$g(i)$ homomorphisms and $g(z)=z^2$ and $g(z)=z^{-1}$ exercise

Question

Exercise:Let $G$ be agroup $(\mathbb{C}\{0\},.)$. Consider $g:G\to G$ to be a homomorphism, and show the possible values of $g(i)$. For each one , give an example of a homomorphism $g:G\to G$ on those conditions.

Solution:

$g(i)=i \:\:\:\text{and}\:\:\: g(z)=z\:\:\:\:\:\:\:\:\:\:\:\:\:g(i)=1\:\:\:\text{and}\:\:\: g(z)=1$

$g(i)=-1 \:\:\:\text{and}\:\:\: g(z)=z^2\:\:\:\:\:\:\:\:\:\:\:\:\:g(i)=-i \:\:\:\text{and}\:\:\: g(z)=z^{-1}$

I understand the $g(i)$ homomorphisms and the fact $g(z)=z $ as well as g(z)=1. However I do not understand $g(z)=z^2$ and $g(z)=z^{-1}$. I tried to think of the orders but I guess I cannot attribute an finite order to $z^2$ or $z^{-1}$,

Question:

Where do $g(z)=z^2$ and $g(z)=z^{-1}$ come from?

Thanks in advance!

Answer 1

Because $g$ is a homomorphism, $i^4=1$ implies $$ 1=g(1)=g(i^4)=(g(i))^4. $$ So we have the above possibilities, i.e., $g(i)\in \{1,-1,i,-i\}$. Of course $g(i)=-1$ would hold for $g(z)=z^2$, since $i^2=-1$. Similarly, with $g(z)=z^{-1}$ we would have $g(i)=i^{-1}=-i$.