Timeline for Is it decidable to check if an element has finite order or not?
Current License: CC BY-SA 3.0
15 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 | vote | accept | Al Tal | ||
| Oct 8, 2016 at 16:38 | |||||
| Oct 8, 2016 at 16:38 | vote | accept | Al Tal | ||
| Oct 8, 2016 at 16:38 | |||||
| Oct 5, 2016 at 18:48 | comment | added | Al Tal | @YCor Indeed, the requirements you mentioned are satisfied in the construction of Higman's embedding theorem. This is is mentioned (and used) in the McCool's paper from Benjamin's answer below. | |
| Oct 5, 2016 at 17:21 | answer | added | Francesco Polizzi | timeline score: 21 | |
| S Oct 5, 2016 at 17:05 | history | suggested | Fan Zheng |
Added two tags.
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| Oct 5, 2016 at 17:02 | answer | added | YCor | timeline score: 15 | |
| Oct 5, 2016 at 16:49 | review | Suggested edits | |||
| S Oct 5, 2016 at 17:05 | |||||
| Oct 5, 2016 at 16:31 | answer | added | Benjamin Steinberg | timeline score: 38 | |
| Oct 5, 2016 at 14:50 | comment | added | Derek Holt | @YCor This does not answer your question, but there is a result by Birget, Olshanskii, Rips, Sapir that a finitely generated group with word problem in NP embed quasi-isometrically as a subgroup of a finitely presented group with polynomial Dehn function. | |
| Oct 5, 2016 at 14:25 | comment | added | YCor | @DerekHolt I forgot to say that I need the image to be a recursive subset, so we would need in addition this, and that the f.p. group itself has a solvable word problem (possibly this follows from the original construction) | |
| Oct 5, 2016 at 14:05 | comment | added | Derek Holt | The question is interesting (and I believe unknown) for automatic groups. But for automatic groups it iseems possible that the answer might be yes, and I would be very surprised if it was yes more generally. | |
| Oct 5, 2016 at 14:02 | comment | added | Derek Holt | @YCor Yes, Boone-Higman, 1974 says a finitely generated group G has solvable word problem if and only if it can be embedded in a simple subgroup of some fintiely presented group. | |
| Oct 5, 2016 at 13:44 | comment | added | YCor | A restatement is whether in a finitely presented group with solvable word problem, the set of finite order elements is recursive (it's clearly recursively enumerable). I'm not sure of the details of Higman's theorem, but doesn't it embed every f.g. group with solvable word problem into a finitely presented one? This would help to get rid of the somewhat constraining finite presentability assumption. | |
| Oct 5, 2016 at 5:41 | history | asked | Al Tal | CC BY-SA 3.0 |