How is $\mathbb Z /m\mathbb Z$ a subgroup of $\mathbb Z / mn \mathbb Z$?

Question

Let $m,n \in \mathbb N$. I have the quotient group $(\mathbb Z / mn \mathbb Z) / (\mathbb Z / m\mathbb Z)$, and I am trying to understand what it looks like. What I can't seem to understand is how $\mathbb Z / m\mathbb Z$ is a subgroup of $\mathbb Z / mn \mathbb Z$. When I try to write out the sets, I get

$\mathbb Z / m\mathbb Z = \{ \bar 0_m, ..., \overline{m-1}_m\} $

and

$\mathbb Z / mn\mathbb Z = \{ \bar 0_{mn}, ..., \overline{m-1}_{mn}\}.$

And that is as far as I get... I feel like I'm missing something super obvious about how these sets are related. I find quotient groups a bit tricky to begin with, and now because of corona virus I have no one to ask for help right now. The problem in my textbook asks me to show that $$(\mathbb Z / mn \mathbb Z) / (\mathbb Z / m\mathbb Z) \cong \mathbb Z / n \mathbb Z,$$ and I'm thinking I will use the third isomorphism theorem to do that.

Hopefully someone here can shed some light on this for me. Thanks.

Answer 1

That is because $n\mathbf Z/mn\mathbf Z$ is a subgroup of $\mathbf Z/mn \mathbf Z$ and we have an isomorphism $$\begin{align} n\mathbf Z/mn\mathbf Z&\tilde{\longrightarrow} \mathbf Z/m\mathbf Z,\\ na+mn\mathbf Z&\longmapsto a+m\mathbf Z. \end{align}$$

Answer 2

As @MatthewTowers correctly points out in his comment, it's not strictly a subgroup, but you have to squint a bit (i.e. use an isomorphism) to see it as a subgroup.

Here is a concrete example that may make it easier to see what is happening: The face of an analog clock represents the group $\mathbb{Z}/12\mathbb{Z}$ (where you have to wear your mathematician's glasses to see that '12' really should be labeled '0').

Since $3$ divides $12$ there should be a copy of $\mathbb{Z}/3\mathbb{Z}$ in $\mathbb{Z}/12\mathbb{Z}$ -- and there is; it's $\{4, 8, 12\}$! (Or $\{0, 4, 8\}$ with your math glasses on.) This is your question with $m=3$ and $mn = 12$.