Timeline for Orders of products of permutations
Current License: CC BY-SA 4.0
28 events
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| Jun 23, 2019 at 14:03 | comment | added | YCor | It's a pity this question (the 2nd one) is so much hidden (with a title seemingly unrelated). I actually once spent a few hours within your questions trying to find it, without success. | |
| Jun 23, 2019 at 10:12 | history | edited | user6976 | CC BY-SA 4.0 |
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| Jun 23, 2019 at 8:03 | history | edited | YCor | CC BY-SA 4.0 |
incorporated 2nd question in title
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| Jun 23, 2019 at 5:31 | history | edited | user64494 | CC BY-SA 4.0 |
A typo in the title is corrected.
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| Jun 23, 2019 at 0:16 | history | edited | user6976 | CC BY-SA 4.0 |
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| Oct 6, 2011 at 15:09 | history | edited | user6976 | CC BY-SA 3.0 |
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| Aug 31, 2011 at 1:12 | history | edited | user6976 | CC BY-SA 3.0 |
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| Sep 21, 2010 at 4:16 | history | edited | user6976 | CC BY-SA 2.5 |
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| Sep 21, 2010 at 1:46 | comment | added | user6976 | @Victor: I just noticed your claim that "The condition that a,b act transitively on a set with at least n elements is equivalent to the condition that the group they generate has order at least n. " That is not true because $S_{n-1}$ is a non-transitive subgroup of $S_n$. | |
| Sep 21, 2010 at 1:37 | history | edited | user6976 | CC BY-SA 2.5 |
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| Sep 13, 2010 at 1:22 | history | bounty ended | CommunityBot | ||
| Sep 6, 2010 at 12:18 | comment | added | user6976 | In any case, yes, positive answer implies existence of infinite 2-generated groups of exponent $p$. But first of all the answer is supposed to be negative, at least I hope so. And second, the infinite Burnside groups of exponent $\ge 665$ do exist. So this observation does not give much. I was thinking about a proof that infinite Burnside groups do not embed into the factor-groups I described above. It is similar to the restricted Burnside problem. But there is no linearization even in the case of $p$-groups, and there is no Hall-Higman theorem reducing everything to the case of $p$-groups. | |
| Sep 6, 2010 at 11:22 | comment | added | Sergei Ivanov | No I meant the following type of an argument. Suppose $B(2,p)$ is finite, say it has $N$ elements. Then it can be defined by finitely many relations (e.g. by the full multiplication table). Each of these relations follows from some finite set of the original relations (those of the form $w^p=e$). Thus there is a finite set of relations of the form $w^p=e$ implying that the group has no more than $N$ elements. Once you replace your $2^p$ by the max length of these relations, you cannot get $n>N$. | |
| Sep 6, 2010 at 10:24 | comment | added | user6976 | Sergei, You mean taking a subgroup of the factor-group of the Cartesian product of the finite groups generated by $a_{i,n_i},b_{i,n_i}$ modulo the direct product? Yes, the subgroup generated by the sequences $(a_{i,n_i}),(b_{i,n_i})$ in that factor−group should be an infinite group of exponent $p$ if $i\to \infty$. Then one would want to apply Zelmanov's solution of the restricted Burnside problem but the factor-group is not necessarily residually finite, unfortunately. | |
| Sep 6, 2010 at 9:44 | comment | added | Sergei Ivanov | If, for every number in place of $2^p$, there is a sequence of examples with $n_i\to\infty$, then a diagonal procedure yields an infinite example for the bounded Burnside problem, isn't it? | |
| Sep 6, 2010 at 9:42 | comment | added | user6976 | Sergei, in fact one can assume that there are counterexamples for, say, $p=673$, any choice of "$2^p$" and any choice of "$2^{2^p}$". Then there is a double indexed sequence of pairs of permutations $a_{i,n}, b_{i,n}$, such that these are permutations of $\{1,...,n\}$, acting transitively, and every word of length $i$ in $a,b$ is of order $p$. I was thinking about passing to some kind of limit, as in the notion of graph limits from combinatorics or asymptotic cones, produce some infinite object which then can be studied by some "infinite mathematics". | |
| Sep 6, 2010 at 9:22 | comment | added | user6976 | Sergei, yes, $2^p$ is a "large number depending only on $p$", and $2^{2^n}$ is "an even larger number". | |
| Sep 6, 2010 at 9:21 | comment | added | user6976 | First of all there is a condition that $p$ does not divide $n$. Otherwise there is an issue with the restricted Burnside problem. If $p\not | n$, then any permutation of order $p$ has fixed points, which can be used. So your reformulation is not "equivalent". As I said, I am sure that any straightforward group theoretic approach to this problem will fail. This is a problem about graphs and, possibly, random walks on them. For example, a conjecture is that for every $a,b$ of order $p$ the random product of length $p^2$ of $a,b$ does no have order $p$ with positive probability. | |
| Sep 6, 2010 at 9:15 | comment | added | Sergei Ivanov | Where does the length bound $2^p$ come from? Does it just stand for "some large number", or is there something special about this constant? | |
| Sep 6, 2010 at 6:39 | comment | added | Victor Protsak | The condition that $a,b$ act transitively on a set with at least $n$ elements is equivalent to the condition that the group they generate has order at least $n.$ One direction is trivial and the other direction is given by the Cayley construction (action on itself by left multiplication). Thus the question can be reformulated as follows: does there exist a 2-generated group of large order $n>2^{2^p}$ in which all short words (length at most $2^p$) have order dividing $p$? | |
| Sep 6, 2010 at 1:01 | history | bounty started | CommunityBot | ||
| Sep 2, 2010 at 2:13 | history | edited | user6976 |
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| Sep 1, 2010 at 13:56 | comment | added | user6976 | It would be great to have a sequence $n_i\to \infty$, and pairs $a_i,b_i$ (for a fixed $p$). It would be even better to prove that such a sequence does not exist. | |
| Sep 1, 2010 at 11:58 | comment | added | Keivan Karai | Does the hypothetical construction have to work for all large enough values of $n$, or would it suffice to give a sequence $n_i \to \infty$ for which the construction works? | |
| Sep 1, 2010 at 9:30 | comment | added | user6976 | It does not apply to the bounded Burnside problem. Together with the solution of Burnside problem it applies to some other group theory problem. Unlike the bounded Burnside problem, this is about finite groups. In fact group theory motivation is misleading here because I do not see how to apply any group theory to this problem. It may be just a combinatorial Olympiad-type problem about graphs or it may belong to a completely different area of mathematics. So in this case, I think, the less you know about Burnside problems the better. | |
| Sep 1, 2010 at 8:21 | comment | added | Victor Protsak | The number 665 in the comment below, no doubt, arises from the negative solution of Burnside's problem. Can you, please, edit the question and indicate, as a motivation, what you know about the connection? | |
| Sep 1, 2010 at 2:58 | answer | added | Robert Bell | timeline score: 6 | |
| Sep 1, 2010 at 2:36 | history | asked | user6976 | CC BY-SA 2.5 |