Order of Element Divides Order of Finite Group

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Theorem

$\forall x \in G: \order x \divides \order G$

Proof

Let $G$ be a group.

Let $x \in G$.

By definition, the order of $x$ is the order of the subgroup generated by $x$.

Therefore, by Lagrange's Theorem, $\order x$ is a divisor of $\order G$.

$\blacksquare$