Timeline for Can you decide whether the commutator subgroup of a f.p. group is f.g?
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 24, 2015 at 19:10 | vote | accept | suitangi | ||
| Mar 23, 2015 at 23:41 | answer | added | YCor | timeline score: 16 | |
| Mar 23, 2015 at 17:46 | review | Close votes | |||
| Mar 23, 2015 at 18:27 | |||||
| Mar 23, 2015 at 17:37 | comment | added | Todd Trimble | I agree with Derek. I think the question is fine for MO, and rather than close it because it's been answered in a comment, it would be better to turn the comment into an actual answer (and I'd be pleased if @YCor does the honors). I spoke further on this here: meta.mathoverflow.net/a/2111/2926 | |
| Mar 23, 2015 at 17:32 | comment | added | Derek Holt | The solution is easy when you see it, but not obvious, so I would not vote to close it. | |
| Mar 23, 2015 at 17:15 | comment | added | YCor | It's undecidable, by a classical argument. Indeed, a finitely generated group $G$ is nontrivial if and only if the commutator subgroup of $G\ast\mathbf{Z}$ is infinitely generated. Hence if we could solve this problem by some machine $X$, we would solve the triviality problem of $G$ by inputting in $X$ the presentation with an additional generator. | |
| Mar 23, 2015 at 17:11 | review | First posts | |||
| Mar 23, 2015 at 17:23 | |||||
| Mar 23, 2015 at 17:06 | history | asked | suitangi | CC BY-SA 3.0 |