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    The question has me stumped I have no idea of how to go about it.

    Question: A group G has an element a with order n, so that a^n=e, where e is the identity . It is given that x is any element of G distinct from a and e

    i)Prove that the order of x^-1ax is n, making clear with group property is being used at each stage.
    ii) express the inverse of x^-1ax in terms of some or all of x,x^-1,a, a^-1 show sufficient working to justify it.
    iii)it is now given that a commutes with every element of G. Prove that a^-1 commutes with every element.

    Help would be appreciated.
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    (Original post by Dualcore)
    The question has me stumped I have no idea of how to go about it.

    Question: A group G has an element a with order n, so that a^n=e, where e is the identity . It is given that x is any element of G distinct from a and e

    i)Prove that the order of x^-1ax is n, making clear with group property is being used at each stage.
    ii) express the inverse of x^-1ax in terms of some or all of x,x^-1,a, a^-1 show sufficient working to justify it.
    iii)it is now given that a commutes with every element of G. Prove that a^-1 commutes with every element.

    Help would be appreciated.
    To start with, can you see what would happen if you worked out
    (x^{-1}ax)(x^{-1}ax) ?
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    (Original post by davros)
    To start with, can you see what would happen if you worked out
    (x^{-1}ax)(x^{-1}ax) ?
    (x^{-1}aax)

    as xx^{-1}=e and ea=a
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    (Original post by Dualcore)
    (x^{-1}aax)

    as xx^{-1}=e and ea=a
    And how else can you write the product aa? What happens if you work out
    (x^{-1}ax)(x^{-1}ax)(x^{-1}ax) ?

    Can you see where this is going?
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    (Original post by davros)
    And how else can you write the product aa? What happens if you work out
    (x^{-1}ax)(x^{-1}ax)(x^{-1}ax) ?

    Can you see where this is going?
    Thanks for your help, i've manage to solve it now
    Even though I've done (ii) and (iii) now, for (iii) would it be sufficient enough to prove that the element a^{-1} commutes with x
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    (Original post by Dualcore)
    Thanks for your help, i've manage to solve it now
    Even though I've done (ii) and (iii) now, for (iii) would it be sufficient enough to prove that the element a^{-1} commutes with x
    Well you have to prove that it commutes with any element x!

    This follows almost immediately if they tell you that ax = xa for all x because you can just pre- and post-multiply by a^{-1} and use associativity.
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    (Original post by davros)
    Well you have to prove that it commutes with any element x!

    This follows almost immediately if they tell you that ax = xa for all x because you can just pre- and post-multiply by a^{-1} and use associativity.
    Yeah that's what I did, thanks again for your help.
 
 
 
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