February 14th, 2012, 08:31 AM  #31  
Global Moderator Joined: Nov 2009 From: Northwest Arkansas Posts: 2,766 Thanks: 4  Re: PERMUTATION GROUPS Quote:
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).  
February 14th, 2012, 08:39 AM  #32 
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233  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! 
February 14th, 2012, 08:42 AM  #33 
Global Moderator Joined: Nov 2009 From: Northwest Arkansas Posts: 2,766 Thanks: 4  Re: PERMUTATION GROUPS
Right. In a proper cycle, everything is necessarily moved, where that need it be the case for arbitrary bijections/permutations.

February 14th, 2012, 08:52 AM  #34 
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233  Re: PERMUTATION GROUPS
So the meaning of (5 1 4 3 2) in your notation is in my notation. 
February 14th, 2012, 08:57 AM  #35 
Global Moderator Joined: Nov 2009 From: Northwest Arkansas Posts: 2,766 Thanks: 4  Re: PERMUTATION GROUPS
yes!

February 14th, 2012, 09:16 AM  #36 
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233  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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