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Math Help - additive and multiplicative groups of a field are not isomorphic

  1. #1
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    additive and multiplicative groups of a field are not isomorphic

    This is from Dummit and Foote Abstract Algebra.

    Prove that the additive and multiplicative groups of a field are never isomorphic.

    Hint from the book:

    Consider three cases: when |\mathbb{F}| is finite, when -1 \ne 1 and when -1=-1 in  \mathbb{F}.

    The finite part is not bad.

    let |\mathbb{F}|=n

    then |(\mathbb{F},+)|=n but |(\mathbb{F}, \cdot )|=n-1

    because 0 does not have an inverse in the multiplicative group.

    For the next case I think contradiction is the way to go so I supposed that there exists an isomorphism \phi : (\mathbb{F},\cdot) \to (\mathbb{F},+) and that -1 \ne 1

    I know that \phi(1)=0 becuase the identity must map to the identity. Also that

    \phi(1)=\phi[(-1)(-1)]=\phi(-1)+\phi(-1)=0

    and for any a \in \mathbb{F}

    \phi[(-1)(a)]=\phi(-1)+\phi(a)
    \phi(1(-a))=\phi(1)+\phi(-a)

    I know that inverses must get mapped to inverses by an isomorphism.

    So I don't see how to get a contradiction So I know I am missing something.

    Thanks

    TES
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    Quote Originally Posted by TheEmptySet View Post
    This is from Dummit and Foote Abstract Algebra.

    Prove that the additive and multiplicative groups of a field are never isomorphic.

    Hint from the book:

    Consider three cases: when |\mathbb{F}| is finite, when -1 \ne 1 and when -1=-1 in  \mathbb{F}.

    The finite part is not bad.

    let |\mathbb{F}|=n

    then |(\mathbb{F},+)|=n but |(\mathbb{F}, \cdot )|=n-1

    because 0 does not have an inverse in the multiplicative group.

    For the next case I think contradiction is the way to go so I supposed that there exists an isomorphism \phi : (\mathbb{F},\cdot) \to (\mathbb{F},+) and that -1 \ne 1

    I know that \phi(1)=0 becuase the identity must map to the identity. Also that

    \phi(1)=\phi[(-1)(-1)]=\phi(-1)+\phi(-1)=0

    and for any a \in \mathbb{F}

    \phi[(-1)(a)]=\phi(-1)+\phi(a)
    \phi(1(-a))=\phi(1)+\phi(-a)

    I know that inverses must get mapped to inverses by an isomorphism.

    So I don't see how to get a contradiction So I know I am missing something.

    Thanks

    TES
    suppose there's a group isomorphism f: (F,+) \longrightarrow F^{\times}.

    1) \text{char} F = 2: then (f(1))^2=f(2)=f(0)=1. the equation x^2=1 in a field of characteristic 2 has the unique solution x = 1. thus f(1)=1=f(0), which is impossible because f is injective.

    2) \text{char} F \neq 2: since f is surjective, there exists a \in F: \ f(a)=-1. but then f(0)=1=(f(a))^2=f(2a). thus 2a=0, and hence a=0. but then -1=f(a)=f(0)=1. contradiction!
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