I heard a talk yesterday on MacIntyre's theorems, which involved a decomposition of ($\omega$-stable) groups into a divisible part and a bounded exponent part. Apparently there is a result of Kaplansky that all abelian groups have a maximal divisible part.
Clearly, if $D$ is a divisible subgroup, then $D=\bigcap_{n\in\mathbb N} nD\subseteq \bigcap_{n\in\mathbb N} nG$. Under certain hypotheses (e.g. $\omega$-saturation of the group), it's easy to show that this subset relation is actually equality. I was told that some hypothesis is necessary to guarantee that, which made me wonder:
Is there an abelian group $G$ such that $\bigcap_{n\in\mathbb N} nG$ is not divisible?
So, there would have to be an element $x$ which is "divisible" (in that you can divide $x$ by $n$ for every $n>0$, since $x\in \bigcap_{n\in\mathbb N}nG$), but in order to divide it by $n$, the result must not be divisible (i.e. for some $n$, and every $y$, if $ny=x$, then $y$ is not "divisible").
Is such a group possible?
Define $G$ by the abelian group presentation $\langle t, x_k\,(k>1) \mid 2t=0,\,kx_k=t\,(k>1)\rangle$. Then $\cap_{n \in {\mathbb N}} \,nG = \langle t \rangle$, a group of order $2$, which is not divisible.