Timeline for Cycles in almost breakable semigroups
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Jul 7, 2023 at 12:46 | history | edited | Salvo Tringali | CC BY-SA 4.0 |
fixed something that had got twisted in my mind
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| Jul 6, 2023 at 14:55 | vote | accept | Salvo Tringali | ||
| Jul 6, 2023 at 2:23 | answer | added | Benjamin Steinberg | timeline score: 1 | |
| Jul 5, 2023 at 21:51 | comment | added | Benjamin Steinberg | This is similar to the argument I had but I made choices for the products. I think this pattern should continue | |
| Jul 5, 2023 at 21:49 | comment | added | Salvo Tringali | Minor remark (I don't know if it can help): If $(x_1,x_2,x_3)$ is a $3$-cycle in an almost breakable semigroup, then $x_1x_2,x_2x_3,x_3x_1\notin\{x_1,x_2,x_3\}$. By symmetry (and the fact that $x_1\ne x_1x_2\ne x_2$), it suffices to check that $x_1x_2\ne x_3$. Assume to the contrary that $x_1x_2=x_3$. Since $x_2x_1\in\{x_1,x_2\}$, we then have that $x_3x_1=x_1x_2x_1\in\{x_1,x_1x_2\}=\{x_1,x_3\}$. This is however impossible, because $(x_3,x_1)$ is an irregular pair. | |
| Jul 5, 2023 at 21:36 | history | edited | Salvo Tringali | CC BY-SA 4.0 |
extended the terminology
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| Jul 5, 2023 at 20:38 | comment | added | Benjamin Steinberg | If my back of the envelope calculation is correct then there cannot be 3-cycles and so I suspect no cycles. The trick is to start the cycle at the largest element of the J-order. | |
| Jul 5, 2023 at 19:29 | comment | added | Salvo Tringali | Maybe it would be better to say "irregular $n$-cycles" (rather than "$n$-cycles"). Graphically, let $G(S)$ be the digraph whose vertex set is (the underlying set of) $S$ and where a node $x$ is joined to a node $y$ by an arc iff $x \ne xy \ne y$. Then I'm asking whether $S$ being almost breakable implies that $G(S)$ is a DAG (a directed acyclic graph). I'm not sure whether this makes the notion any clearer, but... | |
| Jul 5, 2023 at 19:22 | comment | added | Benjamin Steinberg | I'm not sure I've assimilated your notion of cycle but the structure of an almost breakable semigroup is very close to that of a breakable one. The principal ideals must be a chain. Each J-class is either a left or right zero semigroup. The difference seems to be that elements which are J-above need only fix elements below on one side | |
| Jul 5, 2023 at 18:56 | history | asked | Salvo Tringali | CC BY-SA 4.0 |