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Thread: Isomorphism of 2 groups

  1. #1
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    Isomorphism of 2 groups

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

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

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

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

    Hi--

    By $\displaystyle \mathbb{R}+$ 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 $\displaystyle \mathbb{R}$.

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

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

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