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Want to show:

Consider the additive group, $\mathbb{Z}$. Show that the subgroups are of the form $n\mathbb{Z}$, for some $n\in \mathbb{Z}$

Proof:

Note that for $1\in \mathbb{Z}$, $\langle 1\rangle=\mathbb{Z}$, hence $(\mathbb{Z},+)$ is a cyclic group, and therefore every subgroup is cyclic. So let $H \leq \mathbb{Z}$, that means $H = \langle h\rangle = \{ nh\mid n\in \mathbb{Z} \}= h \mathbb{Z}$.

Is the proof correct? May I have feedback, please? Every proof I've seen relies on the division algorithm at some point, isn't the above proof, a more direct proof?

Note: I have considered this to be a corollary to theorem, every subgroup of a cyclic subgroup is cyclic.

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    $\begingroup$ A better proof is given at this duplicate, without using that subgroups of cyclic groups are cyclic. $\endgroup$ – Dietrich Burde May 8 at 21:16
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    $\begingroup$ I'm afraid the the proof of the assertion on subgroups of cyclic groups relies on properties of $\mathbf Z$. $\endgroup$ – Bernard May 8 at 21:16
  • $\begingroup$ @Bernard what does that say about my proof? $\endgroup$ – topologicalmagician May 8 at 21:28
  • $\begingroup$ @DietrichBurde okay, but that doesn't say where my proof fails, may you please elaborate? $\endgroup$ – topologicalmagician May 8 at 21:28
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    $\begingroup$ Your proof has a gap right here: "and therefore every subgroup is cyclic". This statement is true. But you have not justified it. In fact, I would say that the justification of this step is the most important part of the proof. So, your proof is incomplete. $\endgroup$ – Lee Mosher May 8 at 22:04

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