2 1. Hence the cyclic structure is not a group. a Contributed by Mr. Bao (B ) on December 23, c 2017 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 837

14 Cosets a If H is a subgroup of G and a G, theset ah = { a h : h H } is called a (left) coset of H in G. An element of ah is called a coset representative of ah. a Augustin Louis Cauchy ( ), who published more than 800 papers. c 2017 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 838

15 Cosets (continued) ah = H when H is finite. a ah H by definition. If ah < H, thena h i = a h j for some distinct h i,h j H. Butthatimpliesh i = h j by the left-cancellation property, a contradiction. a Contributed by Mr. Kai-Yuan Hou (B ) on June 7, Of course, if G is finite, then H must be, too. c 2017 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 839

16 Cosets (concluded) Similarly, we can also define a right coset of H in G, Ha = { h a : h H }. Let G/H denote the family of all the (left) cosets of H. When G/H is a group, it is called the quotient group G mod H. c 2017 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 840

17 Cosets as Partitions Let H be a subgroup of a group G. For a, b G, eitherah = bh or ah bh =. a Assume ah bh. Let c = a h 1 = b h 2 for some h 1,h 2 H. If x ah, thenx = a h for some h H and x =(b h 2 h 1 1 ) h = b (h 2 h 1 1 h) bh. So ah bh. Similarly, we can prove that bh ah. a Do we need to require that G be finite for this result as the textbook does? Contributed by Mr. Kai-Yuan Hou (B ) on June 7, c 2017 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 841

18 Cosets as Partitions (continued) As a ah for any a G, G can be partitioned by cosets. c 2017 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 842

19 Cosets as Partitions (concluded) e a b H ah bh c 2017 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 843

20 Constructing a Coset Partition of a Finite Group Let G be a group and H a subgroup. 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 c 2017 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 844

21 Lagrange s a Theorem Theorem 123 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. b Hence H divides G. a Joseph Louis Lagrange ( ). b p c 2017 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 845

Joseph Louis Lagrange (1736 1813) c 2017 Prof.

22 Joseph Louis Lagrange ( ) c 2017 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 846

23 Applications of Lagrange s Theorem Suppose G = 16, then the order of its subgroup must be 1, 2, 4, 8, or 16. Suppose G = 18, then the order of its subgroup must be 1, 2, 3, 6, 9, or 18. Suppose G = 11, a prime, then the order of its subgroup must be 1 or 11. c 2017 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 847

24 Index of a Subgroup Let H be a subgroup of G. The index of H in G, denoted by [ G : H ], is the number of cosets of H in G. In other words, the index is the size of G/H, a i.e., [ G : H ] Δ = G/H. When G is finite, because every coset has the same size, b [ G : H ]= G H. (101) a p b Recall the proof of Lagrange s theorem (p. 845). c 2017 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 848

25 Index of a Subgroup (concluded) Corollary 124 If H is a subgroup of a finite group G, then G/H = G H. Corollary 125 If H is a subgroup of a finite group G, then Because G/{ e } = G/H H/{ e }. G/{ e } = G {e } = G H H G/H H/{ e }. {e } c 2017 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 849

26 First Corollary of Lagrange s Theorem a Corollary 126 If G is a finite group and a G, theno(a) divides G. The set generated by a, { a k : k Z }, hassizeo(a) by Lemma 120 (p. 833). Set { a k : k Z } is a subgroup of G by Lemma 121 (p. 835). Lagrange s theorem then implies our claim. a See also Lemma 117 (p. 829). c 2017 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 850

27 The Fermat a -Euler Theorem Theorem 127 If G is a finite group, then every a G satisfies a G = e. By Corollary 126 (p. 850), o(a) divides G. Let G = o(a) k, wherek Z +. Now, a G = a o(a) k =(a o(a) ) k = e k = e. a Pierre de Fermat ( ). c 2017 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 851

Pierre de Fermat (1601 1665) c 2017 Prof.

