Let \[ \ZZ_n[m]=\{x\in \ZZ/n\ZZ : mx=0\} \] be the subgroup of \(m\)-torsion elements of \(\ZZ/n\ZZ\). Prove that \[ \ZZ_n[m]\cong \ZZ/\gcd(m,n)\ZZ. \]
Solution
Set \(d=\gcd(m,n)\).
1. The subgroup \(\ZZ_n[m]\) is generated by the class of \(n/d\).
First, \[ m\frac nd=\frac md\,n, \] so \(m[n/d]_n=0\) in \(\ZZ/n\ZZ\). Thus \[ \langle[n/d]_n\rangle\subseteq \ZZ_n[m]. \]
Conversely, suppose \([x]_n\in\ZZ_n[m]\). Then \(n\mid mx\). Dividing by \(d\) gives \[ \frac nd\mid \frac md x. \] Since \[ \gcd\!\left(\frac nd,\frac md\right)=1, \] Euclid’s lemma yields \(n/d\mid x\). Hence \([x]_n\) lies in the subgroup generated by \([n/d]_n\).
2. The element \([n/d]_n\) has order \(d\).
Its order is the least positive integer \(r\) such that \[ n\mid r\frac nd. \] Cancelling the positive integer \(n/d\) gives \(d\mid r\). Hence the least such \(r\) is \(d\).
3. Therefore\[ \ZZ_n[m]\cong \ZZ/d\ZZ=\ZZ/\gcd(m,n)\ZZ. \]
By 1, the subgroup is cyclic generated by \([n/d]_n\), and by 2 that generator has order \(d\). A cyclic group of order \(d\) is isomorphic to \(\ZZ/d\ZZ\).