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Check that P3*P6=P4

asked 2012-06-12 14:18:09 +0100

bk322 gravatar image

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

The book says

I want to check that P3*P6=P4.

G = SymmetricGroup(3)

BookNumbers = [1, 4, 2, 3, 6, 5]

P = [0]
for i in BookNumbers:
    P.append(sorted(G.list())[i-1])

print (P[3] * P[6]).list(), P[4].list()
print (P[3] * P[6]) == P[4]

it gives:

[3, 2, 1] [2, 1, 3]
False

so they are the same actually. But how do I make sage say True?.

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answered 2012-06-12 14:54:40 +0100

kcrisman gravatar image

The problem is that you are comparing lists, not Sage objects. achrzesz gives an answer which compares two Sage group elements, so they are equal. The lists you give are simple Python ordered lists, and as ordered things, certainly aren't the same. Why didn't you just compare P[3]*P[6] and P[4]?

By the way, they're not the same. But that's a different issue. They're using the notation of the second row of your notation in your original question, not cycle notation.

sage: P[3] * P[6]
(1,3)
sage: P[4]
(1,2)
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answered 2012-06-12 14:30:51 +0100

achrzesz gravatar image
sage: G = SymmetricGroup(3)
sage: P3=G((3,2))
sage: P6=G((3,1,2))
sage: P4=G((2,1))
sage: P3*P6==P4
True
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Comments

But mine are also the same. [3, 2, 1] is same as [2, 1, 3], isn't it?

bk322 gravatar imagebk322 ( 2012-06-12 14:45:50 +0100 )edit

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Asked: 2012-06-12 14:18:09 +0100

Seen: 365 times

Last updated: Jun 12 '12