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 $n$.


1 Answer 1


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 ...?

  • $\begingroup$ 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? $\endgroup$ May 7, 2019 at 13:42
  • 1
    $\begingroup$ @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, 2019 at 14:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.