I find it hard to understand a part of the proof of the existence of any basic subgroup in every abelian torsion group.I'm going to write you the information I think useful.
Let $G$ an abelian torsion group. Let $B=\langle X\rangle$ where $X$ is a maximal pure independent subset. If we prove that $G/B$ is divisible it follows that $B$ is basic and so our thesis.
Suppose by contradiction that $G / B$ is not divisible, hence it has a nontrivial pure cyclic subgroup $\langle g+B\rangle$. By purity of $B$ we have that if $ p^dg$ belongs to $B$, then $p^dg$ belongs to $p^dB$. Hence $p^dg=p^db$ and $p^d(g-b)=0$ where $b\in B$.
Since $(g-b)+B=g+B$ it follows that $g'=g-b$ and $g'+B$ have the same order.
Why is the latter true? How can I prove that $g'=g-b$ and $g'+B$ have the same order? I hope I gave you all relevant information.
Can anyone help me? Thanks.