|November 24th, 2016, 01:28 PM||#1|
Joined: Nov 2016
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;
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.
|April 5th, 2017, 01:26 PM||#2|
Joined: Apr 2017
Do you know how to write in LaTeX? If you know, then please rewrite your question in LaTeX. It would be easier to understand.
|cardinality, element, finding, orbit|
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