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February 14th, 2012, 08:31 AM   #31
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Re: PERMUTATION GROUPS

Quote:
Originally Posted by agentredlum
How do i write this permutation using your notation?


Open parenthesis (
Start with the smallest number, and "track it" until you get back to that number. Close parenthesis)
Then track the next smallest number (that isn't already) in a cycle.
Eventually you will have all numbers.

e.g. we track 1 --> 5 since s(1) = 5. Then we follow 5 --> 2 since s(5) = 2. Then we follow 2 --> 1, since s(2) = 1. Arrived back at 1, we stop and write
(1 5 2).
The smallest number not in this cycle is 3.
s(3) = 4 and s(4) = 3, so we write
(34)

In conclusion, (1 5 2)(3 4).
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February 14th, 2012, 08:39 AM   #32
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Re: PERMUTATION GROUPS

I see...

so it is impossible to write it using 1 set of parenthesis, or as you would say, a product of 1 cycle.

Well, i'm glad we got that cleared up!
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February 14th, 2012, 08:42 AM   #33
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Re: PERMUTATION GROUPS

Right. In a proper cycle, everything is necessarily moved, where that need it be the case for arbitrary bijections/permutations.
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February 14th, 2012, 08:52 AM   #34
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Re: PERMUTATION GROUPS

So the meaning of (5 1 4 3 2) in your notation is



in my notation.

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February 14th, 2012, 08:57 AM   #35
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Re: PERMUTATION GROUPS

yes!
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February 14th, 2012, 09:16 AM   #36
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Re: PERMUTATION GROUPS

Thanx Chaz!

For the original problem, i converted everything to my notation, got the right answer my way, then converted to your notation and posted. Doing it your way is much faster and well worth the effort to understand your way.

I'm glad we had this conversation
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