Determining Missing Images in an Even Permutation

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Homework Statement


Hi, I have been MIA lately due to work, but I am back with questions, and eager to learn! I am self studying, and so I have inconsistencies in my learning which I hope to iron out.

Suppose you are told that the permutation
( 1 2 3 4 5 6 7 8 9
3 1 2 X Y 7 8 9 6)

In S9 the symmetric group, where the images of 5 and 4 have been lost, is an even permutation. What must the images of 5 and 4 be?


Homework Equations





The Attempt at a Solution


I thought X should be 5, and Y should be 4, so that you have 6 transpositions or am I oversimplifying the logic here?
Thank you.
 
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Justabeginner said:

Homework Statement


Hi, I have been MIA lately due to work, but I am back with questions, and eager to learn! I am self studying, and so I have inconsistencies in my learning which I hope to iron out.

Suppose you are told that the permutation
( 1 2 3 4 5 6 7 8 9
3 1 2 X Y 7 8 9 6)

In S9 the symmetric group, where the images of 5 and 4 have been lost, is an even permutation. What must the images of 5 and 4 be?


Homework Equations





The Attempt at a Solution


I thought X should be 5, and Y should be 4, so that you have 6 transpositions or am I oversimplifying the logic here?
Thank you.
Looks good to me.
 
Thank you.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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