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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?

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    $\begingroup$ For the converse, you have to show that if $n|m$ then any element of $m\mathbb Z$ (not only $m$) is in $n\mathbb Z$ $\endgroup$ – J. W. Tanner Mar 20 at 16:18
  • $\begingroup$ @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 $\mathbb Z$. $\endgroup$ – o.h. Mar 20 at 16:18
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    $\begingroup$ @o.h. You're right. I read it as $\Bbb Z_m$ and $\Bbb Z_n$. That's not what this question is about. $\endgroup$ – Arthur Mar 20 at 16:19
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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 $n\mathbb Z$."

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

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  • $\begingroup$ I made a slight correction, @David. $\endgroup$ – Shaun Mar 20 at 21:35

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