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The presentation is a bit prettier if we set $b_i := b^{a^i}$ for $i\in \mathbb{Z}$:

$\langle a, b \mid [b, b_i] \; (i\in \mathbb{Z}), b_0^{c_0}b_1^{c_1}\ldots b_s^{c_s}\rangle$,

Since the subgroup generated by the $b_i$ is abelian, we can even write the last relator additively as $\sum_{i=0}^s c_ib_i$.

These groups occur in a paper Derek Holt and I are writing on subgroups of finitely generated soluble groups. They are all metabelian, and if the $c_i$ are relatively prime, they are also torsion-free.

I would like to call these groups by their proper name if they have one, especially as we have not so far managed to come up with a good name. Since they are fairly straightforward groups, I expect it is possible they have already come up somewhere else and been named, but unfortunately presentations are difficult to search for on MathSciNet.

EDIT: I might as well mention our current name for these groups. We use the notation $G(\mathbb{c})$, where $\mathbb{c} = (c_0,\ldots,c_s)$, for the group with the above presentation. This I am happy with. But our current name for the groups in general is `Gc-groups', which I don't like very much, and so I was hoping to find out that they already have a better name.

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Presumably you've noticed that $\langle a, b \mid [b, b_i] \; (i\in \mathbb{Z}), \rangle\cong\mathbb{Z}\wr \mathbb{Z}$. –  HJRW Feb 1 '11 at 18:42
    
Certainly! I probably should have said, I am assuming $s\geq 1$. –  Tara Brough Feb 1 '11 at 18:46
    
I only included very minimal information about what we know about these groups, since I expect that if somebody knows what they are called, they will have seen them in at least a very similar form before. Of course the $s\geq 1$ assumption doesn't really matter, but it cuts out the easy examples $\mathbb{Z}\wr\mathbb{Z}$ and $C_p\wr\mathbb{Z}$. There are plenty of other 'easy' examples, though: for example if $|c_0| = |c_s| = 1$, then we get a polycyclic group. –  Tara Brough Feb 1 '11 at 18:55
    
I am not quite seeing how they can be torsion free. Since the b_i commute and some product of powers of the b_i is trivial, then the product of those b_i has finite order. –  Tobias Kildetoft Feb 2 '11 at 8:45
    
Oops, just noticed a mistake (a 'not' where there shouldn't have been one) in my reply to Tobias's comment, and then accidentally deleted it. Here is the corrected version: Not so (though you are not the first to have raised this objection). For example, $b_0b_1^{-2}=1$ only says that $b_0=b_1^2$. Why should $b_0b_1 = b_1^3$ have finite order? It's not trivial to show that these groups are torsion-free in general. The proof takes a little over a page in the current draft of our paper. –  Tara Brough Feb 8 '11 at 19:10

1 Answer 1

These groups are virtually wreath products of finitely generated Abelian groups with $\mathbb Z$. More precisely the subgroup $G_s=\langle a^s, b\rangle$ is a wreath product of a 1-related Abelian group (the relation is your product of powers) and $\langle a^s\rangle$, and your group is an extension of $G_s$ by a cyclic group of order $s$. I do not think there exists a special name for these groups. As with every metabelian group which is Abelian-by-cyclic, you can get a lot of properties of it by looking at the corresponding $\mathbb{Z}[b,b^{-1}]$-module.

Update. I just realized that this group is not a virtual wreath product. One can construct the group as follows. Consider the ring $R=\mathbb{Z}[t,t^{-1}]$ and the ideal $I$. there generated by the polynomial $f=c_0+c_1t+\ldots+c_st^s$. The group $\mathbb{Z}$ acts on the Abelian group $R/I$ by left multiplication (which are automorphisms of the additive group of $R/I$). Your group is the corresponding semidirect product $R/I\rtimes \mathbb{Z}$. One can also represent this group by $2\times 2$ matrices using the Magnus embedding. Anyway this does not answer your question about the name of the group. I still do not think it has a name.

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Thank you for your reply. I think you are probably right that these groups have no name. However, I'll leave the question open at least until our paper is accepted, just in case. –  Tara Brough Apr 7 '11 at 15:08
    
By the way, do you actually mean $a^{s+1}$ rather than $a^s$? Otherwise I don't understand what the base group in your expression of $G_s$ as a wreath product would be. –  Tara Brough Apr 7 '11 at 15:10
    
yes, it should be s+1. –  Mark Sapir Apr 7 '11 at 16:12
    
@Tara: What is your paper about? Is it on the arXiv? –  Mark Sapir Apr 7 '11 at 22:26
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The paper shows that every finitely generated soluble group which is not virtually abelian has a subgroup isomorphic to one of a small number of types (one of those being these Gc-groups). It's called 'Finitely generated soluble groups and their subgroups', and yes, it is on the arXiv. –  Tara Brough Apr 7 '11 at 22:40

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