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I want to find a permutation in $S_8$ such that $σ(172)(35)(48)σ^{-1}= (12)(34)(567)$.

To do that i wanted to do $σ(172)σ^{-1} =(567)$ and use the propriety : $σ(i_1, i_2, . . . , i_k)σ^{−1} = (σ(i_1), σ(i_2), . . . , σ(i_k))$.

Since $(172)(35)(48)$ is a disjoint support. But i'm not sure if i can do this?

So i want to know if we can do this for all the cycle we had. i mean if we can do $σ(172)σ^{-1}=(567)$ and find $(σ(1), σ(7), σ(2))=(567)$ and do it again $σ(35)σ^{-1}=(34)$ and find $(σ(3), σ(5)) =(34) σ.(48)σ^{-1}=(12)$ and find $(σ(4), σ(8)) =(12).$

Then conclude $σ=(154)(2768)$?

I'm not sure because it seems we applied $3$ time $σ^{-1}$ and $σ$ Or we had it only one time in the first product.

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    $\begingroup$ Have a look here on MSE, which has several answers, e.g., here. $\endgroup$ – Dietrich Burde Nov 18 '17 at 14:18
  • $\begingroup$ thanks the method explain helped me, and i found the same result, so how can explain that i can applied σ.t.σ^-1 to each cycle t of our permutation ? $\endgroup$ – Farouk Deutsch Nov 18 '17 at 14:44
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Using the formula $$\sigma (i_1 i_2\dots i_r)\sigma^{-1}=\bigl(\sigma(i_1)\sigma(i_2)\dots \sigma(i_r)\bigr), $$ you see that you must have $$\sigma\bigl(\{1,2,7\}\bigr)=\{5,6,7\}\quad\text{and}\quad\bigl\{\sigma\bigl(\{3,5\}\bigr),\sigma\bigl(\{4,8\}\bigr)\bigr\}=\bigl\{\{1,2\},\{3,4\}\bigr\}$$ For instance, we can take the $8$-cycle $$\sigma=(15276843),\quad\sigma^{-1}=(13486725).$$

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  • $\begingroup$ thanks, i understand that there are many possible results, i wanted to know how can explain that i can applied σ.t.σ^-1 to each cycle t of our permutation ? $\endgroup$ – Farouk Deutsch Nov 18 '17 at 14:45
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    $\begingroup$ Because you have initially a decomposition into disjoint (hence commuting) cycles, and they remain so after any automorphism. $\endgroup$ – Bernard Nov 18 '17 at 14:50
  • $\begingroup$ So we can rewrite this σ(172)(35)(48)σ^{-1}= (12)(34)(567) as σσ^{-1}(172)(35)(48)= (12)(34)(567) and finally just (172)(35)(48)= (12)(34)(567) ?? $\endgroup$ – Farouk Deutsch Nov 18 '17 at 15:02
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    $\begingroup$ Certainly not: $\sigma$ does not commute with the cycles of the decomposition. But $\sigma(\tau\tau'\gamma)\sigma^{-1}=(\sigma\tau\sigma^{-1})(\sigma\tau'\sigma^{-1})(\sigma\gamma\sigma^{-1})$. $\endgroup$ – Bernard Nov 18 '17 at 15:13
  • $\begingroup$ yes it is exactly what my question is about. How can we prove it ? $\endgroup$ – Farouk Deutsch Nov 18 '17 at 16:49
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\begin{eqnarray*} (145286) (172)(35)(48) (168254) =(12)(34)(567). \end{eqnarray*} so $ \sigma=\color{blue}{(145286)}$ will do.

In general if you want to find the element that conjugate same shaped cycles, write them one above the other and read off the mapping from top to bottom \begin{eqnarray*} (172)(35)(48) (6) \\ (567)(34)(12)(8) \end{eqnarray*} Giving $\sigma=(154)(2768)$ as you did in the question.

The general theory being used here is \begin{eqnarray*} \sigma (i_1 i_2\dots i_r)\sigma^{-1}=(\sigma(i_1)\sigma(i_2)\dots \sigma(i_r) ). \end{eqnarray*}

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