Timeline for Groups that do not exist
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Jan 14, 2020 at 6:47 | comment | added | Yuji Tachikawa | A bit more on @KConrad's comment: Dummit & Foote analyzes simple groups of order 168=$2^3\cdot 3\cdot 7$ and of order $3^3\cdot 7\cdot 13\cdot 409$ in parallel. The former exists but the latter doesn't. As Dummit & Foote said in their textbook, the simple group of this particular odd order needs to be ruled out as a small part of the proof of the odd order theorem. Dummit & Foote also outlines a character-theoretic proof of the nonexistence of this group on p.898. | |
| Dec 12, 2012 at 20:18 | comment | added | Mark Meckes | @Timothy: I assumed as much, although simply inserting "nonabelian" in the appropriate place strikes me as a better compromise. | |
| Dec 9, 2012 at 0:35 | comment | added | Timothy Chow | It's common practice among group theorists to omit the locution "except for cyclic groups of prime order" for brevity, understanding that the listener will insert it wherever it is necessary. | |
| Dec 8, 2012 at 21:23 | comment | added | Grant Olney Passmore | "Suppose G is a simple group of odd order. ... They were seeking, of course, to demonstrate a contradiction." But, there is no contradiction here, unless G is also specified to not be cyclic of prime order, right? | |
| Dec 8, 2012 at 13:08 | comment | added | Mark Meckes | I'm quite sure I can construct an infinite sequence of simple groups of odd order... | |
| Dec 7, 2012 at 23:20 | comment | added | KConrad | On p. 212 of Dummit & Foote (3rd ed.), the possibility of a finite simple group of order $3^3 \cdot 7 \cdot 13 \cdot 409$ is discussed, with a certain amount of consistent data for such a group being obtained, but ultimately there can't be such a group since there are no simple groups of odd order. | |
| Dec 7, 2012 at 22:33 | comment | added | Mariano Suárez-Álvarez | @Ricky, one possible situation I imagine is that one conjectures there is a group such and such, comes up with a description of its category of modules but which cannot correspond to a group; then it might come from a Hopf algebra, an association scheme, or some other things that resemble groups. | |
| Dec 7, 2012 at 21:45 | comment | added | user5810 | Is there a formal sense in which "there is almost such a group"? $\:$ (e.g., a binary operation on a set such that associativity fails for exactly one ordered triple of elements, which is otherwise a group) | |
| Dec 7, 2012 at 21:36 | comment | added | Nick Gill | Yes `the conjectured to exist' part of your question made me wonder if the FT-theorem was relevant. Still, as Gorenstein tells it, there is a real sense that something awfully like a group was lurking in that final configuration... Maybe a group that wanted to exist but didn't know how :-) | |
| Dec 7, 2012 at 21:26 | comment | added | Mariano Suárez-Álvarez | My question is somewhat vague, because it is difficult to pinpoint exactly what one can mean by «a group was conjectured to exist»; there is a difference between that (whatever it is :-) ) and assuming the group exists to derive a contradiction, as in F&T's proof, though. In the case of the odd order theorem, I think that experts believed the theorem to be true, and had from some time (since Burnside asked the question probably believing the answer was positive) | |
| Dec 7, 2012 at 21:18 | history | answered | Nick Gill | CC BY-SA 3.0 |