Show that $m\mathbb{Z}$ is a subgroup of $n\mathbb{Z} \iff m|n$

Show that $$m\mathbb{Z}$$ is a subgroup of $$n\mathbb{Z} \iff n|m$$

I think my solution for one way of this is correct:

$$\Rightarrow$$ Suppose $$m \mathbb{Z}$$ is a subgroup of $$n\mathbb{Z}$$ , then $$m \mathbb{Z}$$ is a subset of $$n\mathbb{Z}$$

Therefore $$m$$ is an element of $$n\mathbb{Z}$$, $$m=nz$$ for some $$z$$ in $$\mathbb{Z}$$

And so $$n|m$$ as required.

For the converse, am I allowed to do these steps in reverse or is there more I must do?

• For the converse, you have to show that if $n|m$ then any element of $m\mathbb Z$ (not only $m$) is in $n\mathbb Z$ – J. W. Tanner Mar 20 at 16:18
• @Arthur No, since $m\mathbb Z \cong n\mathbb Z$ (which is trivially a subgroup of itself) for all $m,n$, that is not a very interesting statement. The question does make sense since $m\mathbb Z$ must be taken to describe a subgroup of $\mathbb Z$. – o.h. Mar 20 at 16:18
• @o.h. You're right. I read it as $\Bbb Z_m$ and $\Bbb Z_n$. That's not what this question is about. – Arthur Mar 20 at 16:19

Arguably the last statement of the resulting proof -- in which we conclude that $$m\in n\mathbb Z$$ -- should be followed by something of the form: "... and since $$m$$ generates $$m\mathbb Z$$, this shows that $$m\mathbb Z$$ is a subgroup of $$n\mathbb Z$$."
Suppose $$n\mid m$$ which means that $$m=nc$$ for some $$c\in \mathbb{Z}$$. The set $$n\mathbb{Z}$$ is set of all integers of the form $$N=nz$$ for some $$z\in \mathbb{Z}$$. Now let's look on the set $$m\mathbb{Z}$$, it the set of all integers of the form $$M=mz=(nc)z=n(cz)$$ for some $$z,c\in \mathbb{Z}$$.Now we can clearly see evey element $$M\in \mathbb{mZ}$$ is in $$\mathbb{nZ}$$. Hence $$\mathbb{mZ}\subseteq\mathbb{nZ}$$ and we can even say $$\mathbb{mZ}=\mathbb{nZ}$$.