Is the group $\displaystyle (\mathbb{C},\cdot)=\{ z \in \mathbb{C} \ | z \neq 0\}$ isomorphic to $\displaystyle S \times \mathbb{R}$, where $\displaystyle S=\{ z \in \mathbb{C} \ | |z|=1\}$
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.
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: $\displaystyle S\times \mathbb{R}_{>0}\cong \mathbb{C}^*$