28 Pierre de Fermat ( ) c 2017 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 852

29 Euler s Theorem Recall that Z n is the set of positive integers between 1 and n 1 that are relatively prime to n. a Theorem 128 (Euler s theorem) For all a Z n, (Z n, ) is a group. b Z n = φ(n). c a φ(n) 1modn. Apply Theorem 127 (p. 851). a p b p c p c 2017 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 853

30 Fermat s Little Theorem Theorem 129 (Fermat s little theorem) Suppose p is aprime.then a p 1 1modp for all a Z p. By Euler s theorem (p. 853). c 2017 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 854

31 Three Easy Applications The inverse of a in (Z p, ) isa p 2 mod p. a p 2 a = a p 1 1modp by Fermat s little theorem. 3 (n 2 1) when 3 n. The number 3 is a prime. By Fermat s little theorem, n 3 1 = n 2 1mod3. c 2017 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 855

32 Three Easy Applications (concluded) a pn p n 1 1modp n for odd prime p and gcd(a, p) =1. By Euler s theorem (p. 853) and Theorem 61 (p. 418), 1 a φ(pn) a pn p n 1 mod p n. c 2017 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 856

33 Application: The RSA Function a Let n = pq, wherep and q are distinct odd primes. Then φ(n) =(p 1)(q 1) by Theorem 61 (p. 418). Let e be an odd integer relatively prime to φ(n). b The RSA function is defined as where gcd(x, n) =1. E(x) =x e mod n, a Rivest, Shamir, & Adleman (1978). b This e should not be confused with the identity of a group. c 2017 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 857

34 Adi Shamir, Ron Rivest, and Leonard Adleman c 2017 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 858

35 Encryption Using the RSA Function The RSA function is a good candidate for the encryption of message x. The number e is called the encryption key. Furthermore, the prime number theorem (p. 158) guarantees an abundance of primes. c 2017 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 859

36 Decryption and Trapdoor Information To be useful, an efficient algorithm must exist to recover x from E(x). Butthisisanopenproblem(sofar). The way out is the existence of the trapdoor information not available to the enemies. One possible piece of trapdoor information is the factorization of n. Factorization is believed to be hard. a a Numbers can be factorized efficiently by Shor s (1994) quantum algorithm. c 2017 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 860

37 Inversion of the RSA Function Let d be the inverse of e modulo φ(n), that is, ed =1modφ(n). Because gcd(e, φ(n)) = 1, such d exists. d can be found by the extended Euclidean algorithm (p. 792). By Euler s theorem (p. 853), the encrypted message y = Δ E(x) canbedecrypted by y d =(x e ) d = x ed = x 1+kφ(n) = xx kφ(n) = x mod n. So the decryption function is D(y) =y d mod n. c 2017 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 861

38 Sophie Germain Primes How many e s are there such that gcd(e, φ(n)) = 1? The density of numbers between 1 and φ(n) thatsatisfy the above condition is a φ(φ(n)) φ(n) = φ((p 1)(q 1)) (p 1)(q 1). a The 5th edition of Grimaldi s Discrete and Combinatorial Mathematics errs with φ(n)/n on p c 2017 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 862

39 Sophie Germain Primes (concluded) Suppose p =2p +1andq =2q +1,wherep,q are also primes. Such primes are called Sophie Germain primes. The density becomes φ(4p q ) (p 1)(q 1) = 2(p 1)(q 1) 1 4p q 2. c 2017 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 863

40 Sophie Germain ( ) A French mathematician. Gauss on Germain: But when a person of the sex which, according to our customs and prejudices, must encounter infinitely more difficulties than men to familiarize herself with these thorny researches, succeeds nevertheless in surmounting these obstacles and penetrating the most obscure parts of them, then without doubt she must have the noblest courage, quite extraordinary talents and superior genius. c 2017 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 864

41 Second Corollary of Lagrange s Theorem Corollary 130 Every group of prime order is cyclic. Pick any element a e of the group G. a Note that o(a) > 1. As o(a) also divides G, b aprimenumber,o(a) = G. This implies that every b G must be of the form a k for some k Z. a Because a group of prime order has at least 2 elements, such an a exists. b See Corollary 126 (p. 850). c 2017 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 865

42 Criterion for Generators The computational problem of verifying if g is a generator is believed to be hard without the factorization of G. Exhaustive testing is too slow, taking O( G ) time. A better algorithm is based on the next corollary, assuming the factorization of G is available. c 2017 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 866

43 Third Corollary of Lagrange s Theorem Corollary 131 Let G be a finite cyclic group with prime factorization of its order m = p a 1 1 pa 2 2 pa n n.theng Gis a generator of G if and only if for i =1, 2,...,n. Define m i Δ = m/pi. g m/pi e (102) Hence m i = p a 1 1 pa 2 2 pa i 1 i p a n n. Suppose g is a generator. c 2017 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 867

