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Subject : MATHEMATICS Subject : MATHEMATICS Paper 1 : ABSTRACT ALGEBRA Chapter 1 : Direct Product of Groups Module 2 : Internal direct product of groups Anjan Kumar Bhuniya Department of Mathematics Visva-Bharati; Santiniketan West Bengal 1 Internal direct product of groups Learning outcomes: 1. Internal direct product of groups. 2. Necessary and sufficient condition for a group to be an internal direct product. 3. Isomorphism between external and internal direct products. In the previous module we introduced and characterized external direct products of groups, that provides us a formulation to think a family of distinct groups as subgroups of a larger group. To be specific, consider two groups G1 and G2 having the identity elements e1 and e2 respectively. Then N1 = G1 × fe2g ' G1 and N2 = fe1g × G2 ' G2 are two normal subgroups of G1 × G2. In this module we consider the reverse problem, that is, given a group G whether there is a family of subgroups H1;H2; ··· ;Hk of G such that G ' H1 × H2 × · · · × Hk. As we can expect, it is not possible for every group in general. Even if it is possible for a group G, then also the subgroups of which external direct product is isomorphic to G need to satisfy some conditions. Following result gives us an idea on the conditions that the subgroups need to satisfy. Henceforth we use simply multiplicative notation instead of ∗ to mean the group operation of the direct product. Theorem 0.1. Let G1;G2; ··· ;Gn be a family of groups. Denote G = G1 × G2 × · · · × Gn and Hi = f(e1; ··· ; ei−1; ai; ei+1; ··· ; en) j ai 2 Gig for every i = 1; 2; ··· ; n. Then 1. Hi is a normal subgroup of G and Hi ' Gi for every i = 1; 2; ··· ; n; 2. every element of G can be expressed uniquely as h1h2 ··· hn where hi 2 Hi for every i = 1; 2; ··· ; n. Proof. 1. First note that (e1; e2; ··· ; en) 2 Hi which ensures that Hi 6= ;. 2 Now consider a = (e1; ··· ; ei−1; ai; ei+1; ··· ; en); b = (e1; ··· ; ei−1; bi; ei+1; ··· ; en) 2 Hi. Then −1 −1 ab = (e1; ··· ; ei−1; ai; ei+1; ··· ; en)(e1; ··· ; ei−1; bi; ei+1; ··· ; en) −1 = (e1; ··· ; ei−1; ai; ei+1; ··· ; en)(e1; ··· ; ei−1; bi ; ei+1; ··· ; en) −1 = (e1; ··· ; ei−1; aibi ; ei+1; ··· ; en) 2 Hi; −1 since aibi 2 Gi. Thus Hi is a subgroup of G. Similarly for every g = (g1; g2; ··· ; gn) 2 G, −1 −1 gag = (e1; ··· ; ei−1; giaigi ; ei+1; ··· ; en) 2 Hi implies that Hi is a normal subgroup of G. 2. Let a = (a1; a2; ··· ; an) 2 G. Then ai 2 Gi and so hi = (e1; ··· ; ei−1; ai; ei+1; ··· ; en) 2 Hi for every i = 1; 2; ··· ; n are such that a = h1h2 ··· hn. To prove uniqueness of such representation, suppose that a = k1k2 ··· kn where ki 2 Hi. Now ki 2 Hi implies that ki = (e1; ··· ; ei−1; bi; ei+1; ··· ; en) for some bi 2 Gi. Then (a1; a2; ··· ; an) = a = k1k2 ··· kn = (b1; b2; ··· ; bn) implies that ai = bi and hence hi = ki for every i = 1; 2; ··· ; n. Since every element of G is of the form h1h2 ··· hn where hi 2 Hi for every i = 1; 2; ··· ; n, it follows that G = H1H2 ··· Hn. Also Gi ' Hi for every i = 1; 2; ··· ; n implies that G ' H1 × H2 × · · · × Hn. Definition 0.2. Let G be a group and N1;N2; ··· ;Nk be normal subgroups of a group G. Then G is called an internal direct product of N1;N2; ··· ;Nk if every element of G can be expressed uniquely as a = a1a2 ··· ak where ai 2 Ni for all i = 1; 2; ··· ; k. The adjective `internal' emphasizes the facts that all the components Ni are required to be subgroup of a single group G and the product a1a2 ··· ak of the elements of N1;N2; ··· ;Nk is actually a product within the group G. 3 Example 0.3. Consider the Klein's 4-group K4 = fe; a; b; cg. Then N1 = fe; ag and N2 = fe; bg are two normal subgroups of K4. Now N1N2 = fe; agfe; bg = fe; a; b; cg = K4 implies that every element of K4 is expressed as a product of an element of N1 and an element of N2. To check the uniqueness, −1 −1 assume x1; x2 2 N1 and y1; y2 2 N2 such that x1y1 = x2y2. Then x2 x1 = y2y1 2 N1 \ N2 = feg −1 −1 which implies that x2 x1 = e = y2y1 , that is, x1 = x2 and y1 = y2. Hence G is an internal direct product of the subgroups N1 and N2. Example 0.4. Every nonzero complex number z has unique representation