Given that $(\mathbb{Z}, +)$ and $(\mathbb(C), *)$ are groups, show that the function $f: ((\mathbb{Z}, +) \rightarrow (\mathbb{C}, *)$ is a homomorphism of groups with $f(x) = i^x, \forall x \in \mathbb{Z}$.
Based on what I understand, I am to prove that:
- $f(x + y) = f(x) * f(y)$
From which I got:
$$ i^{(x+y)} = i^x * i^y \\ i^{(x+y)} = i^{(x+y)} $$
- $f(e_\mathbb{Z}) = e_\mathbb{C}$ : $e$ - neutral element of the set.
From which I deduced that $e_\mathbb{Z} = 0$ therefore:
$$ f($e_\mathbb{Z}) = f_0 = i^0 = I = e_\mathbb{C} $$
- $f(x^+) = f[(x)]^*$ where $x^+$ if the inverse of $x$.
Having that the inverse of a number for the addition in $\mathbb{Z}$ is it negative value, I have deduced that:
$$ f(-x) = (i^x)^* \\ i^{-x} = (i^x)^{-1} \\ i^{-x} = i^{-x} $$
Would that last part be correct as it relates to the inverse of complex numbers?
Also how do I determine the kernel/nucleus of $f$?
More generally, if $G$ is a group (written multiplicatively) and $g \in G$, then the map $f: \mathbb Z \to G$ given by $n \mapsto g^n$ is a homomorphism. It is the unique homomorphism $\mathbb Z \to G$ with $1 \mapsto g$.
The image of $f$ is the subgroup of $G$ generated by $g$.
The kernel of $f$ is $m\mathbb Z$, where $m$ is the order of $g$, if $g$ has finite order, or $m=0$ if $g$ has infinite order.
In your example, $g=i \in \mathbb C$ has order $4$.