Question about permutation cycles

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The discussion centers on the mathematical concept of permutation cycles, specifically examining the cycle A = (1 4 5)(2 3 6) and its square A^2 = (1 5 4)(2 6 3). The key question is whether there exists a permutation B such that BAB^-1 = A^2. The conclusion drawn is that B must reverse the elements 5 and 4, as well as 6 and 3, to achieve the desired conjugation.

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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?
 

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