Let \(A\) be an abelian group and let \[ A[m]=\{a\in A:ma=0\}. \] Prove that \[ \Hom_{\ZZ}(\ZZ/m\ZZ,A)\cong A[m]. \]
Solution
Define evaluation at the standard generator by \[ \Phi:\Hom_{\ZZ}(\ZZ/m\ZZ,A)\longrightarrow A, \qquad \Phi(f)=f([1]). \]
1. The image of \(\Phi\) is contained in \(A[m]\).
For any homomorphism \(f\), \[ m\Phi(f)=mf([1])=f(m[1])=f([0])=0. \] Thus \(\Phi(f)\in A[m]\).
2. Every element of \(A[m]\) occurs as \(f([1])\) for a unique homomorphism \(f\).
Given \(a\in A[m]\), define \[ f_a([n])=na. \] If \(n\equiv n'\pmod m\), then \(n-n'=qm\) for some integer \(q\), so \[ na-n'a=qma=0. \] Hence \(f_a\) is well defined. It is a group homomorphism and satisfies \(f_a([1])=a\).
Uniqueness follows because \(\ZZ/m\ZZ\) is generated by \([1]\): any homomorphism is determined by the image of this generator.
3. Hence \(\Phi\) is an isomorphism onto \(A[m]\).
By 2, \(\Phi\) is both surjective onto \(A[m]\) and injective. It is plainly a homomorphism of abelian groups, so \[ \Hom_{\ZZ}(\ZZ/m\ZZ,A)\cong A[m]. \]