Preserving cycle structure in transpositions

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Homework Help Overview

The discussion revolves around the properties of permutations in the symmetric group S_n, specifically focusing on the preservation of cycle structure under conjugation by another permutation. The original poster presents a problem involving a cycle r and a permutation ß, seeking to demonstrate that the cycle structure remains intact when applying ß to r and its inverse.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the implications of the conjugation operation on the cycle structure, questioning how elements are transformed under the action of ß and its inverse. There is a focus on understanding the relationship between elements of the cycle and their images under the permutation.

Discussion Status

The conversation has progressed with participants engaging in back-and-forth questioning and clarification. Some have begun to articulate their understanding of the problem, while others are still grappling with the definitions and implications of the operations involved. There is a sense of productive exploration, with hints and questions guiding participants toward deeper insights.

Contextual Notes

Participants note the distinction between permutations and elements being permuted, which is crucial for understanding the operations discussed. The original poster expresses uncertainty about generalizing their examples, indicating a need for clarity on the broader implications of their findings.

PiRGood
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Hi all, long time reader first time poster! Just need a hand on this problem I've been stuck on for a few days

Homework Statement


Let r=(a_1,a_2...a_k) be in S_n. Suppose that ß is in S_n. Show that:
ßrß^-1=(ß(a_1), ß(a_2)...ß(a_k)).



Homework Equations





The Attempt at a Solution


Basically here i need to prove that the cycle structure of transposition is preserved in composition. For example if r=(1,2)(3,4) and ß=(1,2,3,4) then ßrß^1 = (ß(1),ß(2),ß(3),ß(4))=(2,3)(4,1). That is, r is a double transposition and so is ßrß^1.
I've been able to solve plenty when using examples like the one above but i can't seem to do it without loss of generality. Any help would be greatly appreciated :)
 
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What is brb^(-1) of b(a_1)?
 
Dick said:
What is brb^(-1) of b(a_1)?

I apologize, I am not sure if this is a question or a hint. Either way i do not understand.

ß is an element of the group S_n. a_1 is an element of r, which is also a set of transpositions contained in S_n. ß(a_1) is a composition of transpositions, i believe

Does that help?
 
PiRGood said:
I apologize, I am not sure if this is a question or a hint. Either way i do not understand.

ß is an element of the group S_n. a_1 is an element of r, which is also a set of transpositions contained in S_n. ß(a_1) is a composition of transpositions, i believe

Does that help?

It's a question that becomes a hint if you figure out how to answer it. a_1 is an element of {1,...,n}, the set of elements that S_n permutes. So is b(a_1). Now here's a simpler question, what's b^(-1) of b(a_1)? I'm using b instead of your beta because it's easier to write.
 
isn't it simply b^-1(a_1)? Or (b(a_1))^-1?
 
Dick said:
What is brb^(-1) of b(a_1)?

Is it b(a_1)? the problem says b(a_1)b^(-1)=b(a_1)
 
PiRGood said:
Is it b(a_1)? the problem says b(a_1)b^(-1)=b(a_1)

No, it's not b(a_1). b(a_1)b^(-1)=b(a_1) doesn't make much sense. b(a_1) is an element of {1...n} and b^(-1) is a permutation. If you want to find brb^(-1) first find b^(-1)(b(a_1)). b^(-1) is the inverse of b, right? What does that mean?
 
Dick said:
No, it's not b(a_1). b(a_1)b^(-1)=b(a_1) doesn't make much sense. b(a_1) is an element of {1...n} and b^(-1) is a permutation. If you want to find brb^(-1) first find b^(-1)(b(a_1)).

I thought b, (a_1) and b^(-1) were all permutations?

b^(-1) is the inverse of b, right? What does that mean?
It means that b^(-1)b is e, or the identity element
 
PiRGood said:
I thought b, (a_1) and b^(-1) were all permutations?


It means that b^(-1)b is e, or the identity element

b is a permutation, b^(-1) is a permutation. a_1 is NOT a permutation, it's an element in the set that's being permuted. b^(-1)b is e, in the sense that it is the identity permutation. e(x)=x for all x. I think you are kind of confused about what these things really mean. Take a deep breath, think about it, and tell me again what is b^(-1)(b(a_1)).
 
  • #10
tell me again what is b^(-1)(b(a_1)).

Well b will permute at to some number (or set, whichever you like) and b^(-1) will permute it back, so won't it just be a_1?
 
  • #11
PiRGood said:
Well b will permute at to some number (or set, whichever you like) and b^(-1) will permute it back, so won't it just be a_1?

Now you're getting someplace. Sure it's a_1. Next you apply r. So what's r(a_1)?
 
  • #12
Dick said:
Now you're getting someplace. Sure it's a_1. Next you apply r. So what's r(a_1)?

a_2?
 
  • #13
PiRGood said:
a_2?

Right again. So now the answer to the original question, "what is brb^(-1) of b(a_1)"?
 
  • #14
Dick said:
Right again. So now the answer to the original question, "what is brb^(-1) of b(a_1)"?
a_2?

I apologize i think you lost me. you want to know what b(a_1)b^-1 is?

However i thinking like this may have just turned a lightbulb on for me. What i must show is that for any cycle r of length k, and any b in S. brb^-1 is also a cycle of the same length k. Am i getting somewhere?
 
  • #15
PiRGood said:
a_2?

I apologize i think you lost me. you want to know what b(a_1)b^-1 is?

However i thinking like this may have just turned a lightbulb on for me. What i must show is that for any cycle r of length k, and any b in S. brb^-1 is also a cycle of the same length k. Am i getting somewhere?

Yes, you are trying to show if (a_1...a_k) is a k-cycle of r, then (b(a_1),...b(a_k)) is a k-cycle of brb^(-1). But as to the original question, b^(-1) turns b(a_1) into a_1. r turns a_1 into a_2. Finally the b makes it b(a_2), doesn't it? So brb^(-1) maps b(a_1) to b(a_2). Do you see where this is going?
 
  • #16
I do! The lightbulb just went on! thank you so much for bearing with me! :) :)
 

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