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Check that P3*P6=P4

asked 12 years ago

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 12 years ago

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 12 years ago

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

bk322 gravatar imagebk322 ( 12 years ago )

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Asked: 12 years ago

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Last updated: Jun 12 '12