TITLE: An explicit example of a finitely presented group containing a subgroup isomorphic to $(\mathbb Q,+)$.
QUESTION [37 upvotes]: 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?

REPLY [6 votes]: 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I find in this 2012 unpublished paper of A. Fossas "On the number of conjugacy classes of torsion elements on Thompson's group T" that the same counting holds in $T$ (she asserted it for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 $3$...!