$G$ infinite abelian group with $[G:H]$ finite for every non trivial subgroup $H$ , to prove $G$ is cyclic 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.
