By Julia Collins

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Then we define a partial ordering on these sets by (E1 , g1 ) ≤ (E2 , g2 ) if E1 ⊂ E2 and g2 extends g1 . By Zorn’s Lemma there is a maximal element (E0 , g0 ). Suppose for contradiction that E0 = B, and let b ∈ B\E0 . Now consider the Z-module E := {e + nb| e ∈ E0 , n ∈ Z} If nb ∈ / E0 for all nonzero n, then we can extend g0 to g : E → D by defining g (e + nb) = g0 (e). Otherwise we can find a minimal m ∈ N such that mb ∈ E0 , and we put d = g0 (mb). Then we say g (e + nb) = g0 (e) + nx, where x ∈ D is such that mx = d.

E. there is a subset S of M such that every element of M can be written in a unique way 29 as a Z-linear combination of elements of S. The structure theorem for finitely generated abelian groups says that every such group is isomorphic to Zn ⊕ Zp1 ⊕ Zp2 ⊕ · · · ⊕ Zpn , where n ∈ N and the pi are powers of primes. It then follows that every finitely generated torsion-free abelian group is isomorphic to Zn , and is thus free. Note that we are relying on finite generation here, as Q (an abelian group under addition) is torsionfree but not free.

1. Any homomorphism γ : Z → B is completely characterised by what happens to γ(1), as γ(n) = nγ(1). Thus the map γ → γ(1) is an isomorphism between Hom(Z, B) and B. 2. Let B, C be Z-modules, γ : Z → C a homomorphism and ε : B → C a surjective homomorphism. To prove the assertion, we must find a map β : Z → B such that εβ = γ. From (1) we know that γ and β are completely characterised by where they send the element ‘1’. By surjectivity of ε we can find b ∈ B such that ε(b) = γ(1). Let β(1) = b. Then for any n ∈ Z we have ε(β(n)) = ε(nβ(1)) = ε(nb) = nε(b) = nγ(1) = γ(n).