Math 371 Lecture #36 x8.1 (Ed.2), 9.1 (Ed.3): Direct Products We take another step in the classification of groups by way of products of normal sub- groups. Example. The abelian group U15 = f1; 2; 4; 7; 8; 11; 13; 14g has two (normal) subgroups M = f1; 11g and N = f1; 2; 4; 8g. The \product" MN consists of all products mn where m 2 M and n 2 N. We have M ⊆ MN because m = m1 for all m 2 M, and N ⊆ MN because n = 1n for all n 2 N. Do 7; 13; 14 2 MN? Yes, because 11 · 2 = 7, 11 · 8 = 13, and 11 · 4 = 14. Thus U15 = MN, the \product" of two normal subgroups of U15. Notice that every element of U15 can be written uniquely in the form mn for m 2 M and n 2 N. Are the normal subgroups M and N of U15 isomorphic to quotient groups of U15? Since M \ N = f1g and both M and N are normal, we have by the Second Isomorphism Theorem that ∼ ∼ ∼ ∼ M = M=(M \ N) = NM=N = U15=N; N = N=(N \ M) = MN=M = U15=M: ∼ This says that U15=N × U15=M = M × N. ∼ Could it happen that U15 = M × N, the Cartesian product of two normal subgroups? The table for M × N is · (1; 1) (1; 2) (1; 4) (1; 8) (11; 1) (11; 2) (11; 4) (11; 8) (1; 1) (1; 1) (1; 2) (1; 4) (1; 8) (11; 1) (11; 2) (11; 4) (11; 8) (1; 2) (1; 2) (1; 4) (1; 8) (1; 1) (11; 2) (11; 4) (11; 8) (11; 1) (1; 4) (1; 4) (1; 8) (1; 1) (1; 2) (11; 4) (11; 8) (11; 1) (11; 2) (1; 8) (1; 8) (1; 1) (1; 2) (1; 4) (11; 8) (11; 1) (11; 2) (11; 4) (11; 1) (11; 1) (11; 2) (11; 4) (11; 8) (1; 1) (1; 2) (1; 4) (1; 8) (11; 2) (11; 2) (11; 4) (11; 8) (11; 1) (1; 2) (1; 4) (1; 8) (1; 1) (11; 4) (11; 4) (11; 8) (11; 1) (11; 2) (1; 4) (1; 8) (1; 1) (1; 2) (11; 8) (11; 8) (11; 1) (11; 2) (11; 4) (1; 8) (1; 1) (1; 2) (1; 4) Could the function f : M × N ! MN given by f(m; n) = mn be an isomorphism? We will answer this question in a moment in a more general setting which we now describe. The Cartesian product of groups G1;G2;:::;Gn, denoted by G1 × G2 × · · · × Gn, consists of n-tuples (a1; a2; : : : ; an) for ai 2 Gi, equipped with the component operation (a1; a2; : : : ; an)(b1; b2; : : : ; bn) = (a1b1; a2b2; : : : ; anbn): Technically, each Gi is NOT a subgroup of the direct product G1 × G2 × · · · × Gn (but each Gi is isomorphic to the subgroup of G1 × G2 × · · · × Gn). Observe that if all of G1;G2;:::;Gn are finite groups, then the order of G1 ×G2 ×· · ·×Gn is the the product of the orders of G1;G2;:::;Gn. Should G1;G2;:::;Gn happen to be normal subgroups of a group G, we can consider the normal subgroup G1G2 ··· Gn that consists of products of the form a1a2 ··· an in G for ai 2 Gi, i = 1; 2; : : : ; n. Lemma 9.2. Let M and N be normal subgroups of a group G. If M \ N = feg, then ab = ba for all a 2 M and all n 2 N. Proof. Let a 2 M and b 2 N. The goal is to show that ab = ba. Since M is normal, we have that b−1ab 2 M. Since a−1 2 M we have that a−1b−1ab 2 M. Similarly, by the normality of N we have that a−1b−1a 2 N, so that a−1b−1ab 2 N. −1 −1 Since M \ N = feg, we obtain a b ab = e, and hence ab = ba. Theorem 9.1. Let N1;N2;:::;Nk be normal subgroups of a group G such that every element in G can be written uniquely in the form a1a2 ··· ak for ai 2 Ni, i = 1; 2; : : : ; k. Then