Timeline for Calculating presentations for the normal subgroup of a semidirect product
Current License: CC BY-SA 2.5
16 events
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| Apr 9, 2010 at 17:25 | comment | added | HJRW | Ian: For me, your examples confuse rather than illuminate your question. If you're only interested in finite quotients, then as Jack says in his answer, there are well known techniques and the splitting H->G is unimportant. As you've already spotted, you can't hope for a finite answer in general. Do you have any examples with infinite H that you're interested in? | |
| Apr 9, 2010 at 14:38 | answer | added | Jack Schmidt | timeline score: 1 | |
| Apr 9, 2010 at 5:41 | comment | added | S. Carnahan♦ | You seem to be working with a definition of $PSL_2(Z)$ that is different from anything I've seen (although I've just found a rather strange assertion on Wikipedia). Half of your transformations have determinant -1, and are not in the image of anything in $SL_2(Z)$. | |
| Apr 9, 2010 at 5:13 | comment | added | Ian Brown | They do. For PSL_2(Z), the splitting consists of the linear fractional transformations that preserve the set {0,1,infty}. In other words, the 6 linear fractional transformations x, 1/(1-x), (x-1)/x, 1/x, 1-x, and x/(x-1). For SP_4(Z), it is a bit more complicated. I should remark that these splittings are very special to the low-dimensional cases. Neither the maps PSL_n(Z)-->PSL_n(Z/2Z) nor SP_{2n}(Z)-->SP_{2n}(Z/2Z) split for n large. | |
| Apr 9, 2010 at 5:13 | comment | added | Harry Gindi | @Ian Brown: That splitting property has to do with Grp being a semi-abelian category. ncatlab.org/nlab/show/semi-abelian+category Now you don't have to call it a "funny property" any more. | |
| Apr 9, 2010 at 5:05 | comment | added | S. Carnahan♦ | I don't think those splittings exist. | |
| Apr 9, 2010 at 4:54 | history | edited | Ian Brown | CC BY-SA 2.5 |
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| Apr 9, 2010 at 4:47 | comment | added | Ian Brown | Maybe I should say that I'm not interested in whether this is "computable", but in practical ways to compute it. I added another nice example in the text of the question. | |
| Apr 9, 2010 at 4:45 | comment | added | Ian Brown | Yes, but how would you compute that in any natural way (ie other than just starting to enumerate elements of G and then checking which ones map to H)? I suppose that what you say is an answer in some sense, but I'm looking for something along the lines of the Reidemeister-Schreier rewriting process. | |
| Apr 9, 2010 at 4:45 | comment | added | S. Carnahan♦ | I think the real question I should be asking is, what is your model of computation, so that we can distinguish between silly algorithms like mine (assuming I didn't mess up again) and something that would be useful to you? | |
| Apr 9, 2010 at 4:42 | comment | added | S. Carnahan♦ | My mistake. Instead, could you take all elements of G that map to the identity of H as generators, with multiplication in G defining relations? | |
| Apr 9, 2010 at 4:42 | comment | added | Ian Brown | A related comment -- the category of groups has the funny property that a splitting H->G of a short exact sequence 1-->N-->G-->H-->1 does not imply that there is a retract homomorphism G-->N. | |
| Apr 9, 2010 at 4:39 | comment | added | Ian Brown | I'm not sure I follow. There is not necessarily a homomorphism from G to N -- remember, H does not need to be normal. What kinds of relations are you talking about? | |
| Apr 9, 2010 at 4:38 | comment | added | S. Carnahan♦ | If you are allowing infinite presentations, can't you just take elements of G as generators, with relations defined by the multiplication table, together with additional relations sending anything to zero if its image in H is not the identity? | |
| Apr 9, 2010 at 4:33 | history | edited | Ian Brown | CC BY-SA 2.5 |
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| Apr 9, 2010 at 4:24 | history | asked | Ian Brown | CC BY-SA 2.5 |