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Math Help - Algebra, Problems For Fun (9)

  1. #1
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    Algebra, Problems For Fun (9)

    Let G be a group and suppose there exists n \geq 1 such that the map g \mapsto g^n is an automorphism of G. Prove that g^{n-1} \in Z(G) for all g \in G.
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  2. #2
    MHF Contributor Swlabr's Avatar
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    Note that \phi: G \rightarrow G, g \mapsto g^n is a homomorphism and so (gh)^n = g^{n}h^n. Also, as it is onto every h' \in G can be written in this form, h' = h^n for some h \in G.

    Thus, it is sufficient to prove that g^{n-1}h^n = h^{n}g^{n-1}.

    Now, note that g^{n}h^{n} = (gh)^n = g(hg)^{n-1}h
    \Rightarrow g^{n-1}h^{n-1} = (hg)^{n-1} = \underbrace{(hg)(hg)...(hg)}_{n-1} = (hg)^{n}g^{-1}h^{-1} = h^{n}g^{n}g^{-1}h^{-1}.

    Thus, g^{n-1}h^n = (hg)^{n-1}h = h^{n}g^{n}g^{-1}h^{-1}h = h^{n}g^{n-1} as required.

    I hope that's correct - haven't had a chance to check it properly as my wife is badgering me to go through and eat lunch!
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  3. #3
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    Quote Originally Posted by Swlabr View Post
    Note that \phi: G \rightarrow G, g \mapsto g^n is a homomorphism and so (gh)^n = g^{n}h^n. Also, as it is onto every h' \in G can be written in this form, h' = h^n for some h \in G.

    Thus, it is sufficient to prove that g^{n-1}h^n = h^{n}g^{n-1}.

    Now, note that g^{n}h^{n} = (gh)^n = g(hg)^{n-1}h
    \Rightarrow g^{n-1}h^{n-1} = (hg)^{n-1} = \underbrace{(hg)(hg)...(hg)}_{n-1} = (hg)^{n}g^{-1}h^{-1} = h^{n}g^{n}g^{-1}h^{-1}.

    Thus, g^{n-1}h^n = (hg)^{n-1}h = h^{n}g^{n}g^{-1}h^{-1}h = h^{n}g^{n-1} as required.

    I hope that's correct - haven't had a chance to check it properly as my wife is badgering me to go through and eat lunch!
    it's correct. go eat your lunch before you get in trouble! haha
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