5
$\begingroup$

I know several constructions leading to finitely generated non-finitely presented groups, using amalgamated products:

Property: Let $A,B$ be two finitely presented groups. Then $A \underset{C}{\ast} B$ is finitely presented iff $C$ is finitely generated.

using HNN extensions:

Property: Let $A$ be a finitely presented group. Then $\underset{C}{\ast} A$ is finitely presented iff $C$ is finitely generated.

or using wreath products (more difficult result):

Property: Let $A,B$ be two finitely presented groups. Then $A \wr B$ is finitely presented iff $A$ is trivial or $B$ is finite.

However, the only application that I know giving a "nice" group, that is a group with a simple description (not using a presentation of course), is the lamplighter group $L_2= \mathbb{Z}_2 \wr \mathbb{Z}$. Do you know other examples?

$\endgroup$
2
  • $\begingroup$ Upon second reading it is clear enough. It was close to midnight ... $\endgroup$ – Jyrki Lahtonen May 10 '14 at 4:01
  • 2
    $\begingroup$ the wreath products are not defined by a presentation, the lamplighter group is not simpler to describe than the others. Btw another example of a infinitely presented f.g. group is $SL_3(F[t])$ where $F$ is any finite field. This is more difficult than the results you mention. Also, by a result of Bieri-Strebel (1978), a (infinite locally finite)-by-$\mathbf{Z}$ group can never be finitely presented. $\endgroup$ – YCor May 11 '14 at 11:29
4
$\begingroup$

For $n \ge 1$, consider the homomorphism of a direct product $F_2^n$ of $n$ copies of the free group of rank $2$ to ${\mathbb Z}$ in which all generators aret mapped to $1$, and let $K_n$ be its kernel.

Then $K_1$ is not finitely generated, and $K_2$ is finitely generated but not finitely presented. They are finitely presented for $n \ge 3$, but I believe that they satisfy interesting homological conditions: I don't remember the details right now!

$\endgroup$
3
  • 2
    $\begingroup$ The k-th homology has infinite rank for n=k, this is essentially due to Stallings who proved it for k=3. $\endgroup$ – Moishe Kohan May 10 '14 at 2:58
  • $\begingroup$ @DerekHolt: Interesting. Do you have a reference? $\endgroup$ – Seirios Jun 1 '14 at 5:29
  • 1
    $\begingroup$ for completeness, a reference: morse theory and finiteness properties of groups by brady $\endgroup$ – ttt Jul 10 '17 at 5:23
3
$\begingroup$

I have edited this answer to add in a reference to a result of Bieri which essentially proves that the groups in this answer are not finitely presentable. This result is quite general and may be applied in lots of different settings. If I had more time then I would completely rewrite the answer to focus on this result and make Rips' construction an application of it, but alas I do not have the time so instead am writing this preamble...


Examples of finitely generated, non-finitely presentable groups can be found via Rips' construction (E. Rips, Subgroups of small Cancellation Groups, Bull. Lond. Math. Soc., 14 (1982), 45–47 (doi)):

Let $Q$ be the group defined by your favourite presentation. Then there exists a hyperbolic* group $G$ and a short exact sequence $1\rightarrow N\rightarrow G\rightarrow Q\rightarrow 1$ where $N=\langle a, b\rangle$ is finitely generated. You care about this sequence because if gives you subgroups of hyperbolic groups with pathological properties (e.g. take $Q$ to have insoluble word problem, then $N$ has insoluble membership problem, while as the triviality problem is insoluble for groups (so we do not know if $Q$ is trivial) the generation problem is insoluble for hyperbolic groups (we do not know if $G=\langle a, b\rangle=N$)). However:

Fact. $Q$ is infinite if and only if $N$ is finitely presentable.

So the bad subgroups of hyperbolic groups which we found are all non-finitely presentable.

The proof of the fact is non-trivial, and follows from three facts:

  1. In general, if a group $H$ has cohomological dimension two and $K\lhd H$ is a finitely presented normal subgroup of $H$ then $K$ is free or of finite index in $H$. (This is Corollary 8.6 in Bieri, Robert. "Homological dimension of discrete groups." Queen Mary College Mathematics Notes (1976).)

  2. The group $G$ in Rips' construction has cohomological dimension $2$ (as it is torsion-free small cancellation).

  3. The subgroup $N$ is normal but not free.

*In fact $C'(\lambda)$ for your favourite $\lambda\leq1/6$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.