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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?

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    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 $F_2\times F_2$?2011-04-28

2 Answers 2

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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.

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    QPrototank This could be a nice exercise. AS a reference, see Baumslag's book (Topics in Combinatorial Group Theory) Theorem 12 on Page 52.2018-10-01
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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.

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    Thanks - it hadn't occurred to me to use MathJax.2018-04-27