# An explicit example of a finitely presented group containing a subgroup isomorphic to $(\mathbb Q,+)$.

A theorem (I do unfortunately not remember to whom it is due) states that there exists a finitely presented group containing a subgroup isomorphic to the additive group of rational numbers. Can somebody give an explicit construction?

• This is not an answer, but a characterization of subgroups of finitely presentable groups as those groups with consistently recursively enumerable word problem is given by Sacerdote: plms.oxfordjournals.org/cgi/reprint/s3-35/2/193.pdf – S. Carnahan May 3 '10 at 18:38
• I was asked this question by Martin Bridson this March. It seems to be an 'open problem'. – HJRW May 4 '10 at 1:47
• I learned this question about 15 years ago from Pierre de la Harpe. – Roland Bacher May 4 '10 at 9:07
• The theorem, as mentioned in the answers, is Higman's embedding theorem: en.wikipedia.org/wiki/Higman%27s_embedding_theorem. I talked about this question with Laurent Bartholdi. He said that noone has ever constructed an explicit finitely presented group with an explicit embedding of $\mathbb Q$. I seem to remember to have read such a statement also in de la Harpes' book on geometric group theory, but I can't find it right now, so perhaps my memory serves me badly. – Łukasz Grabowski Nov 5 '10 at 12:39
• Oh, I found it :-) III.A.17 "Research problem on embedding of $\mathbb Q$". It starts "Find a <<natural and explicit>> embedding of $\mathbb Q$ in a finitely generated group". – Łukasz Grabowski Nov 5 '10 at 17:37

## 5 Answers

Francesco Matucci, James Hyde and I have just posted an arXiv preprint with a solution to this problem. We prove that $$\mathbb{Q}$$ embeds in the group $$\overline{T}$$ of piecewise-linear homeomorphisms of the real line obtained by lifting Thompson's group $$T$$ through the covering map from the line to the circle. That is, $$\overline{T}$$ consists of all piecewise-linear homeomorphisms $$f$$ of the real line that satisfy the following conditions:

1. Each linear segment of $$f$$ has the form $$f(x) = 2^a x + \dfrac{b}{2^c}$$ for some $$a,b,c\in\mathbb{Z}$$.

2. Each breakpoint of $$f$$ has dyadic rational coordinates.

3. $$f(x+1)=f(x)+1$$ for all $$x\in\mathbb{R}$$.

We also prove that this group $$\overline{T}$$ has a presentation with two generators and four relations.

It follows from this result together with a theorem of Brin that $$\mathbb{Q}$$ embeds in the automorphism group of Thompson's group $$F$$, which is also finitely presented. Similarly, $$\mathbb{Q}$$ embeds in the braided Thompson group $$BV$$ introduced by Brin and Dehornoy, which again is known to be finitely presented.

Our preprint also proves that $$\mathbb{Q}$$ embeds into a finitely presented simple group which we denote $$T\mathcal{A}$$. (None of the other groups listed above are simple.) This is a certain group of homeomorphisms of the circle that are "nearly piecewise-linear" in the sense that they have infinitely many linear pieces that accumulate at finitely many points. We prove that this group $$T\mathcal{A}$$ is two-generated and has type $$\mathrm{F}_\infty$$, and we indicate how an explicit finite presentation of $$T\mathcal{A}$$ could be derived.

• Ten years later! Marvellous. – David Roberts May 6 '20 at 11:29
• Actually the question was on the air even before. The only "natural" examples I knew of finitely generated groups containing $\mathbf{Q}$ were (up to trivial modifications) solvable groups of the form $V\rtimes\Gamma$ constructed by Ph. Hall in the 50s, with $\Gamma=\mathbf{Z}\wr\mathbf{Z}$ and $V$ a simple $\mathbf{Z}\Gamma$-module, isomorphic as group to $\mathbf{Q}^{(\infty)}$. – YCor May 6 '20 at 15:21

This is an extended comment, to make precise the theorem to which Roland was making reference.

