TITLE: free group of finite rank can contain free groups of infinite rank as a subgroup
QUESTION [7 upvotes]: how a free group of rank greater than or equal to 3 contains every free group of countable ranks as a sugroup?

REPLY [3 votes]: There is a very nice theorem which states:
Thm: If $F$ is a free group on $n$ generators and $H$ is a normal subgroup of index $j$ then if both $n$ and $j$ are finite $H$ is a free group on $j(n-1)+1$ generators. If $n$ is infinite and $j$ is finite then $H$ is a free group on infinitely many generators. Finally, if $j$ is infinite, then $H$ may be finitely or infinitely generated; however, if $H$ contains a normal subgroup $N$ of $F$, $N\neq 1$, then $H$ is a free group on infinitely many generators.
This is theorem 2.10 in Magnus, Karrass and Solitar "Combinatorial Group Theory" (and has a name...Schreier's formula?)
Once you have this theorem you can prove, for example, that the commutator subgroup is a free group on infinitely many generators (it contains a normal subgroup and is itself characteristic so you can apply the last line of the theorem). But doing it the way Benjamin Steinberg did it is much neater. I just thought it would be useful to mention this formula!