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How can we prove that: $n|m \Leftrightarrow m\mathbb{Z} \le n\mathbb{Z}$
I understand that if $m\mathbb{Z} \le n\mathbb{Z}$ so that $m=m\cdot 1\in n\mathbb{Z}$ and then $n|m$ but I don't understand why it's true.

Xavi
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    If you know that $m \Bbb{Z}$ and $n \Bbb{Z}$ are subgroups of $\Bbb{Z}$, then all you need to do is prove $m \Bbb{Z} \subseteq n \Bbb{Z}$. I would first suggest you think about what it means to say $n \mid m$. Then think why $mk \in n \Bbb{Z}$ for any $k \in \Bbb{Z}$. – Theo Bendit Nov 11 '21 at 09:03
  • Since this question is a bit old, I'm assuming that the OP understands it now. But in general I think it's useful to indicate (with questions like this) whether the OP lacks the intuition for the proposition, or doesn't know how to represent that intuition with rigor. – Brian Tung Dec 14 '21 at 01:48

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Let $G $ be a group and $A,B \leq G$. If $A \subseteq B$, then $A \leq B$ (Try to prove it if you haven't already). So in this case you just need to prove that $m \mathbb Z \subseteq n\mathbb Z$.

I'll mark the proof as a spoiler just in case you want to try it yourself first:

Let $x \in m\mathbb Z$. Then $x = mh$, for some $h \in \mathbb Z$. But $n|m$, meaning that $m = nq$, for some $q \in \mathbb Z$, so we can conclude that $x = mh = n \cdot qh \in n\mathbb Z$. So $m\mathbb Z\subseteq n \mathbb Z$, emaning that $m \mathbb Z \leq n\mathbb Z$

Eduardo Magalhães
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