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Is there an abelian group $G$ for which the largest divisible subgroup of $G$ (given by the sum of all divisible subgroups) is not the intersection of all subgroups of the form $nG$ over the positive integers $n$?

Note that if $ny=x \in G$, then for any positive integer $m$, if $z \in G$ and $nmz=x$, then of course $nmz=ny$, but one cannot in general "cancel" the $n$ to get $mz=y$. Cancellation is possible only if one knows that $0$ is the only element of $G$ whose order divides $n$, which in particular holds if $G$ is torsionfree.

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I think the following construction works:

Let $F = \bigoplus_{\Bbb N} \Bbb Z$. Define $H \leqslant F$ by $\langle e_1 - 2e_2, e_1 - 3e_3, \ldots \rangle$.
Set $G = F/H$.

Then, $\overline{e_1} \in nG$ for all $n \geqslant 1$.
On the other hand, given any $x \in F$ with $\overline{e_1} = 2 \overline{x}$, there exists $n \in \Bbb N$ such that $x \notin nG$.
This shows that $\overline{e_1}$ cannot be in any divisible subgroup of $G$, even though it is in the intersection of all $nG$.


For the proofs of the above statements, see my answer here.
Also see this answer to note that the biggest divisible subgroup consists of all totally divisible elements, instead of all divisible elements.

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