On Fri, May 15, 2009 at 2:36 AM, <simon.k...@uni-jena.de> wrote: > > Dear Jim, > > On May 15, 4:03 am, jimfar <jamesfar...@mac.com> wrote: >> Thanks, I was confusing myself with the definition of the order of an >> element with order of the cycle. > > Are you really confusing it? > > As much as I understood, you only want those elements that have a > single (!) cycle of length 3. Then, order() is the wrong thing to ask > for! > > sage: G=PermutationGroup([((1,2,3),(4,5,6))]) > sage: G.0 > (1,2,3)(4,5,6) > sage: G.0.order() > 3 > > So, G.0 is not a 3-cycle, but it is of order 3. > > I don't know if there is better way to test the number of cycles, but > you could do: > [x for x in AlternatingGroup(5) if x.order()==3 and len(x.cycles()) > ==1] > > The difference becomes apparent when you do it for AlternatingGroup > (6), rather than AlternatingGroup(5).
The point is that for Alt(5) it is a simple one liner! Of course, for more complicated groups, it is not the same thing as order=3 so you need to use len(x.cycle_tuples())==1. > > Cheers, > Simon > > > > --~--~---------~--~----~------------~-------~--~----~ To post to this group, send email to sage-support@googlegroups.com To unsubscribe from this group, send email to sage-support-unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/sage-support URLs: http://www.sagemath.org -~----------~----~----~----~------~----~------~--~---
