My Math Forum  

Go Back   My Math Forum > College Math Forum > Abstract Algebra

Abstract Algebra Abstract Algebra Math Forum


Reply
 
LinkBack Thread Tools Display Modes
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
princetongirl818 is offline  
 
Reply

  My Math Forum > College Math Forum > Abstract Algebra

Tags
cardinality, element, finding, orbit



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Cardinality of integers equals cardinality of reals BenFRayfield Number Theory 0 February 15th, 2014 03:55 PM
Flashlight in stable orbit must start falling when turned on BenFRayfield Advanced Statistics 0 May 6th, 2013 07:16 AM
Set with one element blabla Applied Math 3 April 12th, 2012 03:05 PM
Related Rates Elliptical Orbit/Satellite Problem SummerRain Calculus 0 November 6th, 2008 05:45 PM
Find Equation of the Orbit symmetry Algebra 1 November 27th, 2007 07:23 PM





Copyright © 2017 My Math Forum. All rights reserved.