Let G be an infinite locally compact abelian group that is isomorphic to its own dual. If H is a closed subgroup of G, is it necessarily true that $H \cong \widehat{G/H}$?
I ask because in the case of the real numbers and p-adic numbers, this seems to be true:
For the real numbers, any non-trivial closed subgroup is going to be like the integers, and its quotient is isomorphic to $\mathbb{S} ^1$, which is the dual of the integers.
In the case of the p-adic numbers $\mathbb{Q} _p$, I'm not sure whether all its closed subgroups are of the form $p^k\mathbb{Z} _p$, but atleast for those particular subgroups, their quotient is isomorphic to the Prufer p-group, which is precisely the dual of the p-adic integers.
Since in the general abelian locally compact case, there is a correspondence between closed subgroups of G and its dual through annihilators, and since for any closed subgroup H of G, there is an isomorphism $\widehat{G/H} \cong A(H)$, it seems tempting to assume this pattern would keep repeating. At the same time, I embarrassingly don't know of any other self-dual groups besides these two and the Adele group of the rationals (whose structure I know basically nothing about), so it's hard for me to potentially cook any counterexamples besides finite abelian groups (which I'm not interested on in this question).
To extend the example by Mariano in the comments, let $A$ be an infinite compact Hausdorff abelian group, such as $S^1$, so its dual group $\widehat{A}$ is a discrete Hausdorff abelian group.
Then $G = A \times A \times \widehat{A} \times \widehat{A}$ is a self-dual group and its subgroup $H = A \times \{0\} \times \{0\} \times \{0\}$ is closed. We'll show $\widehat{G/H} \not\cong H$ as topological groups.
Since $G/H \cong A \times \widehat{A} \times \widehat{A}$ we have $\widehat{G/H} \cong \widehat{A} \times A \times A$. If $\widehat{G/H} \cong H$ as topological groups then $\widehat{G/H} \cong A$, so $\widehat{G/H}$ would be compact and it has the discrete closed subgroup $\widehat{A} \times \{0\} \times \{0\}$. (Actually, in a compact Hausdorff group, every discrete subgroup is closed: a proof is here.) In a compact space all closed subsets are compact, so $\widehat{A}$ is both discrete and compact. Compact discrete spaces are finite, so $\widehat{A}$ would be finite. Then $A$ would be finite by double duality, which contradicts the assumption that $A$ is infinite. Thus $\widehat{G/H} \not\cong H$ as topological groups.