Subgroup of $\Bbb {Z}_m \oplus \Bbb {Z}_n$ where $(m,n)=1$.

Question

Let $m,n>1$, $(m,n)=1$. Prove that every subgroup $H$ of $\Bbb {Z}_m \oplus \Bbb {Z}_n$ is $H=A\oplus B$ where $A=H\cap \Bbb {Z}_n$ and $B=H\cap \Bbb {Z}_m$.

First attempt: $G=\Bbb {Z}_m \oplus \Bbb {Z}_n \cong \Bbb {Z}_{mn}$. Therefore $G$ is cyclic and abelian and finite. $H$ is cyclic and therefore $H=\left<(a,b)\right>$ where $a\in \Bbb {Z}_m$ and $b\in \Bbb {Z}_n$. For $A=\left<a\right>$ and $B=\left<b\right>$ is it admissible?

I can see exactly why this is true (what I have to prove)... How, though, do I show it? I would appreciate your help.

Answer 1

Obviously $$ ( H \cap \mathbb{Z}_n) \oplus ( H \cap \mathbb{Z}_m) \leq H$$ Now suppose $r, s \in \mathbb{Z}$ such that $$sn + rm = 1$$ Suppose $h \in H$, $$h = a + b$$ with $$a \in \mathbb{Z}_n\ \ b \in \mathbb{Z}_m $$Then $$ rmh =rm(a+b)=rma= (rm +sn)a = a$$ This means that $a \in H \cap \mathbb{Z}_n$ because obviously $rmh \in H$. Analogously $b \in H \cap \mathbb{Z}_m$, and so $$H \leq ( H \cap \mathbb{Z}_n) \oplus ( H \cap \mathbb{Z}_m)$$

Answer 2

$$\mathbb{Z}_m\oplus \mathbb{Z}_n=\mathbb{Z}_{mn}.$$ And from here it is easy to show that all the subgroups are $$\mathbb{Z}_k,$$ where $k$ is divisor of $mn$.