Timeline for decidability of group homomorphism existence
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| S Sep 27, 2013 at 8:45 | history | suggested | Sergiy Kozerenko | CC BY-SA 3.0 |
TeX corrections
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| Sep 27, 2013 at 7:42 | review | Suggested edits | |||
| S Sep 27, 2013 at 8:45 | |||||
| Sep 6, 2011 at 23:00 | answer | added | user6976 | timeline score: 8 | |
| Nov 4, 2009 at 3:09 | answer | added | HJRW | timeline score: 3 | |
| Oct 20, 2009 at 23:36 | comment | added | Steven Sam | I think the original poster means finite cyclic instead of cyclic. My understanding of this is that if we have two words x and y in H, and we want to know if they are equal, we just set h = xy^{-1} and see if every generator of a finite cyclic group maps to it. If so, then it has to be the identity element because it has order n for all n. Well, I haven't shown that this question is undecideable for a given finite cyclic group (just that we can't do it for all of them for any given h), so maybe there's a better way to see this. | |
| Oct 19, 2009 at 21:11 | answer | added | Hugh Thomas | timeline score: 1 | |
| Oct 19, 2009 at 14:45 | answer | added | Charles Siegel | timeline score: 0 | |
| Oct 18, 2009 at 21:22 | comment | added | Reid Barton | Maybe I'm missing something but if the problem is undecidable when G is cyclic, how could it not also be undecidable when we know less about G? | |
| Oct 18, 2009 at 16:46 | history | asked | user667 | CC BY-SA 2.5 |