Timeline for Universal group such that every finite group is a quotient
Current License: CC BY-SA 4.0
22 events
| when toggle format | what | by | license | comment | |
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| Nov 20, 2023 at 10:42 | comment | added | YCor | Dominic, I find it weird to accept an answer that doesn't give an answer the question. This might discourage somebody to provide a full answer. | |
| Nov 19, 2023 at 10:16 | answer | added | Immanuel van Santen | timeline score: 1 | |
| Nov 16, 2023 at 10:13 | vote | accept | Dominic van der Zypen | ||
| Nov 20, 2023 at 12:47 | |||||
| Nov 15, 2023 at 20:06 | history | edited | Dominic van der Zypen | CC BY-SA 4.0 |
went back to original question without the wrong bits
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| Nov 15, 2023 at 20:02 | comment | added | Dominic van der Zypen | I'll undo the changes and go back to the old question. Apologies for changing the question so much in the wrong direction. I understand everyone who wants to close it. | |
| Nov 15, 2023 at 19:59 | vote | accept | Dominic van der Zypen | ||
| Nov 15, 2023 at 20:07 | |||||
| Nov 15, 2023 at 18:18 | review | Close votes | |||
| Nov 22, 2023 at 3:04 | |||||
| Nov 15, 2023 at 17:49 | comment | added | Noah Schweber | For your new version of the question with no universality requirement, just take the direct sum (rather than product) of the finite groups. | |
| Nov 15, 2023 at 17:26 | history | edited | Dominic van der Zypen | CC BY-SA 4.0 |
deleted 95 characters in body
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| Nov 15, 2023 at 17:21 | comment | added | Dominic van der Zypen | Sorry for my false statement about $I_\mathbb{N}$. Will take the wrong bit out of the question and rephrase the question altogether. I was wondering about a countable group such that all finite groups are a quotient of $G$, just like $I_\mathbb{N}$ is countable. Thanks @SeanEberhard (and the others here) for putting thought effort into the question, apologies for my false statement and rephrasing | |
| Nov 15, 2023 at 17:07 | comment | added | Sean Eberhard | @AlessandroCodenotti There do not exist just-infinite groups covering all finite groups. Suppose $G$ were one. Then in particular we have surjections $f_n : G \to C_n$ for all $n$ and hence a map $(f_n) : G \to \prod C_n$ with infinite image, hence an embedding, which implies that $G$ is abelian. | |
| Nov 15, 2023 at 13:29 | comment | added | Denis T | @MikhailKatz There's no reference, this is just wrong, as Yves noted above. | |
| Nov 15, 2023 at 12:40 | comment | added | Mikhail Katz | Do you have a reference for the claim that "$I_\mathbb{N}$ has the universal property that if $G$ is such that every finite group can be embedded into $G$, then $I_\mathbb{N}$ embeds into $G$" ? | |
| Nov 15, 2023 at 12:40 | comment | added | Dominic van der Zypen | I corrected the terminology and the typo - thanks to Hendrik and YCor! | |
| Nov 15, 2023 at 12:36 | history | edited | Dominic van der Zypen | CC BY-SA 4.0 |
deleted 16 characters in body
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| Nov 15, 2023 at 12:02 | comment | added | YCor | Concerning the subgroup case, as I already said somewhere else on this site (I don't remember where), there is no group that contains all finite groups and embeds into every group containing every finite group. Indeed, the groups $\bigoplus_n S_n$ and $\ast_n S_n$ both contain every finite group but no infinite group embeds into both. | |
| Nov 15, 2023 at 11:40 | comment | added | YCor | What you call "is almost-identical" is widely known as "has finite support" or "is finitely supported", or "is finitary". | |
| Nov 15, 2023 at 10:38 | answer | added | Sean Eberhard | timeline score: 2 | |
| Nov 15, 2023 at 10:17 | comment | added | Sean Eberhard | I don't think $I_{\mathbb N}$ (which is often denoted something like $\operatorname{FSym}(\mathbb N)$) has the universal property you say. For example every finite group embeds into $\prod_{n=1}^\infty S_n$, but there is no injective homomorphism from $\operatorname{FSym}(\mathbb N)$ to $\prod_{n=1}^\infty S_n$. | |
| Nov 15, 2023 at 9:34 | comment | added | Alessandro Codenotti | There are so called "just-infinite" groups, that is infinite groups all of whose proper quotients are finite (examples are infinite simple groups, $\Bbb Z$, the infinite dihedral group), but I don't know examples of just-infinite groups with all finite groups as quotients. If such examples exist and there's two nonisomorphic ones, this would provide a negative answer | |
| Nov 15, 2023 at 8:51 | comment | added | HenrikRüping | I think there is a Typo and in 2 it should be $\pi:G\to F$. | |
| Nov 15, 2023 at 7:24 | history | asked | Dominic van der Zypen | CC BY-SA 4.0 |