It is known [Ould Houcine, Abderezak. Embeddings in finitely presented groups which preserve the center. J. Algebra 307 (2007), no. 1, 1--23. MR2278040 (2007i:20043)] that there is a finitely presented group which has $$\mathbb Q$$ as center, which is a very nice result! Indeed, the paper shows that (i) every countable group $$G$$ embeds into a finitely generated group $$K$$ such that $$Z(G)=Z(K)$$ and (ii) Every finitely generated recursively presented group $$G$$ embeds into a finitely presented group $$K$$ such that $$Z(G)=Z(K)$$.

I had originally started the answer with the following:

The statement to which you are referring is a consequence of G. Higman's theorem that states that every group having a recursive presentation can be embedded in a finitely presented group. See [Higman, G. Subgroups of finitely presented groups. Proc. Roy. Soc. Ser. A 262 1961 455--475. MR0130286 (24 #A152)]

Since $$\mathbb Q$$ is plainly recursively presentable, there is a group like the one you want... You can follow the construction given by Rotman in the last chapter of his Introduction to the Theory of Groups to obtain a presentation---the end result is not going to be pretty, though... (It is the proof of this result that makes use of the unsettling folded plates that come with the book)

but then retracted it because Rotman actually states (and proves) Higman's theorem for finitely generated recursively presented groups, and $$\mathbb Q$$ is not finitely generated!

Later, though, Jack Schmidt observed than in fact Higman does deal with the countably generated case in the original paper (please refer to his comments below for details) so the retracted text should be unretracted.

• I don't think you had to torch the whole thing -- I think the 2007 article you linked to claims to answer the question affirmatively, as you said. – Cam McLeman May 3 '10 at 18:23
• @Cam: yup, I was just editing, but wanted to kill the wrong part as soon as possible. – Mariano Suárez-Álvarez May 3 '10 at 18:30
• I am delighted at the civility and integrity of this exchange of comments: thanks, Cam and Mariano. – Georges Elencwajg May 3 '10 at 18:42
• After defining recursively presented, Higman refers to his earlier embedding, ams.org/mathscinet-getitem?mr=32641 (HNN49), to embed a recursively presented group (which is countable) into a two-generator group (an iterated HNN extension), and furthermore one can observe that this embedding is effective and that the 2-generator group is recursively presented. Corollary Any recursively presented group can be effectively embedded in a finitely presented group. – Jack Schmidt May 3 '10 at 19:01
• @Henry, the point was to make precise the theorem to which Roland was making reference. – Mariano Suárez-Álvarez May 3 '10 at 19:56

Every recursively presented (even infinitely generated) group can be effectively embedded into a finitely generated recursively presented group either by using HNN extensions (as in Higman-Neumann-Neumann original paper) or by using small cancelation quotients of the free group. Every finitely generated recursively presented group can be effectively embedded into a finitely presented group by the Higman embedding theorem. The finite presentation is explicitly constructed using any Turing machine recognizing the set of relations of the finitely generated group. This means that one can explicitly write down a finite presentation of a group containing $$\mathbb Q$$. The number of generators and the number of relations will depend (linearly) on the number of commands in the Turing machine recognizing the defining relations of $$\mathbb Q$$. The real question is to find a "natural" finitely presented group containing $$\mathbb Q$$. That is not known so far. There are finitely presented groups containing close relatives of $$\mathbb Q$$. The Baumslag-Solitar group $$BS(1,d)$$ contains the group of $$d$$-adic rationals. And the R.Thompson group $$V$$ contains the group $${\mathbb Q}/{\mathbb Z}$$.

Update 1. The first step can be simplified for $$\mathbb Q$$. For every $$n\ge 2$$ take the group $$G_n=BS(1,n)=\langle a_n,b_n\mid b_n^{-1}a_nb_n=a_n^n\rangle$$. The direct product $$\Pi G_n$$ contains $$\mathbb Q$$. Add two generators $$t,s$$ to the presentation of the direct product and all relations $$t^{-1}a_it=a_{i+1}$$, $$s^{-1}b_is=b_{i+1}$$ for all $$i\ge 1$$. That is a finitely generated (by $$a_1,b_1, t,s$$) recursively presented group containing $$\mathbb Q$$. The presentation of that group can be easily recognized by a Turing machine. Then the Higman construction gives a presentation of a finitely presented group containing $$\mathbb Q$$. The presentation will contain something like a 100 generators and 100 relations (I did not compute exact numbers).

