Timeline for Infinitely presented group where every finite sub-presentation is virtually free
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 21, 2019 at 19:58 | comment | added | Ian Agol | I suspect now that such presentations cannot exist. I think at least one should be able to use small cancellation theory to rule out presentations where the power is high, maybe at least 6. | |
| Feb 21, 2019 at 11:29 | comment | added | YCor | @DerekHolt no need of HNN theory to see that it contains $\mathbf{Z}^2$ (generated by $\{a,a^t\}$): instead just map onto $\mathbf{Z}\wr\mathbf{Z}$. | |
| Feb 21, 2019 at 9:37 | comment | added | HJRW | Perhaps it's worth pointing out that it's not terribly difficult to ensure that all the one-relator subpresentations are free. Just add an extra generator $t$ to $S$, add the relation $t$, and change every $r_i$ to $r_it$. Of course, it's much harder to do something similar for larger subsets of relators, but then one gets out of the theory of one-relator groups. | |
| Feb 21, 2019 at 8:53 | comment | added | DavidHume | @IanAgol Thanks, this seems like a very interesting approach in general, I just wondered if the community knew of an example before diving into this. | |
| Feb 21, 2019 at 7:53 | comment | added | Derek Holt | But the easiest argument is that $\langle a,t \mid [a,a^t] \rangle \cong \langle a,b,t \mid [a,b] = 1, b = a^t \rangle$ is an HNN extension of ${\mathbb Z}^2$ and hence has ${\mathbb Z^2}$ as a subgroup and cannot be virtually free. Of course that still requires the theory of HNN extensions. | |
| Feb 21, 2019 at 6:14 | history | answered | Ian Agol | CC BY-SA 4.0 |