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Can someone help on this. I'm stuck on this part. I am trying to find a permutation $\sigma$ such that $$\sigma(1,2)(3,4)\sigma^{-1} = (5,6)(3,1)$$. By a particular theorem, I know I can have this one $$ \sigma (1,2)\sigma^{-1}\cdot \sigma(3,4)\sigma^{-1} =(5,6)(3,1)$$Then, $$(\sigma(1) , \sigma(2)) = (5,6) $$ and $$(\sigma(3) , \sigma(4)) = (3,1) $$ Am I right on doing this? $$\sigma(1) = 5 \qquad \mbox{and} \qquad \sigma(2)=6$$ and $$\sigma(3) = 3 \qquad \mbox{and} \qquad \sigma(4)=1$$ I don't really know what to do next. I thought of making them as one permutation but it doesn't really makes sense to me. Please help.


marked as duplicate by Elaqqad, Community Apr 14 '15 at 5:59

This question was marked as an exact duplicate of an existing question.


Your work is fine so far. The only thing remaining is to define $\sigma(5)$ and $\sigma(6)$. Note that since $\sigma$ is a permutation of $\{1,2,3,4,5,6\}$, you just need to make sure it is bijective. This means either $\sigma(5) = 4, \sigma(6) = 2$, or $\sigma(6) = 4, \sigma(5) = 2$. I leave it to you to write either result in cycle notation.

  • $\begingroup$ does this mean I have to answers? $\endgroup$ – marg_ocruz Apr 14 '15 at 5:04
  • $\begingroup$ There are more possible answers than that, even. I leave it as an exercise to show that there are sixteen possible values of $\sigma$ such that $\sigma (1,2)(3,4) \sigma^{-1}= (5,6)(3,1)$. My guess is that you only need to find one of these. $\endgroup$ – Rolf Hoyer Apr 14 '15 at 5:08
  • $\begingroup$ is there an algorithm for that? because I think that I can also do the same thing when i find the stabilizer.. I am also stuck on such problem.. $\endgroup$ – marg_ocruz Apr 14 '15 at 5:11
  • $\begingroup$ I didn't do anything fancy, I think you just have to be careul thinking about it. Note that there are eight different ways to write $(1,2)(3,4)$ in cycle notation. $\endgroup$ – Rolf Hoyer Apr 14 '15 at 5:15
  • $\begingroup$ oh I see.. :) thanks! $\endgroup$ – marg_ocruz Apr 14 '15 at 5:23

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