# Can an uncountable group have a countable number of subgroups? [closed]

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, TheSimpliFireDec 9 '18 at 19:28

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• I'm frankly a bit surprised at the negative reaction to this question. – Noah Schweber Dec 9 '18 at 3:00
• Possible duplicate of Countable number of subgroups $\implies$ countable group – Carmeister Dec 9 '18 at 10:05
• Then why is it marked as "off-topic" rather than "duplicate"? – C Monsour Dec 9 '18 at 20:42
• I'm voting to reopen. If it should be closed please give the correct reason. – C Monsour Dec 9 '18 at 20:44
• @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. – 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.

• How can we be sure that the subgroups generated by $g$ and $g'$ (for $g\not= g'$) are not the same? – BenjaminH Dec 9 '18 at 8:54
• 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. – Tobias Kildetoft Dec 9 '18 at 9:00
• @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.) – bof Dec 9 '18 at 10:17
• @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). – 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.

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