Determine the following factor group: $\mathbb{C}^* / \mathbb{R}_{>0}$ Using the first isomorphism theorem, determine the following factor group:
$\mathbb{C}^* / P$ with $P = \mathbb{R}_{>0}$
How to find the factor group:
Find a surjective group homomorphism from the respective group whose kernel is the respective subgroup:
$\phi : \mathbb{C}^* \rightarrow G $ with ker$(\phi) = \{x \in \mathbb{C}^*, \phi(x) = $ neutral element of G$ \} =  P = \mathbb{R}_{>0} $
After that use the isomorphism theorem to describe the respective factor group down to isomorphism.
My problem here is that I can't find a suitable group G so that exactly the kernel is equal to P.
Which group G would fit here and what would be the corresponding mapping rule and the neutral element of G?

The obvious answer is the circle group $\Bbb T$, denoted that way because it is a $1$-torus.
$\Bbb T$ is a group under the operation of complex multiplication,  with the role of the identity played by $z=1$.
The homomorphism is $\Bbb C^*\ni z\mapsto \dfrac z{\lvert z\rvert}\in\Bbb T$, and it's kernel is $P$.
This is an important example.
