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Math Help - Sanity check about a homomorphism

  1. #1
    Forum Admin topsquark's Avatar
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    Sanity check about a homomorphism

    (sighs) I need someone to confirm if I'm going insane again.
    Prove that for each fixed nonzero k \in \mathbb{Q} the map from \mathbb{Q} to itself by q \to kq is an automorphism of \mathbb{Q}.
    If it's an automorphism then it's a homomorphism. But:
    \phi _k : \mathbb{Q} \to \mathbb{Q}: q \mapsto kq is not a homomorphism, except for k = 1:

    Let f,g \in \mathbb{Q}.
    \phi _k (f g ) = kfg

    and
    \phi _k (f) \phi_k(g) = (kf) (kg) = k^2fg

    What am I missing this time? (It's just been that kind of day.)

    -Dan
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  2. #2
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    Re: Sanity check about a homomorphism

    Well, it's an automorphism with respect to addition...
    Thanks from topsquark
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