Timeline for Infinite group for which it is still unknown if it is simple
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Apr 15, 2021 at 15:01 | history | became hot network question | |||
| Apr 15, 2021 at 11:31 | history | edited | YCor | CC BY-SA 4.0 |
fixed English, expanded title
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| Apr 15, 2021 at 10:52 | review | First posts | |||
| Apr 15, 2021 at 10:55 | |||||
| Apr 15, 2021 at 9:50 | comment | added | YCor | @ADL this is indeed imprecise but the way the group comes into play is also "to the appreciation of the reader": as group presentation? as group explicitly described as acting by automorphisms on some structure. In any case, I quite like Uri's answer. | |
| Apr 15, 2021 at 9:44 | comment | added | ADL | @YCor Sure, although I'm not sure why you mention simple groups. My point is that any "criterion for determining the simplicity" of finitely presented groups cannot be algorithmically checked on general examples (unless the criterion implies, but is not equivalent to, simplicity - the word "determines" makes the precise situation unclear to me, but that's maybe just me). | |
| Apr 15, 2021 at 8:56 | comment | added | YCor | @ADL sure; this also applies to "trivial groups" — the input here is a finite group presentation. | |
| Apr 15, 2021 at 8:36 | answer | added | Uri Bader | timeline score: 34 | |
| Apr 15, 2021 at 8:35 | comment | added | ADL | Finitely presented simple groups have decidable word problem, and so being simple is a Markov property, and so it is undecidable if a given presentation defines a simple group. (In his survey article "Decision problems for groups: survey and reflections", Miller attributes this to Kuznetsov. He also provides an easy, half-page proof.) | |
| Apr 15, 2021 at 7:26 | comment | added | YCor | I also think I remember there are some diffeomorphism groups ($\mathrm{Diff}^k_0(M)$ for suitable $k$ and manifold $M$, index $0$ meaning isotopic to identity), whose simplicity is unknown — and certainly unrelated to set-theoretic subtleties as in my previous example. | |
| Apr 15, 2021 at 7:23 | comment | added | YCor | The question is very vague, but let me mention that if A=2^N/fin is the quotient Boolean algebra of the power set of N by finite subsets, then it's undecidable in ZFC whether Aut(A) is simple. (But it's a theorem of ZFC, due to Rubin, that its derived subgroup is simple, and it's a theorem of ZFC+CH that Aut(A) is simple.) | |
| Apr 15, 2021 at 7:01 | history | asked | user181168 | CC BY-SA 4.0 |