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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?

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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
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3 Answers 3

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 finitely 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.

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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
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@Cam: yup, I was just editing, but wanted to kill the wrong part as soon as possible. –  Mariano Suárez-Alvarez May 3 '10 at 18:30
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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
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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
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@Henry, the point was to make precise the theorem to which Roland was making reference. –  Mariano Suárez-Alvarez May 3 '10 at 19:56
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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 $\mathbb Q$.

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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.

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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
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