Timeline for Group which "resembles" the free product of a cyclic group of order two and a cyclic group of order three, but isn't.
Current License: CC BY-SA 2.5
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 12, 2010 at 22:25 | vote | accept | Todd Trimble | ||
| Oct 12, 2010 at 21:16 | answer | added | Derek Holt | timeline score: 18 | |
| Oct 12, 2010 at 19:17 | comment | added | user6976 | Todd: R. Thompson's group $V$ is pretty specific. Right? But you would need to check the references (Cannon-Floyd-Parry). | |
| Oct 12, 2010 at 18:56 | comment | added | Todd Trimble | Qiaochu: I haven't thought hard about it. Everyone: it's not that the existence of such things is particularly in question (I fully believe it). But all the examples I've been provided with are handwavy. What I want is a very explicit example, and hopefully a reference. | |
| Oct 12, 2010 at 18:33 | comment | added | Autumn Kent | @Peter Oops. Yeah, all those words should be set equal to 1. | |
| Oct 12, 2010 at 18:31 | comment | added | Qiaochu Yuan | @Todd: have you tried geometric examples, e.g. letting a and b be rotations about two carefully chosen skew lines in R^3? | |
| Oct 12, 2010 at 18:29 | comment | added | HJRW | To perhaps make Mark's answer below a little more explicit: you can quotient your free product by almost any 'sufficiently complicated' element, and you will get another infinite hyperbolic group with the properties you want. 'Sufficiently complicated' means something like 'satisfying a suitable small-cancellation condition'. | |
| Oct 12, 2010 at 18:27 | answer | added | Vagabond | timeline score: 0 | |
| Oct 12, 2010 at 18:27 | comment | added | Peter Tingley | @Richard Are you missing at ``$=1$" in your example? Otherwise I don't think the example you give is a quotient of the group described in the question. | |
| Oct 12, 2010 at 18:24 | answer | added | user6976 | timeline score: 22 | |
| Oct 12, 2010 at 18:23 | answer | added | Autumn Kent | timeline score: 11 | |
| Oct 12, 2010 at 18:17 | comment | added | Autumn Kent | @Steve No. The group $\langle a, b | a^2 = b^3 = (ab)^7 \rangle$ is the fundamental group of a closed hyperbolic $2$-orbifold, and is infinite. | |
| Oct 12, 2010 at 18:07 | comment | added | Steve D | Isn't the free product of $C_2$ and $C_3$ a just-infinite group? | |
| Oct 12, 2010 at 17:30 | history | asked | Todd Trimble | CC BY-SA 2.5 |