Update 2. In Valiev, M. K. Universal group with twenty-one defining relations. Discrete Math. 17 (1977), no. 2, 207–213, Valiev constructed an explicit presentation with 21 defining relations of a group containing all finitely presented groups, hence containing $$\mathbb Q$$ (earlier a 26-relator example was constructed by Boone and Collins). The difference with the example in Update 1 is that it is hard to describe an embedding of $$\mathbb Q$$ in that group. That embedding is defined by the Turing machine describing a presentation of $$\mathbb Q$$.

Here's an answer after reading Jim's answer. Indeed, given the previous knowledge on finite order elements in Thompson's group $$T$$ it seems we were very close to the answer.

First, a general fact:

Proposition. Let $$G$$ be a torsion-free group with central cyclic subgroup $$Z$$. Suppose that $$G/Z$$ is isomorphic to $$\mathbf{Q}/\mathbf{Z}$$. Then $$G$$ is isomorphic to $$\mathbf{Q}$$.

Proof, since $$G$$ is central-by-(locally cyclic), it is abelian. It necessarily has $$\mathbf{Q}$$-rank 1, so has an injective homomorphism $$i$$ into $$\mathbf{Q}$$. This induces an injection $$\bar{i}$$ from $$G/Z\simeq\mathbf{Q}/\mathbf{Z}$$ into $$\mathbf{Q}/i(Z)$$, which is itself isomorphic to $$\mathbf{Q}/\mathbf{Z}$$. Since every injective endomorphism of $$\mathbf{Q}/\mathbf{Z}$$ is an automorphism, we deduce that $$\bar{i}$$ is an isomorphism, and hence so is $$i$$.

Now let us consider finite order elements. First recall that in $$G=\mathrm{Homeo}^+(\mathbf{R}/\mathbf{Z})$$, every element of finite order is conjugate to a (finite order) rotation (=translation $$r\mapsto r+\theta$$); the rotation number $$\theta$$ is a conjugacy invariant. Thus there are $$\varphi(n)$$ conjugacy classes of elements of order $$n$$, and single conjugacy class of cyclic subgroups of order $$n$$.

I find in this 2012 unpublished paper of A. Fossas "On the number of conjugacy classes of torsion elements on Thompson's group T" that the same counting holds in $$T$$ (she asserted it for elements, but the simpler assertion for cyclic subgroups immediately follows). Let me emphasize this:

Theorem. In Thompson's group $$T$$, there is exactly one conjugacy class of cyclic subgroup of order $$n$$, for every integer $$n\ge 1$$.

An immediate corollary is that if $$m|n$$, every cyclic subgroup of order $$m$$ is contained in a cyclic subgroup of order $$n$$. In this way, we can produce an ascending sequence $$(H_n)$$ of cyclic subgroups with $$H_n$$ of order $$n!$$. Hence its union is isomorphic to $$\mathbf{Q}/\mathbf{Z}$$. Now the inverse image $$\tilde{T}$$ of $$T$$ in the group of $$\mathbf{Z}$$-equivariant self-homeomorphisms of $$\mathbf{R}$$ is a torsion-free central extension of $$T$$ and the proposition applies:

Corollary. The finitely presented group $$\tilde{T}$$ contains an isomorphic copy of $$\mathbf{Q}$$.

