Timeline for Nilpotency class of a certain finite 2-group
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
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| Nov 8, 2011 at 16:04 | comment | added | Max Horn | Because the presentation I gave is a consistent polycyclic presentation (verifying that is straight forward), and so the order of each generator is a multiple of its relative order. So the order of $w$ is a multiple of 2 (it's relative order). Since we also know that $w^2=1$, it has in fact exactly order 2. Of course you can also use Derek's argument :). | |
| Nov 7, 2011 at 16:26 | comment | added | Alireza Abdollahi | Why $w$ is non-trivial? | |
| Nov 7, 2011 at 15:42 | comment | added | Max Horn | PS: If you don't know about polycyclic presentations, here is a short writeup: icm.tu-bs.de/ag_algebra/software/polycyclic/htm/CHAP002.htm | |
| Nov 7, 2011 at 15:37 | history | edited | Max Horn | CC BY-SA 3.0 |
Correct presentation (I mixed up the power commutator and power conjugacy presentations in my mind)
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| Nov 7, 2011 at 14:51 | history | answered | Max Horn | CC BY-SA 3.0 |