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We have already seen this question that "Give an example of a noncyclic Abelian group all of whose proper subgroups are cyclic.". Now the easiest answer is $\Bbb Z_2 \times \Bbb Z_2$ and the bit more interesting result(this is interesting in number-theoretic ways and Galois theoretic way as well) is $G=\{a/2^k| a\in \Bbb Z, k\in \Bbb N\}$

Here I have proved that this is a non-cyclic group but the thing I could not understand is why every proper subgroup is cyclic. Now please read this few liner blog. The argument that is given here. "Suppose $H$ is a proper subgroup of $G$ then Since $H$ is a proper subgroup, there exists a smallest $m∈\Bbb N$ such that no element of H has a denominator equal to $2^m$." You can read that because the article was written just after the first question raised in math stack. I could not understand that immediately why there should always exist one such big $m$ because $H$ is proper. I was thinking that what will happen if $(5/2)^n \in H$ for all $n \in \Bbb N$, then we have to calculate a bit to prove(might be!!) that $H=G$ or what will happen if we have some other similar cases. Is the reason obvious? Is there any other way that I am missing.

By the way, another interpretation of $\Bbb Z(p^{\infty})$ is not G in the first math stack post.

Ri-Li
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    Isn't $H = {3a/2^k : a \in {\mathbb Z}, k \in {\mathbb N} }$ a non-cyclic proper subgroup of $G$ of index $3$? – Derek Holt Dec 30 '19 at 10:12
  • To some extent this is my question. I don't know about the authenticity of the blog. $H$ is proper(obvious) it is non cyclic because if some $H=<3a/2^k>$ where a is odd then they can't generate $3/2^{k+1}$ and so on. – Ri-Li Dec 30 '19 at 10:24
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    It seems this blog post is just false. Clearly the subgroup given by @DerekHolt is a counterexample to the claim about the smallest $m$, since all $3/2^m$ are in $H$. – Captain Lama Dec 30 '19 at 11:20

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