This question already has an answer here:

Let $G$ be an abelian group with all proper subgroup finite, then is $G$ finite or at least finitely generated ?


marked as duplicate by user26857, Hurkyl, Community Jan 12 '18 at 16:20

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 1
    $\begingroup$ Have you attempted to approach this problem on your own yet? What have you tried? $\endgroup$ – Andrew Tawfeek Jan 12 '18 at 15:57
  • $\begingroup$ @Andrew Tawfeek: yes ... with all examples I know pointing out that the group is finite ... $\endgroup$ – user495643 Jan 12 '18 at 16:01
  • $\begingroup$ Start taking subgroups generated by non $0$ element if there are. Those are cyclic and finite. Quotient by it and keep on doing that with the result. If eventually you reach the trivial group, you are done, the group is finite. If you can keep on going indefinitely, exclude one of the factors and look at the group generated by all other factors. This one is proper, and infinite. Contradiction. $\endgroup$ – orole Jan 12 '18 at 16:03
  • $\begingroup$ @orole: You overlook the possibility that the excluded "factor" is contained in the group generated by the other "factors". (also, you seem to be assuming that the proper subgroups are actually direct summands of the whole) $\endgroup$ – Hurkyl Jan 12 '18 at 16:08
  • $\begingroup$ @Hurkyl "Abelian" $\endgroup$ – orole Jan 12 '18 at 16:10

Let $G$ be the group of all $z\in\mathbb{C}$ such that $z^{2^n}=1$ for some $n\in\mathbb{Z}^+$ (under the usual multiplication). Then $G$ is infinite, but every proper subgroup is finite.


If $p$ is a prime number, then consider the ring $\mathbb{Z}[\frac{1}{p}]$ of all rational numbers whose denominator is a power of $p$.

Taking quotients of the underlying additive group, the group

$$ \mathbb{Z}\left[\frac{1}{p} \right] / \mathbb{Z} $$

is an example of a group with this property. This group is sometimes denoted as $\mathbb{Z} / p^\infty$, since it is the "limit" (in a suitable sense) of the cyclic groups $\mathbb{Z} / p^k$ as $k \to \infty$.

In the case of $p=2$, this group is isomorphic to the one described in the other answer.