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

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

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 ...?

• 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? – tomatoketchup May 7 '19 at 13:42
• @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? – Arthur May 7 '19 at 14:49