G is isomorphic to the direct product N1 × N2 × · · · × Nk. Proof. We define the map f : N1 × N2 × · · · × Nk ! G by f(a1; a2; : : : ; ak) = a1a2 ··· ak: The map f is surjective because by hypothesis, each element of G can be written in the form a1a2 ··· ak for some ai 2 Ni, i = 1; 2; : : : ; k. The map f is injective because when f(a1; a2; : : : ; ak) = f(b1; b2; : : : ; bk), we obtain that a1a2 ··· ak = b1b2 ··· bk, so that by the uniqueness of writing each element of G as a product, we have a1 = b1, a2 = b2,..., ak = bk. Before we show that f is a homomorphism, we show for i 6= j that Ni \ Nj = feg. To do this we take a 2 Ni \ Nj, which a can be written as product of elements of N1;N2;:::;Nk in two different ways: a = e ··· e a e ··· e e e ··· e """" N1 Ni Nj Nk a = e ··· e e e ··· e a e ··· e """" N1 Ni Nj Nk Uniqueness of writing a as a product of the elements of Ni implies that a = e, so that Ni \ Nj = feg. By Lemma 9.2, we then have that aibj = bjai for all ai 2 Ni and bj 2 Nj when i 6= j. Repeated using this commutativity we show that f is a homomorphism: f((a1; a2; : : : ; ak)(b1; b2; : : : ; bk)) = f(a1b1; a2b2; : : : ; akbk) = a1b1a2b2 ··· akbk = a1a2b1b2 ··· akbk = a1a2 ··· akb1b2 ··· bk = f(a1; a2; : : : ; ak)f(b1; b2; : : : ; bk): Therefore f : N1 × N2 × · · · × Nk ! G is an isomorphism. Definitions. Suppose that normal subgroups N1;N2;:::;Nk of a group G have the property that every element of G can be written uniquely in the form a1a2 ··· ak for ai 2 Ni, i = 1; 2; : : : ; k. Then G = N1N2 ··· Nk and we say that G is an internal direct product of N1;N2;:::;Nk. By Theorem 9.1, the internal direct product N1N2 ··· Nk is isomorphic to N1 ×N2 ×· · ·× Nk, and we say that G is the external direct product of N1;N2;:::;Nk. Each normal subgroup Ni in the internal or external direct product is called a direct factor of G. Theorem 9.3. If M and N are normal subgroups of a group G such that G = MN and M \ N = feg, then G ∼= M × N where M ∼= G=N and N ∼= G=M. Proof. Suppose every element of G is of the form mn for m 2 M and n 2 N. We want to show that this form mn is unique, so suppose mn = m1n1 for some m1 2 M and n1 2 N. −1 −1 −1 −1 −1 −1 Then m1 mn = m1 m1n1 = n1, and so n1n = m1 mnn = m1 m. −1 −1 Since M and N are subgroups, we have m1m 2 M and n1n 2 N, −1 −1 −1 −1 Since m1 m = n1n and M \ N = feg, then m1 m = e and n1n = e, and hence m1 = m and n1 = n. Thus every element of G can be written uniquely as mn for m 2 M and n 2 N. Since M and N are normal subgroups of G, Theorem 9.1 implies that G ∼= M × N. Because M\N = feg, the Second Isomorphism Theorem implies that M ∼= M=(M\N) ∼= ∼ ∼ MN=N = G=N and N = N=(M \ N) = MN=M = G=M. In practice we typically identity the exterior direct product and the interior direct product because they are isomorphic. Example (Continued). The group U15 = f1; 2; 4; 7; 8; 11; 13; 14g has two normal subgroups M = f0; 11g and N = f1; 2; 4; 8g. Because G = MN and M \ N = feg, we have by Theorem 9.3 that G ∼= M × N. ∼ ∼ ∼ Since M = Z2 and N = Z4, we have that U15 = Z2 × Z4. The group Z2 × Z4 is one of five isomorphism classes of groups of order 8..