Let $G$ be a totally disconnected locally compact abelian group. Let $U \leq G$ be open and such that $G/U \cong \mathbb{Z}(p^\infty)$, the Prüfer $p$-group. In general $U$ is not necessarily a direct summand of $G$: for instance when $G = \mathbb{Q}_p$ and $U = \mathbb{Z}_p$, or when $G = \mathbb{Z}[1/p]$ and $U = \mathbb{Z}$.
In both examples, $U$ is not minimal for this property: there exists a proper subgroup $V < U$ which is open and such that $G/V \cong \mathbb{Z}(p^\infty)$.
Is this the only obstruction? That is, if $U$ is minimal for this property, does it follow that it is a direct summand of $G$?
