Timeline for Is there a 0-1 law for the theory of groups?
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21 events
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| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
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| Aug 8, 2014 at 0:20 | comment | added | Gerry Myerson | A search of MathSciNet for zero-one law with MSC Primary 20 gave no results. | |
| Aug 7, 2014 at 22:27 | answer | added | David E Speyer | timeline score: 8 | |
| Dec 6, 2013 at 21:46 | answer | added | Emil Jeřábek | timeline score: 18 | |
| Dec 4, 2013 at 1:55 | comment | added | Jason Rute | From ww2.ii.uj.edu.pl/Schedae/idziak.pdf (2003, so out of date): "there are only a few results on zero–one (or more generally on limit) laws for specific theories $T$. One reason is that for such counting a deep insight into the structure of finite models of $T$ is required. The counting is even more difficult if the language of $T$ contains function symbols. Except for unary functions [...] and Abelian groups (where we completely understand the structure) there are only a few other results on limit laws for algebras to report. The reader may wish to consult [...]" | |
| Dec 3, 2013 at 23:15 | comment | added | Joel David Hamkins | @EmilJeřábek, why not expand that comment into an answer? I find that quite interesting. | |
| Dec 3, 2013 at 22:59 | comment | added | Emil Jeřábek | @Joel: According to math.uwaterloo.ca/~snburris/htdocs/DENSITY/overview.pdf , abelian groups have a limit law, but not a 0–1 law, and he gives an example that the probability of an abelian group having an element of order $2$ is $1-\prod_{n\ge1}(1-2^{-n})$. | |
| Dec 3, 2013 at 22:35 | comment | added | Joel David Hamkins | Can we hope to prove a 0-1 law for the class of finite abelian groups? After all, in abelian groups we have a decision procedure and some understanding of the possible assertions one can make in the first-order language of group theory. | |
| Dec 3, 2013 at 18:12 | comment | added | David E Speyer | @Michael The number of groups of order $p^n$ is $p^{\frac{2}{27} n^3 + O(n^{8/3})}$, see plms.oxfordjournals.org/content/s3-15/1/151.full.pdf . The number of extensions $1 \to C_p^{\lceil n/3 \rceil} \to G \to C_p^{\lfloor 2n/3 \rfloor} \to 1$ is also $p^{\frac{2}{27} n^3 + O(n^{8/3})}$ (with the $O()$ representing a different function of course.) This is the basis for the conjecture that almost all groups are $2$-step extensions, although these bounds aren't tight enough to prove that. The division by $3$ suggests we might be able to detect $n \bmod 3$, but I couldn't figure out how. | |
| Dec 3, 2013 at 17:15 | comment | added | Michael | A follow-up question: is the denominator of $p_N(\theta)$ known? Is there an asymptotic formula for the denominator as $N\to\infty$? | |
| Dec 3, 2013 at 0:19 | comment | added | Marty | @Michael, I think the more typical way to count isomorphism classes is to weight each class by the reciprocal of the cardinality of its automorphism group. So, each cyclic group of order $N$ would be weighted by $1 / \phi(N)$, for example. | |
| Dec 2, 2013 at 22:04 | comment | added | David E Speyer | Can we even figure out whether "extensions of the form $1 \to (\mathbb{Z}/2)^a \to G \to (\mathbb{Z}/2)^b \to 1$" obey a zero-one law? I thought about it a fair bit and came to the tentative conclusion that they do, but I don't think it's easy either way. | |
| Dec 2, 2013 at 21:26 | answer | added | Michael | timeline score: -1 | |
| Dec 2, 2013 at 21:18 | comment | added | Michael | @StefanKohl, I think $p_N(\theta)$ is a legitimate measure, in many way similar to the measure of natural numbers less than $N$. One can associate a natural number $n$ with the cyclic group $Z_n$ to show that natural numbers have "structure" (a.k.a. prime decomposition) and "order" (magnitude). And yet in the questions such as "what's the proportion of primes among natural numbers" one often takes essentially the same measure, the limit of the proportion among the natural numbers under $N$ for $N\to \infty$, as in this question for groups. | |
| Dec 2, 2013 at 20:46 | comment | added | Derek Holt | It's an interesting question, and it would be very surprising indeed if the answer was no (even if it is not true that almost all groups are class 2 nilpotent, then it seems very implausible that half of there are), but unfortunately there seems to be no prospects at all of resolving it. Or rather, it could only be resolved if it could somehow be proved true without needing to know much about the intricate details of the structure of "almost all groups". | |
| Dec 2, 2013 at 19:53 | history | edited | Stefan Kohl♦ |
Added top-level tag.
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| Dec 2, 2013 at 18:11 | comment | added | Stefan Kohl♦ | I think your definition of $p_N(\phi)$ is maybe not really what one wants here, as the order of a group is not a good measure for how much structure it may have: e.g. why should one give the 2328 groups of order $2^7$ in a sense a higher weight than the 9310 groups of order $3^7$, the 34297 groups of order $5^7$, the 113147 groups of order $7^7$, etc.? | |
| Dec 2, 2013 at 18:01 | review | First posts | |||
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| Dec 2, 2013 at 17:57 | comment | added | Emil Jeřábek | I see: Qiaochu Yuan mentions in the comments on MSE that standard conjectures imply that almost all groups are class-2 nilpotent. | |
| Dec 2, 2013 at 17:48 | comment | added | Emil Jeřábek | What about nilpotency class? | |
| Dec 2, 2013 at 17:42 | history | asked | Seirios | CC BY-SA 3.0 |