If $ B^{8}=(234591)$and if $B\in S_{9}$.what is the value of $B$

Question

Basically the cycle is of order 6. So probably $B^{2}=(234591)$.what next do I do? Any hint?

Answer 1

You are correct, $B^8$ has order $6$ and hence $B^8 = B^2$. A permutation admits a square root if and only if, when written as a product of disjoint cycles, the multiplicity of each cycle type of even length is even (which includes zero). The "if" direction has a long proof, but the "only if" is easy to see: the sign of a product of permutations is the product of the signs of its factors, so a square always has an even sign. Hence, permutations with an odd sign do not have square roots.

As a special case of this (just one cycle of length $6$, and one is odd), what we have denoted as $B^2$ cannot be a square of any $B \in S_9$.