Timeline for Is it decidable to check if an element has finite order or not?
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Oct 8, 2016 at 16:38 | vote | accept | Al Tal | ||
| Oct 8, 2016 at 16:38 | |||||
| Oct 6, 2016 at 5:08 | history | edited | Francesco Polizzi | CC BY-SA 3.0 |
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| Oct 6, 2016 at 5:06 | comment | added | Francesco Polizzi | @TannerSwett: oh, right. I meant that the decidability of the world problem does not imply the decidability of the order problem. Fixed, thanks. | |
| Oct 5, 2016 at 23:50 | comment | added | Sophie Swett | @FrancescoPolizzi, don't you mean the answer is no (it is not decidable in general), rather than yes (it is decidable in general)? | |
| Oct 5, 2016 at 18:29 | comment | added | Al Tal | Indeed, Collins uses a special notation. For him, the identity element $1_G$ is a $0$-th power of all elements, so $(x,1)$ all belong to the set of power pairs. Therefore, the power problem may indeed have Turing degree below the order problem. | |
| Oct 5, 2016 at 18:11 | comment | added | Benjamin Steinberg | Maybe his version is not unidirectional | |
| Oct 5, 2016 at 18:01 | comment | added | Francesco Polizzi | At the moment I have no access to the full paper. However, the abstract says: The chapter concludes by proving a lemma that shows that groups with a soluble power problem and an insoluble order problem have an unexpected algebraic property, so it seems that he is actually able to provide examples where the power problem has a lower Turing degree than the order problem. | |
| Oct 5, 2016 at 17:51 | comment | added | Benjamin Steinberg | Unless Collins is using a different definition of the power problem than McCool I think the power problem should have a higher Turing degree than the order problem | |
| Oct 5, 2016 at 17:27 | history | edited | Francesco Polizzi | CC BY-SA 3.0 |
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| Oct 5, 2016 at 17:21 | history | answered | Francesco Polizzi | CC BY-SA 3.0 |