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  • 1 Post By Kuldeep Guha Mazumder
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November 24th, 2016, 12:28 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  
April 5th, 2017, 01:26 PM   #2
Joined: Apr 2017
From: Kolkata

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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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Kuldeep Guha Mazumder is offline  

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