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November 24th, 2016, 01:28 PM  #1 
Newbie Joined: Nov 2016 From: CA Posts: 2 Thanks: 0  Finding the cardinality of an orbit of an element
Let A=(123…n)A=(123…n) be an element of PnPn. So that the group PnPn acts on itself by means of the action of conjugation, for B∈PnB∈Pn, B⋅A=BAB(^−1). Stabilizer of A is a subgroup; {A^c ∣c=0,…,n−1}. I want to find the cardinality of OrbPn(A). So by applying the Orbit stabilizer theorem: Then for A to act in itself by conjugation then I have that OrbPn(A)=ABA^(−1) where A is an element of Pn. Is this correct :O Thanks for reading this. nomad609 
April 5th, 2017, 01:26 PM  #2 
Newbie Joined: Apr 2017 From: Kolkata Posts: 2 Thanks: 1 
Do you know how to write in LaTeX? If you know, then please rewrite your question in LaTeX. It would be easier to understand.


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