# Finding the permutation that shows two permutations are conjugates method?

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.

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 $\sigma\in S_n$.
• 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? – David Fuentes Oct 10 '13 at 0:45
• 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. – David Fuentes Oct 10 '13 at 1:35