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 Post subject: Disjoint cyclesPosted: Mon, 27 Feb 2012 05:29:00 UTC
 S.O.S. Oldtimer

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?

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 Post subject: Re: Disjoint cyclesPosted: 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 ?

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 Post subject: Re: Disjoint cyclesPosted: Mon, 27 Feb 2012 05:59:21 UTC
 S.O.S. Oldtimer

Joined: Sat, 21 Jan 2012 03:59:22 UTC
Posts: 182
disjoint cycles are commutative?

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 Post subject: Re: Disjoint cyclesPosted: 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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 Post subject: Re: Disjoint cyclesPosted: Mon, 27 Feb 2012 06:17:18 UTC
 S.O.S. Oldtimer

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.

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 Post subject: Re: Disjoint cyclesPosted: 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 (for all )?

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 Post subject: Re: Disjoint cyclesPosted: Mon, 27 Feb 2012 06:26:15 UTC
 S.O.S. Oldtimer

Joined: Sat, 21 Jan 2012 03:59:22 UTC
Posts: 182
ai is less than bj? i'm sorry im utterly lost

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 Post subject: Re: Disjoint cyclesPosted: 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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 Post subject: Re: Disjoint cyclesPosted: Mon, 27 Feb 2012 07:02:01 UTC
 S.O.S. Oldtimer

Joined: Sat, 21 Jan 2012 03:59:22 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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 Post subject: Re: Disjoint cyclesPosted: 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 ?

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 Post subject: Re: Disjoint cyclesPosted: Mon, 27 Feb 2012 07:08:50 UTC
 S.O.S. Oldtimer

Joined: Sat, 21 Jan 2012 03:59:22 UTC
Posts: 182
if there are no pairs in common it should not have an effect?

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 Post subject: Re: Disjoint cyclesPosted: 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 ? (Assuming your group acts on the left.)

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 Post subject: Re: Disjoint cyclesPosted: Mon, 27 Feb 2012 07:26:01 UTC
 S.O.S. Oldtimer

Joined: Sat, 21 Jan 2012 03:59:22 UTC
Posts: 182
a1 should equal a1?

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 Post subject: Re: Disjoint cyclesPosted: Mon, 27 Feb 2012 07:28:41 UTC
 S.O.S. Oldtimer

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?

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 Post subject: Re: Disjoint cyclesPosted: Mon, 27 Feb 2012 07:34:34 UTC
 S.O.S. Oldtimer

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?

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