TITLE:  $\mathbb{Z}_m \oplus \mathbb{Z}_n$ isomorphic to $\mathbb{Z}_{mn}$ as $\mathbb{Z}_{mn}$ modules
QUESTION [5 upvotes]: I would like to know if, when $\gcd(m,n)=1$, it is the case that $\mathbb{Z}_m \oplus \mathbb{Z}_n$ is isomorphic to $\mathbb{Z}_{mn}$  as $\mathbb{Z}_{mn}$ modules. I know they are isomorphic as groups but I couldn't show they were isomorphic as modules over the ring $\mathbb{Z}_{mn}$.

REPLY [9 votes]: A $\mathbb{Z}_k$ module is just an abelian group of exponent $k$ (that is, an abelian group in which $kx=0$ for all $x$). In particular:
Theorem. Two abelian groups of exponent $k$ are isomorphic as $\mathbb{Z}_k$ modules if and only if they are isomorphic as groups.
Proof. Let $G$ and $H$ be two abelian groups (written additively) of exponent $k$. If they are isomorphic as $\mathbb{Z}_k$ modules, then there is a bijective $\mathbb{Z}_k$-module isomorphism $\phi\colon G\to H$; this module isomorphism is in particular an abelian group isomorphism, so $G$ and $H$ are isomorphic as abelian groups.
Conversely, suppose that $G$ and $H$ are isomorphic as abelian groups, and let $\psi\colon G\to H$ be an abelian group isomorphism. I claim that it is a $\mathbb{Z}_k$-module isomorphism. To establish this, it suffices to show that $\psi(ax) = a\psi x$ for all $a\in\mathbb{Z}_k$. Since $G$ and $H$ are of exponent $k$, "$ax$" is well defined since if $a\equiv a'\pmod{k}$, then $a=a'+km$ for some $m$, and then $ax = (a'+km)x = a'x + m(kx) = a'x+0 = a'x$. We proceed by induction on $a$: for $a=0$, $\psi(0x) = \psi(0) = 0 = 0\psi(x)$. Assuming the result holds for $a$, we have $\psi((a+1)x) = \psi(ax+x) = \psi(ax)+\psi(x)$ (since $\psi$ is an abelian group homomorphism), and $\psi(ax)+\psi(x) = a\psi(x)+\psi(x) = (a+1)\psi(x)$. Thus, $\psi$ is a $\mathbb{Z}_k$-module homomorphism as well. $\Box$