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February 13th, 2012, 08:40 PM   #21
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Re: PERMUTATION GROUPS

Quote:
 Originally Posted by MarkFL 57 = 3·19... :P

too bad i'm not allowed to swear, i didn't notice it either.

I need more coffee...

I could understand if i mistook 51 for a prime but 57 is inexcusable...

February 13th, 2012, 09:15 PM   #22
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Re: PERMUTATION GROUPS

Quote:
 Originally Posted by The Chaz You can't just pick out the (1 3) from the product.
but isn't that what you did when you picked out (1 5) from the product (1 5)(2 3 4) = (5 3 2 4 1)

BTW I now realize that i copied it wrong when i transferred my calculations from paper to forum post, it should be...

(1 5)(2 3 4) = (5 3 4 2 1)

so yes i was wrong but not wrong for the reason you gave.

Oh man... my brain is fried... this isn't working for me... can you help me figure out what i'm doing wrong?

2 hours ago i was convinced i knew how to multiply cycles, now i'm not sure...

 February 13th, 2012, 09:40 PM #23 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 Re: PERMUTATION GROUPS All right, i think i got it. Disregard my previous example cause it was wrong. Here is example that i think is correct. write (5 1 4 3 2) as a product of disjoint cycles answer: (3 4)(1 5 2) or (1 5 2)(3 4) because disjoint cycles commute or (1 5 2)(4 3) cause the elements in a transposition commute, (3 4) is a transposition or (5 2 1)(3 4) or (2 1 5)(3 4) because they contain the same information obviously both of these commute also and 3 4 can switch positions obviously. Am i right?
 February 14th, 2012, 04:16 AM #24 Global Moderator     Joined: Nov 2009 From: Northwest Arkansas Posts: 2,766 Thanks: 4 Re: PERMUTATION GROUPS No. From the wikipedia page: "the expression of the permutation is unique up to the order of the cycles" So you trying to rewrite a 5-cycle as the disjoint product of a 2-cycle and a 3-cycle will never work. Never. Not ever.
 February 14th, 2012, 06:41 AM #25 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 Re: PERMUTATION GROUPS so what is the right way to write (5 1 4 3 2) as a product of disjoint cycles? I don't see the relevance of the wiki quote and your subsequent conclusion, what's the connection?
February 14th, 2012, 07:00 AM   #26
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Re: PERMUTATION GROUPS

Quote:
 Originally Posted by agentredlum so what is the right way to write (5 1 4 3 2) as a product of disjoint cycles? I don't see the relevance of the wiki quote and your subsequent conclusion, what's the connection?
There are k ways to write a k-cycle, but they just differ on where you choose to start...
(51432) = (14325) = (43251) = (32514) = (25143)
Those are the five ways to write that 5-cycle.

There are no other ways. You seem to be insisting that there be two or more cycles in some alternate way of expressing (51432), and that these two or more cycles will have no element in common. You cannot achieve this. Please stop trying/asking!

A GOOD question would be,
"How do I write [the composition/product of some amount >1 of NON-disjoint cycles] as a product of disjoint cycles?"

The example in the link you gave does exactly this.

My example: Write s = (12)(12)(12)(12)(12)(12)(12)(13)(12)(12) as a product of disjoint cycles on the set {1,2,3}
Solution, since s sends 1 to 3, 2 to 1, and 3 to 2, we can write s = (132) = (321) = (213). There are I other such ways to express s.

Let's try one more round of Q&A. If we can't make any progress, I'll close this.

 February 14th, 2012, 07:35 AM #27 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 Re: PERMUTATION GROUPS Well then how do you explain this? http://www.wolframalpha.com/input/?i=Pe ... C3%2C2%29+ wolfram alpha confirms my calculation of decomposition.
 February 14th, 2012, 07:59 AM #28 Global Moderator     Joined: Nov 2009 From: Northwest Arkansas Posts: 2,766 Thanks: 4 Re: PERMUTATION GROUPS *&$*^(#*(&$@ We are using two different notations!!! I totally forgot about the difference. First, your comment about not writing 1-cycles adds to the confusion by implying that you are using cycle decomposition notation. When you write (51432), what you MEAN is $\begin{pmatrix} 1 &2 & 3 &4 &5 \\ 5 &1 &4 &3 &2 \end{pmatrix}$ This is one way to represent a particular bijection/permutation of {1,2,3,4,5}. Let's call this function s. Since the top line is always 1 2 3 ... n, sometimes people omit the top line and just write the bottom line : (51432). Please stop doing this! Another way to express s is to explicitly list what it does. s(1) = 5 s(2) = 1 s(3) = 4 s(4) = 3 s(5) = 2 Of course, the cycle decomposition of s is (152)(34) = (34)(152) = ... In this context, you cannot use both notations without great risk of confusion.
February 14th, 2012, 08:15 AM   #29
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Re: PERMUTATION GROUPS

Now let's look at the original problem in light of this recent development.

Quote:
 Originally Posted by samyuktha Express the product (2 5 4) (1 4 3) (2 1) as a product of disjoint cycles ..
1. We observe 1, 2, 3, 4, and 5 in this problem, so we consider cycles on supersets of {1, 2, 3, 4, 5}
2. We do NOT see any "cycles" of EXACTLY five numbers, so there is absolutely no reason to use the notation that you introduced. That notation compares position with value, which isn't well-defined for "cycles" of length different than 5.
3. Now we track where each of 1, 2, 3, 4, 5 goes...
(21) sends 1 to 2.
(143) doesn't send 2 anywhere (but to 2, if you want)
(254) sends 2 to 5.

So 1 goes to 5.
Repeat until you get back to 1.
Then repeat until all of 1, 2, 3, 4, 5 have been represented.

 February 14th, 2012, 08:26 AM #30 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 Re: PERMUTATION GROUPS How do i write this permutation using your notation? $\begin{pmatrix} 1 &2 & 3 &4 &5 \\ 5 &1 &4 &3 &2 \end{pmatrix}$

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