> On Thu, May 14, 2009 at 8:57 PM, jimfar 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 p
r of the
> element?
>
> On Thu, May 14, 2009 at 7:53 PM, jimfar 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?
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I can generate a list from any given group, but how would I go about
generating a list of just 3 or 5 cycles?
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Thank you very much
On May 6, 7:47 pm, Robert Bradshaw
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
> > T
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?
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thank you very much
On May 4, 11:57 pm, Michael Welsh 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)?
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