Non finitely presented subgroup

Consider the kernel of the homomorphism from two copies of the free group $F_2 \times F_2$ onto the integers sending every generator to 1. How to see that this subgroup is not finitely presented?

-
Sorry, what's $F_2$? Surely not the field of two elements? – Gerry Myerson Apr 28 '11 at 9:06
Surely the free group of rank 2. – PseudoNeo Apr 28 '11 at 10:30
This looks related: mathoverflow.net/questions/54975/… – mac Apr 28 '11 at 11:14
Ah, I mis-parsed it. It's not, "two copies of (the free group $F_2\times F_2$)," it's that $F_2\times F_2$ is two copies of the free group $F_2$. So, what are the generators of $F_2\times F_2$? – Gerry Myerson Apr 28 '11 at 13:25

What I have done is to calculate an (infinite) presentation of the kernel $K$ in question. Let the two copies of $F_2$ be generated by $\{a,b\}$ and $\{x,y\}$. Then $K$ is generated by $B := ba^{-1}$, $X := xa^{-1}$ and $Y := ya^{-1}$. Using a reasonably straightforward Reidemeister-Schreier calculation, we get the presentation
$\langle\, X, Y, B \mid [B,(YX^{-1})^{X^i}]\: (i \in \mathbb{Z})\, \rangle$
of $K$, which is an HNN-extension of the free group $\langle X,Y \rangle$ with stable letter $B$, where the subgroup generated by $(YX^{-1})^{X^i}$ for $i \in \mathbb{Z}$, which is not finitely generated, is centralized by $B$.
Edited: To show that $K$ is not finitely presentable, it is enough to show that any group $K'$ defined by a presentation using a finite subset of the relators in the above presentation is unequal to $K$. Notice that $K'$ is also an HNN-extension of $\langle X,Y \rangle$ by $B$, but the subgroup of $\langle X,Y \rangle$ centralized by $B$ is finitely generated, and so is a proper subgroup of the subgroup centralized by $B$ in $K$. So $K \ne K'$ and hence $K$ is not finitely presentable.