Preserving cycle structure in transpositions

In summary: 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?
  • #1
PiRGood
14
0
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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  • #2
What is brb^(-1) of b(a_1)?
 
  • #3
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?
 
  • #4
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.
 
  • #5
isn't it simply b^-1(a_1)? Or (b(a_1))^-1?
 
  • #6
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)
 
  • #7
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?
 
  • #8
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
 
  • #9
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! :) :)
 

1. What is the significance of preserving cycle structure in transpositions?

The cycle structure refers to the way in which elements are rearranged in a permutation. In transpositions, only two elements are swapped, while the rest remain in their original positions. Preserving cycle structure means that the resulting permutation will have the same number and length of cycles as the original permutation.

2. Why is preserving cycle structure important in mathematical applications?

In mathematical applications, preserving cycle structure is important because it allows for the analysis and understanding of the underlying structure and patterns within a permutation. It can also help in solving certain problems or equations, as the cycle structure provides information about the relationships between elements.

3. How is cycle structure preserved in transpositions?

In transpositions, the elements being swapped are part of a cycle in the original permutation. This means that the elements have a direct or indirect relationship with each other. By swapping these elements, the cycle structure is preserved as the relationship between the elements remains unchanged.

4. Can cycle structure be preserved in other types of permutations?

Yes, cycle structure can be preserved in other types of permutations such as rotations, reflections, and combinations of transpositions. As long as the underlying relationships between elements are maintained, the cycle structure will remain the same.

5. What is the practical use of preserving cycle structure in transpositions?

Preserving cycle structure in transpositions has practical applications in various fields such as cryptography, computer science, and group theory. It can be used to analyze and solve problems involving permutations, as well as to understand the structure and behavior of different systems and algorithms.

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