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I know that if a group $G$ has a finite number of subgroups then the group $G$ is finite.

But if a group $G$ has countable number of subgroups then is the group countable?

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  • $\begingroup$ Firstly , it is very difficult to count all the subgroups of a given group $\endgroup$ – user210387 Nov 13 '15 at 15:57
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    $\begingroup$ Are you making any set theoretic assumptions, like axiom of choice? Pretty sure you can look at all the cyclic subgroups, each of which can contain at most countable many, so there are uncountable many such groups. But that argument might be trickier without choice (or false). $\endgroup$ – Paul Plummer Nov 13 '15 at 15:57
  • $\begingroup$ Perhaps surprisingly, the converse is not true: math.stackexchange.com/questions/257175/… $\endgroup$ – lhf Nov 13 '15 at 16:20
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Let $S$ be the set of all subgroups of $G$.

Consider the map $\phi: G \to S$ given $\phi(g) = \langle g \rangle$.

If $\phi$ were injective, then we'd be done.

Unfortunately, $\phi$ is not injective, but fortunately we can control how not injective it is.

Indeed, every $H \in S$ has a finite number of pre-images since every cyclic group has finite number of generators (including infinite cyclic groups). Therefore $$ G = \bigcup_{H \in S} \phi^{-1}(H) $$ is a countable union of finite sets and so is countable.

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  • $\begingroup$ Without choice I am not sure it is necessarily true, I know that countable union of pairs do not have to be countable (this situation would come up in a torsion free group), and I don't see any way to actually choose one or the other. It seems somewhat plausible to construct a group using this idea, if you say all these pairs generate distinct cyclic subgroups, and their unions is not countable. (this is related to the Russel's socks "problem": On the number of Russell’s socks or $2+2+2+. . .= ?$ byHorst Herrlich, Eleftherios Tachtsi. $\endgroup$ – Paul Plummer Nov 13 '15 at 16:55
  • $\begingroup$ @Paul: While I'm not saying that it cannot happen, most sets which cannot be well-ordered will also have an uncountable number of finite subsets. So you need to take extra care in which subsets are allowed to span subgroups and which are not. $\endgroup$ – Asaf Karagila Nov 13 '15 at 19:42
  • $\begingroup$ @lhf i understand that every cyclic group has finite number of generators but why is that $H$ has a finite number of pre-images? is it because every group is a union of cyclic groups? $\endgroup$ – Learnmore Nov 14 '15 at 2:24
  • $\begingroup$ @learnmore, if $H$ is not cyclic, then it has no pre-iimages. The image of $\phi$ is the set of cyclic subgroups of $G$. $\endgroup$ – lhf Nov 14 '15 at 10:16
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    $\begingroup$ @JeremyRickard I agree, I figured I would bring up the point since it is interesting. I decided to ask the question as a separate question here so I don't turn this question into a question it wasn't designed to be. $\endgroup$ – Paul Plummer Nov 16 '15 at 17:24

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