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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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What have you tried? –  Qiaochu Yuan Sep 9 '12 at 20:01
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This is harder than I first thought. –  Makoto Kato Sep 9 '12 at 20:53
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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). –  Jack Schmidt Sep 9 '12 at 23:16

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Consider the $n$ such that $A^n := \{ a^n : a\in A \} \neq A$. In general these $A/A^n$ will be modules over the self-injective rings $\mathbb{Z}/n\mathbb{Z}$ so Prüfer/Kulikov's theorem shows you have a finite index subgroup. However if you choose a good $n$, you can just use linear algebra. Simple version under the spoiler, since a few other people were working on it.

If $A$ is not divisible then the set $S = \{ n \in \mathbb{Z} : n > 0, \{ a^n : a \in A \} \neq A \}$ has a least element (more generally, the set of ideals without the Baer property has a maximal element). Let $n$ be the least element of $S$. If $n=ab$ for $a,b > 1$, then $A \neq A^n = A^{ab} = (A^a)^b = A^b = A$, since $a,b < n$ so $a,b \notin A$. Hence $n$ is prime and $A/A^n$ is a vector space over the field $\mathbb{Z}/n\mathbb{Z}$. In particular, by the axiom of choice it has a maximal subspace $B$ and since $B$ has codimension 1, $[A:B] =n$ is a finite index.

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I might be picky, but does it really require choice? I wouldn't be surprised if it really did, but since you invoked it expressis verbis, I thought that maybe you can answer this question. :) –  tomasz Sep 9 '12 at 21:37
    
@tomasz: I'm not sure if it is equivalent to AC, but there are vector spaces in ZF without AC that have no codimension 1 subspaces. mathoverflow.net/questions/80905/… (question#2) I believe that means a non-divisible abelian group without finite index subgroups also exists. However this will just be because there are vector spaces over Z/pZ without finite index subspaces. –  Jack Schmidt Sep 9 '12 at 23:08
    
Thanks for the reply. :) It is not obvious to me that every space occurs as a quotient, but I'm still convinced enough that we do need some choice. :) –  tomasz Sep 9 '12 at 23:34

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