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Math Help - Abstract Algebra: cyclic groups

  1. #1
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    Abstract Algebra: cyclic groups

    Let G be a cyclic group with generator a, and let G' be a group isomorphic to G. If phi: G-> G' is an isomorphism, show that for every x in G, phi(x) is completely determined by the value phi(a). That is, if phi: G -> G' and psi: G -> G' are two isomorphisms such that phi(a) = psi(a) and phi(x) = psi(x) for all x in G.

    I know that it has something to do with how x = a^n for some integer. But I have no idea what to do next! Thanks in advance.
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    Quote Originally Posted by fulltwist8 View Post
    Let G be a cyclic group with generator a, and let G' be a group isomorphic to G. If phi: G-> G' is an isomorphism, show that for every x in G, phi(x) is completely determined by the value phi(a). That is, if phi: G -> G' and psi: G -> G' are two isomorphisms such that phi(a) = psi(a) and phi(x) = psi(x) for all x in G.
    Since \left< a \right> = G it means if x\in G then x = a^k and thus \phi(x) = \phi(a^k) = \phi(a)^k.
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    ok and that makes sense, but where does the psi come into play?
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    Quote Originally Posted by fulltwist8 View Post
    ok and that makes sense, but where does the psi come into play?
    It is saying that if \phi ans \psi are two isomorphisms which agree at the group generator then they agree eveywhere, i.e. it is the same isomorphism.
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