Timeline for A special residually finite group
Current License: CC BY-SA 2.5
13 events
| when toggle format | what | by | license | comment | |
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| Oct 11, 2010 at 10:14 | comment | added | user6976 | @Andy: The fact that torsion linear groups are finite was known long before Tits-Milnor-Wolf. It is a theorem of Burnside (~1900). | |
| Oct 11, 2010 at 2:37 | comment | added | HJRW | Andy - with a 'virtually' in front of 'nilpotent', that's the Milnor--Wolf Theorem. | |
| Oct 11, 2010 at 0:32 | comment | added | Andy Putman | Whoops! Thanks for the correction Mark. Is it true that any non-nilpotent solvable group has a free non-cyclic subsemigroup? | |
| Oct 10, 2010 at 23:59 | comment | added | Mustafa Gokhan Benli | @Pete, thanks for the far better rewriting of my answer. | |
| Oct 10, 2010 at 22:54 | comment | added | user6976 | @Andy: solvable groups can have exponential growth. For example ${\mathbb Z}\wr {\mathbb Z}$ is solvable of class 2 and has exponential growth because it contains a free non-cyclic subsemigroup. | |
| Oct 10, 2010 at 21:09 | comment | added | Andy Putman | By the way, I highly recommend reading the final chapter in Pierre de la Harpe's book "Topics in Geometric Group Theory", which is entirely devoted to the Grigorchuk group. It serves as a sort of "universal counterexample" to conjectures in geometric group theory. | |
| Oct 10, 2010 at 20:54 | comment | added | Andy Putman | It's not hard to prove that the Grigorchuk grp is not linear. For instance, the Tits alternative says that every fg linear grp G either contains a solvable subgroup of finite index or contains a nonabelian free subgroup. This implies that the "growth function" f(n) of G (here f(n) is the number of elements of G of length at most n in a fixed genset) grows either polynomially (if G has a solvable subgrp of finite index) or exponentially (if G contains a nonabelian free subgrp). However, the first major thm about the Grigorchuk grp is that its growth fcn is superpolynomial but subexponential. | |
| Oct 10, 2010 at 20:08 | comment | added | Greg Kuperberg | These two references say that it is known that the Grigorchuk group is not linear. math.uchicago.edu/~may/VIGRE/VIGRE2008/REUPapers/Waddle.pdf bernoulli.epfl.ch/graphs/talks/Buliga.pdf | |
| Oct 10, 2010 at 19:53 | comment | added | Pete L. Clark | Note: I had a little trouble reading the answer (no doubt in large part because I am not a group theorist), so I decided to rewrite it. I hope that is okay. [P.S.: +1.] | |
| Oct 10, 2010 at 19:51 | history | edited | Pete L. Clark | CC BY-SA 2.5 |
rewrote for increased clarity
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| Oct 10, 2010 at 7:34 | history | edited | Mustafa Gokhan Benli | CC BY-SA 2.5 |
added 27 characters in body
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| Oct 10, 2010 at 7:30 | comment | added | Colin Reid | That is, assuming the group itself isn't linear over a field of characteristic zero. The Grigorchuk group works though as it is known not to be linear. | |
| Oct 10, 2010 at 7:02 | history | answered | Mustafa Gokhan Benli | CC BY-SA 2.5 |