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Prove that for the additive group $(\mathbb{Z}, +)$ of integers every subgroup is of the form $k\mathbb{Z}$.

Here is the proof I wrote, and I knew it was off, so I sent it to my professor for some help. His reply: "Sorry, but you missed the point. You needed to show that any subgroup of $\mathbb{Z}$ is of the form $k\mathbb{Z}$ for some $k$. Keep on trying." Which is less than helpful. This is a problem off a practice final.

Can someone give me a hint to get me started?

edit: here's my new attempt. LINK Is this enough? Or is more needed?

Dana Hill
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    You did miss the point. Your argument (correctly) shows that the set $k\mathbb Z$ is a subgroup but that's not what you were asked. You were asked to prove that EVERY subgroup has that form. – lulu Apr 20 '17 at 23:49
  • You proved that $k\mathbb{Z}$ is a subgroup. What you are supposed to prove is that if you are handed a subgroup $H$ it is really just one of the $k\mathbb{Z}$'s. Hint: what is the smallest nonzero element in $k\mathbb{Z}$? – Ethan Bolker Apr 20 '17 at 23:50
  • You start proving that $kZ $ is a subgroup. But you can't do that if uou're yo answer the question. You need to start with the addumptio that $U $ is a subgroup of $(Z,+)$ and shiw that $U=kZ$ for some $k $. – AnyAD Apr 20 '17 at 23:51
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    Hint: see the Lemma in this answer. – Bill Dubuque Apr 21 '17 at 02:16
  • Showing that $k\Bbb Z$ is a subgroup is quite different from show that any $H\leq \Bbb Z$ has the form $k\Bbb Z$ – Zelos Malum Apr 21 '17 at 07:48

1 Answers1

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What you proved is that $k\mathbb{Z}$ is a subgroup for any $k$. But to prove the statement given to you, your proof should begin: "Let $H$ be a subgroup of $\mathbb{Z}$" and conclude with "Therefore $H = k\mathbb{Z}$ for some $k \in \mathbb{Z}$."

If $H$ is a subgroup of $\mathbb{Z}$, try looking at the smallest (in absolute value) element of $H$.

Joshua Ruiter
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  • Ah... you said absolute value. Others just said "what's the smallest value" and for integers, there isn't one, there's always a smaller integer. But if you say absolute value... now you can say 1. Maybe I can start with that. – Dana Hill Apr 21 '17 at 16:42
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    The new attempt at the link is good proof! It would be good to include a reference to where that lemma is coming from, for completeness. – Joshua Ruiter Apr 23 '17 at 18:09