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Let $S_6$ represent the set of all permutations of $\Omega = \{1, 2, \ldots, 6\}$.

Let $\sigma, \tau \in S_6$, such that $\sigma = (1, 2, 3)(4, 5, 6)$ and $\tau = (2, 4, 5)(3, 1, 6)$.

Find an element $\gamma \in S_6$ such that $\sigma = \gamma\tau\gamma ^{-1}$

My initial thought was to use the algebra of permutations to rewrite the expression as to remove the $\gamma^{-1}$ like so: $\sigma\gamma = \gamma\tau\gamma^{-1}\gamma = \gamma\tau\circ e = \gamma\tau$.

Unfortunately, this gave me mappings containing some unknown elements of $\gamma$ (i.e. $\gamma = (\gamma_1, \gamma_2, \ldots, \gamma_6)$) which seems to be a dead end.

From previous knowledge of centralizers I know that for an arbitrary element $a \in G$, where $G$ is a Group, $C(a) = \{g\in G : ga = ag\}$.

Is a centralizer just a special case of the given problem? What parallels can we draw (if any) between the two definitions?

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  • $\begingroup$ If you pair up the $3$-cycles of $\sigma$ with those of $\tau$, then $\gamma$ will take elements of one $3$-cycle to its pair. $\endgroup$ – Michael Burr Jan 5 '18 at 12:59
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The conjugation $\gamma \tau \gamma^{-1}$ acts on $\gamma(1), \gamma(2), \dots, \gamma(6)$ in the same way as $\tau$ acts on $1, 2, \dots, 6$; formally: $$\gamma \tau \gamma^{-1}( \gamma(i) ) = \gamma( \tau(i) ).$$ Since $$\tau = (2,4,5)(3,1,6),$$ this means that $$\gamma\tau\gamma^{-1} = (\gamma(2), \gamma(4), \gamma(5))(\gamma(3), \gamma(1), \gamma(6)).$$ So, for this to be equal to $\sigma = (1,2,3)(4,5,6)$, $\gamma$ could be the permutation that maps $2$ to $1$, $4$ to $2$, $5$ to $3$, $3$ to $4$, $1$ to $5$ and $6$ to $6$, i.e., $\gamma = (2,1,5,3,4)(6)$.

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  • $\begingroup$ Surely you mean to say that $\gamma\tau\gamma^{-1} = (\gamma (2), \gamma (4), \gamma (5))(\gamma (3), \gamma (1), \gamma (6))\gamma^{-1}$? Since I can't get your value for $\gamma$ to work $\endgroup$ – Joel Biffin Jan 5 '18 at 13:59
  • $\begingroup$ can you see this? $\endgroup$ – Joel Biffin Jan 5 '18 at 15:03
  • $\begingroup$ No, I really mean $\gamma\tau\gamma^{-1} = (\gamma(2), \gamma(4), \gamma(5))(\gamma(3), \gamma(1), \gamma(6))$. For instance, using just the definition of $\tau$, $\gamma\tau\gamma^{-1}(\gamma(3)) = \gamma(\tau(3)) = \gamma(1)$. So $\gamma\tau\gamma^{-1}$ maps $\gamma(3)$ to $\gamma(1)$, which is what the cycle notation also says. $\endgroup$ – Magdiragdag Jan 5 '18 at 19:26

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