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Jan 9, 2019 at 18:07 history closed YCor gr.group-theory 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