# Does $\mathbb{Q}$ embed into a finitely generated solvable group?

Does $$\mathbb{Q}$$ embed into a finitely generated solvable group?

I've checked that $$\mathbb{Q}$$ is not a subgroup of any finitely generated metabelian group. I don't know how to show this (or whether it is true) for solvable groups of higher step.

• Tiny note: no one-relator group contains $\mathbb{Q}$ as a subgroup, a result due to B. B. Newman, but this does not stray far at all from what you already have in the metabelian case. – Carl-Fredrik Nyberg Brodda Jan 13 at 20:29

## 1 Answer

Yes, it's due to Ph. Hall. It embeds into a f.g. 3-step solvable group.

Let $$s:\mathbf{Z}\to\mathbf{Q}^*$$ be a map (thought as an bi-infinite word) such that every finite sequence of nonzero rational numbers occurs as subword. Define two automorphisms $$u,v$$ of $$\mathbf{Q}^{(\mathbf{Z})}$$ (vector space over $$\mathbf{Q}$$ with basis $$(e_n)_{n\in\mathbf{Z}}$$) as follows: $$u$$ is the shift $$e_n\mapsto e_{n+1}$$, and $$v$$ is the diagonal map $$e_n\mapsto s(n)e_n$$. Since $$u^nvu^{-n}$$ is also diagonal for every $$n$$ it commutes with $$v$$, and hence the pair $$\langle u,v\rangle$$ generates a quotient of the wreath product $$\mathbf{Z}\wr\mathbf{Z}$$ (actually a copy of it). Moreover, as $$\mathbf{Z}[\langle u,v\rangle$$]-module, $$\mathbf{Q}^{(\mathbf{Z})}$$ is readily seen to be simple. Hence $$\langle u,v\rangle\ltimes \mathbf{Q}^{(\mathbf{Z})}$$ is generated by 3 elements, contains a copy of $$\mathbf{Q}$$ and is 3-step solvable. (Reference of roughly the same construction: Ph. Hall, On the finiteness of certain soluble groups, Proc. London Math. Soc. (3) 9 (1959).)

It was also proved by Neumann-Neumann that every countable $$k$$-step solvable group embeds into a finitely generated $$(k+2)$$-step solvable group (reference: B. H. Neumann, H. Neumann. Embedding Theorems for Groups, J. London Math. Soc. 34 (1959), 465-479).

Edit: Ph. Hall earlier (Finiteness conditions for soluble groups, Proc. London Math. Soc. (3) 4 (1954)) checked that there exists a 3-step solvable finitely generated group, whose center is free abelian of infinite rank (such a group can be viewed as group of upper triangular $$3\times 3$$ matrices over $$\mathbf{Z}[t^{\pm 1}]$$). It is straightforward that every abelian group can be embedded into (the center of) a quotient of the latter group (I guess he was aware of this, and I think I remember that Neumann-Neumann claim to generalize Hall's result but don't have their paper right now).

• Thank you! Do you happen to have the name on hand for the Neumann-Neumann paper? – Josh F Jan 13 at 20:55