Order of Element Divides Order of Finite Group

Theorem

In a finite group, the order of a group element divides the order of its group:

$\forall x \in G: \left|{x}\right| \backslash \left|{G}\right|$

Proof

Let $G$ be a group.

Let $x \in G$.

By Order of Subgroup Generated by Single Element, the order of the subgroup generated by $x$ equals the order of $x$.

Therefore, by Lagrange's Theorem, $\left|{x}\right|$ is a divisor of $\left|{G}\right|$.

$\blacksquare$

Also see

• Element to the Power of Group Order in which it is shown that $x^{\left|{G}\right|} = e$.

Sources

• J.A. Green: Sets and Groups (1965)... (previous)... (next): $\S 6.4$: Example $117$
• Seth Warner: Modern Algebra (1965)... (previous)... (next): $\S 25$: Theorem $25.7$
• Richard A. Dean: Elements of Abstract Algebra (1966): $\S 1.9$: Theorem $12$ Corollary
• Allan Clark: Elements of Abstract Algebra (1971)... (previous)... (next): $\S 41$
• Thomas A. Whitelaw: An Introduction to Abstract Algebra (1978)... (previous)... (next): $\S 44.1$
• John F. Humphreys: A Course in Group Theory (1996): $\S 5$: Corollary $5.12$