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 25th, 2009, 06:12 AM   #1
Guest
 
Joined:

Posts: n/a
Thanks:

prove permutation inverse

This should be rather straight forward, but I thought I would ask for some inout from those better an group theory than me.

Let . Prove that

What would be a good way to go about this?. I am just learning some group theory and thought this was a good problem involving cycles.


I was thinking Let , where Then,

because



and keep going until we get to e. I have more, but it is too much Latexing with all those sigmas
 
 
November 25th, 2009, 08:26 AM   #2
Newbie
 
Joined: Aug 2008

Posts: 10
Thanks: 0

Re: prove permutation inverse

To start with , holds for all bijections (not just permutations -- on the other hand it holds for all groups, either way works). Splitting any cycle into a product of 2-cycles, and knowing that 2-cycles and their own inverse will be useful -- but it's important to know which 2-cycles a cycle is split into. You should be able to prove the theorem just using these three facts: If so, then it's worth going back to prove those lemmas.
vicky is offline  
November 26th, 2009, 04:50 AM   #3
Guest
 
Joined:

Posts: n/a
Thanks:

Re: prove permutation inverse

Thanks.
 
Reply

  My Math Forum > College Math Forum > Abstract Algebra

Tags
inverse, permutation, prove



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
prove there exists an inverse function annakar Calculus 1 December 18th, 2012 08:57 AM
Prove that the element n-1 is its own inverse restin84 Number Theory 5 April 25th, 2012 05:17 PM
find an inverse of a permutation cycle rayman Abstract Algebra 5 January 29th, 2012 01:26 AM
find inverse of inverse function jaredbeach Algebra 1 November 17th, 2011 11:58 AM
prove prove prove. currently dont know where to post qweiop90 Algebra 1 July 31st, 2008 06:27 AM





Copyright © 2019 My Math Forum. All rights reserved.