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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.

No. Suppose $G$ is an uncountable group. Every element $g$ of $G$ belongs to a countable subgroup of $G$, namely the cyclic subgroup $langle grangle$. 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.

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 $G$, $langle Arangle$, 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.