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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).

KCd
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Pedro Lourenço
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  • Oh that answers it yeah lol. Thanks – Pedro Lourenço Aug 13 '23 at 22:32
  • Every nonzero proper closed subgroups of $\mathbf Q_p$ is $p^k\mathbf Z_p$ for some integer $k$. Proof: Let $H$ be a nonzero closed subgroup containing a nonzero $x$. Due to $H$ being closed, $H$ contains $x\mathbf Z_p$, which is $p^n\mathbf Z_p$ where $n = {\rm ord}p(x)$. So if $H$ is unbounded then $H = \mathbf Q_p$, while if $H$ is bounded and $m = {\rm min}{x \in H}({\rm ord}_p(x))$ then $|x|_p \leq 1/p^m$ for all $x \in H$ and equality is achieved at some $x$, so $H = p^m{\mathbf Z}_p$. – KCd Aug 13 '23 at 23:14
  • I accidentally removed my upvote but that makes sense too, appreciate it. – Pedro Lourenço Aug 14 '23 at 12:24

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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.

KCd
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