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 $4$:

$\begin{array}{r|rrrr} +_4 & 0 & 1 & 2 & 3 \\ \hline 0 & 0 & 1 & 2 & 3 \\ 1 & 1 & 2 & 3 & 0 \\ 2 & 2 & 3 & 0 & 1 \\ 3 & 3 & 0 & 1 & 2 \\ \end{array}$

$\blacksquare$


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