# Thread: Isomorphism of 2 groups

1. ## Isomorphism of 2 groups

Is the group $(\mathbb{C},\cdot)=\{ z \in \mathbb{C} \ | z \neq 0\}$ isomorphic to $S \times \mathbb{R}$, where $S=\{ z \in \mathbb{C} \ | |z|=1\}$

2. 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 $(f) =
\left\{(1,0)\right\}$
. After that apply the isomorphism-theorem.

3. ## R doesnt mean that

Hi--

By $\mathbb{R}+$ imean the set of all positive real numbers.

4. 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: $S\times \mathbb{R}_{>0}\cong \mathbb{C}^*$