z = reiθ where r = jzj and θ = amp z. Then r 2 R+, the group of all positive real numbers and eiθ 2 S1 = fu 2 C∗ j juj = 1g. Hence C∗ is the internal direct product of R+ and S1. Example 0.5. The group Z of all integers can not be represented as an internal direct product (sum) of two subgroups. Note that for any two distinct nonzero integers m and n, mn 2 mZ \ nZ implies that mZ \ nZ 6= f0g. Following is a necessary and sufficient condition for a group to be an internal direct product of a given family of normal subgroups. To prove this result, we need the following lemma: Lemma 0.6. Let G be a group and N1;N2; ··· ;Nk normal subgroups of G. If Ni\(N1N2 ··· Ni−1Ni+1 ··· Nk) = feg for every i = 1; 2; ··· ; k, then aiaj = ajai for every ai 2 Ni; aj 2 Nj, i 6= j. −1 −1 −1 Proof. Consider ai 2 Ni and aj 2 Nj. Then aiajai aj 2 (aiNjai )Nj ⊆ Nj, since Nj is −1 −1 −1 −1 a normal subgroup of G. Similarly aiajai aj 2 Ni. Hence aiajai aj 2 Ni \ Nj ⊆ Ni \ (N1N2 ··· Ni−1Ni+1 ··· Nk) = feg, and so aiaj = ajai. Theorem 0.7. Let G be a group and N1;N2; ··· ;Nk normal subgroups of G. Then G is an internal direct product of N1;N2; ··· ;Nk if and only if (i) G = N1N2 ··· Nk; and (ii) Ni \ (N1N2 ··· Ni−1Ni+1 ··· Nk) = feg for every i = 1; 2; ··· ; k. Proof. First assume that G is an internal direct product of the normal subgroups N1;N2; ··· ;Nk. Then for every a 2 G there are unique ai 2 Ni such that a = a1a2 ··· ak, and so a 2 N1N2 ··· Nk. Hence G ⊆ N1N2 ··· Nk. Also N1N2 ··· Nk ⊆ G, since every Ni is a subgroup of G. Thus 4 G = N1N2 ··· Nk. Now consider a 2 Ni \ (N1N2 ··· Ni−1Ni+1 ··· Nk). Then a 2 Ni and a = a1 ··· ai−1ai+1 ··· ak for some aj 2 Nj, j = 1; ··· ; i − 1; i + 1; ··· ; k implies that e ··· eae ··· e = a = a1 ··· ai−1eai+1 ··· ak: Since every such representation is unique, it follows that a = e. Thus Ni\(N1N2 ··· Ni−1Ni+1 ··· Nk) = feg for every i = 1; 2; ··· ; k. Conversely, assume that the family of normal subgroups N1;N2; ··· ;Nk satisfies the given conditions. Then G = N1N2 ··· Nk implies that every a 2 G can be represented as a = a1a2 ··· ak where ai 2 Ni. To prove uniqueness of such representations, suppose that a also has another representation a = b1b2 ··· bk where bi 2 Ni. Then a1a2 ··· ak = b1b2 ··· bk implies, by Lemma 0.6, that −1 −1 −1 −1 −1 aibi = (b1a1 ) ··· (bi−1ai−1)(bi+1ai+1) ··· (bkak ) 2 Ni \ (N1N2 ··· Ni−1Ni+1 ··· Nk) = feg: Thus ai = bi for every i = 1; 2; ··· ; k, forcing uniqueness of the representation of every a 2 G in the desired form. The following is a special case of the above result, but is used frequently. Corollary 0.8. Let G be a group, and H and K two normal subgroups of G. Then G is an internal direct product of H and K if and only if G = HK and H \ K = feg. The discussion preceding the definition of internal direct product can be stated as: If G = G1 × G2 then G is an internal direct product of two normal subgroups N1 ' G1 and N2 ' G2. Conversely every internal direct product can also be realized as an external direct product up to isomorphism. Theorem 0.9. If G is the internal direct product of the normal subgroups N1;N2; ··· ;Nk, then G ' N1 × N2 × · · · × Nk: Proof. Define f : G −! N1 × N2 × · · · × Nk by: f(a1a2 ··· ak) = (a1; a2; ··· ; ak) 5 for every (a1; a2; ··· ; ak) 2 G. Since every element of G has a unique representation as a1a2 ··· ak for ai 2 Ni, so f is well defined. Also it follows directly that f is both one-to-one and onto. Now, for every a = a1a2 ··· ak; b = b1b2 ··· bk 2 G, f(ab) = f(a1a2 ··· akb1b2 ··· bk) = f(a1b1a2b2 ··· akbk) (by Lemma 0.6) = (a1b1; a2b2; ··· ; akbk) = (a1; a2; ··· ; ak)(b1; b2; ··· ; bk) = f(a)f(b) shows that f is an isomorphism. I wonder whether any reader, after going through this note, thinks that the condition of normality of the subgroups in the definition of internal direct product is superfluous. Check, if it is possible to prove the above result without the condition of normality of the subgroups! Since we wanted to get a representation of an external direct product as a product of subgroups, so we had to define the new product of subgroups in such a way that the above isomorphism follows.
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