Arbitrary Cyclic Group of Order 4
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Theorem
Let $S = \set {1, 2, 3, 4}$.
Consider the algebraic structure $\struct {S, \circ}$ given by the Cayley table:
- $\begin{array}{r|rrrr} \circ & 2 & 3 & 4 & 1 \\ \hline 2 & 2 & 3 & 4 & 1 \\ 3 & 3 & 4 & 1 & 2 \\ 4 & 4 & 1 & 2 & 3 \\ 1 & 1 & 2 & 3 & 4 \\ \end{array}$
Then $\struct {S, \circ}$ is a group.
Specifically, $\struct {S, \circ}$ is the cyclic group of order $4$.
Proof
Let $S' = \set {0, 1, 2, 3}$.
Let $\phi: S \to S'$ be the bijection:
\(\ds \map \phi 2\) | \(=\) | \(\ds 0\) | ||||||||||||
\(\ds \map \phi 3\) | \(=\) | \(\ds 1\) | ||||||||||||
\(\ds \map \phi 4\) | \(=\) | \(\ds 2\) | ||||||||||||
\(\ds \map \phi 1\) | \(=\) | \(\ds 3\) |
By inspection, the Cayley table presented above is in the same form as the Cayley table for the cyclic group of order $4$:
- $\begin{array}{r|rrrr} +_4 & 0 & 1 & 2 & 3 \\ \hline 0 & 0 & 1 & 2 & 3 \\ 1 & 1 & 2 & 3 & 0 \\ 2 & 2 & 3 & 0 & 1 \\ 3 & 3 & 0 & 1 & 2 \\ \end{array}$
$\blacksquare$
Sources
- 1967: George McCarty: Topology: An Introduction with Application to Topological Groups ... (previous) ... (next): Chapter $\text{II}$: Groups: Exercise $\text{B v}$