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Group thoery question.

  1. Aug 24, 2007 #1
    1. The problem statement, all variables and given/known data

    For any elements a and b from a group and any integer n, prove that (a^(-1)*ba)^n = a^(-1) b^n *a


    2. Relevant equations

    There are no equations given for this particular problem




    3. The attempt at a solution

    by law of exponents and the distributive property(a^-1 *b*a)= a^-n *b^n *a^n=a^(-n +n)*b^n = a^0 * b^n = b^n

    Likewise, a^(-1) * b^n * a^1 = a^(-1+1)* b^n

    since (a^-1 *b *a)^n =b^n and since a^-1 * b^n *a^1 = b^n , then (a^-1 * b*a) = (a^-1 *b^n *a^1)
     
  2. jcsd
  3. Aug 24, 2007 #2

    matt grime

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    Groups are not commutative. You can't just say, as you have, that (xy)^n=x^ny^n, that is just not true. So what is the definition of x^n? It is x*x*x*..*x with n x's. End of story.
     
  4. Aug 24, 2007 #3

    dextercioby

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    [tex] \left(a^{-1}ba\right)^{n}= \left(a^{-1}ba\right) \left(a^{-1}ba\right) \left(a^{-1}ba\right)... [/tex]

    Now use associativity.
     
  5. Aug 25, 2007 #4
    Assiociative property: ((a^-1 *b)*a)^n= (a^-1 *(b*a))^n : that would be how you would apply the assiociative property to this problem.

    I know the assiociative property is one of the 3 properties, along with the identity and inverse property to prove that a particular non-empty set is a group.

    But how is the assiociative property related to what you wrote : ((a^-1 *b)*a)^n=(a^-1 *b)*a)*(a^-1 *b)*a)*(a^-1 *b)*a)
     
  6. Aug 25, 2007 #5

    matt grime

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    Associativity means that one does not need brackets.
     
  7. Aug 25, 2007 #6
    What do you mean associativity does not need brackets?

    Assiociativity is based on this fundamental operation: (ab)c=a(bc) for all a,b, c in G
     
  8. Aug 25, 2007 #7
    Yes, and there is no doubt at all that matt knows this. Let's take it one step at a time. Can you calculate/simplify (a-1ba)2=(a-1ba)(a-1ba)?

    Hint: apply the associative property.
     
  9. Aug 25, 2007 #8
    yes.
     
  10. Aug 25, 2007 #9
    Ok... And what is it?
     
  11. Aug 25, 2007 #10
    Why do I need to simplify (a-1ba)^2=(a-1ba)^1 *(a-1ba)^1 again when you already simplfied the equation for me? I thought you wanted me to regconize whether the particular equation you wrote was simplified or not.
     
  12. Aug 25, 2007 #11
    I don't recall doing this, I asked you if you could simplify it yourself and you replied yes, so what did you come up with?

    The point of that rather simple question was that if you cannot do that, there is little hope of solving the more general problem in your first post.

    So shall we try this again if we have (a-1ba)2=(a-1ba)(a-1ba)

    Can you simplify this? Your first post says that (a-1ba)n=(a-1bna) for all natural n. So in other words (a-1ba)2 should equal (a-1b2a).

    Now can you show this?
     
  13. Aug 26, 2007 #12

    matt grime

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    That is not an operation. Now, it is an identity, which means we can unambiguously write abc, and we know it has a unique meaning.

    Perhaps you would prefer it, and find it helpful, if I pointed out that associatvity means we can insert and move brackets around at will in an expression. Now, please, simplify

    [tex]aba^{-1}aba^{-1}[/tex]

    Notice how I'm omitting the brackets?
     
    Last edited: Aug 26, 2007
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