# Group whose Order equals Order of Element is Cyclic

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## 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.

$\blacksquare$

## Sources

- 1964: Walter Ledermann:
*Introduction to the Theory of Finite Groups*(5th ed.) ... (previous) ... (next): Chapter $\text {I}$: The Group Concept: $\S 9$: Cyclic Groups