Write (13257)(23)(47512) as a product of disjoint cycles

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Homework Statement


Write (13257)(23)(47512) as a product of disjoint cycles. Each bracket is a permutation of seven elements written in cycle notation.

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The Attempt at a Solution


This isn't too hard of a problem. One easy way would be to evaluate the entire product, and then write that product in cycle notation. However, is there an easier, faster way of doing this, just be looking at the expression (13257)(23)(47512) directly?
 
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Not that I know of. And ##2## occurs in all three cycles, so any possible "rule" is likely more complicated than simply multiply them.
 
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Also, one quick related question. I need to calculate the order of (125)(34). Is there a quick way to do this, or do I have to literally keep composing the permutation with itself until I get the identity permutation?
 
If they are disjoint, they commute. So ##(ab)^n=a^nb^n## and the order is the least common multiple of both orders. And a cycle of length ##n## is of order ##n##.
 
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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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