$N>4$. Consider the symmetric group $S(N)$. Does the equation $X^3 = (1,2,3)$ have a solution in $S(N)$? Since $(1,2,3)$ is an even permutation, any solution (if any exist) must also be an even permutation. I also know that if $z$ is a cycle of length $k$ in $S(N)$ and $m$ is a positive integer, then $z^m$ is a product of $d$ disjoint cycles of length $k/d$ each (where $d$ is the GCD of $k$ and $m$). I have not been able to get anywhere beyond this. Can I get a few hints? Since I don't have a very good knowledge of rings and fields, I would be happy to get hints that use undergraduate group theory. Thanks.
solution to $X^3=(1,2,3)$
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symmetric-groups
1 Answers
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It is enough to search for a solution that is a cycle. (why?) If $z$ is such a cycle then, using $z^3 = (1,2,3)$ and the facts you wrote about the powers of cycles, what can you deduce about the length $k$ of $z$?