Understanding the group product of $\mathbb{Z_n}$ and $\mathbb{Z_m}$, $G=\mathbb{Z_m} \mathbb{Z_n}$ $G=\mathbb{Z_m} \mathbb{Z_n} = \{[a][b] : [a] \in \mathbb{Z_m}, [b] \in \mathbb{Z_n}\}$
Specifically when $gcd(m,n)=1$. Can somebody show me what $G$ will equal as a set, and if you could go to far as to open up the equivalence classes of $[a],[b]$ and show me what exactly is going on, I'd appreciate it. I know that when $gcd(m,n)=1$ this product will be isomorphic to $\mathbb{Z_{mn}}$ and am just looking for a bit of details to clarify. Thanks!
 A: Algebraically it would most likely be easier to start with a group $H = \mathbb{Z}_n \times \mathbb{Z}_m$ with $\gcd(m,n) = 1$. Then it's very straightforward to see that $\mathbb{Z}_n \mathbb{Z}_m \cong H$ (and many proofs of the Chinese Remainder Theorem include this line of reasoning; recall the Chinese Remainder Theorem says essentially that $\mathbb{Z}_n \times \mathbb{Z}_m \cong \mathbb{Z}_{nm}$ exactly when $\gcd(m,n) = 1$).
But you seem to want a more hand-on understanding. So let's produce an example, say with $6 = 2 \cdot 3$. And we'll begin by starting with the group $H = \mathbb{Z}_6$, which I'll recognize as $(\mathbb{Z}/7\mathbb{Z})^\times$, the multiplicative units mod $7$.
(As a small note: writing $[n]$ for equivalence classes all the time is annoying. But essentially integer that appears below that is a member of a group is an equivalence class in that group).
Let $A = \langle 2 \rangle = \{ 2, 4, 1 \} \cong \mathbb{Z}_3$ be the subgroup generated by $2$ and let $B = \langle 6 \rangle = \{6, 1 \} \cong \mathbb{Z}_2$ be the subgroup generated by $6$.
Then your question asks what $AB$ looks like as a subset of $H$. Of course, we know that $AB = H$, so really I suppose we're asking if we can gain an understanding of how $AB = H$.
So let's do it. The product $AB$ consists of the following $6$ elements (coming from $3$ choices from $A$ and $2$ choices from $B$):
$$\begin{array}{lllll}
2 \cdot 6 = 5 && 4 \cdot 6 = 3 && 1 \cdot 6 = 6 \\
2 \cdot 1 = 2 && 4 \cdot 1 = 4 && 1 \cdot 1 = 1
\end{array}$$
Notice that each element $1, 2, 3, 4, 5, 6$ is represented, and thus (as sets) it's clear that $AB = H$. From this it's immediately clear that this is true as groups as well, knowing that a subgroup of $H$ with the same (finite) size as $H$ must be $H$.
Further, note that each element is represented in a unique way! This is also part of the statement of the Chinese Remainder Theorem.
