Let $\text{gcd}(m,n)=1$. Prove that $\mathbb{Z}_{mn}\cong \mathbb{Z}_m \times \mathbb{Z}_n$.
Hint: This is a "strong" restatement of the Chinese Remainder Theorem.
I know that I can build a map from $\mathbb{Z}_{mn}$ directly and check that the map is well-defined, but I think it may be easier to instead build a map on $\mathbb{Z}$ and study its kernel. I need help though because I don't know how to do this. Any help would be great, thanks.
On
Is true that "$\mathbb{Z}_{m}\times \mathbb{Z}_{n}$ have order $mn$ iff $(m,n)=1$" so you can prove like this: Suppose $(m$,n)=1, let $ k=ord((1,1))$ then you have $(1,1)^k=(0,0)$ then $m\mid k$ and $n\mid k$, but if there is ${k}´\in\mathbb{Z}$ such that $m\mid k´$ and $n\mid k´$, $(1,1)^{k^´}=(0,0)$ then $k=\left [ m,n \right ]$ but it was $mn$ so $\mathbb{Z}_{m}\times \mathbb{Z}_{n}$
Consider the isomorphism $f: \mathbb{Z} \rightarrow \mathbb{Z}_m \times \mathbb{Z}_n, f(g)=(g\mod m, g\mod n)$. Then check that $ker(f)=(mn) \mathbb{Z}$. The desired result follows from the first ismorphism theorem.