Timeline for Is there a 0-1 law for the theory of groups?
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Dec 3, 2013 at 19:21 | comment | added | Joel David Hamkins | I agree with Emil, and had objected in my comment only because the question mentions the first-order language of groups. | |
| Dec 3, 2013 at 12:30 | comment | added | Emil Jeřábek | Joel is quite right, but I think it should be mentioned that while the question asks specifically about first-order logic, where cyclicity is indeed not expressible, there is nothing particularly fundamental about this choice. People have studied 0–1 laws for various extensions of FO that can express cyclicity, like FO(TC), FO(IFP), or MSO. Even more importantly, when we discuss a 0–1 law for a certain logic relative to a class of structures, there is no need for this class to be itself definable in said logic. It makes perfect sense to consider an FO 0–1 law on the class of all cyclic groups. | |
| Dec 2, 2013 at 22:48 | comment | added | Joel David Hamkins | To achieve a 0-1 law, which are truly amazing when they are true, one should have a precise formal language. | |
| Dec 2, 2013 at 22:48 | comment | added | Joel David Hamkins | The language of groups allows you to form terms using the group operation, to express equalities and inequalities, to use logical connectives and to quantify over the elements of the group ("there is a group element $x$ such that..." or "for every element $x$ in the group..."). You may not generally quantify over natural numbers, over subsets of the group, over homomorphisms etc. (Those are expressible in more powerful languages, such as the language of set theory or suitable category-theoretic languages, but not in the language of groups.) | |
| Dec 2, 2013 at 22:25 | comment | added | Michael | @JoelDavidHamkins, why the existence of surjective homomorphism from $Z$ to $G$ is not a statement in the language of groups? | |
| Dec 2, 2013 at 21:52 | comment | added | Joel David Hamkins | Another problem with this answer is that being cyclic is not expressible in the language of groups. So the assertion $\theta$ that is discussed is not in the desired language. | |
| Dec 2, 2013 at 21:49 | history | edited | Michael | CC BY-SA 3.0 |
added 157 characters in body
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| Dec 2, 2013 at 21:46 | comment | added | Michael | Oops, you are right, I meant $p(\theta)/p(being_{ }cyclic)=6/\pi^2$. Need to think whether the answer is recoverable for the property of not having square roots of unity alone, w/out being cyclic. | |
| Dec 2, 2013 at 21:35 | comment | added | Stefan Kohl♦ | This answer is wrong since $p(\theta) = 0 \in \{0,1\}$. -- If it would be that easy, the question wouldn't be interesting. | |
| Dec 2, 2013 at 21:26 | history | answered | Michael | CC BY-SA 3.0 |