Natural example of finitely generated group with $\mathbb{Q}$ as a subgroup.

Question

I know from an answer to this that every countable group embeds into a finitely generated group, but I'm curious if there is a nice explicit example that demonstrates this is possible with the rationals. I was able to exhibit the dyadic rationals as a subgroup of a finitely generated semi-direct product of them and the integers, but my method for that fails to generalize to $\mathbb{Q}$ due to the existence of infinitely many primes.

Answer 1

See this question and the answer by Jim Belk. The group is one of the "relatives" of the R.Thompson group $T$. This one is obtained by lifting Thompson's group $T$ through the covering map from the line to the circle. It is a natural finitely generated (even finitely presented) group, and it has been considered many times before by Ghys, Sergiescu and others.

Answer 2

Most of the "natural" examples of finitely generated groups I know are linear, which implies that they are residually finite by Malcev's theorem. Residual finiteness passes to subgroups, but $\mathbb{Q}$ is not residually finite (any map from $\mathbb{Q}$ to a finite group is zero by divisibility), so $\mathbb{Q}$ can't be a subgroup of a residually finite group and hence can't be a subgroup of a finitely generated linear group.

Note that this argument doesn't rule out $\mathbb{Z} \left[ \frac{1}{2} \right]$, which is residually finite, and in fact it doesn't rule out $\mathbb{Z}_{(p)}$ ($\mathbb{Z}$ localized at all primes not equal to $p$). I'm not sure one way or the other whether $\mathbb{Z}_{(p)}$ can be constructed as a subgroup of a "natural" finitely generated group.

Answer 3

The oldest "natural" example is probably the one due to Philip Hall, 1959. It's very elementary, here is it:

Let $u=(u_n)_{n\in\mathbf{Z}}$ be a bi-infinite sequence of nonzero rationals. Assume (one can ask less but nevermind) that every finite sequence of nonzero rationals appears as subword of $u$.

Let $V=\mathbf{Q}^{(\mathbf{Z})}$ be the vector space over $\mathbf{Q}$ with basis $(e_n)_{n\in\mathbf{Z}}$. Define two operators of $V$: the shift $s(e_n)=e_{n+1}$, and the diagonal map $d_u(e_n)=u_ne_n$. Define $\Gamma=\langle s,d_u\rangle$. Note that $s^nd_us^{-n}=d_{u'}$ where $u'$ is the $n$-shift of $u$, in particular $s^nd_us^{-n}$ and $d_u$ commute, so $\Gamma$ is naturally a quotient of the wreath product $\mathbf{Z}\wr\mathbf{Z}$ (actually, it is isomorphic to it). Next, observe that $V$ is a simple $\mathbf{Z}\Gamma$-module, that is, $V\neq 0$ and the only $\Gamma$-invariant subgroups of $V$ are $\{0\}$ and $V$ (easy from the assumptions).

Hence, the semidirect product $V\rtimes\Gamma$ is generated by 3 elements (the 2 generators of $\Gamma$ and any nonzero element of $V$), and this is a finitely generated (solvable) group containing a copy of $\mathbf{Q}$ (and of even of $\mathbf{Q}^{(\mathbf{Z})}$).

Actually, here one just needs $V$ to be generated by $e_O$ as $\mathbf{Z}\Gamma$-module, so it's enough to assume that the $u_n$ and $u_n^{-1}$ generate $\mathbf{Q}$ as additive group. This holds, for instance, if $u_n=\max(1,|n|)$.