44 The Proof (continued) Because o(g) =m, g m/pi e for all i. Conversely, assume inequality (102). We proceed to show that g must be a generator. Let o(g) =j so g j = e. Because j divides m by Lagrange s theorem (p. 845), m = dj for some d 1. c 2017 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 868

45 The Proof (concluded) Let j = p b 1 1 pb 2 2 pb n n, where 0 b i a i for i =1, 2,...,n. What if j<m? Then b i <a i for some i. But then j divides m i. This implies that g m i (102). = e, contradicting inequality We must conclude that j = m and g is a generator. c 2017 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 869

46 Algorithm for Testing If g Is a Generator of G 1: m := p a 1 1 pa 2 2 pa n n ; 2: for i =1, 2,...,n do 3: if g m/p i = e then 4: return g is not a generator ; 5: end if 6: end for 7: return g is a generator ; Note that n = O(log 2 m). So the number of steps is polynomial in log 2 m. In contrast, the exhaustive method takes m steps. c 2017 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 870

47 Number of Generators in Finite Cyclic Groups Lemma 132 Let G be a finite cyclic group with order m and g be a generator of G. Then the generators are where 1 i<mand gcd(i, m) =1. Hence the number of generators is φ(m), Euler s phi function (p. 418). g i, Suppose 1 i<mis relatively prime to m. Let j = o(g i ). So g ij = e. As g is a generator, m divides ij by Lemma 117 (p. 829). c 2017 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 871

48 The Proof (concluded) As m cannot divide i by assumption, m divides j. As 1 j m, wemusthavej = m and g i is a generator. Next assume 1 i<mbut gcd(i, m) =d>1. Define j = m/d. Now, 0 <j<m. By the Fermat-Euler theorem (p. 851), (g i ) j = g ij = g im/d = g (i/d)m =(g m ) i/d = e. So g i is not a generator. c 2017 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 872

49 Number of Generators in (Z n, ), IfAny Theorem 133 If (Z n, ) has a generator, then it has φ(φ(n)) generators. a Recall Euler s phi function (p. 418). If (Z n, ) has a generator, then it is a finite cyclic group. Lemma 132 (p. 871) then implies the theorem. a A common mistake is to answer φ(n). Is it easy to calculate φ(φ(n)) even if one knows the factorization of n? c 2017 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 873

50 Group Homomorphism and Isomorphism Let (G, ) and(h, ) be 2 groups. A function f : G H is a homomorphism if for all x, y G. f(x y) =f(x) f(y) It is called an isomorphism if f is a bijection. G and H are said to be isomorphic (written as G = H) if an isomorphism exists between them. c 2017 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 874

51 Group Homomorphism and Isomorphism (concluded) Isomorphic groups have the same multiplication table (up to relabeling by f). When ambiguity is can be an issue, Write e G for the identity of G. Write e H for the identity of H. c 2017 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 875

52 All Cyclic Groups Are Isomorphic Lemma 134 Cyclic groups of the same order are isomorphic. Let G =( g, g )andh =( h, h ) be 2 cyclic groups of the same order. Define f : G H by f(g i )=h i. f is a one-to-one correspondence between G and H because f(g) =h clearly generates H. For all x = g i G and y = g j G, f(x g y)=f(g i+j )=h i+j = h i h h j = f(x) h f(y). c 2017 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 876

53 All Cyclic Groups Are Isomorphic (concluded) Corollary 135 Every cyclic group of order m>1 is isomorphic to (Z n, +). (Z n, +) is a cyclic abelian group. a Lemma 134 (p. 876) then implies this corollary. a p c 2017 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 877

54 Permutations a Let function f : { 1, 2,...,n} {1, 2,...,n} be one-to-one and onto. f must be a permutation of { 1, 2,...,n}. Write f as I = 1 2 n f(1) f(2) f(n) 1 2 n 1 2 n, theidentity permutation. a Lagrange (1770); Ruffini (1799); Cauchy (1815). Recall p c 2017 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 878

55 Permutations (concluded) So permutations are functions. We are mainly interested in permutations of a finite set. But permutations can be defined as bijections on an infinite set. c 2017 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 879

