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Math Help - Homomorphisms/Isomorphisms

  1. #1
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    Homomorphisms/Isomorphisms

    a) The inclusion Z -> R is a homomorphisms of additive groups.

    b) The subgroup {0} of Z is isomorphic to the subgroup {(1)} of S5.

    These are both true statements, but I must know how to prove them. Can someone show me please? Thanks.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Re: Homomorphisms/Isomorphisms

    Quote Originally Posted by jzellt View Post
    a) The inclusion Z -> R is a homomorphisms of additive groups.

    b) The subgroup {0} of Z is isomorphic to the subgroup {(1)} of S5.

    These are both true statements, but I must know how to prove them. Can someone show me please? Thanks.
    Let me ask you a question, if \iota:\mathbb{Z}\hookrightarrow \mathbb{R} is our inclusion what is \iota(x+y)? What about \iota(x)+\iota(y)?
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  3. #3
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    Re: Homomorphisms/Isomorphisms

    I'm not sure. If f is our function f: Z -> R, doesn't it have to be defined for me to continue?
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    Re: Homomorphisms/Isomorphisms

    Quote Originally Posted by jzellt View Post
    I'm not sure. If f is our function f: Z -> R, doesn't it have to be defined for me to continue?
    We have defined the function, it's the inclusion map. Let's pause for a second, did you take the time to fully understand all the words in the problem statement? Do you know what an inclusion map is?
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  5. #5
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    Re: Homomorphisms/Isomorphisms

    nevermind see next post...
    Last edited by jzellt; October 22nd 2011 at 08:01 PM.
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    Re: Homomorphisms/Isomorphisms

    So I googled it and it looks like the inclusion map of function f is defined by f(x) = x. So to show that f: Z->R is an homomorphism, I must show that f(x+y) = f(x) + f(y). But f(x+y) = x+y and f(x) + f(y) = x+y. So were done. COrrect?
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  7. #7
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    Re: Homomorphisms/Isomorphisms

    Quote Originally Posted by jzellt View Post
    So I googled it and it looks like the inclusion map of function f is defined by f(x) = x. So to show that f: Z->R is an homomorphism, I must show that f(x+y) = f(x) + f(y). But f(x+y) = x+y and f(x) + f(y) = x+y. So were done. COrrect?
    Exactly.
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