Proof on Order of Elements in a Group

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Discussion Overview

The discussion revolves around proving that the order of the element \( bab^{-1} \) in a group \( G \) is equal to the order of the element \( a \). Participants explore the implications of group properties, particularly in non-abelian contexts, and engage in mathematical reasoning related to group automorphisms and the definition of order.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant expresses confusion about proving the equality of orders without assuming the group is abelian, questioning the rearrangement of elements.
  • Another participant suggests examining the expression \( (bab^{-1})^2 \) as a potential approach to understanding the order of \( bab^{-1} \).
  • A participant notes that if \( n \) is the order of \( a \), then \( (bab^{-1})^n = ba^nb^{-1} \) but seeks clarification on how this demonstrates that the orders are equal.
  • There is a discussion about the implications of automorphisms and whether the proof of \( b \cdot b^{-1} \) being an automorphism relates to the original problem.
  • Participants inquire about the order of the composition \( ba^nb^{-1} \) and its relationship to the order of \( a \). Some express that they believe it remains \( n \) due to the nature of \( b \) and \( b^{-1} \).
  • There is a request for clarification on whether \( (ab)^{x} = 1 \) is equivalent to \( (ab)^{x} = e \), indicating a potential exploration of definitions and properties of group elements.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the proof or the implications of their findings. Multiple viewpoints and uncertainties remain regarding the relationship between the orders of \( a \) and \( bab^{-1} \), as well as the nature of automorphisms in this context.

Contextual Notes

Participants express uncertainty about the definitions and properties involved, particularly in non-abelian groups. There are unresolved questions about the implications of their mathematical manipulations and the assumptions underlying their reasoning.

jeffreydk
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I'm trying to figure out how to prove the following...

If [tex]a, b \in G[/tex] where G is a group, then the order of [tex]bab^{-1}[/tex] equals the order of [tex]a[/tex].

I'm rather stumped because the group is not necessarily abelian and it seems like it would have to be in order to directly show that you can rearrange b and b's inverse to get rid of them. I'm confused party because I'm not sure if those properties still hold when you're working with the order of the elements. Any help is greatly appreciated, thanks.
 
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What is (bab-1)n? And given b, and k in a group G when is it true that bkb-1 is equal to the identity?
 
As a hint, note that [tex](bab^{-1})^2 = (bab^{-1})(bab) = ba(b^{-1}b)ab^{-1} = baeab^{-1} = baab^{-1} = ba^2b^{-1}[/tex], where [tex]e[/tex] is the identity in [tex]G[/tex].
 
Oh ok, it just hit me it makes perfect sense. I wasn't thinking hard enough about the actual definition of order. Thank you both for your help.
 
let f:G-->G be an automorphism of G. if x has order n, prove f(x) also has order n.
 
But in order to demonstrate that b . b^-1 is an automorphism, you would be basically doing the very proof shown above though?
 
doing, and understanding WHAT you are doing, are two different things.
 
I can show
[tex](bab^{-1})^n = ba^nb^{-1}[/tex]

But how does this show the orders are equal?

Any help would be great please!
 
If n is the order of a, what is [itex]ba^nb^{-1}[/itex] ?
 
  • #10
matt grime said:
If n is the order of a, what is [itex]ba^nb^{-1}[/itex] ?

are you asking what the order of [itex]ba^nb^{-1}[/itex] is?

would I be right in saying that it's n, because no matter what n is the b and b^-1 stay the same?
 
  • #11
Is there some sort of theorem for the order of a composition of more than one element?
 
  • #12
Firepanda said:
are you asking what the order of [itex]ba^nb^{-1}[/itex] is?


No, I'm asking you what it is. Let me try to make it even more clear: if n is the order of a, what is a^n? Now, what is ba^nb^-1?
 
  • #13
matt grime said:
No, I'm asking you what it is. Let me try to make it even more clear: if n is the order of a, what is a^n? Now, what is ba^nb^-1?

Ah i gotcha! Kk it all made sense now.

Would you be able to help me here also? :

https://www.physicsforums.com/showthread.php?t=181745

is saying [tex](ab)^{x} = 1.[/tex] the same as saying [tex](ab)^{x} = e[/tex]?

:)
 

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