CHAPTER 8 External Direct Products Definition and Examples Definition (External Direct Product). Let G1, G2, . , Gn be a finite col- lection of groups. The external direct product of G1, G2, . , Gn is G1 G2 G = (g , g , . , g ) g G ⊕ ⊕ · · · ⊕ n { 1 2 n | i 2 i} where (g1, g2, . , gn)(g10 , g20 , . gn0 ) = (g1g10 , g2g20 , . gngn0 ) and each gigi0 is performed with the operation of Gi. Corollary. Let G1, G2, . , Gn be a finite collection of groups. Then G1 G2 G is a group. ⊕ ⊕ · · · ⊕ n Proof. 1 1 1 1 The identity is (e1, e2, . , en) and (g1, g2, . , gn)− = (g1− , g2− , . , gn− ). Associativity carries through componentwise. ⇤ Example. R2 = R R and R3 = R R R with the operation being componentwise (vector)⊕addition. ⊕ ⊕ Example. U(5) = 1, 2, 3, 4 and U(12) = 1, 5, 7, 11 . So { } { } U(5) U(12) = (1, 1), (1, 5), (1, 7), (1, 11), (2, 1), (2, 5), (2, 7), (2, 11), ⊕ (3, 1), (3, 5), (3, 7), (3, 11), (4, 1), (4, 5), (4, 7), (4, 11) . (2, 7)(3, 11) = (6 mod 5, 77 mod 12) = (1, 5). 106 8. EXTERNAL DIRECT PRODUCTS 107 Example. Z2 Z5 = (0, 0), (0, 1), (0, 2), (0, 3), (0, 4), (1, 0), (1, 1), (1, 2), (1, 3), (1, 4) . ⊕ Consider (1, 1) . Since the operation is addition, h i (1, 1) = (1, 1), 2(1, 1) = (0, 2), 3(1, 1) = (1, 3), 4(1, 1) = (0, 4) h i 5(1, 1) = (1, 0), 6(1, 1) = (0, 1), 7(1, 1) = (1, 2) 8(1, 1) = (0, 3), 9(1, 1) = (1, 4), 10(1, 1) = (0, 0), so Z2 Z5 = (1, 1) is cyclic. ⊕ h i Are all groups of the form Zn Zm cyclic? ⊕ Example. Consider Z2 Z2 = (0, 0), (0, 1), (1, 0), (1, 2) . ⊕ Except for (0, 0), each element has order 2, so Z2 Z2 is the Klein 4-group, so is not cyclic. ⊕ Properties of External Direct Products Theorem (8.1 — Order of an Element in a Direct Product). The order of an element in a direct product of a finite number of finite groups is the lcm of the orders of the components of the elements: (g , g , . , g ) = lcm( g , g , . , g ). | 1 2 n | | 1| | 2| | n| Proof. First consider the case where the direct product has 2 factors. s Let (g1, g2) G1 G2. Let s = lcm( g1 , g2 ) and let t = (g1, g2) .T hen(g1, g2) = (gs, gs) = (e,2e) =⊕ (Theorem 4.1, Corollary| | | | 2) t s. Thus| t s.| 1 2 ) | But (gt, gt) = (g , g )t = (e, e) = g t and g t. 1 2 1 2 ) | 1| | 2| Thus t is a common multiple of g and g = s t since s = lcm( g , g ). | 1| | 2| ) | 1| | 2| Thus s = t and (g , g ) = lcm( g , g ). | 1 2 | | 1| | 2| 108 8. EXTERNAL DIRECT PRODUCTS For the general case, suppose the result holds for G1 G2 Gn 1. But ⊕ ⊕ · · · ⊕ − G1 G2 Gn = G1 G2 Gn 1 Gn, ⊕ ⊕ · · · ⊕ ⊕ ⊕ · · · ⊕ − ⊕ so applying the previous argument, the result holds for G G G 1⊕ 2 ⊕ · · · ⊕ n by induction. ⇤ Example. Determine the number of elements of order 7 in Z49 Z7. ⊕ Solution. For each such element, 7 = (a, b) = lcm( a , b ) by Theorem 8.1. There are three mutually exclusive cases:| | | | | | (1) a = 7 and b = 7. There are 6 choices for a (7, 14, 21, 28, 35, 42) and 6 for b| (1| , 2, 3, 4, 5,|6)| for a total of 36. (2) a = 7 and b = 1. We have 6 choices for a and 1 for b, a total of 6. | | | | (3) a = 1 and b = 7. Another 6 choices. | | | | Thus Z49 Z7 has 48 elements of order 7. ⊕ ⇤ Example. Find the number of cyclic subgroups of order 14 in Z196 Z49. ⊕ Solution. We count the elements (a, b) of order 14: (1) a = 14 and b = 7 or 1. Z196 has a unique cyclic subroup of order 14 and this|has| φ(14) = |6|generators (Theorem 4.4), so we have 6 choices for a and 7 for b, a total of 42 for (a, b). (2) a = 2 and b = 7. By Theorem 4.4, since 2 196, there is φ(2) = 1 subgroup| | of order|2,| and thus element of order 2. There| are 6 choices for b, so 6 overall. Thus we have 48 elements of order 14. Because each cyclic group has 6 elements of order 14, and no two of the cyclic groups