Section 1.6. Quotient Groups 55 It is easy to check that the direct product H × K is a group [the identity is (1, 1) and (h, k)−1 = (h−1,k−1)]. We now apply the First Isomorphism Theorem to direct products. Proposition 1.85. Let G and G be groups, and let K G and K G be normal subgroups. Then (K × K ) (G × G ), and there is an isomorphism (G × G )/(K × K ) ∼ = (G/K) × (G /K ). Proof. Let π : G → G/K and π : G → G /K be the natural maps. It is easy to check that f : G × G → (G/K) × (G /K ), given by f : (g, g ) → (π(g),π (g )) = (gK, g K ), is a surjective homomorphism with ker f = K × K . The First Isomorphism Theo- rem now gives the desired isomorphism. • Proposition 1.86. If G is a group containing normal subgroups H and K with H ∩ K = {1} and HK = G, then G ∼ = H × K. Proof. We show first that if g ∈ G, then the factorization g = hk, where h ∈ H and k ∈ K, is unique. If hk = h k , then h −1 h = k k−1 ∈ H ∩ K = {1}. Therefore, h = h and k = k. We may now define a function ϕ: G → H × K by ϕ(g) = (h, k), where g = hk, h ∈ H, and k ∈ K. To see whether ϕ is a homomorphism, let g = h k , so that gg = hkh k . Hence, ϕ(gg ) = ϕ(hkh k ), which is not in the proper form for evaluation. If we knew that hk = kh for h ∈ H and k ∈ K, then we could continue: ϕ(hkh k ) = ϕ(hh kk ) = (hh , kk ) = (h, k)(h , k ) = ϕ(g)ϕ(g ). Let h ∈ H and k ∈ K. Since K is a normal subgroup, (hkh−1)k−1 ∈ K since H is a normal subgroup, h(kh−1k−1) ∈ H. But H ∩ K = {1}, so that hkh−1k−1 = 1 and hk = kh. Finally, we show that the homomorphism ϕ is an isomorphism. If (h, k) ∈ H × K, then the element g ∈ G, defined by g = hk, satisfies ϕ(g) = (h, k) hence ϕ is surjective. If ϕ(g) = (1, 1), then g = 1, so that ker ϕ = 1 and ϕ is injective. Therefore, ϕ is an isomorphism. • Remark. We must assume that both subgroups H and K are normal. For example, S3 has subgroups H = (1 2 3) and K = (1 2) . Now H S3, H ∩ K = {1}, and HK = S3, but S3 ∼ = H × K (because the direct product is abelian). Of course, K is not a normal subgroup of S3. Theorem 1.87. If m and n are relatively prime, then Imn ∼ = Im × In. Proof. If a ∈ Z, denote its congruence class in Im by [a]m. The reader can show that the function f : Z → Im × In, given by a → ([a]m, [a]n), is a homomorphism. We claim that ker f = mn . Clearly, mn ⊆ ker f. For the reverse inclusion, if a ∈ ker f, then [a]m = [0]m and [a]n = [0]n that is, a ≡ 0 mod m and a ≡ 0 mod n that is, m | a and n | a. Since m and n are relatively prime, mn | a (FCAA, Exercise 1.60), and so a ∈ mn , that is, ker f ⊆ mn and ker f = mn . The First

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