Timeline for Have any publications been made in this area of group theory?
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24 events
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| Apr 27, 2012 at 8:23 | history | rollback | user20886 |
Rollback to Revision 5
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| Apr 26, 2012 at 13:44 | history | edited | user20886 | CC BY-SA 3.0 |
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| Feb 1, 2012 at 19:35 | vote | accept | user20886 | ||
| Jan 26, 2012 at 19:44 | vote | accept | user20886 | ||
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| Jan 26, 2012 at 19:43 | history | edited | user20886 | CC BY-SA 3.0 |
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| Jan 26, 2012 at 19:38 | vote | accept | user20886 | ||
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| Jan 26, 2012 at 19:35 | vote | accept | user20886 | ||
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| Jan 26, 2012 at 19:35 | vote | accept | user20886 | ||
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| Jan 26, 2012 at 19:22 | comment | added | Benjamin Steinberg | Colin, I give the proof in my second answer. | |
| Jan 26, 2012 at 19:04 | answer | added | Benjamin Steinberg | timeline score: 3 | |
| Jan 26, 2012 at 16:32 | answer | added | Colin Reid | timeline score: 0 | |
| Jan 26, 2012 at 16:28 | comment | added | Colin Reid | @Benjamin Steinberg: '...the resulting group is finite'. This is not obvious to me, even given restricted Burnside. Reference/proof? | |
| Jan 26, 2012 at 12:47 | comment | added | Benjamin Steinberg | Also it follows from Zelmanov's solution to the restricted Burnside problem that there are only finitely many equivalence classes of such mappings on G for all k iff G had bounded exponent and each finitely generated subgroup of G is finite. | |
| Jan 26, 2012 at 12:15 | comment | added | Benjamin Steinberg | If you want to identify all positive words that agree on all k-tuples of group elements you get the free monoid in the variety generated by G on k generators. If you allow all group words you get the free group in the variety generated by G on k-generators. If G is finite these two guys are the same and the resulting group is finite. | |
| Jan 26, 2012 at 10:19 | comment | added | Alain Valette | Syzygy? ams.org/notices/200604/what-is.pdf | |
| Jan 26, 2012 at 8:40 | comment | added | Colin Reid | If your group G has infinite exponent then yes, it matters whether you allow negative powers. (The positive words only form a monoid in general.) If the group you start with has finite exponent, then you have inverses for free and you are effectively asking about certain quotients of Burnside groups defined by identities. | |
| Jan 26, 2012 at 1:07 | history | edited | user20886 | CC BY-SA 3.0 |
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| Jan 26, 2012 at 0:38 | answer | added | Benjamin Steinberg | timeline score: 2 | |
| Jan 26, 2012 at 0:35 | answer | added | Igor Rivin | timeline score: 6 | |
| Jan 26, 2012 at 0:31 | comment | added | Colin Reid | Your space (if I understand correctly) is the free group on k letters modulo all identities satisfied by G. Groups defined by certain sets of identities have been extensively studied, although it can be difficult to understand them in general: see for instance the theory of Burnside groups or Engel groups. | |
| Jan 26, 2012 at 0:29 | history | edited | Yemon Choi |
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| Jan 25, 2012 at 23:41 | history | edited | user20886 | CC BY-SA 3.0 |
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| Jan 25, 2012 at 23:34 | comment | added | Ryan Budney | In a vague sense it sounds like you're looking not just at the relations in a space, but relations among the relations. In many ways that's what group cohomology is about. | |
| Jan 25, 2012 at 23:32 | history | asked | user20886 | CC BY-SA 3.0 |