can have an element of order 14 in common, there are 48/6 = 8 cyclic subgroups of order 14. ⇤ 8. EXTERNAL DIRECT PRODUCTS 109 Example. Since 8 = 6 in Z48 and 4 = 4 in Z16, 8 4 is a subgroup | | | | h i ⊕ h i of order 24 in Z48 Z16. ⊕ Theorem (8.2 — Criterion for G H to be Cyclic). Let G and H be finite cyclic groups. Then G H is cyclic⊕ G and H are relatively prime. ⊕ () | | | | Proof. Let G = m and H = n, so G H = mn. | | | | | ⊕ | (= ) Assume G H is cyclic. Suppose gcd(m, n) = d and (g, h) is a generator of )G H. Now ⊕ ⊕ mn m n n m mn (g, h) d = (g ) d , (h ) d = (e, e) = mn = (g, h) = d = 1. ) | | d ) Thus G and H are relatively prime. | | | | ( =) Suppose G = g and H = h , and gcd(m, n) = 1. Then ( h i h i (g, h) = lcm(m, n) = mn = G H , | | | ⊕ | so (g, h) is a generator of G H = G H is cyclic. ⊕ ) ⊕ ⇤ Corollary (1 — Criterion for G G G to be Cyclic). An 1 ⊕ 2 ⊕ · · · ⊕ n external direct product G1 G2 Gn of a finite number of finite cyclic groups is cyclic ⊕G and⊕ · ·G· ⊕are relatively prime when i = j. () | i| | j| 6 Proof. Theorem 8.2 and induction. ⇤ Corollary (Criterion for Zn n n Zn Zn Zn ). 1 2··· k ⇡ 1 ⊕ 2 ⊕ · · · ⊕ k Let m = n1n2 nk. Then Zm Zn1 Zn2 Znk ni and nj are relatively prime· · ·when i = j. ⇡ ⊕ ⊕ · · · ⊕ () 6 110 8. EXTERNAL DIRECT PRODUCTS Example. Z2 Z2 Z5 Z7 Z2 Z10 Z7 Z2 Z70, ⊕ ⊕ ⊕ ⇡ ⊕ ⊕ ⇡ ⊕ and Z2 Z2 Z5 Z7 Z2 Z10 Z7 Z2 Z5 Z2 Z7 Z10 Z14. ⊕ ⊕ ⊕ ⇡ ⊕ ⊕ ⇡ ⊕ ⊕ ⊕ ⇡ ⊕ Thus Z2 Z70 Z10 Z14, but Z2 Z70 Z140 ⊕ ⇡ ⊕ ⊕ 6⇡ U(n) as an External Direct Product If k n, let | Uk(n) = x U(n) x mod k = 1 . Example. { 2 | } U(21) = 1, 2, 4, 5, 8, 10, 11, 13, 16, 17, 19, 20 , so { } U (21) = 1, 4, 10, 13, 16, 19 and 3 { } U (21) = 1, 8 . 7 { } Lemma. If k n, U (n) U(n). | k Proof. For a, b U (n), a mod k = 1 and b mod k = 1 = 2 k ) (ab) mod k = (a mod k)(b mod k) = 1 1 = 1, · so Uk(n) is closed under multiplication mod k. Since 1 U (n), U (n) = , U (n) U(n) by the Finite Subgroup test. 2 k k 6 ; k ⇤ 8. EXTERNAL DIRECT PRODUCTS 111 The next theorem requires two preliminary results: Lemma (1 — Page 24 #17). Let a, b, s, t Z. If a mod st = b mod st, a mod s = b mod s and a mod t = b mod2t. The converse is true if gcd(s, t) = 1. Proof. (= ) Since st a b, s a b and t a b = ) | − | − | − ) a mod s = b mod s and a mod t = b mod t. ( =) Assume gcd(s, t) = 1, s a b and t a b. Then lcm(s, t) a b. But ( | − | − | − st = lcm(s, t) gcd(st) = lcm(st) 1 = lcm(st), · · so st a b = a mod st = b mod st. | − ) ⇤ Lemma (2 — Page 24 # 19). gcd(a, bc) = 1 gcd(a, b) = 1 and gcd(a, c) = 1. () Proof. By the Fundamental Theorem of Arithmetic, a = p p p , b = q q q , c = r r r 1 2 · · · k 1 2 · · · l 1 2 · · · m where the pi, qi, ri are primes. gcd(a, bc) = 1 p = q and p = r for any i, j, h. () i 6 j i 6 h gcd(a, b) = 1 p = q for any i, j. () i 6 j gcd(a, c) = 1 p = r for any i, h. () i 6 h The result clearly follows. ⇤ 112 8. EXTERNAL DIRECT PRODUCTS Theorem (8.3 — U(n) as an Exterior Direct Product). Suppose s and t are relatively prime. Then U(st) U(s) U(t). Also, Us(st) U(t) and U (st) U(s). ⇡ ⊕ ⇡ t ⇡ Proof. We are given gcd(s, t) = 1. If x U(st), gcd(x, st) = 1 = (Lemma 2) gcd(x, s) = 1 and gcd(x, t) = 1, so define2 ) φ : U(st) U(s) U(t) by φ(x) = (x mod s, x mod t). ! ⊕ [Since elements in U(st), U(s), and U(t) have multiple representations, we need to show φ is well-defined.] Suppose x = y mod st. By Lemma 1, x = y mod s and x = y mod t, so φ(x) = (x mod s, x mod t) = (y mod s, y mod t) = φ(y), so φ is well-defined. [To show φ is 1–1.] Suppose x, y U(st) and φ(x) = φ(y). Then 2 (x mod s, x mod t) = (y mod s, y mod t) = ) x = y mod s and x = y mod t = ) (by Lemma 1 since gcd(s, t) = 1) x = y mod st. Thus φ is 1–1. [To show φ is onto.] Let (a, b) U(s) U(t).