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For $x = (12)(34)$ and $y = (56)(13),$ find a permutation $a$ such that $a^{-1}xa = y$.

I written $a^{-1}xa = y$ as $xa = ay$ thus $(12)(34)a = a(56)(13)$ but I can't find the $a$?

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Conjugation affects cycles like so: $\sigma(a_1~a_2~\cdots~a_r)\sigma^{-1}=(\sigma(a_1)~\sigma(a_2)~\cdots~\sigma(a_r))$.

So, how does conjugation affect a permutation's disjoint cycle representation?

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  • $\begingroup$ Anon I still don't know how to solve this problem. Can you explain it more please? $\endgroup$ – user104235 Nov 13 '13 at 3:56
  • $\begingroup$ And this chapter was before conjugacy, so I don't think we can apply that here. $\endgroup$ – user104235 Nov 13 '13 at 4:02
  • $\begingroup$ @user104235 If you're not allowed to talk about conjugation, even without using the word "conjugation," then you're not allowed to do the problem at all, plain and simple, because the problem is about a particular conjugation. The reason I ask questions in my answers is so that people can come to understand problems in multiple stages (rather than just doing the whole thing for others). However you have not told me any of your thoughts on the pedagogical question I posed, so I don't know where I need to break the problem up into simpler steps. $\endgroup$ – anon Nov 13 '13 at 4:27
  • $\begingroup$ In particular: (1) do you understand what I've said so far? (2) can you extrapolate to figure out how conjugation explicitly affects a product of (disjoint) cycles? (3) so then (rearranging $a^{-1}xa=y$ $\Leftrightarrow x=aya^{-1}$) what does $a(56)(13)a^{-1}$ look like? and finally (4) what can you select $a$ to be to achieve $(12)(34)$? $\endgroup$ – anon Nov 13 '13 at 4:32
  • $\begingroup$ $a^{-1}(12)(34)a = (56)(13)$. I am stuck in finding such an a. $\endgroup$ – user104235 Nov 13 '13 at 4:32

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