56 Permutation Groups Let f and g be two permutations of { 1, 2,...,n}. Then f g is defined as 1 2 n g(f(1)) g(f(2)) g(f(n)). (103) Note that f is applied first. The alternative of applying g first is more consistent with function composition on p But our convention is more convenient in calculations. Either convention works. c 2017 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 880

57 Permutation Groups (continued) For example, = In general, permutations can work on any finite set, not just { 1, 2,...,n}. The finite set can even be a set of permutations! When a set of permutations forms a group under, we have a permutation group. In general, is not abelian. c 2017 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 881

58 Permutation Groups (concluded) A key result of Cayley (p. 901) says every group is isomorphic to a permutation group! But the permutation perspective has one unique advantage over groups: Permutations are functions! Under this perspective, notations like g(x), x X, makes sense for a group element g G that acts on X. This idea was used to construct interconnection networks for parallel computers. a a Annexstein, Baumslag, & Rosenberg (1990). c 2017 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 882

59 Permutation Group as a Multiplication Table g 1 g 2 g 3 g 4 g 5 g 6 c 2017 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 883

60 The Symmetric Group There are n! permutations of { 1, 2,...,n}. These permutations form a group (verify it). This group S n is called the symmetric group of degree n. S n = n!. A permutation group is thus a subgroup of S n. And by Cayley (p. 901), every group is a subgroup of a symmetric group. In general, S X denotes the set of all permutations of a set X. c 2017 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 884

61 Cycles Call a cycle (i 1 i 2 i m )anm-cycle, where i 1,i 2,...,i m are distinct. It represents the permutation i 1 i 2 i 3 i m 1 i m other fixed points i 2 i 3 i 4 i m i 1 other fixed points. The order of an m-cycle g is m because the identity permutation. g m = I, c 2017 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 885

62 A 1-cycle is a fixed point. The inverse of a cycle: Cycles (concluded) (i 1 i 2 i m ) 1 =(i m i n 1 i 1 ). Because (i 1 i 2 i m )(i m i m 1 i 1 )=(i 1 )(i 2 ) (i m ). c 2017 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 886

63 Cycle Decomposition of Permutations A permutation like represented as (1 3)(2 4)(5). There are 3 disjoint cycles above. can be 5 is a fixed point; it is invariant under the permutation. In general, a permutation is either a cycle or a product of disjoint cycles. c 2017 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 887

64 Cycle Decomposition of Permutations (concluded) Now, (1 3)(2 4)(5) = (1 3)(2 4). So fixed points drop out. A cycle decomposition a of a permutation is a product of disjoint cycles that contains a 1-cycle for every invariant element. A cycle decomposition can be calculated efficiently. a Also called a complete factorization. c 2017 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 888

65 Another Cycle Decomposition = (1 2 3)(4 5)(6). There are 3 disjoint cycles above. Equivalent cycle decompositions: (3 1 2)(5 4)(6), (4 5)(1 2 3)(6),. The cycle decomposition is essentially unique. c 2017 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 889

66 Transpositions 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 not necessarily disjoint transpositions. c 2017 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 890

67 Order of a Permutation Theorem 136 Let g S n.ifg = g 1 g m is a product of disjoint cycles, a then where g i is an r i -cycle. o(g) =lcm(r 1,r 2,...,r m ), We knew o(g i )=r i (p. 885). Suppose o(g) =M. Clearly, g M =(g 1 g m ) M = g M 1 g M m because the g i s are disjoint and hence commute. a It is not necessarily a cycle decomposition of g because 1-cycles may be suppressed here. c 2017 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 891

68 Order of a Permutation (concluded) The disjointness of the g i s also implies g M i = I for each i. Then r i M by Lemma 117 (p. 829), 1 i m. Hence lcm(r 1,r 2,...,r m ) M as well. But g lcm(r 1,r 2,...,r m ) = I. Trivially, r i divides lcm(r 1,r 2,...,r m ). So g lcm(r 1,...,r m ) = g lcm(r 1,...,r m ) 1 g lcm(r 1,...,r m ) m = I. Hence lcm(r 1,r 2,...,r m )=M. Corollary 137 Let g S n.ifp is a prime, then o(g) =p if and only if g is a p-cycle or a product of disjoint p-cycles. c 2017 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 892

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

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