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$G=\mathbb{Z_m} \mathbb{Z_n} = \{[a][b] : [a] \in \mathbb{Z_m}, [b] \in \mathbb{Z_n}\}$

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!

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    $\begingroup$ Do you mean the direct product $G=\mathbb{Z}_m \times \mathbb{Z}_n$ ? $\endgroup$ – lhf Feb 1 '19 at 18:08
  • $\begingroup$ No, I don't actually. Like, if you have two subgroups of $G$, say $H,K$, then you can describe $HK$ as a set, and this set is a subgroup of $G$ if either $K$ or $H$ is normal in $G$. I'm looking for someone to explain to me why $\mathbb{Z_n} \mathbb{Z_m} = \mathbb{Z_{mn}}$ when $gcd(m,n)=1$, and specifically I would like to "open up the equivalence classes" and see why this makes sense in a number theoretic sort of way $\endgroup$ – Mathematical Mushroom Feb 1 '19 at 18:17
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    $\begingroup$ So $\mathbb Z_n$, $\mathbb Z_m$ are subgroups of which group? $\endgroup$ – eduard Feb 1 '19 at 18:24
  • $\begingroup$ Ah, right, sorry. They are isomorphic to subgroups of $\mathbb{Z_n} \times \mathbb{Z_m}$ $\endgroup$ – Mathematical Mushroom Feb 1 '19 at 18:33
  • $\begingroup$ Can you explain your attempt? Do you feel comfortable with Isomorphism theorems? $\endgroup$ – eduard Feb 1 '19 at 18:58
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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.

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  • $\begingroup$ Thanks. I wanted to see that $\mathbb{Z_n} \mathbb{Z_m}=\mathbb{Z_{mn}}$ when $gcd(m,n)=1$ in order to conclude that $\mathbb{Z_n} \times \mathbb{Z_m}=\mathbb{Z_{mn}}$ $\endgroup$ – Mathematical Mushroom Feb 4 '19 at 1:29

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