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Let $G$ be generated by the elements $g$ and $\{e_i\}$ for $i\in\mathbb{Z}$, having the relation $ge_ig^{-1}=e_{i+1}$.

It seems to me like there is a subgroup $\langle e_i,e_{i+1}\rangle$ for example, that is the free group on two generators. But I also know that this group is solvable (although I don't know how to prove it) and that the free group on two generators is not solvable. Thus cannot be a subgroup. Is there a direct argument to see that the group generated by two of the $e_i$'s is not the free group on two generators?

Edit: It turns out this group is only solvable when $e_i e_j = e_j e_i$

  • How certain are you that this group is solvable? I don't see it. In fact, this looks to me like an HNN extension of a free group generated by the $e_i$'s. – Andreas Blass Jan 25 '20 at 00:39
  • Well, I suppose it might not be... My professor asked us to prove that it is solvable on a Homework. It is supposed to be an example of a group that is solvable but is not polycyclic. –  Jan 25 '20 at 00:42
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    Perhaps you have left off some relations? Are $e_i,e_j$ required to commute? – Lee Mosher Jan 25 '20 at 00:47
  • Well perhaps my professor did. I will check and update my question. –  Jan 25 '20 at 00:49
  • During class he mentioned that this was a "wreath product" and an answer to this question: https://math.stackexchange.com/questions/7896/subgroups-of-finitely-generated-groups-are-not-necessarily-finitely-generated

    gives a description that seems to be the same as the description of the group in my homework.

     –  Jan 25 '20 at 00:54
  • Related: https://math.stackexchange.com/a/308229/30382 – Servaes Jan 25 '20 at 01:04
  • If it is a wreath product, then the $e_i$ commute; what you have is the restricted wreath product of the infinite cyclic group by itself, $\mathbb{Z}\wr_r\mathbb{Z}$. But in the wreath product $A\wr_r B$, you take a direct sum (restricted direct product) of $|B|$ copies of $A$, and then $B$ acts on this sum by acting on the indices. So if your group is supposed to be a wreath product, then you are missing the relations $e_ie_j=e_je_i$ for all $i,j$. – Arturo Magidin Jan 25 '20 at 01:14
  • @pictorexcrucia: The difference between the description given by Andreas in the question you link to and yours is that he explicitly says the base group is abelian (“coproduct in the category of abelian groups”) whereas your description does not say that $\langle e_i\mid i\in\mathbb{Z}\rangle$ is abelian. – Arturo Magidin Jan 25 '20 at 01:17

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