TITLE: Show that $m\mathbb{Z}$ is a subgroup of $n\mathbb{Z} \iff m|n $
QUESTION [6 upvotes]: 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?

REPLY [2 votes]: 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 $\mathbb{mZ}=\mathbb{nZ}$.