Preserving cycle structure in transpositions

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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