Timeline for Groups with triple system of self-normalizing subgroups
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Jun 15, 2020 at 7:27 | history | edited | CommunityBot |
Commonmark migration
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| Jul 1, 2011 at 10:37 | vote | accept | Tom De Medts | ||
| Jun 24, 2011 at 21:49 | answer | added | Geoff Robinson | timeline score: 16 | |
| Jun 24, 2011 at 20:55 | comment | added | Max Horn | Just a minor remark: For searching the literature, it might be helpful to know that that when $AB=G$, $A\cap B=1$, then $G$ is also called an Zappa–Szép product, knit product or bicrossed product of $A$ and $B$. The group $G$ then also is referred to as a permutable group, or "group that admits an exact factorization". | |
| Jun 24, 2011 at 20:16 | answer | added | Andreas Thom | timeline score: 1 | |
| Jun 24, 2011 at 18:18 | comment | added | Steve D | Another comment in the finite case: Obviously we have $|A|=|B|=|C|$, and so $|G|$ is a perfect square, and each of $A,B,C$ is core-free. Even just the existence of an $A$ and a $B$ is equivalent to: $G$ is a transitive permutation group of degree $n$, order $n^2$, and has a self-normalizing regular subgroup. I don't know a lot about permutation groups, but I suspect this limits the structure quite a bit. If we could somehow show $|A^b\cap A|$ is independent of $b\in B$, we could even conclude $G$ is primitive. | |
| Jun 24, 2011 at 15:37 | comment | added | Geoff Robinson | A tiny remark is that for finite $G$, this can't happen with $A,B,C$ all nilpotent. If all were nilpotent, then $G$ is solvable by Kegel-Wielandt, and then $G$ has a unique conjugacy class of self-normalizing nilpotent subgroups (the Carter subgroups). Thus $A,B,C$ are all conjugate. But a group is never a product of two proper conjugate subgroups. | |
| Jun 24, 2011 at 11:55 | history | edited | Tom De Medts | CC BY-SA 3.0 |
edited title
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| Jun 24, 2011 at 8:41 | history | asked | Tom De Medts | CC BY-SA 3.0 |