2.4. DIRECT PRODUCTS AND FINITELY GENERATED ABELIAN GROUPS31 2.4 Direct Products and Finitely Generated Abelian Groups 2.4.1 Direct Products 4 Def 2.43. The Cartesian product of sets S1,S2, ··· ,Sn is the set of all ordered n-tuples (a1, ··· , an) where ai ∈ Si for i = 1, 2 ··· , n. n Y Si = S1 × S2 × · · · × Sn i=1 := {(a1, a2, ··· , an) | ai ∈ Si for i = 1, 2, ··· , n}. Caution: The notation (a1, a2, ··· , an) here has different meaning from the cycle notation in permutation groups. Thm 2.44. If G1,G2, ··· ,Gn are groups with the corresponding multiplica- Qn tions, then i=1 Gi is a group under the following multiplication: (a1, a2, ··· , an)(b1, b2, ··· , bn) := (a1b1, a2b2, ··· , anbn) Qn for (a1, a2, ··· , an) and (b1, b2, ··· , bn) in i=1 Gi. Proof. Qn 1. Closed: ai ∈ Gi, bi ∈ Gi ⇒ aibi ∈ Gi ⇒ (a1b1, ··· , anbn) ∈ i=1 Gi. 2. Associativity: By the associativity in each Gi. 3. Identity: Let ei be the identity in Gi. Then e := (e1, e2, ··· , en) is the identity of G. −1 −1 4. Inverse: The inverse of (a1, ··· , an) is (a1 , ··· , an ). Ex 2.45. R2, R3 are abelian groups. (Every finite dimensional real vector space is an abelian group that is iso- morphic to certain Rn.) Ex 2.46. Z2 = {(a, b) | a, b ∈ Z} is a subgroup of R2. Qn Qn (If Hi ≤ Gi for i = 1, ··· , n, then i=1 Hi ≤ i=1 Gi). 41st HW: 1, 2, 14, 34, 46 32 CHAPTER 2. PERMUTATIONS, COSETS, DIRECT PRODUCTS Ex 2.47. If ei is the identity of Gi for i = 1, ··· , n. Then n Y Gj '{e1} × {e2} × · · · × {ej−1} × Gj × {ej+1} × · · · × {en} ≤ Gi. i=1 Qn So every Gj is isomorphic to a subgroup of i=1 Gi for j = 1, ··· , n. Ex 2.48. The Klein 4-group V is isomorphic to Z2 ×Z2, which is not cyclic. Ex 2.49. Z2 × Z3 ' Z6 is cyclic, and (1, 1) is a generator. Thm 2.50. Let m and n be positive integers. 1. If gcd(m, n) = 1 (i.e. m and n are relative prime), then Zm × Zn is cyclic and is isomorphic to Zmn, and (1, 1) is a generator of Zm × Zn. 2. If gcd(m, n) 6= 1, then Zm × Zn is not cyclic. Proof. 5 1. Suppose gcd(m, n) = 1. The sum of k copies of (1, 1) is equal to (k mod m, k mod n). The smallest positive integer k that makes (k mod m, k mod n) = (0, 0) is k = mn because m and n are rel- ative prime. So the cyclic subgroup of (1, 1) in Zm × Zn has the order mn = |Zm ×Zn|. So (1, 1) generates Zm ×Zn. It implies that Zm ×Zn is a cyclic group isomorphic to Zmn. 2. Suppose gcd(m, n) = d 6= 1. Given (a, b) ∈ Zm × Zn, the sum of k copies of (a, b) is (ka mod m, kb mod n). Notice that both m and n mn n m divide d = m · d = d · n. Then mn mn ( a mod m, b mod n) = (0, 0). d d So the cyclic subgroup generated by (a, b) has an order no more than mn d < mn = |Zm × Zn|. Thus Zm × Zn can not be generated by any one of its elements, which implies that Zm × Zn is not cyclic. Qn Cor 2.51. The group i=1 Zmi is cyclic and is isomorphic to Zm1m2···mn if and only if the numbers mi for i = 1, ··· , n are pairwise relative prime, that is, the gcd of any two of them is 1. 51st HW: 1, 2, 14, 34, 46 2.4. DIRECT PRODUCTS AND FINITELY GENERATED ABELIAN GROUPS33 Proof. (For reference only) If mi for i = 1, ··· , n are pairwise relative prime, then n n ! n ! Y Y Y Zmi ' Zm1m2 Zmi ' Zm1m2m3 Zmi '···' Zm1m2···mn . i=1 i=3 i=4 Qn Qn Conversely, if i=1 Zmi ' Zm1m2···mn , then i=1 Zmi contains a subgroup isomorphic to Zmi × Zmj for 1 ≤ i < j ≤ n. Every subgroup of the cyclic Qn group i=1 Zmi ' Zm1m2···mn is cyclic. Hence Zmi × Zmj is cyclic. So gcd(mi, mj) = 1 for 1 ≤ i < j ≤ n. Cor 2.52. 