Timeline for Countable group with uncountable number of subgroups $< 2^{\aleph_0}$ [duplicate]
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7 events
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Jan 9, 2019 at 18:07 | history | closed | YCor gr.group-theory Users with the gr.group-theory badge or a synonym can single-handedly close gr.group-theory questions as duplicates and reopen them as needed. | Duplicate of Finitely generated group with $\aleph_0<X_G<2^{\aleph_0}$ normal subgroups? | |
Jun 18, 2015 at 6:38 | vote | accept | Dominic van der Zypen | ||
Jun 17, 2015 at 23:58 | comment | added | user35370 | YCor basically answers in this question, also there is some discussion of whether or not quotients could be uncountable but not $2^{\aleph_0}$. There seems, a priori, that a quotient could have $\aleph_1$ many elements, as pointed out by Emil, although I am not sure if there have been proofs that there are groups like that, (The question is not exactly the same, but YCor points out it works for subgroups) | |
Jun 17, 2015 at 15:17 | answer | added | Emil Jeřábek | timeline score: 26 | |
Jun 17, 2015 at 14:56 | comment | added | Joel David Hamkins | Emil, why not post as an answer? Although the set theorists are accustomed to thinking of the class of countable groups as a Polish space, this might be less familiar to the group theorists, and so this may be a good opportunity to explain this perspective. | |
Jun 17, 2015 at 14:43 | comment | added | Emil Jeřábek | No, the set of subgroups of $G$ is closed as a subset of the Polish space $2^G$. | |
Jun 17, 2015 at 14:22 | history | asked | Dominic van der Zypen | CC BY-SA 3.0 |