I've seen the 'standard' proof where we essentially construct a generator and use the division method.
I was thinking through the problem attempting to do this using isomorphism's and was looking for some help, both in terms of the 'idea' and practice. I'm not sure if I'm going down the correct route.
Case 1: Let $G$ be a finite group of order $n$ and $H \leq G, |H| = m$ and we write $n =am$ due to langrange's theorem.
I know that $G \cong \mathbb{Z} / n \mathbb{Z}$. I want to essentially show that $|H|$ is also isomorphic to some structure relating $\mathbb{Z}$ and addition that is also cyclic.
Can I claim that $H = m\mathbb{Z} / n\mathbb{Z}$? Is this true / if so is this a cyclic group? Further, how would I prove this? Also is it equal to $\mathbb{Z}_a$ ?
Case 2: $G$ is an infinite group thus $G \cong \mathbb{Z}$. Is it hence the case that $|H| \cong k\mathbb{Z}$ which is a cyclic group of generator $k$?
Thanks
Edit:
Does this argument follow? (Case 1): $G \cong \mathbb{Z} / n \mathbb{Z}$. Every subgroup of $\mathbb{Z} / n \mathbb{Z}$ is of the form $m \mathbb{Z} / n \mathbb{Z}$ where $m | n$, in this case let $m = an$. Then this subgroup has the form $\left\{0, m . . . m^{a-1}\right\}$ which is cyclic. Now if I recall there is a theorem that if $H \leq G$ and and $\phi: G \rightarrow G'$ then $\phi(H) \leq G'$. Since we can form an isomorphism $G \rightarrow \mathbb{Z} / n \mathbb{Z}$ then take a subgroup $H \leq G$, then $\phi(G) = m \mathbb{Z} / n \mathbb{Z}$ for some m | n. Hence $H$ is cyclic.
There is a less standard proof that could be what you're looking for. The proof is based on the following important and well known result.
Theorem. If $H$ is a subgroup of $\mathbb{Z}$, then $H=a\mathbb{Z}$ for a unique $a\ge 0$.
Now let's prove the general theorem.
Theorem. If $G$ is a cyclic group and $H$ is a subgroup of $G$, then $H$ is cyclic.
Proof. Let $G$ be a cyclic group and let $H$ be a subgroup. Consider a surjective homomorphism $\varphi\colon\mathbb{Z}\to G$, which exists by hypothesis. Then $H=\varphi(\varphi^{-1}(H))$ and, since $\varphi^{-1}(H)=a\mathbb{Z}$ is cyclic, also $H$ is cyclic, because it's the homomorphic image of a cyclic group. QED