If $\mathbb{Z}_m \times \mathbb{Z}_n \cong \mathbb{Z}_{m'} \times \mathbb{Z}_{n'}$ with $m|n$ and $m'|n'$, then does $m=m'$ and $n=n'$?

Question

Suppose $\mathbb{Z}_m \times \mathbb{Z}_n $ is isomorphic to $\mathbb{Z}_{m'} \times \mathbb{Z}_{n'}$ as groups, where $m$ divides $n$ and $m'$ divides $n'$. Does that mean $m=m'$ and $n=n'$?

Answer 1

Yes (assuming $m,n,m',n'$ are positive integers). The group isomorphism produces an isomorphism of $\mathbb Z$-modules $$ \mathbb Z_m\oplus\mathbb Z_n\cong\mathbb Z_{m'}\oplus\mathbb Z_{n'}. $$ Thus for any positive integer $c$ we have $$ \mathbb Z_c\otimes_{\mathbb Z}(\mathbb Z_m\oplus\mathbb Z_n)\cong \mathbb Z_c\otimes_{\mathbb Z}(\mathbb Z_{m'}\oplus\mathbb Z_{n'}). $$ But $$ \mathbb Z_c\otimes_{\mathbb Z}\mathbb Z_m\cong\mathbb Z_{\gcd(c,m)}, $$ and similarly for the other components. Considering the cardinality of both sides in the above isomorphism, $$ \gcd(c,m)\gcd(c,n)=\gcd(c,m')\gcd(c,n'). $$ Recall that $c$ was arbitrary. Setting $c=m$ gives $m|m'$ and $c=m'$ gives $m'|m$, so $m=m'$. Now setting $c=n$ gives $n|n'$ and $c=n'$ gives $n'|n$, so $n=n'$.

For a more general result see the structure theorem for finitely generated modules