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By definition, group $G$ is divisible if for any $g\in G$ and natural number $n$ there is $h\in G$ such that $g=h^n$. Let $A$ be abelian group with no proper subgroups of finite index. How can I prove that $A$ is divisible?

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    This is harder than I first thought.2012-09-09
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    @MakotoKato: Since $\mathbb{Z}/n\mathbb{Z}$ is such a nice ring, its modules have submodule lattices that are self-dual (I believe $\operatorname{Hom}(-,\mathbb{Z}/n\mathbb{Z})$ is the duality), and so a finite subgroup does imply a finite index subgroup (here you leverage that $\mathbb{Z}/n\mathbb{Z}$ can define $n$ module-theoretically, so also “finite”). However, things are much simpler if you take $n$ prime. $\mathbb{Z}$ is not a nice ring, and so you really do need to use something special about $A/nA$ (not just that it is torsion, but that it is bounded).2012-09-09

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