Moreover when we produce the embedding $$H_{n}\subset H_{n+1}$$ with $$n\ge 1$$, we readily see that there exists an element in $$T$$ centralizing $$H_n$$ but not normalizing $$H_{n+1}$$. Conjugating by this element, we see that we have at least two choices for $$H_{n+1}$$. Hence we obtain $$2^{\aleph_0}$$ choices of such sequences, and hence $$2^{\aleph_0}$$ subgroups of $$\tilde{T}$$ isomorphic to $$\mathbf{Q}/\mathbf{Z}$$, and, pulling back, $$2^{\aleph_0}$$ subgroups of $$T$$ isomorphic to $$\mathbf{Q}$$ (as obtained by Belk, Hyde and Matucci). In particular, these copies are not all conjugate.

In addition let me mention that it is easy to find one copy of $$\mathbf{Q}$$ with solvable membership problem, just being careful when choosing roots (of course at most countably many of those copies have solvable membership problem).

Note: Fossas' short paper on uniqueness up to conjugation is unpublished. I have double checked (in my own way) and am convinced it's true.

I'm far from an overview of the past knowledge on classifying finite order elements in Thompson's group $$T$$. Let me mention this 2008 paper by Liousse "Rotation numbers in Thompson-Stein groups and applications", which already includes (see Cor. 1) the fact that $$T$$ admits elements of all possible orders.

All this retrospectively looks very easy. But I have to confess that I hadn't even realized so far (before Jim's answer) that $$T$$ has a element of order $$3$$...!

• It is quite obvious that $T$ has elements of order 3, since $T$ contains $\mathrm{PSL}(2,\mathbb{Z})$ (this follows from Thurston's realization of $T$ as a group of piecewise $\mathrm{PSL}(2,\mathbb{Z}$ homeomorphisms). The first non-trivial case is order 5... – Michele Triestino May 14 '20 at 8:05
• @MicheleTriestino it's obvious modulo having this observation in mind. Once you have an explicit element of order 3, it's obvious that it exists, and actually the whole recpe to produce elements of arbitrary odd order is very elementary... Anyway this last sentence was a bit meta-mathematical: I meant that I didn't have the right picture in mind. Nevertheless your observation might give the oldest reference for this observation that there are nontrivial elements of odd order in $T$. – YCor May 14 '20 at 8:20
• I have just realized that the proof of Proposition III.2.1 of Ghys-Sergiescu eudml.org/doc/140083 contains a construction of torsion elements of every order in $T$ – Michele Triestino Jun 25 '20 at 7:47
• @MicheleTriestino Thanks! that proposition states that for every rational mod $\mathbf{Z}$ is rotation number of some element of $T$, and the proof starts, for every $n$, exhibiting an element $g_n$ of order $n+2$ (explicit: $0\mapsto 1-2^{-n-1}$, $1/2\mapsto 1$, $1-2^{-n-1}\mapsto 1-2^{-n}$, affine in between). They credit anonymous handwritten notes for $g_n$. – YCor Jun 25 '20 at 7:58

I believe this is unknown, but mostly for metamathematical reasons.

The major result in this area is Higman's embedding theorem that a finitely generated and recursively presented group is embeddable in a finitely presented group. While $\mathbb{Q}$ is certainly recursively presented, it is not finitely generated, so this doesn't apply.

My main reason for thinking it unknown is that in Johnson's "Embedding Some Recursively Presented Groups" Groups St. Andrews, 1997 in Bath, Volume 2, the author states specifically that they could not find such a group.

Edit: Ah, I see Mariano's answer shows mine as incorrect/incomplete. I'll leave my answer up just for the observation that you need a little more Higman's original embedding theorem to get the conclusion.

• I confirm that Johnson's paper contains that passage, but Higman's paper also specifically says (p. 456 paragraph following the corollary) that the rational numbers (indeed a countable restricted direct product of copies of the rationals) can be "effectively embedded" in a finitely presented group. Presumably what Johnson meant is exactly what the OP is asking though: effective as in "Ok, and what are the relators?". – Jack Schmidt May 3 '10 at 18:26
• Ah, great. Thanks for coming to the rescue, Jack -- I was getting pretty turned around. – Cam McLeman May 3 '10 at 20:55
• I suspect Higman means 'effectively' in the sense that 'there's an algorithm to compute it'. – HJRW May 4 '10 at 1:47