TITLE: The set of subgroups of $F_2$
QUESTION [10 upvotes]: This question came up in our algebraic topology class and our Professor didn't know the answer. I also couldn't find an answer so far.

What is the cardinality of the set of subgroups of $F_2$?

Here $F_2 = \mathbb Z * \mathbb Z$ denotes the free group on two generators. The cardinality of the set of subgroups is clearly bounded below by $\aleph_0$ (as $F_2$ contains subgroups of all countable ranks) and above by $2^{\aleph_0}$.

REPLY [21 votes]: It is clear that there are $2^{\aleph_0}$ subgroups of the free group $F_\infty$ on countably many generators (because each subset of a free generating set generates a different subgroup). In addition, it is well known that $F_2$ contains a subgroup that is not finitely generated. Since all subgroups of a free group are free, this subgroup is isomorphic to $F_\infty$. Then every subgroup of this $F_\infty$ is also a subgroup of $F_2$, which means that $F_2$ has $2^{\aleph_0}$ subgroups. (However, this simple argument does not prove Andy Putman's stronger statement that there are $2^{\aleph_0}$ normal subgroups.)