Timeline for Is there a nonabelian free group inside a group of positive rank gradient?
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 2, 2015 at 5:29 | comment | added | Andreas Thom | @YiftachBarnea: Thanks for pointing this out! | |
| Sep 1, 2015 at 18:33 | comment | added | Yiftach Barnea | It is worth mentioning that Jan-Christoph Schlage-Puchta proved a similar result independently, see his paper: A p-group with positive rank gradient. J. Group Theory 15 (2012), no. 2, 261–270. You might also like to look at a joint paper with me: On p-deficiency in groups. J. Group Theory 16 (2013), no. 4, 497–517. (Both can be found on the Arxiv.) | |
| Sep 1, 2015 at 15:33 | comment | added | Joël | Thanks, Benjamin and Ycor, and Derek. Things are clear now. | |
| Sep 1, 2015 at 15:26 | comment | added | YCor | @DerekHolt: no, many torsion-free groups are residually-$p$ (= residually a finite $p$-group). The correct interpretation of Andreas's post is a $p$-group which is also residually finite (and hence residually-$p$) | |
| Sep 1, 2015 at 15:11 | comment | added | Benjamin Steinberg | p-group means each element has order a p-power. It need not be finite. Like the Grigorchuk group | |
| Sep 1, 2015 at 14:56 | comment | added | Joël | I am still confused. What is a $p$-group? For me it is a finite group of order a power of $p$. | |
| Sep 1, 2015 at 14:46 | comment | added | Derek Holt | @Joel But I think "(residually finite) $p$-group" is equivalent to "residually (finite $p$-group)". | |
| Sep 1, 2015 at 14:37 | comment | added | Joël | I am not completely sure how to parse the statement of the theorem. Should "residually finite $p$-group" be read together, meaning that the intersection of subgroups of index a power of $p$ is trivial ? | |
| Sep 1, 2015 at 13:59 | vote | accept | Pablo | ||
| Sep 1, 2015 at 13:53 | history | answered | Andreas Thom | CC BY-SA 3.0 |