Timeline for Finiteness of the Burnside Group
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
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| S Apr 28, 2023 at 17:37 | history | suggested | mathreader | CC BY-SA 4.0 |
correct spelling of `Finiteness'
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| Apr 28, 2023 at 15:47 | review | Suggested edits | |||
| S Apr 28, 2023 at 17:37 | |||||
| Aug 22, 2015 at 23:27 | comment | added | Leandro Vendramin | Related MO question: mathoverflow.net/questions/188968/… | |
| Aug 22, 2015 at 22:08 | history | edited | Yiftach Barnea | CC BY-SA 3.0 |
added 413 characters in body
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| Aug 20, 2015 at 21:08 | answer | added | YCor | timeline score: 6 | |
| Aug 20, 2015 at 21:04 | comment | added | Yiftach Barnea | If I understand correctly, then, in other words, if the group is finite, you can determine it, and determine its size. But what I am asking is a different question. The RBP gives an upper bound on the possible size, so can we determine that the group is larger than this size, thus, infinite. In general even for finitely presented groups this is impossible as there are finite presentations where you cannot even say whether the group is trivial, let alone larger than $k$. | |
| Aug 20, 2015 at 20:46 | comment | added | YCor | What I said is general about recursive presentations. If you have a recursive group presentation on finitely many generators which turns out to be that of a group of order $\le k$, then for some $n$ you get enough equalities in the $n$-ball to ensure it has $\le k$ elements, and enough equalities to ensure that all elements in the $(n+1)$-ball are in the $n$-ball: the algorithm just consisting in computing consequences of relations. But this is just an algorithm that stops if the group has cardinal $\le k$. | |
| Aug 20, 2015 at 20:38 | comment | added | Yiftach Barnea | Yves I am not sure I understand your claim. Isn't what you are saying is exactly what I am asking? If I understand correctly, you claim that for each $k$ you can check whether the size of $B(d,n)$ is k. Why? | |
| Aug 20, 2015 at 11:13 | comment | added | YCor | Btw, as you probably observed, the set of $(n,d)$ such that $B(n,d)$ has cardinal $\le k$ is obviously recursively enumerable (with an explicit algorithm). So whether we can test the other inequality is equivalent to determining whether $(n,d)\mapsto |B(n,d)|$ is computable. | |
| Aug 20, 2015 at 11:09 | comment | added | YCor | If $d$ is fixed, there is in principle an algorithm whose input is $n$ and the output is the cardinal of $B(d,n)$. This is just because this sequence is eventually infinite, hence computable, but this does not say what the sequence (nor the algorithm) is. | |
| Aug 20, 2015 at 11:09 | comment | added | Yiftach Barnea | The triple $(n,d,k)$. | |
| Aug 20, 2015 at 11:07 | comment | added | YCor | What is the input? the pair $(n,d)$? the triple $(n,d,k)$? | |
| Aug 20, 2015 at 11:03 | history | edited | Yiftach Barnea | CC BY-SA 3.0 |
edited title
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| Aug 20, 2015 at 10:56 | history | asked | Yiftach Barnea | CC BY-SA 3.0 |