# Order of element divides order of group

This article states a result of the form that one natural number divides another. Specifically, the (order of an element) of a/an/the (element of a group) divides the (order of a group) of a/an/the (group).
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## Statement

Let $G$ be a finite group and $g \in G$ be an element. Let $m$ be the order of the element $g$: the smallest positive integer $m$ such that $g^m$ is the identity element. Then, $m$ divides the order of $G$. In particular, we have, for any $g \in G$, that: $g^{|G|} = e$.

## Facts used

1. Lagrange's theorem

## Related facts

### Converse

• Cauchy's theorem: This states that for every prime dividing the order of a group, there is an element whose order equals that prime number.
• Sylow's theorem: This asserts the existence of $p$-Sylow subgroups for every prime $p$ dividing the order of the group.
• Exponent of a finite group has precisely the same prime factors as order: This is a consequence of Cauchy's theorem.

## Proof

The proof follows from Lagrange's theorem, along with the observation that the order of the element $g$ equals the order of the cyclic subgroup gneerated by $g$, which is therefore a subgroup of $G$.