Can an uncountable group have a countable number of subgroups?

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.

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.

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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.