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S3's elements

asked 2012-06-10 06:41:13 -0500

bk322 gravatar image

Here's elements of Symmetric group of 6th order: S3:

The book says

I want to get the same in Sage. So I do:

sage: G = SymmetricGroup(3)
sage: G.list()
[(), (2,3), (1,2), (1,2,3), (1,3,2), (1,3)]

Now I can't find (1,3,2) element in the book. As far as I understand:

P1 -> ()
P2 -> (1,2,3)
P3 -> (2,3)
P4 -> (1,2)
P5 -> (1,3)
P6 -> ??? also (1,2,3) ???

So my question is to set the correct map from sage to my book...

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answered 2012-06-10 08:26:56 -0500

DSM gravatar image

updated 2012-06-10 08:30:08 -0500

The results of

sage: G = SymmetricGroup(3)
sage: G.list()
[(), (2,3), (1,2), (1,2,3), (1,3,2), (1,3)]

are given in cycle notation. So the book's equivalent of (1,3,2) is one where 1 becomes 3, 3 becomes 2, and 2 becomes 1, which is P6, if I'm reading correctly.

If you want to match the bottom three elements of your matrix, you can simply convert each to a list, or maybe a dict would make the mapping more explicit:

sage: for g in G:
....:     print g, g.list(), g.dict()
....:     
() [1, 2, 3] {1: 1, 2: 2, 3: 3}
(2,3) [1, 3, 2] {1: 1, 2: 3, 3: 2}
(1,2) [2, 1, 3] {1: 2, 2: 1, 3: 3}
(1,2,3) [2, 3, 1] {1: 2, 2: 3, 3: 1}
(1,3,2) [3, 1, 2] {1: 3, 2: 1, 3: 2}
(1,3) [3, 2, 1] {1: 3, 2: 2, 3: 1}
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That's very cool. I did this: `for e in sorted(G): print '{0:>8s}{1:>10s}{2:>19s}'.format(e, e.list(), e.dict())` (I type more then, but it inserts >...)

bk322 gravatar imagebk322 ( 2012-06-12 05:52:40 -0500 )edit

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Asked: 2012-06-10 06:41:13 -0500

Seen: 243 times

Last updated: Jun 10 '12