TITLE: Non finitely presented subgroup
QUESTION [8 upvotes]: 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?

REPLY [4 votes]: 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.