Timeline for How to get Latin squares from a finite group and a subgroup
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Aug 24, 2017 at 5:11 | comment | added | Anupam Ah | you mean their product in $S_5$? like you said in the very first comment the unique element of $ X\cap xyH$? | |
| Aug 24, 2017 at 5:04 | comment | added | Colin Reid | As the multiplication table of the operation $*$. | |
| Aug 24, 2017 at 4:46 | comment | added | Anupam Ah | I am actually confused of the construction of Latin squares out of transversal's. Let me start with the latin square I said in en.wikipedia.org/wiki/Latin_square. It is a reduced Latin square. I read out the rows as functions, then I will be getting a collection $X$ of permutations in $S_5$ satisfying for $f,g \in X$ distinct $f(x) \neq g(x) \forall x$. Using these representative how can we construct the latin square back? | |
| Aug 23, 2017 at 21:05 | comment | added | Colin Reid | I would guess that if you start with a reduced Latin square (taking $S_4$ to be the stabilizer of $1$ in $S_5$) and read off the rows as functions, you get your original Latin square back. | |
| Aug 23, 2017 at 12:41 | comment | added | Anupam Ah | I need a particular latin square which does not arise from a group. It exists only in order 5. see it in en.wikipedia.org/wiki/Latin_square. For that which I should choose? | |
| Aug 23, 2017 at 11:56 | comment | added | Colin Reid | Sure, but the loop you get in that case will just be the cyclic group $C_5$. What do you want to do with the resulting Latin square? | |
| Aug 23, 2017 at 11:09 | comment | added | Anupam Ah | I need just one set only. You are telling powers of (1,2,3,4,5) will work. | |
| Aug 23, 2017 at 11:00 | comment | added | Colin Reid | In the special case of $S_{n-1}$ inside $S_n$, a loop transversal amounts to picking a set of permutations $X$ such that distinct elements $f,g \in X$ satisfy $\forall x: f(x) \neq g(x)$. There will be many such choices in general (the most obvious is to take the powers of an $n$-cycle), but it's not easy to tell which will give isomorphic loops. | |
| Aug 23, 2017 at 10:37 | vote | accept | Anupam Ah | ||
| Aug 23, 2017 at 10:32 | comment | added | Anupam Ah | Is it possible to construct one such set in particular for $G=S_5$ and $H=S_4$ the symmetric groups? | |
| Aug 22, 2017 at 8:41 | history | edited | Colin Reid | CC BY-SA 3.0 |
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| Aug 22, 2017 at 8:01 | history | answered | Colin Reid | CC BY-SA 3.0 |