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Math Help - is f(f(x)) = f(x) f(x) , f is automorphism f^2 is identity

  1. #1
    MHF Contributor Amer's Avatar
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    is f(f(x)) = f(x) f(x) , f is automorphism f^2 is identity

    let G be a finite group
    if sigma is automorphism from G to G such that

     \sigma (g) = g \text{ if and only if }\;\; g =1

    given that  \sigma^2 is the identity automorphism

    prove that G is abelian ( hint prove that every element of G can be written as  x^{-1}\sigma (x) apply sigma to such expression


     x = x^{-1} \sigma (x)

     \sigma (x) = \sigma (x^{-1} \sigma (x))

     \sigma (x) = \sigma (x^{-1}) \sigma (\sigma (x))

     \sigma (x) \sigma (x) = \sigma (x) \sigma (x^{-1}) \sigma (\sigma (x))

     ( \sigma (x) )^2 = \sigma (\sigma (x))


    as you can see I end up with what I am asking about, did I made something wrong?

    I work in it in another way

     \sigma ^2 (x) = x since sigma^2 is the identity

    let  \sigma (x) = x*

     \sigma (\sigma (x)) = \sigma (x*) \Rightarrow \;\; x = \sigma (x*)

    so if x image is x* then x* image is x
    Last edited by Amer; November 14th 2011 at 10:08 AM.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Re: is f(f(x)) = f(x) f(x) , f is automorphism f^2 is identity

    Quote Originally Posted by Amer View Post
    let G be a finite group
    if sigma is automorphism from G to G such that

     \sigma (g) = g \text{ if and only if }\;\; g =1

    given that  \sigma^2 is the identity automorphism

    prove that G is abelian ( hint prove that every element of G can be written as  x^{-1}\sigma (x) apply sigma to such expression


     x = x^{-1} \sigma (x)

     \sigma (x) = \sigma (x^{-1} \sigma (x))

     \sigma (x) = \sigma (x^{-1}) \sigma (\sigma (x))

     \sigma (x) \sigma (x) = \sigma (x) \sigma (x^{-1}) \sigma (\sigma (x))

     ( \sigma (x) )^2 = \sigma (\sigma (x))


    as you can see I end up with what I am asking about, did I made something wrong?

    I work in it in another way

     \sigma ^2 (x) = x since sigma^2 is the identity

    let  \sigma (x) = x*

     \sigma (\sigma (x)) = \sigma (x*) \Rightarrow \;\; x = \sigma (x*)

    so if x image is x* then x* image is x
    I think you are making the mistake of interpreting (\sigma^2)(x)=(\sigma(x))^2 and not \sigma(\sigma(x)), no?
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  3. #3
    MHF Contributor Amer's Avatar
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    Re: is f(f(x)) = f(x) f(x) , f is automorphism f^2 is identity

    I think you are making the mistake of interpreting (\sigma^2)(x)=(\sigma(x))^2 and not \sigma(\sigma(x)), no?
    Thanks for your respond
    I think it is wrong but I reached it from the given hint tell me if I did something wrong in my work

    I want to give the whole question as it is in the book

    Let G be a finite group which possesses an automorphism \sigma such that \sigma (g) = g if and only if g=1 .
    if \sigma ^2 is the identity map from G to G,
    Prove that G is abelian (such an automorphism called fixed point free of order 2 ).

    [show that every element of G can be written in the form x^{-1}\sigma (x) and apply \sigma to such expression ]

    Page 38, Question 23. TextBook D.S.Dummit and R.M Foote, Abstract Algebra, 3^{rd} edition, prentice-Hall, 2004. QS162.D85
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  4. #4
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    Re: is f(f(x)) = f(x) f(x) , f is automorphism f^2 is identity

    the multiplication in Aut(G) is functional composition, not the "element-wise" multiplication of G.
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  5. #5
    MHF Contributor Amer's Avatar
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    Re: is f(f(x)) = f(x) f(x) , f is automorphism f^2 is identity

    the multiplication in Aut(G) is functional composition, not the "element-wise" multiplication of G.
    so what I did wrong in this
    the question gave me a hint which said show that every element in G can be written as

     x^{-1} \sigma (x)

    so I suppose that

     x = x^{-1} \sigma (x) then i apply sigma to both sides

    \sigma (x) = \sigma (x^{-1}\sigma (x))

    \sigma (x) = \sigma (x^{-1}) \sigma (\sigma (x)) since sigma is homo
    then I multiply both sides by sigm(x)

    \sigma (x) \sigma (x) = \sigma (x) \sigma (x^{-1}) \sigma (\sigma (x))

    ( \sigma (x))^2 = \sigma (\sigma (x))

    I cant see a mistake :S I know that the multiplication in Auto G is composition of functions
    Thanks
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  6. #6
    MHF Contributor Swlabr's Avatar
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    Re: is f(f(x)) = f(x) f(x) , f is automorphism f^2 is identity

    Quote Originally Posted by Amer View Post
    so what I did wrong in this
    the question gave me a hint which said show that every element in G can be written as

     x^{-1} \sigma (x)

    so I suppose that

     x = x^{-1} \sigma (x) then i apply sigma to both sides
    No - every element can be written as x^{-1}\sigma(x), but the x may vary. Basically, for every y\in G there exists an x\in G such that y=x^{-1}\sigma(x).
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  7. #7
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    Re: is f(f(x)) = f(x) f(x) , f is automorphism f^2 is identity

    Quote Originally Posted by Amer View Post
    let G be a finite group
    if sigma is automorphism from G to G such that

     \sigma (g) = g \text{ if and only if }\;\; g =1

    given that  \sigma^2 is the identity automorphism

    prove that G is abelian ( hint prove that every element of G can be written as  x^{-1}\sigma (x) apply sigma to such expression
    define the map f:G \longrightarrow G by f(x) = x^{-1}\sigma(x) for all x \in G. then f is one-to-one because if f(x)=f(y), then \sigma(xy^{-1})=xy^{-1} and so x= y, by the first condition you gave us. thus f is onto because G is finite. now let g \in G and choose x \in G such that g=f(x)=x^{-1}\sigma(x). then \sigma(g)=\sigma(x^{-1})x = g^{-1}. hence \sigma(g)=g^{-1} for all g \in G and so G is abelian.
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  8. #8
    MHF Contributor Amer's Avatar
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    Re: is f(f(x)) = f(x) f(x) , f is automorphism f^2 is identity

    Swlabr, NonCommAlg Thanks very much it is very clear now
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