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November 24th, 2016, 01:28 PM   #1
Joined: Nov 2016
From: CA

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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,


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.

princetongirl818 is offline  

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cardinality, element, finding, orbit

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