Very strange permutation problem

betty2301
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Homework Statement


~


Homework Equations



product of permutation

The Attempt at a Solution


i have difficulty understanding this q.
why in \sigma there are b1b2...bn new element?
why can i insert \tau in the permutation "matrix"?\tauitself is a "matrix"??
 

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It's not a matrix. It is the two-line notation for a permutation. See here for more details:

http://en.wikipedia.org/wiki/Permutation

This is not a strange problem; in fact it's one of the most standard and fundamental facts about permutations.
 
jbunniii said:
It's not a matrix. It is the two-line notation for a permutation. See here for more details:

http://en.wikipedia.org/wiki/Permutation

This is not a strange problem; in fact it's one of the most standard and fundamental facts about permutations.

i know this is not a matrix that's why i write "matrix".
help!
 
is \tau a transposition or any permutation?
 
betty2301 said:
is \tau a transposition or any permutation?

\tau can be any permutation. (That is exactly what \forall \tau \in S_n means.)

Remember that \tau \sigma \tau^{-1} is a composition of three permutations, in the following order: first \tau^{-1}, then \sigma, then \tau. If the "input" set of elements is \tau(a_1), ..., \tau(a_n), then just follow what each stage of the composition does to these elements. Try not to overthink it - this is actually a very easy problem.
 
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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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