# $\mathbb{Z}_m \oplus \mathbb{Z}_n$ isomorphic to $\mathbb{Z}_{mn}$ as $\mathbb{Z}_{mn}$ modules

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 $\mathbb{Z}_{mn}$.

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I would mention the relatively prime bit at the beginning! – Dylan Moreland Sep 21 '11 at 15:18
@ Dylan Thanks, good thought – user9352 Sep 21 '11 at 15:21

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. $\Box$

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First, as abelian groups $\mathbb{Z}_m \times \mathbb{Z}_n$ and $\mathbb{Z}_{mn}$ are only isomorph if $(m,n) = 1$. Otherwise is $$\mathbb{Z}_m \times \mathbb{Z}_n \cong \mathbb{Z}_{[m,n]}.$$

That said, what's the structure of $\mathbb{Z}_{mn}$-module on $\mathbb{Z}_m \times \mathbb{Z}_n$? Probably you refer to the following $$[h] ([x], [y]) = ([hx], [hy]) = h ([x], [y]).$$ Where the multiplication on the left is that of the $\mathbb{Z}_{mn}$-module and that on the right is the multiplication of th $\mathbb{Z}$-module.

Now, given an isomorphism of $\mathbb{Z}$-modules $\Phi: \mathbb{Z}_m \times \mathbb{Z}_n \to \mathbb{Z}_{[m,n]}$ we have $$\Phi([h] ([x], [y])) = \Phi(h ([x], [y])) = h \Phi([x], [y]) = [h] \Phi([x], [y])$$

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