Order of an element divides the order of the group

Remarks

The order of $g$ is the order of the cyclic subgroup $\langle g \rangle$, by this fact, and Lagrange's theorem applies to $\langle g \rangle$. For example, in a group of order $10$, each element has order $1, 2, 5$ or $10$.

Going deeper

For a student

The order of an element $g$ is the smallest $n \ge 1$ with $g^n = e$. The result says that this number divides the order of the group.

The proof reduces the result to Lagrange's theorem:

1. The powers of $g$ form the cyclic subgroup $\langle g \rangle$.
2. This subgroup has exactly $\operatorname{ord}(g)$ elements, namely $e, g, \dots, g^{n-1}$ with $n = \operatorname{ord}(g)$.
3. By Lagrange's theorem, $|\langle g \rangle|$ divides $|G|$.

A worked example in the group of units of $\mathbb{Z}/7\mathbb{Z}$, which has $6$ elements:

1. Take $g = 2$. Then $2^1 = 2$, $2^2 = 4$ and $2^3 = 8 = 1$.
2. Thus $\operatorname{ord}(2) = 3$, and $3$ divides $6$.
3. Take $g = 3$. The powers are $3, 2, 6, 4, 5, 1$, so $\operatorname{ord}(3) = 6$.

A useful consequence is that $g^{|G|} = e$ for each $g \in G$. Applied to the units of $\mathbb{Z}/p\mathbb{Z}$, this gives Fermat's little theorem: $a^{p-1} \equiv 1 \pmod p$ when $p$ does not divide $a$.

A common error is to think that each divisor of $|G|$ is the order of some element. The Klein four-group has order $4$, but it has no element of order $4$. Cauchy's theorem gives a partial converse: each prime divisor of $|G|$ is the order of some element.