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Math Help - Isomorphism of 2 groups

  1. #1
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    Isomorphism of 2 groups

    Is the group (\mathbb{C},\cdot)=\{ z \in \mathbb{C} \ | z \neq 0\} isomorphic to S \times  \mathbb{R}, where S=\{ z \in \mathbb{C} \ | |z|=1\}
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  2. #2
    Senior Member Dinkydoe's Avatar
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    I take it by \mathbb{R} you mean (\mathbb{R}, +).

    Then on S\times \mathbb{R} we have a group operation: (z_1,r_1)(z_2,r_2) = (z_1z_2,r_1+r_2).

    Then: (z,r)\mapsto e^rz is a homomorphism S\times \mathbb{R}\to \mathbb{C}^*. Work the details out for yourself.

    You can even show that the kernel of this homomorphism is trivial. ker (f) = <br />
\left\{(1,0)\right\}. After that apply the isomorphism-theorem.
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  3. #3
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    R doesnt mean that

    Hi--

    By \mathbb{R}+ imean the set of all positive real numbers.
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  4. #4
    Senior Member Dinkydoe's Avatar
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    Am I supposed to guess that?

    you couldve meant anything with \mathbb{R}.

    (\mathbb{R}, +) is a group,  (\mathbb{R}\setminus \left\{0\right\}, \cdot ) is a group. And yes (\mathbb{R}_{>0}, \cdot) is a group too. Provide these details next time.

    Anyway, yes these groups are isomorphic:
    (z,r) \mapsto rz with inverse z\mapsto (\frac{z}{|z|}, |z|) is a homomorphism: S\times \mathbb{R}_{> 0} \to \mathbb{C}^*. The kernel is of this homomorphism is trivial, namely:
    \left\{(1,1)\right\}.

    The isomorphism theorem gives: S\times \mathbb{R}_{>0}\cong \mathbb{C}^*
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