Timeline for Is the universal inverse semigroup of a commutative semigroup an embedding?
Current License: CC BY-SA 3.0
9 events
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| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
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| Jul 2, 2012 at 19:56 | comment | added | Thomas Klimpel | @Fernando: It seemed to me, because I had only checked it for upper triangular matrices with at most one non-zero entry per row and column. Now it's obvious to me that these matrices were much too closely related to partial one-to-one transformations. I admit that the question is more appropriate for math.stackexchange, because finding a counter-example was easy once somebody told me that there is one. However, I really like the paper from B. Schein... | |
| Jul 2, 2012 at 19:12 | vote | accept | Thomas Klimpel | ||
| Jul 2, 2012 at 15:54 | comment | added | user6976 | In fact if you read Schein's paper (see my answer, the paper is available online), you will see that the question of whether all commutative semigroups embed into inverse semigroups was considered non-trivial. Even an example of a semigroup with commuting idempotents that does not embed into an inverse semigroup was not known for some time. | |
| Jul 2, 2012 at 14:42 | comment | added | Fernando Muro | Sorry, I was thinking of the associated group. | |
| Jul 2, 2012 at 14:40 | comment | added | user6976 | @Fernando: This semigroup (with addition, I assume) embeds into an inverse semigroup. | |
| Jul 2, 2012 at 13:52 | comment | added | Fernando Muro | Why does it seems to you? This question is indeed more appropriate for math.stackexchange. Take $S=[0,+\infty]$ for a counterexample. | |
| Jul 2, 2012 at 3:15 | answer | added | user6976 | timeline score: 6 | |
| Jul 1, 2012 at 21:36 | history | asked | Thomas Klimpel | CC BY-SA 3.0 |