S.O.S. Mathematics CyberBoard

Your Resource for mathematics help on the web!
It is currently Wed, 22 May 2013 18:48:00 UTC

View unanswered posts | View active topics


Board index » Advanced Mathematics » Algebra and Number Theory

All times are UTC [ DST ]



Post new topic Reply to topic  Page 1 of 2
  [ 20 posts ]  Go to page 1, 2  Next
  Print view Previous topic | Next topic 
Author Message
DgrayMan
 Post subject: Disjoint cycles
PostPosted: Mon, 27 Feb 2012 05:29:00 UTC 
Offline
S.O.S. Oldtimer
User avatar

Joined: Sat, 21 Jan 2012 03:59:22 UTC
Posts: 182
A and B are disjoint cycles, s and r are positive integers.

A=(a1 a1 ... as) , B=(b1 b2 ... br)

How do i prove if AB = e , then A=e and B=e ?


Now this is reminiscent of the theorem if a and b are elements of group G then ab=e implies a=b^-1 and b=a^-1

Why doesn't this rule apply to disjoint cycles?
Top
 Profile  
 
outermeasure
 Post subject: Re: Disjoint cycles
PostPosted: Mon, 27 Feb 2012 05:48:48 UTC 
Offline
Moderator
User avatar

Joined: Mon, 29 Dec 2008 17:49:32 UTC
Posts: 6007
Location: 127.0.0.1, ::1 (avatar courtesy of UDN)
DgrayMan wrote:
A and B are disjoint cycles, s and r are positive integers.

A=(a1 a1 ... as) , B=(b1 b2 ... br)

How do i prove if AB = e , then A=e and B=e ?


Now this is reminiscent of the theorem if a and b are elements of group G then ab=e implies a=b^-1 and b=a^-1

Why doesn't this rule apply to disjoint cycles?


The cycles are disjoint, so what do you know about a_i,b_j?

_________________
\begin{aligned} Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\ Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\ Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\ Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\ Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\ Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4 \end{aligned}
Top
 Profile  
 
DgrayMan
 Post subject: Re: Disjoint cycles
PostPosted: Mon, 27 Feb 2012 05:59:21 UTC 
Offline
S.O.S. Oldtimer
User avatar

Joined: Sat, 21 Jan 2012 03:59:22 UTC
Posts: 182
disjoint cycles are commutative?
Top
 Profile  
 
outermeasure
 Post subject: Re: Disjoint cycles
PostPosted: Mon, 27 Feb 2012 06:09:51 UTC 
Offline
Moderator
User avatar

Joined: Mon, 29 Dec 2008 17:49:32 UTC
Posts: 6007
Location: 127.0.0.1, ::1 (avatar courtesy of UDN)
DgrayMan wrote:
disjoint cycles are commutative?


True but irrelevant, and you haven't answered my question.

_________________
\begin{aligned} Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\ Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\ Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\ Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\ Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\ Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4 \end{aligned}
Top
 Profile  
 
DgrayMan
 Post subject: Re: Disjoint cycles
PostPosted: Mon, 27 Feb 2012 06:17:18 UTC 
Offline
S.O.S. Oldtimer
User avatar

Joined: Sat, 21 Jan 2012 03:59:22 UTC
Posts: 182
I'm not sure. i know every permutation is the product of disjoint cycles and i know e is the product of a even number of transpositions. But that sounds irrelevant too.
Top
 Profile  
 
outermeasure
 Post subject: Re: Disjoint cycles
PostPosted: Mon, 27 Feb 2012 06:20:05 UTC 
Offline
Moderator
User avatar

Joined: Mon, 29 Dec 2008 17:49:32 UTC
Posts: 6007
Location: 127.0.0.1, ::1 (avatar courtesy of UDN)
DgrayMan wrote:
I'm not sure. i know every permutation is the product of disjoint cycles and i know e is the product of a even number of transpositions. But that sounds irrelevant too.


I'll repeat my question: what do you know about a_i,b_j (for all 1\leq i\leq s, 1\leq j\leq r)?

