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
- 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $8$: The Homomorphism Theorem: Example $8.4$