Talk:Wreath product
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It seems like all we need is that H acts on the set U. Is it necessary that H act transitively, or that it be a permutation group?
Also, if U is infinite, I have an inkling that we should use the direct product rather than the direct sum of copies of G. Although I don't know why. 199.17.238.92 22:42 Nov 13, 2002 (UTC)
- All my references seem to require H to be a transitive permutation group for the unrestricted product; but it appears that the generalized wreath product does not require transitivity; I don't have a written reference, but you can try [1] (http://www.mathe2.uni-bayreuth.de/frib/html/book/hyl00_63.html#6) for example (it seems pretty dense going).
- If H is not a permutation group, but only a transformation group, then I think we just find that if we define N = {h in H : h.u = u for all u in U}, then we essentially get G wr (H,U) = (G wr (H/N, U)) × N, with (H/N, U) a permutation group; so it's typically not very interesting.
- I think the reason for the restriction to transitive H is one of applications. We certainly never need to worry about transitivity when talking about G wr K for ordinary groups (i.e., taking K as the permutation group (K, K)); and one writer [2] (http://www.math.uni.wroc.pl/~kisiel/permutat.ps) seemed to imply that the wreath product was originally developed as a tool to describe the structure of permutation groups, where intransitive groups are just direct products or semidirect products of transitive subgroups. But the real answer is - I dunno!
- As regards direct sum, we actually want the external direct sum; which is a subset of the direct product over U, including only those elements whose components are eG except for a finite number of components. Particularly for uncountable U, this allows us to ensure that limits of a product exist, etc. The construction is not unlike the Tychonoff product of topologies. Chas zzz brown 00:48 Nov 14, 2002 (UTC)
Minor correction to the article proposed: I think the example should be C2 wr (Sn, n) instead of C2 wr Sn, if we want to keep the notation consistent. Could anyone of the experts in the subject change it or explain why not? Regards, Alex.
- Alex - I don't think there is any ambiguity in this case since n is given to specify that H is a subgroup of <math>S_n<math>, which in the case <math>H = S_n<math> is obvious. - Gauge 22:00, 23 Feb 2005 (UTC)