[math-fun] uniquely factored groups
JHC> The only candidate for an isomorphism that these invariants leave is between 2xQ8 and 4 :^3 4, which is prohibited by the unique factorisation theorem for direct products, since it's easy to see that the latter of these two isn't a non-trivial direct product. John, could you state precisely the unique factorization theorem? I've never seen it written down. There seem to be some subtleties, such as checking that AxB = AxC implies B=C. (For groups, of course, we interpret "=" as isomorphism.) Is it also true for infinite groups? Rich
On Tue, 23 Sep 2003, Richard Schroeppel wrote:
John, could you state precisely the unique factorization theorem? I've never seen it written down. There seem to be some subtleties, such as checking that AxB = AxC implies B=C. (For groups, of course, we interpret "=" as isomorphism.) Is it also true for infinite groups?
The easiest (because weakest) statement is that every finite group is a direct product of indecomposable groups, and that the isomorphism types of these groups are uniquely determined up to order. ["Indecomposable" means "not a direct product of two non-trivial groups"]. There are stronger forms, but I can never remember them, and find it dangerous to try, because so many possible strengthenings are wrong. For example the groups C2 and D8 (the latter meaning the dihedral group of order 8) are indecomposable, but there are four (or maybe 8) different pairs of subgroups of types C2 and D8 in C2 x D8 of whioch it's the direct product. Regards, John Conway
Quoting John Conway <conway@Math.Princeton.EDU>:
The easiest (because weakest) statement is that every finite group is a direct product of indecomposable groups, and that the isomorphism types of these groups are uniquely determined up to order. ["Indecomposable" means "not a direct product of two non-trivial groups"].
I hope that I haven't blundered into a conversation which I don't understand, but since I have been struggling with this while trying to identify reversible cellular automata, I thought I'd express my doubts to see where my thinking has gone wrong. The version I have is that every finite group is a subgroup of a wreath product of some factor group and its kernel. Of course, if there is no normal subgroup there is no decomposition. The definition of indecomposable above implies <two> normal subgroups, not <one>, but then that would make an awful lot of groups indecomposable that otherwise have a perceptible structure. Like semidirect products. Does the "weakness" alluded to above mostly mean that abelian groups, which have lots of normal subgroups, are being classified? Can anyone reccommend a nice, introductory, treatment of wreath products and Krohn-Rhodes decompositions? - hvm ------------------------------------------------- Obtén tu correo en www.correo.unam.mx UNAMonos Comunicándonos
participants (3)
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John Conway -
mcintosh＠servidor.unam.mx -
Richard Schroeppel