Timeline for "Embedding" functions in groups
Current License: CC BY-SA 2.5
13 events
| when toggle format | what | by | license | comment | |
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| Feb 9, 2010 at 18:01 | answer | added | Gerhard Paseman | timeline score: 1 | |
| Dec 29, 2009 at 9:13 | answer | added | Dinesh | timeline score: 1 | |
| Dec 29, 2009 at 4:34 | comment | added | Jason DeVito - on hiatus | @Barton: Yes, you're right. I wish there was a good way to edit comments! | |
| Dec 29, 2009 at 3:59 | comment | added | Reid Barton | @Dinesh: can you explain a little why you are interested in this question? "Smallest abelian group" is kind of a strange notion, and maybe if we knew the motivation we could get a better perspective on this question. | |
| Dec 29, 2009 at 3:57 | comment | added | Reid Barton | @Jason: Don't you need Z_3, not Z_2, since otherwise x + x is a constant not depending on x? | |
| Dec 29, 2009 at 3:12 | comment | added | Jason DeVito - on hiatus | @Figueroa-O'Farrill, requiring that f is symmetric is neccesary, but it is also sufficient: let G be the free Z_2 module with basis S. Define g to be the obvious inclusion. Define h(x+y) = f(x,y) and define h arbitrarily on all the other elements of G. (Well definedness of h is equivalent to symmetry of f). Thus, we can "embed" f into an abelian group of size 2^|S|. | |
| Dec 29, 2009 at 2:50 | comment | added | José Figueroa-O'Farrill | Clearly not every function will be embeddable. I think you'll need at least to demand that it be symmetric: $f(x,y) = f(y,x)$. | |
| Dec 29, 2009 at 2:38 | history | edited | Dinesh | CC BY-SA 2.5 |
Included Latex for subscripts
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| Dec 29, 2009 at 2:36 | comment | added | Dinesh | By "embeddable", I mean that the function's action should be mimicked by the group operation of an Abelian group (with a possible re-mapping of the outcome of the group operation). If you can phrase the question better, please go ahead. My own conception is nebulous and I might not be describing the problem in its clearest terms. | |
| Dec 29, 2009 at 2:34 | history | edited | Qiaochu Yuan | CC BY-SA 2.5 |
added 257 characters in body
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| Dec 29, 2009 at 2:30 | comment | added | Qiaochu Yuan | Oh, I see; f is required to be the group operation. Do you mind if I rewrite your question to be a little clearer? | |
| Dec 29, 2009 at 2:28 | comment | added | Qiaochu Yuan | I don't understand what you mean by "embeddable." Is f required to satisfy some kind of symmetry with respect to the group or what? (Also, does f have a codomain which is the same finite set from which x and y are taken?) | |
| Dec 29, 2009 at 2:20 | history | asked | Dinesh | CC BY-SA 2.5 |