February 13th, 2012, 08:40 PM  #21  
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233  Re: PERMUTATION GROUPS Quote:
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  
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233  Re: PERMUTATION GROUPS Quote:
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 5cycle as the disjoint product of a 2cycle and a 3cycle 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  
Global Moderator Joined: Nov 2009 From: Northwest Arkansas Posts: 2,766 Thanks: 4  Re: PERMUTATION GROUPS Quote:
(51432) = (14325) = (43251) = (32514) = (25143) Those are the five ways to write that 5cycle. 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 NONdisjoint 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 1cycles adds to the confusion by implying that you are using cycle decomposition notation. When you write (51432), what you MEAN is 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  
Global Moderator Joined: Nov 2009 From: Northwest Arkansas Posts: 2,766 Thanks: 4  Re: PERMUTATION GROUPS
Now let's look at the original problem in light of this recent development. Quote:
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 welldefined 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? 

Tags 
groups, permutation 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Abelian XGroups and Noetherian (Abelian) XGroups  Math Amateur  Abstract Algebra  0  October 29th, 2013 03:46 PM 
Proof on product of cycles (permutation groups)  Kappie  Abstract Algebra  1  March 13th, 2012 09:30 PM 
About minimal normal groups and subnormal groups  Sheila496  Abstract Algebra  0  October 20th, 2011 09:45 AM 
Homomorphism of Lie groups as groups but not as manifolds  lime  Abstract Algebra  3  October 25th, 2010 07:30 AM 
permutation groups, proof  MarkyMark  Abstract Algebra  2  January 15th, 2009 08:34 PM 