An equation with permutations, x^2 = sigma.

Reid
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Homework Statement



x^2 = sigma.
The permutation sigma = (1 2 6 7 5 3 4), a cycle of length seven.

Determine x!

The Attempt at a Solution



I have tried a few times but my attempts are totally wrong. I don't know where to start! :S

I found a similar problem in the textbook. They could, by just lookin at the problem, say that there is no x for that (in the book not the one above) particular type of permutation. How? :S
 
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I don't know if this is the best way, but it definitely works:
First determine what is the order of sigma. This can be used to solve the question. For example: suppose that the order is 3. Then sigma^3 = 1, so sigma^4 = sigma. From this, you can immediately find x.
If the order would be 4, on the other hand, then you would get sigma^4 = 1 so sigma^5 = sigma. The left hand side is not a square of anything, so there is no solution.
 
I'm not sure if i understood this correct. Let's see, with your reasoning...

I know that sigma^7=1, therefore sigma^8=sigma. Then x=sigma^4. Right?
 
Yes, that's exactly right. I didn't quite comprehend what CompuChip meant by "suppose that the order is 3" since the order was clearly 7! I'm glad you did.
 
I tried not to give away the answer completely. Didn't work :smile:
 
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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