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Math Help - Finite group

  1. #1
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    Finite group

    Let G be a finite group which possesses an automorphism \sigma such that \sigma(g)=g iff. g = 1.

    (\Rightarrow)
    Suppose |G| = n and \sigma(g)=g.

    First, I don't get why this would only work if g is 1.
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  2. #2
    Super Member girdav's Avatar
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    Re: Finite group

    I don't understand what we have to show.
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  3. #3
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    Re: Finite group

    Quote Originally Posted by girdav View Post
    I don't understand what we have to show.
    Then you aren't alone.
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  4. #4
    Super Member girdav's Avatar
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    Re: Finite group

    In your first post, you say that we have a finite group which possesses an automorphism \sigma such that \sigma (g)=g if and only if g=1.
    Then, what do we have to do?
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  5. #5
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    Re: Finite group

    Quote Originally Posted by girdav View Post
    In your first post, you say that we have a finite group which possesses an automorphism \sigma such that \sigma (g)=g if and only if g=1.
    Then, what do we have to do?
    Show it is a homomorphism, monic, and epi.
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Re: Finite group

    Quote Originally Posted by dwsmith View Post
    Let G be a finite group which possesses an automorphism \sigma such that \sigma(g)=g iff. g = 1.

    (\Rightarrow)
    Suppose |G| = n and \sigma(g)=g.

    First, I don't get why this would only work if g is 1.
    If I had to guess, you are asking about the common problem: "If G is a finite group and \sigma\in\text{Aut}(G) such that \sigma^2=\text{id} and \sigma possesses no non-identity fixed points, then \sigma is the inverse map and G is abelian."
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