A subgroup $H$ of an abelian group $G$ is called pure if $nH=nG\cap H$ for all $n$. We always have $nH\subset nG\cap H$, so if $H$ is impure then $nH$ is a proper subgroup of $nG\cap H$ for some $n$.
Is there an abelian group $G$, a subgroup $H$, and an $n$, such that $H\cap nG$ has finite index in $G$, but $nH$ has infinite index?