# Groups that are not amalgams

It is clear that cyclic groups have the property that they cannot be written in a non-trivial way as an amalgamated free product or as an HNN-extension.

Can someone please provide us examples of torsion-free 2-generated groups having this property? Any comments related to this question are welcome!

[I'm aware of Serre's Theorem 15 in Trees.]

Edit: The term "in a non-trivial way" means the following. For an amalgamated free product $G=H*_C K$ it is required $H\ne C\ne K$. For an HNN-extension $G=HNN(H,A,B,t)$ it is required $A\ne H\ne B$.

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Nilo, if you insist that $A\neq H\neq B$ then you're not just allowing $\mathbb{Z}$, but also ascending HNN-extensions, including Thompson's group F (if I recall correctly). In your comment below my answer, you seemed to want to exclude Thompson's group F. You can't have it both ways. – HJRW Jan 25 '11 at 3:59
You are definitely right. – Nilo Jan 25 '11 at 5:12
So what's the answer? Are you looking for a two-generator torsion-free group that doesn't split non-trivially in the usual sense, or for one that doesn't split non-trivially in your sense? Because if the latter, then $\mathbb{Z}^2$ works. – HJRW Jan 25 '11 at 21:13
Thanks, Henry! I missed that completely. I would be glad to know what is the usual sense... Thanks again. – Nilo Jan 26 '11 at 0:17
I've added an explicit example to my answer below. – HJRW Jan 27 '11 at 16:18

Actually, your first statement is false for the infinite cyclic group:

$\mathbb{Z}=1*_1$

the trivial HNN extension of the trivial group. The point is that, by the ideas described in Trees, a splitting of a group is the same thing as an action of that group on a tree, and there is an obvious action of $\mathbb{Z}$ on the real line, which is a tree.

A group that does not split is said to have Property FA. There are many constructions of such groups, though some criteria involve torsion, so your torsion-free restriction makes the question more interesting.

One source of examples is given by random groups, which can be two-generated. Dahmani, Guirardel and Przytycki showed that random groups have Property FA.

The following is missing some details; the gaps can be filled in from Serre's Trees and from Peter Scott's article The Geometries of 3-manifolds'. I'll fill in some more details if you ask for them.

Here's an explicit example. You probably know that triangle groups

$T(p,q,r)\cong \langle a,b\mid a^p=b^q=(ab)^r=1)\rangle$

have Property FA. This is quite easy to see by thinking about actions on trees. Because they are all of finite order, $a,b$ and $ab$ each have to fix vertices, and the only way this can happen is if they all fix the same vertex. These groups are 2-generator, but clearly not torsion-free.

Now here's a torsion-free example, which is the fundamental group of a certain Seifert-fibred space. For details on Seifert-fibred spaces, I recommend Peter Scott's article The Geometries of 3-manifolds'.

Let $\mathrm{gcd}(p,p')=\mathrm{gcd}(q,q')=\mathrm{gcd}(r,r')=1$. The group in question is

$\Gamma=\langle a,b,z\mid a^p=z^{p'}, b^q=z^{q'}, (ab)^r=z^{r'}, [a,z]=[b,z]=1\rangle~.$

Note that if $\mathrm{gcd}(p',q',r')=1$ then $\Gamma$ is 2-generated; $\Gamma$ is also torsion-free. $\Gamma$ can be written as a non-split central extension

$1\to\langle z\rangle\to \Gamma\to T(p,q,r)\to 1~.$

Suppose that $\Gamma$ acts on a tree $T$. Then, using some standard lemmas about actions on trees and the fact that it is normal, either $Z=\langle z\rangle$ fixes $T$ pointwise (in which case we are done), or $T$ can be taken to be a line which $z$ translates.

In this latter case, translation distance gives a homomorphism $f:\Gamma\to\mathbb{Z}$ with $f(z)$ non-trivial. But the existence of this homomorphism implies that

$p'/p+q'/q=r'/r$

which of course won't be true for most choices of parameters.

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Since I'm not familiar with this paper's lingo, I understood that such groups are very likely to have property (FA). Maybe you can you mention a specific example of a different kind. Thanks. – Nilo Jan 25 '11 at 0:47
What about Thompson's group F? Though that is an infinitely iterated HNN extension. – user641 Jan 25 '11 at 1:51
@Steve: Sounds interesting, but it seems to me that F is an HNN-extension with infinitely generated base group. – Nilo Jan 25 '11 at 2:32
in the previous comment the word ascending before HNN is missing - sorry. – Nilo Jan 25 '11 at 5:09
@Henry: Many many thanks! – Nilo Jan 29 '11 at 15:46