Order of element divides order of group

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) uses::Lagrange's theorem

Related facts

 * Lagrange's theorem

Applications

 * Exponent divides order

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 $$G$$.