# 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!

• Do you mean the direct product $G=\mathbb{Z}_m \times \mathbb{Z}_n$ ? – lhf Feb 1 '19 at 18:08
• 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 – Mathematical Mushroom Feb 1 '19 at 18:17
• So $\mathbb Z_n$, $\mathbb Z_m$ are subgroups of which group? – eduard Feb 1 '19 at 18:24
• Ah, right, sorry. They are isomorphic to subgroups of $\mathbb{Z_n} \times \mathbb{Z_m}$ – Mathematical Mushroom Feb 1 '19 at 18:33
• Can you explain your attempt? Do you feel comfortable with Isomorphism theorems? – eduard Feb 1 '19 at 18:58

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.

• 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 $\mathbb{Z_n} \times \mathbb{Z_m}=\mathbb{Z_{mn}}$ – Mathematical Mushroom Feb 4 '19 at 1:29