Math 210 A : Algebra , Homework 6

Solution. Let α : Z/nZ → Z/mnZ be defined by α(1) = m. Since 〈1〉 = Z/nZ, this determines the entire homomorphism. Let x ∈ kerα. Then mx ≡ 0, which implies mn | mx so n | x. But since 0 ≤ x < n, we must have x = 0. Therefore the kernel of α is trivial, so α is an injection. Since m | mn, by the fundamental theorem on cyclic groups Z/mnZ has a subgroup of order m isomorphic to Z/mZ. Let β : Z/mnZ→ Z/mZ be defined by projection onto that subgroup, which is a surjective homomorphism. Further, let x ∈ Z/nZ. Them α(x) = mx and β(mx) = 0, so imα ⊂ ker β. Since | imα| = | ker β| = n, imα = ker β. Thus this is an exact sequence. Recall that a sequence is split if and only if the middle term is the direct sum of the outer terms. In our case, we need Z/mnZ ∼= Z/mZ×Z/nZ. By the Chinese Remainder Theorem, this is true if and only if (m,n) = 1.