[sage-support] How would I list just the 3 cycles in AlternatingGroup()
I can generate a list from any given group, but how would I go about generating a list of just 3 or 5 cycles? --~--~-~--~~~---~--~~ 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 -~--~~~~--~~--~--~---
[sage-support] Re: How would I list just the 3 cycles in AlternatingGroup()
I can find the order of the element, but I am looking to generate a list of all of the 3 cycles in something like AlternatingGroup(5) where the list will not go on for too long. On May 14, 5:04 pm, David Joyner wdjoy...@gmail.com wrote: I must be missing something. Why can't you just check the order of the element? On Thu, May 14, 2009 at 7:53 PM, jimfar jamesfar...@mac.com wrote: I can generate a list from any given group, but how would I go about generating a list of just 3 or 5 cycles? --~--~-~--~~~---~--~~ 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 -~--~~~~--~~--~--~---
[sage-support] Re: How would I list just the 3 cycles in AlternatingGroup()
Thanks, I was confusing myself with the definition of the order of an element with order of the cycle. On May 14, 6:52 pm, David Joyner wdjoy...@gmail.com wrote: Why doesn't the obvious 1-liner [x for x in AlternatingGroup(5) if x.order()==3] work? Again, am I missing something? On Thu, May 14, 2009 at 8:57 PM, jimfar jamesfar...@mac.com wrote: I can find the order of the element, but I am looking to generate a list of all of the 3 cycles in something like AlternatingGroup(5) where the list will not go on for too long. On May 14, 5:04 pm, David Joyner wdjoy...@gmail.com wrote: I must be missing something. Why can't you just check the order of the element? On Thu, May 14, 2009 at 7:53 PM, jimfar jamesfar...@mac.com wrote: I can generate a list from any given group, but how would I go about generating a list of just 3 or 5 cycles? --~--~-~--~~~---~--~~ 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 -~--~~~~--~~--~--~---
[sage-support] How do I find the inverse of an element in an Alternating Group?
After generating sage: B=AlternatingGroup(5) and verifying an element is in it, sage: c=(1,2,3) sage: c in B True How do I find the inverse of c in B? --~--~-~--~~~---~--~~ 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 -~--~~~~--~~--~--~---
[sage-support] Re: How do I find the inverse of an element in an Alternating Group?
Thank you very much On May 6, 7:47 pm, Robert Bradshaw rober...@math.washington.edu wrote: On May 6, 2009, at 7:44 PM, jimfar wrote: After generating sage: B=AlternatingGroup(5) and verifying an element is in it, sage: c=(1,2,3) sage: c in B True How do I find the inverse of c in B? Your c here isn't really a permutation element, it's just a tuple. To create (and manipulate) the actual permutation element do: sage: c = B((1,2,3)); c (1,2,3) sage: c^-1 (1,3,2) sage: c^2 (1,3,2) sage: parent(c) Alternating group of order 5!/2 as a permutation group - Robert --~--~-~--~~~---~--~~ 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 -~--~~~~--~~--~--~---
[sage-support] How do I show a permutation group is the alternating group?
I have generated a group using, sage: B=PermutationGroup(['(1,2,4,5,3)','(2,3,1,4,5)']) And I know I can generate a list of the elements and determine the order, but how do I show that this is actually sage: AlternatingGroup(5). Is there a command to verify that B=AlternatingGroup(5)? --~--~-~--~~~---~--~~ 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 -~--~~~~--~~--~--~---
[sage-support] Re: How do I show a permutation group is the alternating group?
thank you very much On May 4, 11:57 pm, Michael Welsh yom...@yomcat.geek.nz wrote: sage: B=PermutationGroup(['(1,2,4,5,3)','(2,3,1,4,5)']) sage: B == AlternatingGroup(5) True sage: B == AlternatingGroup(7) False On 5/05/2009, at 6:47 PM, jimfar wrote: I have generated a group using, sage: B=PermutationGroup(['(1,2,4,5,3)','(2,3,1,4,5)']) And I know I can generate a list of the elements and determine the order, but how do I show that this is actually sage: AlternatingGroup(5). Is there a command to verify that B=AlternatingGroup(5)? --http://yomcat.geek.nz --~--~-~--~~~---~--~~ 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 -~--~~~~--~~--~--~---
