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Math Help - Isomporhisms preserve orders - help with proof

  1. #1
    Super Member Bernhard's Avatar
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    Isomporhisms preserve orders - help with proof

    I am a math hobbyist studying Joseph Gallian's "Contemorary Abstract Algebra" (Fifth Edition).

    Part of Theorem 6.2 on page 124 states that if  \phi :G \rightarrow G' is an isomorphism then the order of an element a  \in G is equal to the order of  \phi (a) ; that is  \ \mid a \mid \ = \ \mid \phi (a) \mid (that is isomorphisms preserve orders).

    Gallian's proof is essentially as follows:

    a has order n  \ \ \Longrightarrow a^n = e

     \ \Longrightarrow \ \phi (a^n) \ = \phi (e) \ = \ e'

     \ \Longrightarrow \ [\phi (a)]^n \ \ = \ e'

    Gallian seems to conclude the proof at this point, but to me he has only shown (using his Th 4.1 Corollary 2) that the order of  \phi (a) divides n

    How do you show that the order of  \phi (a) actually equals n?
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by Bernhard View Post
    How do you show that the order of  \phi (a) actually equals n?

    If ( \phi (a))^m=e' for m<n then ... (absurd).
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    Super Member Bernhard's Avatar
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    Fernando, why exactly is it absurd ... obviously I am missing something

    Can you be more explicit?

    [I am aware that the equality is not necessarily true for homomorphisms so it must be something to do with 1-1 and onto]

    Bernhard
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    \phi^{-1} is an isomorphism so if (\phi(a))^{m}=e' taking \phi^{-1}:


    \phi^{-1}[\;\phi(a)^m\;]=\phi^{-1}[\;e'\;]\Rightarrow a^m=e


    The order of a wouldn't be n .
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  5. #5
    Super Member Bernhard's Avatar
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    Thanks Fernando
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  6. #6
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by Bernhard View Post
    Thanks Fernando

    You are welcome.
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  7. #7
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    put another way, φ(a^n) = (φ(a))^n. now if (φ(a))^m = φ(a^m) = e', that means both e = a^n, and a^m ≠ e are in the kernel of φ, meaning φ is not injective, and thus could not be an isomorphism.

    in fact, φ does not need to be an isomorphism for this to be true, only a monomorphism (injective homomorphism), that is, it is injectivity (1-1) that preserves orders. here's an example:

    consider the map φ:Z2 -->Z4 given by φ(k) = 2k. φ is injective. 0 has order 1 in Z2, and φ(0) = 0 has order 1 in Z4. 1 has order 2 in Z2, and φ(1) = 2 has order 2 in Z4. in this example φ preseves orders even though φ is not an isomorphism of Z2 with Z4.
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