Let \(m\ne0\). Prove that \[ \operatorname{Hom}_{\ZZ}(\ZZ/m\ZZ,\ZZ)=0. \]
Solution
Let \[ f:\ZZ/m\ZZ\to\ZZ \] be a \(\ZZ\)-module homomorphism.
1. The map \(f\) is determined by \(f([1]_m)\).
The cyclic group \(\ZZ/m\ZZ\) is generated by \([1]_m\), and for every integer \(k\), \[ f([k]_m)=k f([1]_m). \]
2. One has \(f([1]_m)=0\).
Since \([m]_m=[0]_m\), \[ 0=f([0]_m)=f([m]_m)=m f([1]_m). \] Because \(m\ne0\) and \(\ZZ\) is torsion-free, this forces \(f([1]_m)=0\).
3. Hence \(f=0\).
Apply 1 and 2.
Therefore the Hom group is zero.