_________________
\begin{aligned} Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\ Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\ Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\ Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\ Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\ Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4 \end{aligned}
Top
 Profile  
 
DgrayMan
 Post subject: Re: Disjoint cycles
PostPosted: Mon, 27 Feb 2012 06:26:15 UTC 
Offline
S.O.S. Oldtimer
User avatar

Joined: Sat, 21 Jan 2012 03:59:22 UTC
Posts: 182
ai is less than bj? i'm sorry im utterly lost
Top
 Profile  
 
outermeasure
 Post subject: Re: Disjoint cycles
PostPosted: Mon, 27 Feb 2012 06:50:01 UTC 
Offline
Moderator
User avatar

Joined: Mon, 29 Dec 2008 17:49:32 UTC
Posts: 6007
Location: 127.0.0.1, ::1 (avatar courtesy of UDN)
DgrayMan wrote:
ai is less than bj? i'm sorry im utterly lost


No, your set need not come with a total order. Think about what the word "disjoint" means.

_________________
\begin{aligned} Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\ Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\ Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\ Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\ Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\ Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4 \end{aligned}
Top
 Profile  
 
DgrayMan
 Post subject: Re: Disjoint cycles
PostPosted: Mon, 27 Feb 2012 07:02:01 UTC 
Offline
S.O.S. Oldtimer
User avatar

Joined: Sat, 21 Jan 2012 03:59:22 UTC
Posts: 182
i know that both permutations are independent of each other, they have no pairs in common
Top
 Profile  
 
outermeasure
 Post subject: Re: Disjoint cycles
PostPosted: Mon, 27 Feb 2012 07:05:26 UTC 
Offline
Moderator
User avatar

Joined: Mon, 29 Dec 2008 17:49:32 UTC
Posts: 6007
Location: 127.0.0.1, ::1 (avatar courtesy of UDN)
DgrayMan wrote:
i know that both permutations are independent of each other, they have no pairs in common


So what is the effect of B on the element a_1?

_________________
\begin{aligned} Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\ Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\ Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\ Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\ Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\ Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4 \end{aligned}
Top
 Profile  
 
DgrayMan
 Post subject: Re: Disjoint cycles
PostPosted: Mon, 27 Feb 2012 07:08:50 UTC 
Offline
S.O.S. Oldtimer
User avatar

Joined: Sat, 21 Jan 2012 03:59:22 UTC
Posts: 182
if there are no pairs in common it should not have an effect?
Top
 Profile  
 
outermeasure
 Post subject: Re: Disjoint cycles
PostPosted: Mon, 27 Feb 2012 07:15:22 UTC 
Offline
Moderator
User avatar

Joined: Mon, 29 Dec 2008 17:49:32 UTC
Posts: 6007
Location: 127.0.0.1, ::1 (avatar courtesy of UDN)
DgrayMan wrote:
if there are no pairs in common it should not have an effect?


So what is the effect of AB on a_1? (Assuming your group acts on the left.)

_________________
\begin{aligned} Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\ Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\ Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\ Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\ Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\ Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4 \end{aligned}
Top
 Profile  
 
DgrayMan
 Post subject: Re: Disjoint cycles
PostPosted: Mon, 27 Feb 2012 07:26:01 UTC 
Offline
S.O.S. Oldtimer
User avatar

Joined: Sat, 21 Jan 2012 03:59:22 UTC
Posts: 182
a1 should equal a1?
Top
 Profile  
 
DgrayMan
 Post subject: Re: Disjoint cycles
PostPosted: Mon, 27 Feb 2012 07:28:41 UTC 
Offline
S.O.S. Oldtimer
User avatar

Joined: Sat, 21 Jan 2012 03:59:22 UTC
Posts: 182
but why couldnt it just be an inverse of A if nothing in B will have an effect on A. Wouldn't that be the same thing?
Top
 Profile  
 
DgrayMan
 Post subject: Re: Disjoint cycles
PostPosted: Mon, 27 Feb 2012 07:34:34 UTC 
Offline
S.O.S. Oldtimer
User avatar

Joined: Sat, 21 Jan 2012 03:59:22 UTC
Posts: 182
ah because theyre disjoint. they can't have anything in common to begin with right? so B cannot be an inverse of A because that would mean A and B are not disjoint right?
Top
 Profile  
 
Display posts from previous:  Sort by  
Post new topic Reply to topic  Page 1 of 2
  [ 20 posts ]  Go to page 1, 2  Next

Board index » Advanced Mathematics » Algebra and Number Theory

All times are UTC [ DST ]

Who is online

Users browsing this forum: No registered users

You cannot post new topics in this forum
You cannot reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum

Search for:
Jump to:  
Contact Us | S.O.S. Mathematics Homepage
Privacy Statement | Search the "old" CyberBoard

users online during the last hour
Powered by phpBB © 2001, 2005-2011 phpBB Group.
Copyright © 1999-2013 MathMedics, LLC. All rights reserved.
Math Medics, LLC. - P.O. Box 12395 - El Paso TX 79913 - USA