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Math Help - Proving that there aren't any more homomorphisms

  1. #1
    Junior Member beebe's Avatar
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    Proving that there aren't any more homomorphisms

    I was asked to find all homomorphisms \Phi :\mathbb{Z}\to\mathbb{Z}

    I'm fairly convinced in my own mind that all of them are \Phi (x)=cx \ \forall \ c \in \mathbb{Z}

    I can show that these are homomorphisms by claiming that \Phi is defined as above and:

    1. \Phi (a+b)=\Phi(a)+\Phi(b) (definition of homomorphism)
    2. c(a+b)=c(a)+c(b) (1, my def. of \Phi)
    3. ca+cb=ca+cb (2, distributive prop.)

    So, how do I show now that there are no other homomorphisms?
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  2. #2
    Super Member ILikeSerena's Avatar
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    Re: Proving that there aren't any more homomorphisms

    Hi beebe!

    Let's make an arbitrary choice for \Phi(1).
    Suppose \Phi(1)=c.
    Now what does \Phi(x) have to be (and why)?
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  3. #3
    MHF Contributor

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    Re: Proving that there aren't any more homomorphisms

    more generally, if G is any cyclic group, with generator x, and φ:G→G is a homomorphism, you can show that φ is completely determined by φ(x).

    (because \varphi(x^k) = (\varphi(x))^k). so if \varphi(x) = x^m, we have \varphi:x^k \to x^{km}.

    now \mathbb{Z} is cyclic, with generator 1, so....

    (recall that in \mathbb{Z}, x^k = x+x+\dots+x\ \text{(k times)}= kx).
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