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Let $m, n$ be positive integers. Show that $m\mathbb{Z}$ is a subgroup of $n\mathbb{Z}$ if and only if $n$ divides $m$.

($\mathbb{Z}$ = set of integers)

I know that for $m\mathbb{Z}$ to be a subgroup of $n\mathbb{Z}$ every element of $m\mathbb{Z}$ must be in $n\mathbb{Z}$ also but not sure how to connect this to $m$ dividing $n$.

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Hint: "every element of $m\mathbb{Z}$ must be in $n\mathbb{Z}$" This includes $m$ itself. So if $m\in n\Bbb Z$, then ...?

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  • $\begingroup$ Could this then be equated to LaGrange's theorem as both mZ and nZ are cyclic. the order of mZ generated by m must divide the order of nZ generated by n? $\endgroup$ – tomatoketchup May 7 '19 at 13:42
  • $\begingroup$ @tomatoketchup It's related, but the orders are both infinite, so Lagrange's theorem doesn't apply. But you seem to overthink this. What is the definition of $n\Bbb Z$? What kind of numbers are in that set? And what does it then mean that $m$ is in there? $\endgroup$ – Arthur May 7 '19 at 14:49

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