It is well known that every subgroup $H$ of a finite abelian group $G$ is isomorphic to a quotient of $G$. I'm wondering whether there is a counterpart for profinite groups.

For example is it true that every open subgroup $H$ of an abelian profinite group $G$ isomorphic to a quotient?

what about the weaker statement that every open subgroup contains an open subgroup which is isomorphic to a quotient?

If G is a direct product of finite groups then the second weaker statement is true because every open subgroup $H$ contains a cylinder neighborhood of the origin $H'$ for which $G=H'\times K$ where $K$ is a finite product of finite groups.


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    $\begingroup$ you are right. I removed that comment. I added another comment instead. $\endgroup$ – user106317 May 19 '17 at 20:14
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    $\begingroup$ By Pontryagin duality, the main question is equivalent to: let $G$ be an locally finite abelian group, and let $H$ be the quotient of $G$ by a finite subgroup. It it true that $H$ is isomorphic to a subgroup of $G$?. The "weaker question" is the same but with the question replaced with It is true that $H$ has a finite subgroup $N$ such that $H/N$ is isomorphic to a subgroup of $G$? $\endgroup$ – YCor May 19 '17 at 22:01
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    $\begingroup$ Another remark: it's false for closed subgroups instead of open. Indeed, $\mathbf{Z}_p$ is isomorphic to a closed subgroup of $\prod_n\mathbf{Z}/p^n\mathbf{Z}$, but not to a quotient (because it's torsion-free while in this product, the torsion elements form a dense subgroup, which passes to quotients). $\endgroup$ – YCor May 19 '17 at 22:03
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    $\begingroup$ Interesting. Is every profinite group embedded in a direct product of finite groups? If so we can use this embedding and conclude the second weaker statement. $\endgroup$ – user106317 May 21 '17 at 16:48
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    $\begingroup$ Another note, this statement is also true for non-profinite groups like $S^1$ as every open subgroup of $S^1$ is $S^1$ which isomorphic to the trivial quotient. $\endgroup$ – user106317 Jun 1 '17 at 19:51

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