TITLE: Commutator subgroup of rank-2 free group is not finitely generated.
QUESTION [19 upvotes]: I'm having trouble with this exercise:

Let $G$ be the free group generated by $a$ and $b$. Prove that the commutator subgroup $G'$ is not finitely generated.

I found a suggestion that says to prove $G'$ is generated by the collection $\{[x^m,y^n]\mid m,n\in\mathbb{Z}\}$. I don't know how to prove this, and how it helps.
If you could please provide me with some hints... Thanks!

REPLY [6 votes]: The answers given thus far are great, but no one has posted my favorite proof of this result yet.  Consider the homomorphism $\phi\colon F_2\to \mathbb{Z}^2$ defined on the generators by $a\mapsto (1,0)$ and $b\mapsto (0,1)$.  This is just the quotient map $F_2\to F_2/[F_2,F_2]$, but we're going to reinterpret it in a slightly different way.  For any $w\in F_2$,
$$ \phi(w)=\left(\#\{a\text{'s in }w\}-\#\{\bar{a}\text{'s in }w\},\#\{b\text{'s in }w\}-\#\{\bar{b}\text{'s in }w\}\right)$$
Don't think of $\phi$ as the quotient by the commutator subgroup, just think of it as a funny map (you should check that it's a homomorphism) defined on the Cayley graph of $F_2$ as described above.  Let's think about the $G=\ker \phi$ and suppose $G$ were finitely generated $G=\langle w_1,\ldots,w_k\rangle$.  Think about the values $\phi$ takes on as you write out each $w_i$ in terms of $a$'s and $b$'s (or equivalently as you walk from the identity to $w_i$ in the Cayley graph).  $\phi$ will increase, then decrease, then increase, then decrease, and so on until you reach $w_i$ where $\phi(w_i)=(0,0)$.  The point is that in each coordinate of $\mathbb{Z}^2$, there will be some maximum value above which you never passed.  This is true for each generator individually, so there is some big compact set in $\mathbb{Z}^2$ which you never leave when writing an individual generator.
Now write a word in your finite generating set of $G$, $w=w_{i_1}w_{i-2}\cdots w_{i_j}$, then write out that word in $a$'s and $b$'s and think about the values $\phi$ takes along the way.  You will be stuck inside the same compact set in $\mathbb{Z}^2$ because each generator only makes it so far then returns to the origin.
To finish, you just say $\phi([a^n,b^n])\in G$, but to write this out in the Cayley graph of $F_2$, you must pass through vertices where $\phi$ is arbitrarily large.  Thus, the finite set of $w_i$'s did not actually generate $G$.