**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.



Potential counterexamples to Bass' trace conjecture
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 $\mathbf{Q}$).




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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](https://arxiv.org/abs/2005.02036)).


([The other answer](https://mathoverflow.net/a/417332/14094) 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.)


