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:

$\forall x \in G: \order x \divides \order G$

Let $G$ be a group.

Let $x \in G$.

By definition, the order of $x$ is the order of the subgroup generated by $x$.

Therefore, by Lagrange's Theorem, $\order x$ is a divisor of $\order G$.

$\blacksquare$

1965: J.A. Green: Sets and Groups ... (previous) ... (next): $\S 6.4$. Lagrange's theorem: Example $117$
1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {IV}$: Rings and Fields: $25$. Cyclic Groups and Lagrange's Theorem: Theorem $25.7$
1966: Richard A. Dean: Elements of Abstract Algebra ... (previous) ... (next): $\S 1.9$: Theorem $12$ Corollary
1967: George McCarty: Topology: An Introduction with Application to Topological Groups ... (previous) ... (next): Chapter $\text{II}$: Groups: Problem $\text{GG}$
1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $2$: Subgroups and Cosets: $\S 41$
1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 44.1$ Some consequences of Lagrange's Theorem
1982: P.M. Cohn: Algebra Volume 1 (2nd ed.) ... (previous) ... (next): $\S 3.4$: Cyclic groups
1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $5$: Cosets and Lagrange's Theorem: Corollary $5.12$