Let $G$ be an infinite abelian group such that for any non-trivial subgroup $H$ of $G$ , $[G:H ]$ is finite ; then how to prove that $G$ is cyclic ? Please don't use any structure theorem of abelian groups . This question has been asked before $G/H$ is a finite group so $G\cong\mathbb Z$ , but I do not understand the second answer ( which does not use any structure theorem) . Thanks


Here is an expansion of that answer : Let $g\in G$ and consider $H = \langle g\rangle$. By hypothesis, $H$ is normal in $G$ and $n:= |G/H|< \infty$. So the map $$ \varphi: G\to H \text{ given by } x \mapsto x^n $$ maps $G$ into $H$ (Why?)

Now we claim that this map is injective. If not, then $\exists x\in G$ such that $x^n=e$, and so take $K = \langle x\rangle$. Now $|K| < \infty$ and by hypothesis, $|G/K| < \infty$. This would imply that $|G| < \infty$ - a contradiction.

Hence the map $\varphi$ is injective, and so $G \cong \varphi(G)$, which is a subgroup of a cyclic group - hence cyclic.

  • $\begingroup$ Have you fixed $g \in G$ ? If so then you have also fixed $n$ , then how come $φ(G):= \{ x^n : x \in G \} $ is cyclic ?? $\endgroup$ – Souvik Dey Oct 8 '14 at 7:09
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    $\begingroup$ Yes, $g$ and $n$ are both fixed. $\varphi(G)$ is a subgroup of $H$, which is cyclic. Also, kindly refrain from downvoting unless the answer is wrong. It is bad form, and people will avoid answering your questions in the future. $\endgroup$ – Prahlad Vaidyanathan Oct 8 '14 at 7:33
  • $\begingroup$ how come $φ(G)$ is a subgroup of $H$ ? To be specific , let $x \in G$ \ $H$ , then can $x^n $ be in $H$ ?? If it can then give me some $m$ such that $ x^n=g^m \in H$ $\endgroup$ – Souvik Dey Oct 8 '14 at 7:38
  • $\begingroup$ $x+H \in G/H$ and $[G:H] = n$, so what can you say about $(x+H)^n$? $\endgroup$ – Prahlad Vaidyanathan Oct 8 '14 at 8:11

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