Homomorphism from Group of Cube Roots of Unity to Itself

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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 $a, b \in U_3$:

\(\ds \map \phi a \map \phi b\) \(=\) \(\ds a^2 b^2\)
\(\ds \) \(=\) \(\ds a b a b\) Definition of Abelian Group
\(\ds \) \(=\) \(\ds \paren {a b}^2\)
\(\ds \) \(=\) \(\ds \map \phi {a b}\)

showing that $\phi$ is a group homomorphism.

$\blacksquare$


Sources

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