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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.

Shaun
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Beren Gunsolus
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    If you just want an explicit example and don't care whether it's "natural," here are three proofs of Higman's embedding theorem, the first one of which at least produces a pretty explicit group: https://jaywillmath.wordpress.com/2013/08/16/three-ways-of-embedding-a-countable-group-in-a-2-generated-group/ – Qiaochu Yuan Sep 18 '20 at 22:42

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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.

markvs
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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 $u_n=\max(1,|n|)$.

YCor
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  • @JCAA It's the whole point of the word "natural", which is subjective. When Ph. Hall did this result, there was already the 1949 HNN construction, and also the 1959 NN alternative approach using a double (unrestricted) wreath product. You can perfectly consider the HNN initial construction, which is explicit, as natural. – YCor Sep 19 '20 at 16:51
  • The following is an easier way to embed any countable group into a 2-generated group. Let $G={g_i | i=1,2,...}$ be a countable group. Let $X={w_{g_i} | i=1,2,...}$ be a set of words in a two-letter alphabet $a, b$ satisfying the small cancelation condition $C"(1/12)$. Let $H=\langle a, b | w_gw_h= w_{gh}\rangle$. Then $G$ naturally embers into $H$. – markvs Sep 19 '20 at 16:55
  • @JCAA easier is a matter of taste, as it's based on small cancelation, to which not everybody is familiar. You'd better post an answer than try to compare the easiness of the various constructions; I'd be happy to upvote it. – YCor Sep 19 '20 at 17:02
  • I have posted an answer. All these are non-answers. – markvs Sep 19 '20 at 17:06
  • @JCAA I meant an answer with this small cancelation argument. Thanks for your useful comments, Mark. – YCor Sep 19 '20 at 17:20
  • Compared to the other answers, it's nice that this group is solvable. Would finitely presented and solvable be possible? – Sean Eberhard Jan 31 '25 at 09:33
  • @SeanEberhard I don't actually remember whether it's known that $\mathbf{Q}$ embeds into a fp solvable group (regardless of the subjective "natural" requirement). But for sure, no "finitely presented" analogue of the above construction is known. I don't even know if there is a fp solvable group $G$ with a simple $\mathbf{Z}G$-module that is torsion-free as abelian group. – YCor Jan 31 '25 at 10:45
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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.

Qiaochu Yuan
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    This does not answer the question. As is well-known, $\Bbb{Z}[1/p]$ is the derived subgroup of the Baumslag-Solitar group $BS(1,p)$. – markvs Sep 18 '20 at 23:33
  • @JCAA more precisely, the derived subgroup of $BS(1,p)$ is isomorphic to $\mathbf{Z}[1/p]$, but is not the obvious copy (unless $p=2$): namely, the derived subgroup equals $(p-1)\mathbf{Z}[1/p]$. – YCor Sep 19 '20 at 13:25
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    @QiaochuYuan as mentioned at several places, $\mathbf{Q}$ and hence $\mathbf{Z}{(p)}$ are subgroups of several "natural" groups. Nevertheless, this suggests the following question: whether $\mathbf{Z}{(p)}$ can be realized as subgroup of a "natural" residually finite finitely generated group. – YCor Sep 19 '20 at 13:27
  • @JCAA If I say that two subgroups $H,K$ of a group are equal, I mean that they contain the same elements (I don't think this is a very exotic convention!). For instance $\mathbf{Z}\neq 2\mathbf{Z}$. Also in $BS(1,p)$, $p\ge 2$, there's a canonical copy of $\mathbf{Z}[1/p]$ (the right-hand factor in the decomposition as $\mathbf{Z}\ltimes\mathbf{Z}[1/p]$). – YCor Sep 19 '20 at 15:11
  • The group that you claim is equal to the derived subgroup is not a subset of $BS(1,p)$. – markvs Sep 19 '20 at 15:14
  • @JCAA You're maybe misunderstanding, but certainly $(p-1)\mathbf{Z}[1/p]\subset\mathbf{Z}[1/p]\subset \mathbf{Z}\ltimes\mathbf{Z}[1/p]$ and $(p-1)\mathbf{Z}[1/p]$ is a subgroup (an ideal) of index $p-1$ in $\mathbf{Z}[1/p]$. To say things differently, the abelianization of $BS(1,p)$ is isomorphic to $\mathbf{Z}\times\mathbf{Z}/(p-1)\mathbf{Z}$ (and not to $\mathbf{Z}$, except for $p=2$), as you can check directly from the presentation. – YCor Sep 19 '20 at 16:31
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    Also to clarify if necessary, I'm identifying $BS(1,p)$, $p\ge 2$, to rational matrices $M(n,x)=\begin{pmatrix} p^n& x\0 & 1\end{pmatrix}$ where $n$ ranges over $\mathbf{Z}$, $x$ ranges over $\mathbf{Z}[1/p]$, and I'm identifying $\mathbf{Z}[1/p]$ to the set of matrices of the form $M(0,x)$. And $(p-1)\mathbf{Z}[1/p]$ is the set of rational numbers of the form $(p-1)a/p^b$ for $a,b\in\mathbf{Z}$. – YCor Sep 19 '20 at 16:34