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

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

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

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 $|G|$.