Timeline for Cancelable commutative monoids with finite maximal subgroups
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 24, 2023 at 18:14 | vote | accept | Nate Ackerman | ||
| Mar 24, 2023 at 17:44 | answer | added | Benjamin Steinberg | timeline score: 4 | |
| Mar 24, 2023 at 17:44 | history | became hot network question | |||
| Mar 24, 2023 at 17:05 | comment | added | Benjamin Steinberg | @Ycor, as yes I forgot 0. Sorry | |
| Mar 24, 2023 at 14:34 | answer | added | R. van Dobben de Bruyn | timeline score: 9 | |
| Mar 24, 2023 at 13:36 | comment | added | YCor | @BenjaminSteinberg the group of "units" is then the group of eventually zero sentences. | |
| Mar 24, 2023 at 12:33 | comment | added | Benjamin Steinberg | @YCor doesnt the eventually nonnegative sequences have trivial group of units and hence what the OP wants is true because the congruence is trivial? | |
| Mar 24, 2023 at 10:58 | comment | added | Benjamin Steinberg | This shouldn't be true. I don't think much about cancellative commutative monoids but I don't see why they should have a submonoid transversal to Green's H-relation. | |
| Mar 24, 2023 at 8:49 | comment | added | YCor | If $G$ is not a direct factor in the group envelope of $M$ this is probably not the case. For instance take $M$ the additive monoid of sequences $\mathbf{N}\to\mathbf{Z}$ that are eventually non-negative. | |
| Mar 24, 2023 at 8:32 | history | asked | Nate Ackerman | CC BY-SA 4.0 |