Timeline for Subgroups and quotients of an abelian pro-finite group
Current License: CC BY-SA 3.0
15 events
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| Jun 1, 2017 at 19:51 | comment | added | user106317 | 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. | |
| May 21, 2017 at 20:47 | history | edited | user106317 | CC BY-SA 3.0 |
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| May 21, 2017 at 20:01 | history | edited | user106317 | CC BY-SA 3.0 |
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| May 21, 2017 at 19:49 | history | edited | user106317 | CC BY-SA 3.0 |
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| May 21, 2017 at 18:04 | comment | added | YCor | Every profinite group embeds as a closed subgroup of a product of finite groups, almost by definition. But I don't think this helps for the "second weaker statement". | |
| May 21, 2017 at 17:55 | history | edited | user106317 | CC BY-SA 3.0 |
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| May 21, 2017 at 16:48 | comment | added | user106317 | 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. | |
| May 20, 2017 at 18:01 | history | edited | user106317 | CC BY-SA 3.0 |
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| May 19, 2017 at 22:03 | comment | added | YCor | 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). | |
| May 19, 2017 at 22:01 | comment | added | YCor | 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 $G$? | |
| May 19, 2017 at 21:56 | history | edited | YCor |
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| May 19, 2017 at 20:14 | history | edited | user106317 | CC BY-SA 3.0 |
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| May 19, 2017 at 20:14 | comment | added | user106317 | you are right. I removed that comment. I added another comment instead. | |
| May 19, 2017 at 18:16 | comment | added | Jeremy Rickard | I don't understand the last comment. $G=C_4$ has an (open) subgroup $H=C_2$, but doesn't split as $H\times K$ for any $K$. | |
| May 19, 2017 at 17:37 | history | asked | user106317 | CC BY-SA 3.0 |