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 $k$?



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.

  • $\begingroup$ The answers to the "Related" at the right margin give many arguments about the proof, I think. So I don't see a new alternative. $\endgroup$ – Dietrich Burde Jan 31 '16 at 21:17
  • $\begingroup$ $G$ cyclic means that there is an epimorphism $\phi\colon \Bbb Z\to G$. For $0\ne H\le G$ consider $\phi^{-1}(H)$ and notice that it is $k\Bbb Z$ (and hence $\phi\circ(x\mapsto kx)$ is an epimorphiusm $\Bbb Z\to H$) with $k$ the minimal element of its intersection with $\Bbb N$. $\endgroup$ – Hagen von Eitzen Jan 31 '16 at 21:21

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

  • $\begingroup$ Thanks. So this addresses the infinite order case. Could you help me with the finite order case? $\endgroup$ – Kevin Jan 31 '16 at 22:11
  • $\begingroup$ @Kevin I reformatted my answer to make the argument clearer. The first theorem (not proved here) is the result about infinite cyclic groups, which can be applied for the general case. $\endgroup$ – egreg Jan 31 '16 at 22:20
  • $\begingroup$ I got to the part about $H = \varphi (\varphi^{-1}(H))$, why is $\varphi^{-1} (H) = a \mathbb{Z}?$ Is it because $H$ is a subset of $G$ and hence maps to some subset of $Z$ (aZ)? I apologise because I'm not really that confident with homomorphisms and would prefer if you could present a proof using just isomorphisms? Thanks $\endgroup$ – Kevin Jan 31 '16 at 22:32

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