Arbitrary Cyclic Group of Order 4

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:

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}$