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Proof on Order of Elements in a Group

  1. Jul 20, 2008 #1
    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.
     
  2. jcsd
  3. Jul 20, 2008 #2
    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?
     
  4. Jul 20, 2008 #3
    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].
     
  5. Jul 20, 2008 #4
    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.
     
  6. Jul 21, 2008 #5

    mathwonk

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    let f:G-->G be an automorphism of G. if x has order n, prove f(x) also has order n.
     
  7. Jul 22, 2008 #6
    But in order to demonstrate that b . b^-1 is an automorphism, you would be basically doing the very proof shown above though?
     
  8. Jul 23, 2008 #7

    mathwonk

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    doing, and understanding WHAT you are doing, are two different things.
     
  9. May 11, 2009 #8
    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!
     
  10. May 12, 2009 #9

    matt grime

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    If n is the order of a, what is [itex]ba^nb^{-1}[/itex] ?
     
  11. May 12, 2009 #10
    are you asking what the order of [itex]ba^nb^{-1}[/itex] is?

    Woudl I be right in saying that it's n, because no matter what n is the b and b^-1 stay the same?
     
  12. May 12, 2009 #11
    Is there some sort of theorem for the order of a composition of more than one element?
     
  13. May 12, 2009 #12

    matt grime

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    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?
     
  14. May 12, 2009 #13
    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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