Timeline for A question on simultaneous conjugation of permutations
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 27, 2022 at 8:40 | comment | added | darij grinberg | For the record: This is proved in Theorem 4.9 of arXiv:1810.08971v4 by Junyao Pan. (I have not checked the proof.) | |
| Aug 8, 2018 at 16:48 | comment | added | MTyson | Consider the polygonal mesh with vertices $\{1,\dots,n\}$, directed edges $i\to a(i)$ and $i\to b(i)$ for each $i$, and for each (possibly trivial) cycle $(k\;\cdots\;k+m-1)$ of $[a,b]$ a face with boundary $[a,b]^m$ starting at $k$. The genus depends on the cycle type of $[a,b]$. Lifting to the universal cover gives a tessellation of the Euclidean or hyperbolic plane. It suffices to find a symmetry of the plane sending each $a$ (resp $b$) edge to an $a$ (resp $b$) edge with opposite orientation. In the commutative case this is realized by a $180^\circ$ rotation around any square face. | |
| Jun 16, 2018 at 21:55 | comment | added | darij grinberg | Have you tried using Rutilo Moreno, Luis Manuel Rivera, Blocks in cycles and $k$-commuting permutations, SpringerPlus, December 2016, 5:1949? (A preprint also appears as arXiv:1306.5708.) Corollary 1 characterizes when the commutator of two permutations has at least $n-k$ fixed points (the authors say that these two permutations $k$-commute). | |
| S Apr 26, 2017 at 16:33 | history | bounty ended | CommunityBot | ||
| S Apr 26, 2017 at 16:33 | history | notice removed | CommunityBot | ||
| Apr 20, 2017 at 15:45 | history | edited | darij grinberg |
edited tags
|
|
| Apr 18, 2017 at 15:44 | comment | added | Peter Mueller | @SamHopkins I don't think so. I think a reason why many fixed points matter is as follows: Suppose $a^{-1}b^{-1}ab=x$. If $z$ inverts $a$ and $b$, then $x^z=a^{-z}b^{-z}a^zb^z=aba^{-1}b^{-1}$, so $x^{zab}=x$. Thus the $z$ we are looking for is in the coset $C_G(x)b^{-1}a^{-1}$. Note that $C_G(x)$ is $S_{n-k}$ times a subgroup of $S_k$, where $k\le4$. I believe the reason for the existence of $z$ is the hugeness of $C_G(x)$. | |
| Apr 18, 2017 at 15:19 | comment | added | Sam Hopkins | Totally naive comment, but would the 4 vs 5 thing have anything to do with when A_n is simple? | |
| Apr 18, 2017 at 15:16 | comment | added | Derek Holt | Note also that it is very easy to find examples in which $[a,b]$ has $n-5$ fixed points and there is no element $z$ inverting both $a$ and $b$. | |
| S Apr 18, 2017 at 14:45 | history | bounty started | Peter Mueller | ||
| S Apr 18, 2017 at 14:45 | history | notice added | Peter Mueller | Draw attention | |
| Apr 14, 2017 at 19:11 | comment | added | Derek Holt | After doing lots of random experiments I am starting to believe that this might be true. If so, then I am afraid that it might not be easy to prove! | |
| Apr 14, 2017 at 11:14 | review | First posts | |||
| Apr 14, 2017 at 11:17 | |||||
| Apr 14, 2017 at 11:11 | history | asked | Danny Neftin | CC BY-SA 3.0 |