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?
 A: This is actually a standard example of a finitely generated but not finitely presented group. I seem to remember that there is a homological proof. I have been trying to think of a reasonably straightforward group-theoretical proof, but I am running out of time.
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.
A: Euler characteristics of classifying spaces give a way to see this.  The Euler characteristic of the free group $F_2$ is -1, because the figure of 8 
has two 1-cells and one 0-cell.  Euler characteristics multiply for direct products, so the Euler characteristic of $F_2 \times F_2$ is 1.  The Euler characteristic of the infinite cyclic group is 0 (= the Euler characteristic of the circle). 
In general, if $N$ is a normal subgroup of $G$ with quotient $Q=G/N$, there is a product formula: the Euler characteristic of $G$ is the product of those of $N$ and $Q$, whenever all three Euler characteristics are defined.  
Going back to our example, since there is no solution to $1=0.x$, we see that the kernel cannot have an Euler characteristic.  We know that the kernel is finitely generated, and we know that it has a 2-dimensional classifying space, so the only way it can fail to have an Euler characteristic is if the classifying space needs infinitely many 2-cells, or equivalently if the kernel is not finitely presented.  
