Let's consider
ϕ:C∗→R+
z↦∣z∣
This function is a surjective homomorphism.
For every x∈R+ , we can view this number as an element of C∗ , x∈C∗,∣x∣=x and thus it is surjective.
It is a homomorphism by the properties of complex numbers: ϕ(z1z2)=∣z1z2∣=∣z1∣∣z2∣=ϕ(z1)ϕ(z2)
We can also see that Ker(ϕ)={z∈C∗:ϕ(z)=∣z∣=1}=S . Thus by the isomorphism theorem ϕˉ:C∗/S→R+ is an isomorphism and so (C∗/S,⋅)≃ϕ(R+,⋅) .
Comments