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September 9th, 2014, 07:35 AM  #1 
Newbie Joined: May 2014 From: Singapore Posts: 11 Thanks: 0  Doubts from Group theory and permutations
Hello everybody, Please take a look at the attachment from the Algebra and Geometry textbook in this post (They are both from the same page!). I understand that the orbits O(Ki) are pairwise disjoint sets and where each of these sets is cyclically permuted by p. This means that if p = (1 2 3 4 5 6 7) (5 7 2 1 4 3 6) The permuatations are (1 5 4) and (2 7 6 3). they are also the orbits of elements 1 and 2? only these two orbits are pairwise disjoint sets. the first orbit set has 3 elements and the second has 4 elements then how they be pairwise checked?? 1 to 2 5 to 7 4 to 6 and 3 to what??? Do the orbits of other elements come from cyclically permuting orbits 1 and 2?? NOW...... what does mutually disjoint orbits mean??? and what is pj, the cycle associated to orbit O(kj) so in the above example how to differentiate between orbits and cycles??? from my understanding cycle and orbits are the same? Theorem 1.3.7 states that p is a product of disjoint cycles p1. p2 . . . .pm. so in the above example is p equal to 5 7 2 1 4 3 6 which is also the unique function. and p1 and p2 are (1 5 4) and (2 7 6 3)?? what does "p = p1. p2 . . . .pm that was derived from the orbit decomposition (1.3.3) is unique up to the order of the 'factors' pj" mean?? what does "number m of factors in this product mean"?? Thank you!!! 
September 9th, 2014, 09:29 PM  #2  
Senior Member Joined: Mar 2012 Posts: 294 Thanks: 88  Quote:
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$1 \to 5$ $2 \to 7$ $3 \to 2$ $4 \to 1$ $5 \to 4$ $6 \to 3$ $7 \to 6$ (You only indicated the imagea function is domain, map AND image) which has cycle decomposition: $(1\ 5\ 4)(2\ 7\ 6\ 3)$. Your assessment of $\rho_1,\rho_2$ is correct. Quote:
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