Timeline for Finitely generated group with $\aleph_0<X_G<2^{\aleph_0}$ normal subgroups?
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| Jan 12, 2015 at 19:26 | history | edited | YCor | CC BY-SA 3.0 |
Added second paragraph
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| Mar 19, 2014 at 18:44 | comment | added | Emil Jeřábek | That’s right, see e.g. mathunion.org/ICM/ICM1994.1/Main/icm1994.1.0269.0276.ocr.pdf . The original result is due to Burgess ams.org/journals/proc/1978-069-02/S0002-9939-1978-0476524-6 . | |
| Mar 19, 2014 at 18:03 | comment | added | YCor | @Emil: do you say that for an analytic equivalence relation (say on the Cantor set), it holds in ZFC that the number of classes, if uncountable, is always $\aleph_1$ or $2^{\aleph_0}$? (e.g. cannot be $\aleph_2$ if $\aleph_2<2^{\aleph_0}$) what's a reference for this? | |
| Mar 19, 2014 at 13:46 | comment | added | Emil Jeřábek | However, this argument does not apply to countable infinitely generated groups, as then the classes may have cardinality $2^\omega$. The equivalence relation is analytic in this case, so a priori the number of equivalence classes can be countable, $\aleph_1$, or $2^\omega$. I wonder whether $\aleph_1$ is possible (assuming $\neg$CH, obviously). | |
| Mar 18, 2014 at 22:21 | comment | added | Emil Jeřábek | Ah, yes, you are right, I missed that. | |
| Mar 18, 2014 at 21:07 | comment | added | YCor | @Emil: it's even simpler, since this relation has countable classes. | |
| Mar 18, 2014 at 18:25 | vote | accept | CommunityBot | moved from User.Id=35370 by developer User.Id=69903 | |
| Mar 18, 2014 at 18:20 | comment | added | Emil Jeřábek | I believe the same also holds for the second question. For $G$ finitely generated, $N\sim M$ is a Borel equivalence relation on a Polish space, and as such it has countably many or $2^\omega$ equivalence classes. | |
| Mar 18, 2014 at 17:54 | history | answered | YCor | CC BY-SA 3.0 |