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I'm currently working in the following excercise:

Find all subgroups of $(\Bbb Z,+)$

I know a subgroup of $(\Bbb Z,+)$ is the multiple of a Natural number $n $ and it has the form: $n\Bbb Z$={$na|n \in \Bbb N, a \in \Bbb Z$}.

But is that the form to describe all subgroups of $(\Bbb Z,+)$? Am I missing something?

Thanks in advance for any hint or help and for taking the time to read my question.

mraz
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  • don't forget the zero subgroup – J. W. Tanner May 27 '19 at 19:44
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    Is $0\in\Bbb N$? And a caveat: ${,na\mid n\in\Bbb N,a\in\Bbb Z,}$ does not mean what you intend. -- Finally, I suspect that what you say after "I know ..." is not considered known in your current course; so a proof of tha fcat may be what is actually asked for – Hagen von Eitzen May 27 '19 at 19:45

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There is obviously the zero subgroup. If $H\leq\mathbb{Z}$ is not the zero subgroup then it must contain positive integers. Let $n=\min\{k\in H: k>0\}$. Then obviously $n\mathbb{Z}\leq H$ because $H$ is closed under the operation of $\mathbb{Z}$ and under taking inverses. Also $H\leq n\mathbb{Z}$ because if $h\in H$ then we can divide $h$ by $n$ with remainder. There are $q,r\in\mathbb{Z}$ such that $h=qn+r,0\leq r<n$. Then $r=h-qn\in H$. Since $n$ is the smallest positive integer in $H$ we conclude that $r=0$ and hence $h=nq\in n\mathbb{Z}$. So $H=n\mathbb{Z}$.

Mark
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Yes. {$0$} can be viewed as $0\Bbb Z$. See here for a complete proof: https://proofwiki.org/wiki/Subgroups_of_Additive_Group_of_Integers

J. W. Tanner
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