Timeline for An ubiquitous pattern of questions
Current License: CC BY-SA 2.5
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 17, 2010 at 4:14 | comment | added | François G. Dorais | I thought some more about how much of Cayley's Theorem you could get by general principles like the above. Robinson's Theorem comes closest, but the only conclusion I could get is that there are first order pure group theory sentences such that every group without a faithful action satisfies one of them and no permutation group satisfies any of them. This is not completely trivial assuming you're not aware of Cayley's Theorem. (The same reasoning applies perhaps less trivially to other representations of groups.) | |
| Feb 16, 2010 at 8:34 | vote | accept | Hans-Peter Stricker | ||
| Feb 16, 2010 at 7:18 | comment | added | Sridhar Ramesh | Ah, alright. Just curious. No worries on not seeing my reply when you deleted your comment; quite possibly, I was still in the middle of composing it. | |
| Feb 16, 2010 at 7:06 | comment | added | François G. Dorais | I retracted my comment because it didn't work, as you observed. (I didn't see you had replied when I deleted it.) You can do a similar trick as with 3 & 4, but again the result is completely trivial. In order to apply Beth's Theorem, you really want phi(x) to say something about the objects of interest. In this case, x should be a group, which makes the process trivial again. | |
| Feb 16, 2010 at 6:48 | comment | added | Sridhar Ramesh | Hm, I appear to have replied to a comment that no longer exists. It was something about how to cast 1 (Cayley's theorem) into this formalism. Have you retracted it because you no longer believe this can be done in a non-trivial way (using Beth's definability theorem to obtain a non-trivial result about groups which substantially anticipates Cayley's theorem) or have you retracted it for some other reason? If the latter, I would still appreciate help seeing the details written out, as I was not able to understand from your previous comment quite what everything should be, as noted above. | |
| Feb 16, 2010 at 6:41 | comment | added | Sridhar Ramesh | Well, suppose T is just the theory of a group and phi is the proposition stating that the action is faithful. Then we won't satisfy the condition that models which agree on the group structure also automatically agree on phi. Alternatively, T could be the theory of groups + the condition that the action is faithful. Then all we get out of this is in the end is that T proves that phi is equivalent to phi_0; i.e., that the two are equivalent for all groups with a faithful action; but this tells us nothing about what kinds of groups actually have faithful actions. Sorry, I still need help. | |
| Feb 16, 2010 at 6:34 | comment | added | François G. Dorais | I had misread 3 when I wrote the above comment. However, 3 & 4 are trivially realized by taking L to be a suitable extension of arithmetic, but then the result is utterly trivial. | |
| Feb 16, 2010 at 6:15 | comment | added | Sridhar Ramesh | No, wait, actually, I can't see how to cast the first one into this formalism; what I was thinking of doing doesn't work quite right. Could you perhaps illustrate exactly what T, L, phi, and phi_0 should be? | |
| Feb 16, 2010 at 6:10 | comment | added | Sridhar Ramesh | Oh, I see now that 1 says "isomorphic to a subgroup of the symmetric group on G" and not just "isomorphic to a subgroup of some symmetric group". I can see how to cast that first one into this formalism. But I couldn't see how to cast the latter. | |
| Feb 16, 2010 at 6:02 | comment | added | François G. Dorais | 1 & 3 are pretty straightforward. (Though not very interesting.) | |
| Feb 16, 2010 at 5:38 | comment | added | Sridhar Ramesh | Hm... Are any of the examples from the question easily cast into this formalism? | |
| Feb 16, 2010 at 4:05 | history | edited | François G. Dorais | CC BY-SA 2.5 |
fixed a small error
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| Feb 16, 2010 at 3:04 | history | answered | François G. Dorais | CC BY-SA 2.5 |