Permutation Conjugation and Order

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Homework Statement


Prove that there is no such permutation a such that
(a-inverse)*(1,2)*(a) = (3,4)(1,5)


The Attempt at a Solution


Does it have something to do with the order of (1,2)? I know the order is 2, so if we square (a-inverse)*(1,2)*(a), then we get the identity...how else can I think about it?
 
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Do you know anything about cycle types? In particular, can you prove that they are invariant under conjugation?

If not, notice that 1->5 on the RHS. Convince yourself that (1,2) must send a(1) to a(5). Do the same for 3.
 
By a(1) and a(5) do you mean the numbers 1 and 5 in whatever the permutation a is?
 
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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