Units of Gaussian Integers form Group
Theorem
Let $U_\C$ be the set of units of the Gaussian integers:
- $U_\C = \set {1, i, -1, -i}$
where $i$ is the imaginary unit: $i = \sqrt {-1}$.
Let $\struct {U_\C, \times}$ be the algebraic structure formed by $U_\C$ under the operation of complex multiplication.
Then $\struct {U_\C, \times}$ forms a cyclic group under complex multiplication.
Proof 1
By definition of the imaginary unit $i$:
\(\ds i^2\) | \(=\) | \(\ds -1\) | ||||||||||||
\(\ds i^3\) | \(=\) | \(\ds -i\) | ||||||||||||
\(\ds i^4\) | \(=\) | \(\ds 1\) |
thus demonstrating that $U_\C$ is generated by $i$.
Thus $\struct {U_\C, \times}$ is by definition a cyclic group of order $4$.
$\blacksquare$
Proof 2
From Gaussian Integer Units are 4th Roots of Unity:
- $\left\{{1, i, -1, -i}\right\}$ constitutes the set of the $4$th roots of unity.
The result follows from Roots of Unity under Multiplication form Cyclic Group.
$\blacksquare$
Proof 3
From Units of Gaussian Integers, $U_\C$ is the set of units of the ring of Gaussian integers.
From Group of Units is Group, $\left({U_\C, \times}\right)$ forms a group.
It remains to note that:
\(\ds i^2\) | \(=\) | \(\ds -1\) | ||||||||||||
\(\ds i^3\) | \(=\) | \(\ds -i\) | ||||||||||||
\(\ds i^4\) | \(=\) | \(\ds 1\) |
thus demonstrating that $U_\C$ is cyclic.
$\blacksquare$
Sources
- 1964: Walter Ledermann: Introduction to the Theory of Finite Groups (5th ed.) ... (previous) ... (next): Chapter $\text {I}$: The Group Concept: $\S 6$: Examples of Finite Groups: $\text{(i)}$
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 34$. Examples of groups: $(6) \ \text{(i)}$
- 1982: P.M. Cohn: Algebra Volume 1 (2nd ed.) ... (previous) ... (next): $\S 3.2$: Groups; the axioms: Exercise $(1)$