Finitely generated free group has finite rank Initially I was trying to prove that the commutator subgroup of $F_2$ (=free group of rank $2$) is not finitely generated. It seems possible to prove that it is indeed free of infinite rank. To get to a contradiction I have to prove that if a free group is finitely generated then it has finite rank but I can't find a way, despite it seems trivial.
 A: If we would have a free group $F$ with rank $|X|$ where $|X|$ is infinite, then we can see $F$ as the group of all words over $X\cup X^{-1}$ with the usual concatenation (the 'standard' free group). Suppose we would have a finite generating set $\{g_1,\dots,g_k  \}$. Then every $g_i$ is a word $x_{i_1}^{\pm}\dots x_{i_{n_i}}^{\pm}$ which uses only finitely many letters of $X$, so there will always be an element $x \in X$ not appearing in any of the generators. This implies that $x$ is not in the subgroup generated by $\{g_1,\dots,g_k\}$ so no such generating set can exist. 
Alternatively, you could prove that the abelianization of the free group $F$ of rank $|X|$ is the abelian free group of rank $|X|$, i.e. $$\bigoplus_{x\in X} \mathbb{Z} $$
using the universal property. Tensoring with $\mathbb{Q}$ gives you a vector space of dimension $|X|$, which has clearly no finitely generating set (because such a set should contain a basis). 
A: The commutator group is the kernel of the canonical surjection $F_2\to Z^2$ given by abelianization. It is therefore isomorphic to the Cayley graph of $Z^2$ endowed with a standard set of generators. From this, drawing a picture of this Cayley graph in the plane, one sees that it is freely generated by the elements $a^nb^m[a,b](a^nb^m)^{-1}$, corresponding to all squares in the plane. Note that these elements are commutators $=[a^nb^ma(a^nb^m)^{-1},a^nb^mb(a^nb^m)^{-1}]$. Geometrically, if an element belongs to the commutator group, it defines a closed curve in the graph, and this curve decomposes into squares of size 1. One can translate this into an algebraic proof that these elemnts is a free basis.  
A: The commutators $[x,y^n]=xy^nx^{-1}y^{-n}$ for $n=1,2,3,\ldots$
generate freely an infinite
rank subgroup of the free group with generators $x$ and $y$.
