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How can I go about showing that $f=(6\;9)(1\;3\;4)(2\;5\;7\;8)$ and $g=(1\;7)(2\;3\;5)(4\;9\;6\;8)$ are conjugate in $S_9$ (the set of permutations on 9 symbols)? I need to do this without using the Cauchy Theorem.

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Permutations are conjugate in the full symmetric group if and only if they have the same cycle structure. This is because conjugation corresponds to renumbering; that is, you get the cycle structure of $ghg^{-1}$ by permuting the numbers in the cycle structure of $h$ according to $g$. Since the full symmetric group always contains the required permutation $h$, any two permutations with the same cycle structure are conjugate in it.

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    @Mike: You're doing it the wrong way around; with the usual notation conventions the permutations should be applied right to left. That explains the problem with $f$ but not the one with my example, where you should get the same result since $(12)^{-1}$ is $(12)$. To find, say, which number $1$ is mapped to, apply $(12)^{-1}$ to get $2$, then $(13)(24)$ to get $4$, then $(12)$ to get $4$. To find which number $4$ is mapped to, apply $(12)^{-1}$ to get $4$, then $(13)(24)$ to get $2$, then $(12)$ to get $1$. Thus the result contains the cycle $(14)$. It works the same for $2$ and $3$.2012-03-02