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