Timeline for Second homology group of free nilpotent p-group
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 3, 2013 at 11:51 | comment | added | Ricardo Andrade | Dear @Tony: It might still be a good idea to add a correction to the question, as not everyone reads the comments. | |
| Jul 4, 2013 at 21:45 | answer | added | Yassine Guerboussa | timeline score: 2 | |
| Mar 6, 2012 at 17:57 | comment | added | Tony | @Derek : Thanks for the correction! You explained it so well that it seems silly to edit the question to include this. | |
| Mar 5, 2012 at 23:37 | comment | added | Derek Holt | I don't know the answer to your question, but I don't think that your comments about these groups being universal are quite correct. The quotients in your series are all elementary abelian $p$-groups. You need to define the $p$-nilpotency class of a group using central series with elementary abelian $p$-groups as factors. Your groups are then universal with respect to groups of $p$-nilpotency class $k$. | |
| Mar 5, 2012 at 21:17 | comment | added | Tony | I tried. The Hopf formula seems to give me the answer for the quotients of the free group by the ordinary lower central series, but I couldn't make it work for the mod p version (for instance, the relations do not live in the commutator subgroup). | |
| Mar 5, 2012 at 16:41 | comment | added | Chris Gerig | Did you try the Hopf Formula? Assuming you know the generators/relations. | |
| Mar 5, 2012 at 16:14 | comment | added | Tony | Edited it because I realized that I asked about general $H_k$ when I really only care about $H_2$ (as indicated in the title). | |
| Mar 5, 2012 at 16:13 | history | edited | Tony | CC BY-SA 3.0 |
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| Mar 2, 2012 at 21:34 | history | asked | Tony | CC BY-SA 3.0 |