Let $G$ be an infinite abelian group such that every proper non-trivial subgroup of $G$ is infinite cyclic ; then is $G$ cyclic ? ( The only characterization I know for infinite abelian groups to be cyclic is that every non-trivial subgroup has finite index . But I am not getting anywhere with the above statement . Please help . Thanks in advance )
NOTE : Here is what I have thought for finitely generated infinite abelian ; if $G$ is infinite and finitely generated , then $G \cong \mathbb Z_{n_1} \times ... \times \mathbb Z_{n_k} \times \mathbb Z^r$ , where $r \ne 0$ and $n_1|n_2|...|n_k ; n_1>1$ ; now if $k=0$ then for $r=1$ , $G$ is cyclic ; if $k=0 ; r>1$ then since for $\mathbb Z^r$ has a non-cyclic infinite non-trivial subgroup (a copy of ) $\mathbb Z \times 2\mathbb Z$ violating the hypothesis . So then let $k \ne 0$ ; then a copy of $\mathbb Z_{n_1} \times ... \times \mathbb Z_{n_k} \times 2\mathbb Z$ is contained in $G$ which is non-trivial , infinite and non-cyclic , violating the hypothesis again . So the only possibility is $G \cong \mathbb Z$ i.e. $G$ is cyclic ; am I correct here ?
The answer is yes. Here is an elementary argument.
Suppose firstly that for each $n\in\mathbb N$ we have $nG=G$. Since $G$ is clearly torsion free, it follows that $G$ is a $\mathbb Q$-vector space, immediately violating the hypothesis.
Thus for some $n\in\mathbb N$ we have $nG<G$, so by hypothesis $nG\cong\mathbb Z$. However $G\cong nG$ since it is torsion free: the map $G\rightarrow G$, $x\mapsto nx$ is injective and has image $nG$.
The group $\mathbb Z\!\left [\frac12\right]$ is not a counterexample: there are plenty of proper non-cyclic subgroups, for example $3\mathbb Z\!\left [\frac12\right]$.