How to Calculate π^σ If We Have 2 Disjoint Cycles?

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Suppose we have 2 disjoint cycles π and σ. How can one calculate π^σ?
I know how to calculate σ^2 or σ^3 but I can't figure out how to solve that.
 
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Are you talking about cycles in the sense of cyclic permutations? The first question would be how we define [itex]\pi^\sigma[/itex].
 
Yes cyclic permuations. For example if[itex]\pi[/itex]= (147)(263859) and σ=(16789)(2345) how can we calculate [itex]\pi[/itex]^σ
 
FoxMCloud said:
Yes cyclic permuations. For example if[itex]\pi[/itex]= (147)(263859) and σ=(16789)(2345) how can we calculate [itex]\pi[/itex]^σ

How did you define ##\pi^\sigma##? Did you define it as ##\sigma \circ \pi \circ \sigma^{-1}##?
 
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I think it's something along the lines of σ^(-1)*[itex]\pi[/itex]*σ. But then again, I'm not seeing how can I calculate σ^(-1).
 
FoxMCloud said:
I think it's something along the lines of σ^(-1)*[itex]\pi[/itex]*σ. But then again, I'm not seeing how can I calculate σ^(-1).

If you have one cycle, then you can find the inverse by reversing the cycle. So if [tex]\sigma = (1 ~2~5~3)[/tex], then [tex]\sigma^{-1} = (3~5~2~1)[/tex]

Then if you have a more general form, then you can calculate the inverse by the formula ##(\sigma\tau)^{-1}= \tau^{-1}\sigma^{-1}##.

For example, if you have ##(1~4~6)(3~2)##, then the inverse is ##(2~3)(6~4~1)##.

So now you can find ##\sigma^{-1}## and thus also ##\pi^\sigma##. However, for a general theorem which makes this a LOT easier: http://www.proofwiki.org/wiki/Cycle_Decomposition_of_Conjugate
 
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