An equation with permutations, x^2 = sigma.

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Homework Help Overview

The discussion revolves around an equation involving permutations, specifically x^2 = sigma, where sigma is a permutation represented as a cycle of length seven. Participants are tasked with determining the value of x! based on the properties of the permutation.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the order of the permutation sigma and how it relates to finding x. Some express confusion about the implications of the order on the equation, while others attempt to clarify the reasoning behind determining x based on the order of sigma.

Discussion Status

The discussion is active, with participants exploring different interpretations of the problem. Some guidance has been offered regarding the relationship between the order of the permutation and the potential solutions for x, but there is no explicit consensus on the final approach.

Contextual Notes

Participants note the complexity of determining x based on the properties of permutations, and there is mention of a textbook example that suggests certain types of permutations may not yield a solution. This raises questions about the assumptions underlying the problem.

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:
 

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