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Math Help - Group proof by induction

  1. #1
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    Group proof by induction

    prove by induction that for elements x and g of a group,

    (x^-1.g.x)^n=x^-1.g^n.x

    My working.

    I've done the easy parts, showing true for n=1, assuming true for n=k but now given that, I need to show true for n=k+1.

    I worked with LHS, expanding the braket and trying to manpiulate to get an expression I already knew but I'm pretty stumped.

    Thanks for any help.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Use:

    (x^{-1}gx)^{k+1}=(x^{-1}gx)(x^{-1}gx)^k

    Now, apply the induction hypothesis.


    Fernando Revilla
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  3. #3
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    Quote Originally Posted by FernandoRevilla View Post
    Use:

    (x^{-1}gx)^{k+1}=(x^{-1}gx)(x^{-1}gx)^k

    Now, apply the induction hypothesis.

    Fernando Revilla

    so I obtain (x^-1.g^k.x)(x^-1g.x) but I can't assume commutativity so how do I simplify to x^-1.g^k+1.x.
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    You needn't apply commutativity:


    (x^{-1}gx)^{k+1}=(x^{-1}gx)(x^{-1}gx)^k=(x^{-1}gx)(x^{-1}g^kx)=\ldots=x^{-1}g^{k+1}x



    Fernando Revilla
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  5. #5
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    Quote Originally Posted by FernandoRevilla View Post
    You needn't apply commutativity:


    (x^{-1}gx)^{k+1}=(x^{-1}gx)(x^{-1}gx)^k=(x^{-1}gx)(x^{-1}g^kx)=\ldots=x^{-1}g^{k+1}x



    Fernando Revilla
    can I just multiply the middle x and x^-1?
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  6. #6
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by Plague01 View Post
    can I just multiply the middle x and x^-1?

    Of course you can. Use the associative property and xx^{-1}=1 .


    Fernando Revilla
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