Since $H$ is a pure subgroup of $G$, for any $h \in H$ and any $n \in \mathbb{Z}$, if $nh \in G$, then $nh \in H$. Now, let $hS \in H / S$ and $n(hS) \in G / S$. Then $nhS \in G / S$, which means that $nh \in G$. Since $H$ is a pure subgroup of $G$, we have $nh
Show more…