Can an uncountable group have only a countable number of subgroups?

Please give examples if any exist!

Edit: I want a group having uncountable cardinality but having a countable number of subgroups.

By countable number of subgroups, I mean the collection of all subgroups of a group is countable.


closed as off-topic by Shaun, Brian Borchers, user10354138, Cesareo, TheSimpliFire Dec 9 '18 at 19:28

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  • 5
    $\begingroup$ I'm frankly a bit surprised at the negative reaction to this question. $\endgroup$ – Noah Schweber Dec 9 '18 at 3:00
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    $\begingroup$ Possible duplicate of Countable number of subgroups $\implies $ countable group $\endgroup$ – Carmeister Dec 9 '18 at 10:05
  • $\begingroup$ Then why is it marked as "off-topic" rather than "duplicate"? $\endgroup$ – C Monsour Dec 9 '18 at 20:42
  • $\begingroup$ I'm voting to reopen. If it should be closed please give the correct reason. $\endgroup$ – C Monsour Dec 9 '18 at 20:44
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    $\begingroup$ @CMonsour It was closed because there is no context. It's just a problem statement with no details. The closure reason is correct. While it might be better that it be listed as a duplicate, I don't see that as a reason to reopen. $\endgroup$ – jgon Dec 9 '18 at 21:10

No. Suppose $G$ is an uncountable group. Every element $g$ of $G$ belongs to a countable subgroup of $G$, namely the cyclic subgroup $\langle g\rangle$. Thus $G$ is the union of all of its countable subgroups. Since a countable union of countable sets is countable, $G$ must have uncountably many countable subgroups.

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    $\begingroup$ How can we be sure that the subgroups generated by $g$ and $g'$ (for $g\not= g'$) are not the same? $\endgroup$ – BenjaminH Dec 9 '18 at 8:54
  • $\begingroup$ With a bit of refinement, this argument can be turned into the group being a countable union of finite sets. I forgot if this makes it countable without choice. $\endgroup$ – Tobias Kildetoft Dec 9 '18 at 9:00
  • $\begingroup$ @TobiasKildetoft The axiom of choice is needed to prove that a countable union of two-element sets is countable. (If the union were countable, you could prove the axiom of choice for two-element sets.) $\endgroup$ – bof Dec 9 '18 at 10:17
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    $\begingroup$ @BenjaminH : You can't, for a specific pair $g,g'$. However, both $\langle g\rangle$ and $\langle g'\rangle$ are countable, so there are indeed uncountably many such subgroups (otherwise a countable union of countable sets would exhaust $G$, which is not possible). $\endgroup$ – MPW Dec 9 '18 at 13:23

EDIT: bof's answer is the right one, but the construction below - while completely pointless overkill - is still an example of a useful technique, so I'm leaving this answer up.

No, this cannot happen.

Suppose $G$ is a group and $A$ is a countable subset of $G$. Then the closure of $A$ under the group operations ($*$ and $^{-1}$) of $G$, $\langle A\rangle$, is again countable - this is a good exercise (HINT: the set of finite strings from a countable set is countable).

With this in hand, if $G$ is an uncountable group we can build an uncountable chain of subgroups of $G$, as follows:

  • We will define a countable subgroup $A_\delta$ for every countable ordinal $\delta$. There are uncountably many of these, so if we can do this we'll be done.

  • We let $A_0$ be the trivial subgroup.

  • Having defined $A_\eta$ for every $\eta<\delta$, we let $a$ be some element of $G$ not in $\bigcup_{\eta<\delta}A_\eta$ - which exists, since this is a countable union of countable subgroups, and $G$ is uncountable - and let $A_\delta=\langle (\bigcup_{\eta<\delta}A_\eta)\cup\{a\}\rangle$.

  • It's easy to prove by transfinite induction that $(A_\delta)_{\delta<\omega_1}$ is a strictly increasing chain of countable subgroups of $G$, so we're done.

  • $\begingroup$ Both answers use the axiom of choice. In ZF, can there be an uncountable group with only a countable number of subgroups? $\endgroup$ – bof Dec 9 '18 at 3:43
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    $\begingroup$ @bof This question that I asked and Noah answered a while ago should answer your question $\endgroup$ – Paul Plummer Dec 9 '18 at 4:51

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