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DgrayMan
 Post subject: Disjoint cycles
PostPosted: Mon, 27 Feb 2012 05:29:00 UTC 
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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?
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outermeasure
 Post subject: Re: Disjoint cycles
PostPosted: Mon, 27 Feb 2012 05:48:48 UTC 
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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?

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\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}
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DgrayMan
 Post subject: Re: Disjoint cycles
PostPosted: Mon, 27 Feb 2012 05:59:21 UTC 
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disjoint cycles are commutative?
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outermeasure
 Post subject: Re: Disjoint cycles
PostPosted: Mon, 27 Feb 2012 06:09:51 UTC 
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DgrayMan wrote:
disjoint cycles are commutative?


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

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\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}
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DgrayMan
 Post subject: Re: Disjoint cycles
PostPosted: Mon, 27 Feb 2012 06:17:18 UTC 
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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.
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outermeasure
 Post subject: Re: Disjoint cycles
PostPosted: Mon, 27 Feb 2012 06:20:05 UTC 
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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)?

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\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}
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DgrayMan
 Post subject: Re: Disjoint cycles
PostPosted: Mon, 27 Feb 2012 06:26:15 UTC 
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ai is less than bj? i'm sorry im utterly lost
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outermeasure
 Post subject: Re: Disjoint cycles
PostPosted: Mon, 27 Feb 2012 06:50:01 UTC 
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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.

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\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}
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DgrayMan
 Post subject: Re: Disjoint cycles
PostPosted: Mon, 27 Feb 2012 07:02:01 UTC 
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i know that both permutations are independent of each other, they have no pairs in common
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outermeasure
 Post subject: Re: Disjoint cycles
PostPosted: Mon, 27 Feb 2012 07:05:26 UTC 
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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}
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DgrayMan
 Post subject: Re: Disjoint cycles
PostPosted: Mon, 27 Feb 2012 07:08:50 UTC 
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if there are no pairs in common it should not have an effect?
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outermeasure
 Post subject: Re: Disjoint cycles
PostPosted: Mon, 27 Feb 2012 07:15:22 UTC 
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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.)

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\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}
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DgrayMan
 Post subject: Re: Disjoint cycles
PostPosted: Mon, 27 Feb 2012 07:26:01 UTC 
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a1 should equal a1?
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DgrayMan
 Post subject: Re: Disjoint cycles
PostPosted: Mon, 27 Feb 2012 07:28:41 UTC 
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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?
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DgrayMan
 Post subject: Re: Disjoint cycles
PostPosted: Mon, 27 Feb 2012 07:34:34 UTC 
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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?
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