If a given permutation in S_n has a given cycle type, describe sgn(sig).

Edellaine
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Homework Statement


5.4: If sigma in S_n has cycle type n_1,...,n_r, what is sgn(sig)? (sgn is the sign homomorphism)

Homework Equations


sgn(sigma) = 1 if sigma is even. sgn(sigma) = -1 is sigma is odd
cycle type is the length of the cycle type. If n_2 = 2, sigma has two 2-cycles.

The Attempt at a Solution


I know what cycle type is (if n_i = j, there are j cycles of length i), and sgn(sigma) is easy. How would I go about expanding (sig) to find how many two cycles I have, if that's even what I should be doing. It doesn't seem that worthwhile to generalize this for cycle type.

I was thinking of writing sgn(sig) as a product of sgn(sig_i) where (sig_i) is an individual cycle of the product of cycles forming (sig), but I don't think that exactly accounts for multiples.

I'm also not sure how to come up with conditions that will say, depending on r, if sgn(sig) = 1, or sgn(sig) = -1.
 
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I'm not quite following your notation here. But to get the sign you just take (-1)^(number of even length cycles), right?
 
Yup. But since you can write any n-cycle as a product of 2-cycles, how do I account for the cycles of odd length.
 
Cycles of even length break into an odd number of two cycles, cycles of odd length break into an even number. I don't see the problem.
 
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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