This answer has been arranged with the help of @Derek and @Conrad.
Definition: $S is pure in $G$ if and only if $x=ng$ for $n\in\mathbb N$ and $y\in G$ then we can always find $s\in S$ such that $x=ns$.
So a subgroup $S$ of $G$ is pure if all elements of $S$ which are $n$-divisible in $G$ are $n$-divisible in $S$.
Theorem: A subgroup $N$ of a torsion-free group $G$ is pure if and only if $G/N$ is torsion-free.
We consider abelian torssion-free group $G=\mathbb Z\times\mathbb Z$, subgroups $H=\langle (3,2)\rangle$ and $K=\langle (1,2)\rangle$. $G/H$ and $G/K$ both are infinite and isomorphic to $\mathbb Z$, so since $\mathbb Z$ is torsion-free then these subgroups are pure in $G$.
Now we consider $S=\langle (1,2),(3,2)\rangle=\{(k+3k’,2k+2k’)|\exists k,k’\in\mathbb Z \}\leq \mathbb Z\times\mathbb Z$. $S$ is not pure in $G$. In fact $(2,0), (1,0)\in G\;\; \text{and}\;\;(2,0)=2(1,0)$ but these two ordered pairs are not not connected as $(2,0)=n(1,0)$ in $S$ because $(2,0)\in S$ and $(1,0)\notin S$. Hence, $H\leq_{pure} G, \;K\leq_{pure} G$ but $\langle H,K\rangle$ is not pure in $G$.
Thanks for step by step hitting me.