Homomorphism from Group of Cube Roots of Unity to Itself

Theorem
Let $\struct {U_3, \times}$ denote the multiplicative group of the complex cube roots of unity.

Here, $U_3 = \set {1, \omega, \omega^2}$ where $\omega = e^{2 i \pi / 3}$.

Let $\phi: U_3 \to U_3$ be defined as:


 * $\forall z \in U_3: \map \phi z = \begin{cases} 1 & : z = 1 \\ \omega^2 & : z = \omega \\ \omega & : z = \omega^2 \end{cases}$

Then $\phi$ is a group homomorphism.

Proof
It is noted that
 * $\paren {\omega^2}^2 = \omega$

and so $\phi$ is the square function.

By Roots of Unity under Multiplication form Cyclic Group and Cyclic Group is Abelian, $U_3$ is abelian.

Thus for all $a, b \in U_3$:

showing that $\phi$ is a group homomorphism.