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September 9th, 2014, 07:35 AM   #1
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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!!!
Attached Images
 alan.jpg (81.1 KB, 2 views) alan21.jpg (91.1 KB, 2 views)

September 9th, 2014, 09:29 PM   #2
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Joined: Mar 2012

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Quote:
 Originally Posted by AspiringPhysicist 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?
The orbits are the SETS $\{1,4,5\}$ and $\{2,3,6,7\}$. Sets are usually "unordered".

Quote:
 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???
Since we have only two orbits, in this case "pariwise disjoint" means we only check that the two orbits have null intersection.

Quote:
 Do the orbits of other elements come from cyclically permuting orbits 1 and 2??
No, any element contained in the orbit of 1 has the same orbit as 1. Likewise for 2.

Quote:
 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?
The index $j$ is just to keep track of how many orbits we have. "Mutually disjoint" means none of the orbits overlap, they form a PARTITION of the set $\{1,2,3,4,5,6,7\}$. The order MATTERS in the cycle: (1 5 4) is not the same as (1 4 5), but these two different permutations have the same orbits.

Quote:
 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)??
No the function is:

$1 \to 5$
$2 \to 7$
$3 \to 2$
$4 \to 1$
$5 \to 4$
$6 \to 3$
$7 \to 6$ (You only indicated the image-a 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:
 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??
We can write (1 5 4)(2 7 6 3) or (2 7 6 3)(1 5 4), these are both the same. If we had more cycles, any re-arrangement of the cycles themselves would yield the same function.

Quote:
 what does "number m of factors in this product mean"?? Thank you!!!
The number of distinct cycles. In this case, m = 2.

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