Is it consistent that there is a countable group $G$ such that the cardinality of the set of subgroups of $G$ is uncountable, but strictly less than $2^{\aleph\_0}$?



Countable group with uncountable number of subgroups $< 2^{\aleph_0}$
Subsets $H\subseteq G$ can be identified with their characteristic functions $\chi\_H\colon G\to\{0,1\}$, which we can view as elements of the Cantor space $2^G$.


In this perspective, subgroups of $G$ form a closed subset of $2^G$: if $H\subseteq G$ is not a subgroup, then $1\notin H$, or $ab^{-1}\notin H$ for some $a,b\in H$, hence
$$\{f\in2^G:f(1)=0\}$$
or
$$\{f\in2^G:f(a)=f(b)=1,f(ab^{-1})=0\}$$
is a basic open neighbourhood of $\chi\_H$ that excludes the characteristic functions of all subgroups.


Thus, $G$ can only have countably many or $2^\omega$ subgroups by the Cantor–Bendixson theorem.


There is nothing special about groups, the same argument applies to substructures of any countable algebraic structure $G$.


