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Inspired by this question, although I don't think it was the OP's intention, hence this separate question: Is there a group $G$ with countably many subgroups, but is a not a countable group itself in $\mathrm{ZF}$?

In $\mathrm{ZFC}$ we can look at the cyclic subgroups of $G$ and "estimate" the number of elements in the group, to conclude that $G$ is countable. But this ends up not going through in $\mathrm{ZF}$ since a countable union of finite sets does not have to be countable, in particular it is known that a countable union of two-element sets does not have to be countable.

So a possible way to construct such a uncountable group (although I am not saying this is a good way to go, I have no idea) is start with a collection $\{ A_i \mid i \in \mathbb{N} \}$ where $A_i$ are pairs, whose union is not a countable set and note that every torsion-free cyclic group has two natural generators, so conceivably there could be a torsion-free group with $A_i$ the natural generating set for a cyclic groups ("$1,-1$" but we could not actually define such a function all $A_i$ without the axiom of choice). Then the constructions would have to make sure there are only countable many subgroups (this seems difficult and would take a lot of care)

An interesting paper "On the number of Russell's socks or $2+2+2+\cdots = ?$" by Horst Herrlich, Eleftherios Tachtsis discusses some of the ideas around countable unions of pairs.

  • A countable union of finite sets is countable. In fact, a countable union of countable sets is countable. – J126 Nov 16 '15 at 17:31
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    @JoeJohnson126: only with the axiom of choice. – mercio Nov 16 '15 at 17:32
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    @JoeJohnson126 In ZFC that is true, but you can not arrive to that conclusion in ZF, as I said in the post. –  Nov 16 '15 at 17:32
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    Sorry about that. My brain added a "C" at the end of "ZF". – J126 Nov 16 '15 at 17:36
  • With the axiom of choice, the easiest proof (that I know) shows that an uncountable group has uncountably many cyclic subgroups. So a related question, that also seems interesting, but might be easier to think about, is whether, without the axiom of choice, an uncountable group must have uncountably many cyclic subgroups. – Jeremy Rickard Nov 16 '15 at 18:13
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    If you had the described injection $S_{\mathbb N}\hookrightarrow S_A$, wouldn't that give you a sock-choosing function? Choose an element of $A_0$; this entails a choice of an element of each $A_i$ by considering the image of the permutation that swaps $0$ with $i$. – hmakholm left over Monica Jan 06 '16 at 19:58
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    And there certainly is an injection $f:X\hookrightarrow S_X$ for every set $X$, without choice: If $X$ is empty then this is trivial; otherwise choose a fixed $x_0\in X$ and let $f(x)$ be the transposition $(x;x_0)$. – hmakholm left over Monica Jan 06 '16 at 20:08
  • @HenningMakholm Ah yes, you are right on all counts. I seemed to have missed that you would have to define these permutations on the elements, and just thought of pairs when I was writing this. (I removed the material from the post) –  Jan 06 '16 at 21:04
  • @PaulPlummer Does my answer seem to work? (Looking at it now, I believe it does.) – Noah Schweber Jan 22 '16 at 22:35
  • @NoahSchweber I apologizes for taking so long, I was meaning to crack open Jech to check/learn the details (my set theory knowledge falls short of most of the answer making complete sense), but it doesn't look like that will be happening anytime soon... The group theory seems to work, and I assume you know what you are doing set theory wise, seems like a fairly standard way to get these sort of choice defying results. I will accept it. Thank you for the answer! –  Jan 23 '16 at 00:50
  • @darij: I admit that I find your edit comment a bit confusing. – Asaf Karagila Nov 27 '19 at 07:41
  • I was just quipping at an alternative (mis)interpretation of the unhyphenated "two element". – darij grinberg Nov 27 '19 at 15:52

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I believe the answer is yes, as follows:

Start with a model of ZF+atoms, $M$, with a set of atoms $A$ which forms a group isomorphic to $\mathbb{D}/\mathbb{Z}$, where $\mathbb{D}$ is the set of dyadic fractions: $\mathbb{D}=\{{p\over 2^k}: p, k\in\mathbb{Z}\}$. Let $G$ be the group of automorphisms of $A$, and consider the symmetric submodel $N$ of $M$ corresponding to the filter of finite supports on $G$. Then in $N$, $A$ is no longer countable, since there are nontrivial automorphisms of $\mathbb{D}/\mathbb{Z}$ fixing arbitrary finite sets; but the only subgroups of $\mathbb{D}/\mathbb{Z}$, other than the whole thing, are those of the form $$\{x: 2^kx=0\}$$ for some fixed $k\in\mathbb{N}$. This provides an explicit bijection - in the original universe, $M$ - between the subgroups of $A$ and $\omega$. Now, passing to $N$, we get no additional subgroups of $A$, and the map described above is symmetric; so in $N$, $A$ has only countably many subgroups.

Meanwhile, the statements "$A$ is uncountable" and "$A$ has countably many finitely generated subgroups" are each "bounded," so we may apply the Jech-Sochor theorem to push this construction into the ZF-setting.

Noah Schweber
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