Order of Element Divides Order of Group

Order of Element Divides Order of Group: Let $G$ be a finite group (so $|G|$ denotes its cardinality), and let $g\in G$ . The order of $g$ , written $\operatorname{ord}(g)$ , is the least integer $n\ge 1$ such that $g^n=e$ , where $e$ is the identity element of $G$ . Then

\operatorname{ord}(g)\ \mid\ |G|.

Examples

In the symmetric group $S_3$ (which has $|S_3|=6$ ), the 3-cycle $(123)$ has $\operatorname{ord}((123))=3$ , and $3\mid 6$ .
In the additive group $\mathbb{Z}/12\mathbb{Z}$ , the element $4$ has order $3$ because $3\cdot 4\equiv 0\pmod{12}$ and no smaller positive multiple is $0$ ; indeed $3\mid 12$ .
The identity element always has order $1$ , and $1$ divides every integer $|G|$ .

Equivalent characterizations

Equivalently, if $\langle g\rangle$ denotes the cyclic subgroup generated by $g$ , then $|\langle g\rangle|=\operatorname{ord}(g)$ and $|\langle g\rangle|\mid |G|$ .

Remarks

This is an immediate corollary of Lagrange's theorem applied to the subgroup $\langle g\rangle\le G$ .