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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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Sorry, what's $F_2$? Surely not the field of two elements? –  Gerry Myerson Apr 28 '11 at 9:06
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Surely the free group of rank 2. –  PseudoNeo Apr 28 '11 at 10:30
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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
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1 Answer 1

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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"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." This is not obvious to me. –  Grumpy Parsnip Apr 29 '11 at 13:17
    
Ths is basically a lemma of B. Neumann: If a group is finitely presented, then given any presentation with infinitely many relators one can remove all but finitely many relators to obtain a finite presentation of the same group. –  Mustafa Gokhan Benli Apr 21 '12 at 12:32
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