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

Question

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?

Answer 1

Yes, in this specific context, running through your arguments in reverse order actually provides a proof of the converse statement. (Obviously this is not generally the case with "if and only if" proofs.)

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 $n\mathbb Z$."

Answer 2

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 $\mathbb{mZ}=\mathbb{nZ}$.