Timeline for Number of commuting pairs in p-group
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Oct 15, 2019 at 13:22 | answer | added | Geoff Robinson | timeline score: 4 | |
| Oct 14, 2019 at 22:20 | review | Close votes | |||
| Oct 31, 2019 at 3:05 | |||||
| Oct 14, 2019 at 18:48 | comment | added | Derek Holt | Geoff Robinson has said already in a previous comment that $k(U) = |G:N_G(U)|$ is false in general. In just computed another example. When $n=p=3$, we have $k(U)=11$ and $|G:N_G(U)| = 52$. | |
| Oct 14, 2019 at 18:44 | comment | added | Nourr Mga | The number of commuting pairs in $U$ is $|U|k(U)$. I want to know if $k(U)=[G:N_{G}(U)]$ or not. More precisely, is the number of commuting pairs in $U$ is $|U|.[G:N_{G}(U)]$?. | |
| Oct 14, 2019 at 18:31 | history | edited | Martin Sleziak |
Removed the deprecated (abstract-algebra) tag - see the tag info: https://mathoverflow.net/tags/abstract-algebra/info (if there are some other suitable tags, choose them instead.)
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| Oct 14, 2019 at 18:19 | comment | added | Nourr Mga | Yes thanks. But which of the two options is correct to compute the number of commuting pairs in U. | |
| Oct 14, 2019 at 18:08 | comment | added | Geoff Robinson | OK, but when $n = 2$, we have $[G:N_{G}(U)] = p+1,$ while $k(U) = p$, so the first option (with "it" $=N_{G}(U)$) does not hold for $n = 2$ (and for lots of other cases). | |
| Oct 14, 2019 at 17:39 | comment | added | Nourr Mga | Thank you sir. I mean in the first option that $k(U)=[G:N_{G}(U)]$ which is easy to calculate. | |
| Oct 14, 2019 at 17:29 | comment | added | Geoff Robinson | The second option does not give a precise formula for $k(U)$. But, as far as I know, Higman's conjecture is still open, as seems clear from the linked paper. As for the first option, do you mean to ask whether $[G:U] = k(U)$ or whether $[G:B] = k(U)$, where $B = N_{G}(U)$? When $n = 2$, neither of these options seems to hold, since $k(U) \leq |U|$ and $[G:B] > |U|$, so I am a little unclear what you mean to ask. | |
| Oct 14, 2019 at 17:20 | history | asked | Nourr Mga | CC BY-SA 4.0 |