Group whose Order equals Order of Element is Cyclic

Theorem
Let $G$ be a finite group of order $n$.

Let $g \in G$ have order $n$.

Then $G$ is a cyclic group which is generated by $g$.

Proof
The subgroup of $G$ generated by $g$ is:


 * $\gen g = \set {g^0, g^1, g^2, \ldots, g^{n - 1} }$

This contains $n$ elements.

Thus:
 * $\gen g = G$

and the result follows by definition of cyclic group.