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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 $2^{\aleph_0}$.

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    $\begingroup$ There are uncountably many non-isomorphic $2$-generator groups. The kernels of surjections from $F_2$ to these groups give uncountably many distinct subgroups of $F_2$. $\endgroup$
    – Andy Putman
    Commented Jul 13, 2016 at 16:33
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    $\begingroup$ To sharpen the statement: the cardinality is the continuum $2^{\aleph_0}$. Source: books.google.com/… $\endgroup$
    – Todd Trimble
    Commented Jul 13, 2016 at 17:14
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    $\begingroup$ I think if you look at the space of infinite index subgroups as a subset of $2^{F_2}$ with the product topology, you will find that it is closed and contains no isolated points and hence is a Cantor set. $\endgroup$
    – Benjamin Steinberg
    Commented Jul 13, 2016 at 17:36
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    $\begingroup$ Actually the set of fully charateristic subgroups (= stable under endomorphisms) itself has continuum cardinal. Since such subgroups parameterize varieties of groups, they are quite important. $\endgroup$
    – YCor
    Commented Jul 13, 2016 at 19:56
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    $\begingroup$ @BenjaminSteinberg indeed, but if the OP didn't know that there are uncountably many subgroups, he might also not know why it has no isolated point (this is based on the LERF property, the Marshal Hall theorem, which is well-known to specialists but far from trivial). $\endgroup$
    – YCor
    Commented Jul 13, 2016 at 23:10

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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.)

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  • $\begingroup$ There are explicit 2-generated groups of $3 \times 3$-matrices over $(\mathbb Z/p\mathbb Z) [t,t^{-1}]$ whose center is isomorphic to $\oplus_{i=1}^{\infty} (\mathbb Z/p\mathbb Z)$. Thus, these groups (and hence $F_2$) have $2^{\aleph_0}$ normal subgroups. $\endgroup$
    – Andreas Thom
    Commented Jul 14, 2016 at 8:16

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