My Math Forum  

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

Abstract Algebra Abstract Algebra Math Forum


Thanks Tree1Thanks
  • 1 Post By Deveno
Reply
 
LinkBack Thread Tools Display Modes
June 9th, 2014, 08:46 AM   #1
Member
 
Joined: May 2014
From: None of your business

Posts: 31
Thanks: 0

Permutations of a finite set

I'm trying to figure out this problem:

Let alpha and Beta be cycles of odd length (not disjoint). Prove that if alpha squared equals beta squared, then alpha equals beta.

What I know thus far is that... since l is an odd length, alpha squared and beta squared are cycles, and not a product of disjoint cycles. Both alpha and beta can be expressed as an even number of transpositions. I also know that I can express alpha as a product of the cycles alpha1 gamma alpha2, and that this product is associative.

Since the problem states that alpha and beta are not disjoint, then I can reason that there is ai in the alpha cycle that is equal to some bj in the beta cycle.

Now I tried expressing beta squared as a product of beta1 (bj bj+2) beta2, and since it is associative, rewrote it as (bj bj+2) beta1 beta2, then multipied by (bj+2 bj) on the left, also multiplying alpha squared by (bj+2 bj) on the left.

With examples, I continued in this fashion until I had the identity permuation on the right hand side of the equation.

Now I'm stuck. I'm thinking that I only showed that the product of beta squared inverse and alpha squared is the identity (not really illuminating).

Please help.
Awesomo is offline  
 
June 9th, 2014, 04:41 PM   #2
Senior Member
 
Joined: Mar 2012

Posts: 294
Thanks: 88

Since $\alpha,\beta$ are not disjoint, they share at least one element in their respective cycles. Hence we may write:

$\alpha = (a_1\ a_2\ \dots\ a_n)$

$\beta = (a_1\ b_2\ \dots\ b_m)$

Thus:

$\alpha^2 = (a_1\ a_3\ \dots a_n\ a_2\ a_4\ \dots a_{n-1})$ (since $n$ is odd)

$\beta^2 = (a_1\ b_3\ \dots b_m\ b_2\ b_4\ \dots b_{m-1})$ (since $m$ is odd).

Since $\alpha^2 = \beta^2$ it follows that $m = n$, and $b_j = a_j$.

To see why we have the stipulation that $n,m$ be odd, consider:

(1 2) and (1 3). Then (1 2)(1 2) = (1 3)(1 3) = e, but (1 2) does not equal (1 3).

**********

The trick here is to see it does not matter which element of a cycle we start it with:

(1 2 3) = (2 3 1) = (3 1 2),

and that two functions $f,g$ are equal iff their domains are equal, their co-domains are equal, and they take the same values in the co-domain for every value in the domain.

(Recall that the cycle $(a\ b\ c)$ is the function:

$f:\{1,2,\dots,n\} \to \{1,2,\dots,n\}$ with:

$f(a) = b$
$f(b) = c$
$f(c) = a$
$f(k) = k$, for all other elements besides $a,b,c$).

So once we prove that $a_3 = \alpha^2(a_1) = \beta^2(a_1) = b_3$, we compare $a_5 = \alpha^2(a_3) = \beta^2(a_3) = \beta^2(b_3) = b_5$, and so on, until we come "full cycle".
Thanks from Awesomo
Deveno is offline  
Reply

  My Math Forum > College Math Forum > Abstract Algebra

Tags
finite, permutations, set



Search tags for this page
Click on a term to search for related topics.
Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Permutations... :S princevisram Advanced Statistics 6 April 17th, 2012 07:41 AM
Star-finite and Locally Finite Sets Hooman Real Analysis 0 February 20th, 2012 09:34 PM
Permutations koolkidx45 Advanced Statistics 1 February 20th, 2012 11:09 AM
Permutations Help bureddogs44 Advanced Statistics 3 September 18th, 2011 12:58 PM
Help please about permutations sinanbasar Advanced Statistics 0 April 16th, 2010 10:37 AM





Copyright © 2019 My Math Forum. All rights reserved.