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.



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> What is the cardinality of the set of subgroups of $F\_2$?
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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 $2^{\aleph\_0}$.



The set of subgroups of $F_2$
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.)


