Timeline for Do these conditions on a semigroup define a group?
Current License: CC BY-SA 2.5
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| Jun 15, 2020 at 7:27 | history | edited | CommunityBot |
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| Jun 12, 2010 at 18:38 | vote | accept | Arturo Magidin | ||
| Jun 10, 2010 at 18:39 | comment | added | Arturo Magidin | In Herstein's "Topics"(in Spanish, 1988); the exercises in Section 2.3 include: prove that a semigroup with right identity and right inverses is a group (problem 12); prove that right identity and left inverses do not yield a group (problem 13); prove that a finite semigroup with both cancellation laws is a group (problem 14); show that a single cancellation law does not suffice (problem 16); prove that if we drop "finite" in problem 14 then the cancellation laws do not suffice (problem 17). Hungerford (section I.2) probem 15 is the same as 14 and 17 from Herstein. No thers I can see. | |
| Jun 10, 2010 at 18:32 | answer | added | Arturo Magidin | timeline score: 16 | |
| Jun 10, 2010 at 18:20 | comment | added | Gerhard Paseman | The cex would be not just infinite, but torsion-free. This sounds vaguely like a homework problem in Hungerford or Herstein. Does anyone know the difference between those homework problems and this formulation? Gerhard "Ask Me About System Design" Paseman, 2010.06.10 | |
| Jun 10, 2010 at 18:18 | comment | added | Homology | oops, sorry then! | |
| Jun 10, 2010 at 18:14 | comment | added | Yemon Choi | @Homology - the original question observes that any counterexample, if it exists, would have to be infinite | |
| Jun 10, 2010 at 17:56 | comment | added | Homology | If $S$ is finite, there exists $e$ s.t. $e^n=e$ for some $n>1$ (by iterating $x \mapsto x^2$), so $e^{n-1}a=a$ for every $a$ in $S$, and you can finish in the same way. So if a counterexample exists, it has to be infinite... | |
| Jun 10, 2010 at 16:59 | history | asked | Arturo Magidin | CC BY-SA 2.5 |