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Problem: Given $\sigma=(12)(34)$ and $\gamma=(56)(13)$ find $\tau\in S_6$ with $\tau^{-1}\sigma\tau=\gamma$

Attempt: I'm kind of new to this but from what I understanding find $\tau$ that satisfies this will show that $\sigma$~$\gamma$ right? This means that they are conjugates of each other. I started off by writing out what the permutations are in $S_6$ but I was not seeing anything. I also rewrote what we are trying to prove as $\sigma\tau$=$\tau\gamma$ by left multiplying by $\tau$.

Question: My main question is whether or not there's a method to solving these types of problems and if there is how can it be applied to this one? Any step in the right direction is appreciated, thank you.

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Hint: for any cycle $(i_1\,i_2\,\ldots\,i_k)$, we have $$ \sigma (i_1\,i_2\,\ldots\,i_k)\sigma^{-1}=(\sigma(i_1)\,\sigma(i_2)\,\ldots\,\sigma(i_k)) $$ for all $\sigma\in S_n$.

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  • $\begingroup$ I thought a permutation times its inverse gives back the original set. So for 1...6 if I do $\sigma\sigma^{-1}$ that gives back 1...6? $\endgroup$ – David Fuentes Oct 10 '13 at 0:45
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    $\begingroup$ I've clarified above. $\endgroup$ – Ian Coley Oct 10 '13 at 0:50
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    $\begingroup$ Ah okay, I understand now, thank you. I believe this also implies that we want $\tau$ such that $\sigma=((\tau(5)\tau(6))(\tau(1)\tau(3))$ and you find a permutation that matches those conditions. $\endgroup$ – David Fuentes Oct 10 '13 at 1:35

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