countable group, uncountably many distinct subgroup? I need to know whether the following statement is true or false?
Every countable group $G$ has only countably many distinct subgroups.
I have not gotten any counter example to disprove the statement but an vague idea to disprove like: if it has uncountably many distinct subgroup then It must have uncountable number of element?
 A: There is another one: $F_\infty$, the free group on countably many letters $x_1,x_2, \ldots$.
It is countable because each of its elements is a finite string of symbols from a countable alphabet. At the same time, for every subset $A\subset \mathbb{N}$, there is a subgroup $H_A$ generated by the set $\{x_i | i \in A\}$. These subgroups are distinct, so we get uncountably many of them.
Also, $F_\infty$ embeds into $F_2$, so we can get these uncountably many subgroups within a finitely presented group!
A: Let $(\mathbb{Q},+)$ be the group of the rational numbers under addition.  For any set $A$ of primes, let $G_A$ be the set of all rationals $a/b$ (in lowest terms) such that every prime factor of the denominator $b$ is in $A$.  It is clear that $G_A$ is a subgroup of $\mathbb{Q}$, and that $G_A = G_{A'}$ iff $A = A'$.  Since there are uncountably many sets of primes, this produces uncountably many distinct subgroups of the countable group $\mathbb{Q}$.
A: One example is the group consisting of all finite subsets of $\mathbb N$, with the group operation being symmetric difference. The group is countably infinite, but for each finite or infinite $A\subseteq \mathbb N$ there's a subgroup consisting of the finite subsets of $A$.
