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Motivation: The following is a theorem of Berrick-Hesselholt (essentially also due to Linnell, though not in this form):

Let $G$ be a group. Suppose that for every subgroup of $G$ isomorphic to $\mathbb Q$, $G$ has a quotient in which the image of this subgroup is central and nontrivial. In this case the Bass trace conjecture holds for $G$.

I can add it for context, but for my question it is not important to know what this conjecture is - I just state this as motivation.

Question : What are some examples of finitely presented groups which do not have this property ?

That is, they have a subgroup isomorphic to $\mathbb Q$ such that for each quotient of $G$, its image is central only if it is trivial.

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    $\begingroup$ AFAIK it’s far from east to construct any f.p. group with a subgroup isomorphic to Q. See mathoverflow.net/questions/23352/… $\endgroup$
    – Alon Amit
    Commented Mar 3, 2022 at 8:24
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    $\begingroup$ @AlonAmit : ah, thanks for pointing this out ! It seems that the answer there actually answers my question, I should have paid more attention: Jim Belk seems to say that $\mathbb Q$ embeds in a f.p. simple group. In particular, for every quotient of this group, the image is trivial or not central $\endgroup$ Commented Mar 3, 2022 at 8:31
  • $\begingroup$ Haha, I was hoping that group might serve but didn’t even have time to check :) great! $\endgroup$
    – Alon Amit
    Commented Mar 3, 2022 at 8:47
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    $\begingroup$ The inverse image itself of Thompson's $T$ was explicitly constructed and used at many places before observing that it contains a copy of $\mathbf{Q}$ (and actually continuum many). This group is not simple but already answers the question (since the image of $\mathbf{Q}$ in every nontrivial quotient is non-central). $\endgroup$
    – YCor
    Commented Mar 3, 2022 at 10:58
  • $\begingroup$ @YCor : if you could add those details (or references for them) as an answer, that would be great too ! $\endgroup$ Commented Mar 3, 2022 at 11:01

2 Answers 2

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a) There are old results which directly imply the existence of such groups:

(1) Boone Higman 1972: every f.g. group with solvable word problem embeds into a simple subgroup of a finitely presented group.

(2) Every countable group with solvable word problem embeds into a f.g. group with solvable word problem (reference? the original HNN construction directly works since it consists of explicit amalgams — alternatively here Ph. Hall produced in the 50s an explicit 3-generated metabelian group with solvable word problem, with copies of $\mathbf{Q}$).


b) A more explicit example is the group $\tilde{T}$ obtained as the set of self-homeomorphisms of $\mathbf{R}/\mathbf{Z}$ commuting with $\sigma:n\mapsto n+1$, that are piecewise affine with dyadic slopes and breakpoints.

The center of $\tilde{T}$ is the infinite cyclic group $\langle\sigma\rangle$ and the quotient is naturally identified with Thompson's group $T$, which is a finitely presented simple group. The normal proper subgroups of $\tilde{T}$ are precisely the subgroups of $\langle\sigma\rangle$.

If $Q$ is any copy of $\mathbf{Q}$ in $\tilde{T}$, it follows that the image of $Q$ in $T$ is non-central (since $\mathbf{Q}$ has no nontrivial cyclic quotient). Hence the same holds in every nontrivial quotient of $\tilde{T}$.

Finally, that $\tilde{T}$ contains a copy of $\mathbf{Q}$ (and even continuum many such copies) is an original observation of Belk, Matucci, Hyde (arXiv).

(The other answer is closely related as it refers to a more complicated finitely presented simple group containing $\tilde{T}$. The group $\tilde{T}$ itself is not finitely presented but has few enough normal subgroups for the condition to hold.)

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    $\begingroup$ Right, but as far as I understand these do not provide (concrete) examples, they prove the existence of examples, or am I mistaken ? $\endgroup$ Commented Mar 3, 2022 at 10:50
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    $\begingroup$ @MaximeRamzi You're right. The HNN construction is totally explicit (you need an explicit enumeration of a generating subset $\mathbf{Q}$, but $(1/n)_{n\ge 1}$ is such an enumeration). On the other hand the Boone-Higman embedding is possibly badly non explicit (maybe it outputs an explicit presentation, but which is not really helpful). $\endgroup$
    – YCor
    Commented Mar 3, 2022 at 10:54
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    $\begingroup$ Thanks for the edit ! $\endgroup$ Commented Mar 3, 2022 at 17:37
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As explained in the comments, Jim Belk's answer to this question answers mine as well : he points ou that the group $T\mathcal A$ he constructed with his coauthors is finitely presented, simple and contains a copy of $\mathbb Q$.

In particular, it has no quotient where $\mathbb Q$ can be central and non trivial, so this answers the question.

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