Question about permutation cycles

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If A is a cycle, and A=(1 4 5) (2 3 6). Is there a B such that BAB^-1=A^2. I found
A^2=(1 5 4) (2 6 3), but I'm not really sure where to go from there.
 
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You know they are conjugate because they have the same cycle structure. If you need to find a B, then looking at your results, I'd say you want a B that reverses 5 and 4 and reverses 6 and 3. Can you think of one?
 

Thread 'Finding the nth roots of a complex number'

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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