Timeline for Group of exponential growth always contains a free sub-group?
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 22 at 19:27 | comment | added | Stefan Kohl♦ | @JosephAdams Consider the wreath product structure of the Lamplighter group. All you need to do is to choose $i$ and $j$ in such a way that $a^ib^j$ lies in the base group. | |
| May 22 at 12:21 | comment | added | Joseph Adams | Why is it that if every you have two elements of the Lamplighter group of infinite order you can arrange then in a way such that the square is the identity? Is this easily derived from its presentation? | |
| May 1, 2022 at 14:10 | comment | added | Ben Wieland | A semidirect product $\mathbb Z^2\rtimes \mathbb Z$ is solvable and finitely presented. If the action is hyperbolic, as is generically true, it has exponential growth. | |
| May 1, 2022 at 13:52 | comment | added | Derek Holt | The solvable Baumslag-Solitar groups $\langle x,y \mid y^{-1}xy=x^k \rangle$ for $|k| > 1$ are examples. | |
| May 1, 2022 at 13:50 | comment | added | Matt Zaremsky | For a finitely presented group with exponential growth and no non-abelian free subgroups you can use Thompson's group F. I think there should also be solvable (i.e., easier) examples too. | |
| May 1, 2022 at 12:28 | comment | added | user44143 | Do you know if the OP’s idea holds for finitely presented groups? | |
| May 1, 2022 at 12:18 | history | answered | Stefan Kohl♦ | CC BY-SA 4.0 |