6 If a positive integer n is factorized as a product of powers of distinct prime numbers: n1 n2 nr n = (p1) (p2) ··· (pr) , then n n n Zn ' Z(p1) 1 × Z(p2) 2 × · · · × Z(pr) r . n Note: Each Z(pi) i can not be further decomposed into a product of proper nontrivial subgroups. 2 2 Ex 2.53. (Z5) Z52 . The group (Z5) is not cyclic, but Z52 = Z25 is cyclic. Def 2.54. The least common multiple (abbreviated lcm) of positive integers r1, r2, ··· , rn is the smallest positive integer that is a multiple of each ri for i = 1, 2, ··· , n. If M is a multiple of ri for i = 1, ··· , n, then M is a multiple of lcm(r1, ··· , rn). Qn Thm 2.55. Let (a1, a2, ··· , an) ∈ i=1 Gi. If ai is of finite order ri in Gi, Qn then the order of (a1, a2, ··· , an) in i=1 Gi is lcm(r1, r2, ··· , rn). Proof. k k k k (a1, a2, ··· , an) = (a1, a2, ··· , an). So m m m (a1 , a2 , ··· , an ) = (e1, e2, ··· , en) if and only if m is a multiple of ri for i = 1, 2, ··· , n, if and only if m is a multiple of lcm(r1, r2, ··· , rn). The smallest such positive integer is lcm(r1, r2, ··· , rn). So the order of (a1, a2, ··· , an) is lcm(r1, r2, ··· , rn). Ex 2.56. The order of (8, 4, 10) in Z12 × Z60 × Z24. 62nd HW: 13, 15, 32, 47, 50 34 CHAPTER 2. PERMUTATIONS, COSETS, DIRECT PRODUCTS 2.4.2 The Structure of Finitely Generated Abelian Groups Thm 2.57 (Fundamental Theorem of Finitely Generated Abelian Groups). Every finitely generated abelian group G is isomorphic to a direct product of cyclic groups in the form r r r Z(p1) 1 × Z(p2) 2 × · · · × Z(pn) n × Z × Z × · · · × Z where the pi are primes, not necessarily distinct, and the ri are positive integers. The direct product is unique except for possible rearrangement of the factors; that is, the number (Betti number of G) of factor Z is unique r and the prime powers (pi) i are unique. The proof (omitted) is challenging. In particular, every finite abelian group is isomorphic to a product of finite cyclic groups. (Caution: finite group 6= finitely generated group) Ex 2.58. Decompose the following three groups: 2 2 Z4 × Z12 × Z15 ' Z22 × (Z22 × Z3) × (Z3 × Z5) ' (Z22 ) × (Z3) × Z5, 2 2 Z3 × Z4 × Z60 ' Z3 × Z22 × (Z22 × Z3 × Z5) ' (Z22 ) × (Z3) × Z5, 2 Z20 × Z36 ' (Z22 × Z5) × (Z22 × Z32 ) ' (Z22 ) × Z32 × Z5. ∼ Therefore, Z4 × Z12 × Z15 = Z3 × Z4 × Z60 Z20 × Z36. Ex 2.59 (Ex 11.13, p.109). Find all abelian groups, up to isomorphism, of order 360. 2.4.3 Applications Def 2.60. A group G is decomposable if it is isomorphic to a direct product of two proper nontrivial subgroups. Otherwise G is indecomposable. Thm 2.61. The finite indecomposable abelian groups are exactly the cyclic groups with order a power of a prime. Thm 2.62. If G is a finite abelian group, then G has a subgroup of order m for every m that divides |G|. Ex 2.63. If m is a square-free integer, then every abelian group of order m is isomorphic to Zm and is cyclic. 2.4. DIRECT PRODUCTS AND FINITELY GENERATED ABELIAN GROUPS35 2.4.4 Homework, II-11, p.110-p.113 1st 1, 2, 14, 34, 46 2nd 13, 15, 32, 47, 50 3rd 18, 26, 36, 53 (opt) 29, 37, 38, 39, 52, 54.