Timeline for Automorphism group of $UT(n,p)$, the group of unitriangular matrices over the field $\mathbb{F}_p$
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Sep 12, 2017 at 8:19 | vote | accept | mayer_vietoris | ||
| Sep 11, 2017 at 19:28 | answer | added | Alexander Chervov | timeline score: 3 | |
| Sep 11, 2017 at 13:35 | comment | added | Derek Holt | Yes that is what I meant. I did the calculation in Magma by computing orbits of the outer automorphism group on the conjugacy classes of subgroups of order $p^3$. I can provide more detailed information on this necessary. | |
| Sep 11, 2017 at 13:30 | comment | added | mayer_vietoris | Thank you very much for all the effort you have put on this. One question only, when you mean "...and the automorphism group has respectively 19, 21 and 23 orbits on these subgroups.", you mean that the $Aut(UT(4,p))$-action on the set of subgroups of order $p^3$ yields these 19,21 and 23 orbits respectively or something different? Because it confuses me a little bit. | |
| Sep 11, 2017 at 12:56 | comment | added | Derek Holt | I did some calculations on the orbits of the automorphism group of ${\rm UT}(4,p)$ on subgroups of order $p^3$ for $p=3,5,7$. At this stage, computations are starting to take longer, but I could go a bit further if necessary. In summary, for $p=3,5,7$, there are respectively $109$, $391$, and $953$ conjugacy classes of subgroups of order $p^3$, and the automorphism group has respectively $19$, $21$ and $23$ orbits on these subgroups. | |
| Sep 11, 2017 at 12:45 | history | edited | mayer_vietoris | CC BY-SA 3.0 |
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| Sep 11, 2017 at 12:11 | comment | added | mayer_vietoris | Derek thank you very much for your reply. For $n=3$ I know that this is the case indeed, since I can write down explicitly what's going on. For $n=4$ though the situation becomes really hard and what you wrote is helpful indeed. What I am trying to understand is mainly how the automoprhisms permute the subgroups of order $p^3$ (for $n=4$ at least) and because I don't have access to any computational system at the moment is difficult to do any computation by hand. | |
| Sep 11, 2017 at 12:04 | comment | added | Derek Holt | I haven't given this any thought at all, but computer calculations with $3 \le n \le 8$ strongly suggest that the order of the automorphism group of ${\rm UT}(n,p)$ is $p^3(p-1)^2(p+1)$ when $n=3$, and $2p^e(p-1)^{n-1}$ with $e = (n^2+n-4)/2$, when $n > 3$. So, for $n=4$, this would give $2p^8(p-1)^3$. | |
| Sep 11, 2017 at 11:00 | history | asked | mayer_vietoris | CC BY-SA 3.0 |