Any two basic subgroups of $G$ are isomorphic ($B_1 \cong B_2$), but I am looking for a counterexample to show that the same is not true of the quotients $G/B_1$ and $G/B_2$. Here, the definition that I am using is that a subgroup $B$ is basic in a torsion abelian group $G$ when:
- $B$ is a direct sum of cyclic groups
- $B$ is pure in $G$ ($B \cap nG = nB$ for all positive integers $n$)
- $G/B$ is divisible ($nG/B = G/B$ for all positive integers $n$)
For context, I believe an incorrect claim is made by Exercise 10.41 of Rotman's An Introduction to the Theory of Groups, which claims that a counterexample is given by $G = \sum_{n = 0}^{\infty} \langle a_n \rangle$ where $a_n$ has order $p^{n + 1}$, and $B = \langle pa_1, a_n - pa_{n + 1}, n \geq 1 \rangle$. In particular, I believe here that $B$ is not actually pure in $G$, since $pa_1 \in B \cap pG$ but $pa_1 \notin pB$. (As well, $G/B$ is not actually divisible, since $a_0 + B$ is not divisible by $p$.)
I think your objection is valid. To fix this, we can replace $B$ with $\langle a_n - pa_{n+1}, n \geq 0 \rangle$.
Now $a_0 + B$ is divisible by $p$, because $p(a_1 + B) = a_0 + B$.