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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.

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    $\begingroup$ 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. $\endgroup$ – Carl-Fredrik Nyberg Brodda Jan 13 at 20:29
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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).

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

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