Is it consistent that there is a countable group $G$ such that the cardinality of the set of subgroups of $G$ is uncountable, but strictly less than $2^{\aleph_0}$?
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asked Jun 17, 2015 at 14:22
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9$\begingroup$ No, the set of subgroups of $G$ is closed as a subset of the Polish space $2^G$. $\endgroup$– Emil JeřábekCommented Jun 17, 2015 at 14:43
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5$\begingroup$ 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. $\endgroup$– Joel David HamkinsCommented Jun 17, 2015 at 14:56
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2$\begingroup$ 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) $\endgroup$– user35370Commented Jun 17, 2015 at 23:58
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Subsets $H\subseteq G$ can be identified with their characteristic functions $\chi_H\colon G\to\{0,1\}$, which we can view as elements of the Cantor space $2^G$.
In this perspective, subgroups of $G$ form a closed subset of $2^G$: if $H\subseteq G$ is not a subgroup, then $1\notin H$, or $ab^{-1}\notin H$ for some $a,b\in H$, hence $$\{f\in2^G:f(1)=0\}$$ or $$\{f\in2^G:f(a)=f(b)=1,f(ab^{-1})=0\}$$ is a basic open neighbourhood of $\chi_H$ that excludes the characteristic functions of all subgroups.
Thus, $G$ can only have countably many or $2^\omega$ subgroups by the Cantor–Bendixson theorem.
There is nothing special about groups, the same argument applies to substructures of any countable algebraic structure $G$.
answered Jun 17, 2015 at 15:17
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1$\begingroup$ There is an exercise attributed to Burris and Kwatinetz at the end of Chapter 1 in the book "Algebras, Lattices, Varieties Volume I" which mentions (for countable algebras of countable type) similar results on the size of the automorphism group and endomorphism monoid and congruence lattice as well. $\endgroup$ Commented Jun 18, 2015 at 1:39
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2$\begingroup$ I think that questions of the form "Does $|X|\ge 2^{\aleph_0}$, i.e., is there a 1-1 map from the Cantor space into $X$?" should always be expanded to "Does $X$ contain a perfect set in some natural T2 topology", or equivalently "Is there a CONTINUOUS 1-1 map from the Cantor space into $X$?". It turns out that a ZFC-answer of "yes" to the first question almost always (in particular: here) is proved by showing that even the second question has a positive answer. The perfect set answer is more interesting because it shows a structural result about $X$, more than a mere cardinality estimate. $\endgroup$ Commented Jun 18, 2015 at 11:54
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$\begingroup$ @Goldstern: Indeed. Though I gather that in case of a simple question like here, the only nonobvious part is to realize it is connected to topology in the first place; once you know that, you already have the answer. $\endgroup$ Commented Jun 18, 2015 at 12:50