Timeline for A finite group that has no decomposition of given cardinality
Current License: CC BY-SA 4.0
15 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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| Nov 26, 2018 at 12:37 | comment | added | Taras Banakh | @FrançoisBrunault The approximative version has been considered by Kozma and Lev, e.g. link.springer.com/article/10.1007%2FBF01190111 | |
| Nov 26, 2018 at 7:18 | comment | added | François Brunault | It may be easier to look at an approximate version of this question: given $0 <a<1$ and $\varepsilon>0$, it it true that for a finite group $G$, there exist subsets $A,B$ with $|A| \ll |G|^{a+\varepsilon}$ and $|B| \ll |G|^{1-a+\varepsilon}$, such that $G=AB$? | |
| Nov 26, 2018 at 0:51 | comment | added | François Brunault | Decomposability makes sense for general Latin squares, but does not hold in general: there is a Latin square of order 9 with no $3 \times 3$ submatrix having distinct entries, see Covers and partial transversals of Latin squares doi.org/10.1007/s10623-018-0499-9 (discussion after Theorem 14). | |
| Nov 25, 2018 at 21:14 | comment | added | François Brunault | It is also easy to see that if $G$ admits two subgroups $H \subset K$ with $(K:H)$ equals to $a$ or $b$, then $G$ is $a \times b$-decomposable. | |
| Nov 25, 2018 at 21:09 | comment | added | Derek Holt | I have no intuition about this at all, but I think it might be a very hard problem to resolve. | |
| Nov 25, 2018 at 20:59 | comment | added | Taras Banakh | So, what is the intuitive expectation concerning the general problem? Is each group $a{\times}b$-decomposable? | |
| Nov 25, 2018 at 20:24 | comment | added | Derek Holt | Questions 1 and 2 have been answered in the affirmative, and the answer to Question 3 is also yes. There is a triple factorization ${\rm PSL}_2(13) = ABC$ into subgroups with $|A|=7$, $|B|=12$, $|C|=13$, with $B \cong A_4$, and you can write $B$ as a product of subgroups of order $3$ and $4$ to get the required $21 \times 52$ factorization. | |
| Nov 25, 2018 at 20:16 | answer | added | Ilya Bogdanov | timeline score: 5 | |
| Nov 25, 2018 at 19:17 | answer | added | François Brunault | timeline score: 5 | |
| Nov 25, 2018 at 9:50 | comment | added | Jeremy Rickard | Also: mathoverflow.net/questions/155986/factor-subset-of-finite-group | |
| Nov 25, 2018 at 9:46 | comment | added | Jeremy Rickard | Related question: mathoverflow.net/questions/177747/… | |
| Nov 25, 2018 at 6:52 | history | edited | Taras Banakh | CC BY-SA 4.0 |
Totally rewrote the problem.
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| Nov 25, 2018 at 2:28 | comment | added | bof | @mathworker21 $A=\{0,1\}$, $B=\{0,2\}$. ($A,B$ are subsets, not necessarily subgroups.) | |
| Nov 25, 2018 at 1:16 | history | asked | Taras Banakh | CC BY-SA 4.0 |