If a=(1 2)(3 4)(5 6)(7 8)(9 10), determine whether there is an n-cycle 'b',( n>=10) with a=b^k for some integer k.
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From Wikipedia: A cycle of length L =k·m, taken to the k-th power, will decompose into k cycles of length m: For example (k=2, m=3), (1 2 3 4 5 6)^2 = (1 3 5) (2 4 6). So yes, it is easy to find a 10-cycle whose 5th degree is a.