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Why all $3$-cycles are conjugate in $\mathfrak S_5$ ? Is it the case for all $n$ or only on $\mathfrak S_n$ ? I mean, in $\mathfrak S_n$ all $3$ cycle are conjugate or not ? (it doesn't look to be the case for $n=3$, but in the doubt, I prefer to ask).

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In $\mathfrak S_n$, any cycles with the same length: $(i_1\,i_2\,\dots\, i_r)$ and $(j_1\,j_2\,\dots\, j_r)$ are conjugate.

Indeed, if $\sigma=(i_1\,j_1)(i_2\,j_2)\dots(i_r\,j_r)$, one has $$\sigma(i_1\,i_2\,\dots\, i_r)\sigma^{-1}=\bigl(\sigma(i_1)\,\sigma(i_2)\,\dots\, \sigma(i_r)\bigr)=(j_1\,j_2\,\dots\, j_r).$$ As a consequence, two permutations are conjugate if and only if their decompositions as a product of disjoint cycles have the same cycles structure.

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  • $\begingroup$ This is fine, as long as you note that there may be values of $k$ for which $i_{k} = j_{k}$, in which case, $(i_{k}j_{k})$ is not really a $2$-cycle, and should be omitted. $\endgroup$ – Geoff Robinson Jan 8 '16 at 13:19
  • $\begingroup$ That was implicit, of course. $\endgroup$ – Bernard Jan 8 '16 at 13:20

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