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Math Help - Abstract Algebra

  1. #1
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    Abstract Algebra

    Show that  (\mathbb{R}, \; +) is not isomorphic to  (\mathbb{R}^*, \; \cdot) , where  \mathbb{R}^* is the group of units in  \mathbb{R} . i.e.  \mathbb{R}^* = \mathbb{R}\backslash \{0\}
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  2. #2
    MHF Contributor red_dog's Avatar
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    Suppose that (\mathbf{R},+) is isomorphic to (\mathbf{R}^*,\cdot)

    Let \mathbf{R},+)\to (\mathbf{R}^*.\cdot)" alt="f\mathbf{R},+)\to (\mathbf{R}^*.\cdot)" /> an isomorphism.

    Then f(x+y)=f(x)f(y), \ \forall x,y\in\mathbf{R}

    For y=x\Rightarrow f(2x)=(f(x))^2\Rightarrow f(x)=\left(f\left(\frac{x}{2}\right)\right)^2\Righ  tarrow f(x)\geq 0, \ \forall x\in\mathbf{R}

    Then f is not a surjection.
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