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Math Help - Automorphism - question

  1. #1
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    Automorphism - question

    G - Finite Group
    T - Automorphism of G
    For all x in G, T(x) = x IFF x=1
    Also, T(T(x))=x for all x in G.

    Prove G is abelian.

    Struggling a lot with this. I suspect T(x) = x^{-1}.
    And once I establish this, proving G is abelian is simple.

    But not going anywhere with the hunch of T(x) = x^{-1}.

    Any pointers please?
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  2. #2
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    Quote Originally Posted by aman_cc View Post
    G - Finite Group
    T - Automorphism of G
    For all x in G, T(x) = x IFF x=1
    Also, T(T(x))=x for all x in G.

    Prove G is abelian.

    Struggling a lot with this. I suspect T(x) = x^{-1}.
    And once I establish this, proving G is abelian is simple.

    But not going anywhere with the hunch of T(x) = x^{-1}.

    Any pointers please?
    define the function f: G \longrightarrow G by f(g)=g^{-1}T(g). now if f(g_1)=f(g_2), then T(g_1g_2^{-1})=g_1g_2^{-1} and thus g_1=g_2, because we're given that T(x)=x \Longrightarrow x = 1. so f is injective and thus

    surjective, because G is finite. now let x \in G. then x=f(g)=g^{-1}T(g), for some g \in G. thus T(x)=T(g^{-1}T(g))=T(g^{-1})T(T(g))=(T(g))^{-1}g=(g^{-1}T(g))^{-1}=x^{-1}. hence G is abelian.
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  3. #3
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    Quote Originally Posted by NonCommAlg View Post
    define the function f: G \longrightarrow G by f(g)=g^{-1}T(g). now if f(g_1)=f(g_2), then T(g_1g_2^{-1})=g_1g_2^{-1} and thus g_1=g_2, because we're given that T(x)=x \Longrightarrow x = 1. so f is injective and thus

    surjective, because G is finite. now let x \in G. then x=f(g)=g^{-1}T(g), for some g \in G. thus T(x)=T(g^{-1}T(g))=T(g^{-1})T(T(g))=(T(g))^{-1}g=(g^{-1}T(g))^{-1}=x^{-1}. hence G is abelian.
    Thanks !
    I could never have guessed it - to define f(g)=g^{-1}T(g) and then do it !!!
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