Subgroups of the integers

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.

• A better proof is given at this duplicate, without using that subgroups of cyclic groups are cyclic. – Dietrich Burde May 8 at 21:16
• I'm afraid the the proof of the assertion on subgroups of cyclic groups relies on properties of $\mathbf Z$. – Bernard May 8 at 21:16
• @Bernard what does that say about my proof? – topologicalmagician May 8 at 21:28
• @DietrichBurde okay, but that doesn't say where my proof fails, may you please elaborate? – topologicalmagician May 8 at 21:28
• 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. – Lee Mosher May 8 at 22:04