The order $|G|$ of a finite group and the order $\operatorname{ord}(g)$ of an element, with $\operatorname{ord}(g) = \infty$ when no positive power of $g$ is the identity

The order of a finite group. Let $G$ be a group (Group and abelian group) whose underlying set is finite (Finite, countably infinite, countable, uncountable), so that $G \approx n$ for some $n \in \mathbb{N}$ (Equinumerous sets, $A \approx B$ and $A \preceq B$). That natural number is unique: if $G \approx n$ and $G \approx n'$ then $n \approx n'$, since $\approx$ is symmetric and transitive, and then $n = n'$ by claim 3 of The pigeonhole principle on $\mathbb{N}$. The order of $G$ is that unique natural number, written $|G|$. A group is infinite when its underlying set is not finite, and $|G|$ is then not defined.

The order of an element. Let $G$ be any group and $g \in G$, with natural powers as in Powers $g^{n}$: natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$. Put

$S_g \;:=\; \{\, k \in \mathbb{N} \;:\; k \ge 1 \text{ and } g^{k} = e \,\} \;\subseteq\; \mathbb{N}.$

• If $S_g \ne \varnothing$, the order of $g$ is its least element,

  $\operatorname{ord}(g) \;:=\; \min S_g \;\in\; \mathbb{N},$

  which exists by the well-ordering principle (The well-ordering principle): every nonempty subset of $\mathbb{N}$ has a least element, and that element is unique, being $\le$ every element of $S_g$ and a member of it. We then say $g$ has finite order.

• If $S_g = \varnothing$ we say $g$ has infinite order and write $\operatorname{ord}(g) = \infty$, where $\infty$ is a symbol reserved for this case and is not a natural number. No arithmetic is performed with it here.

By construction $\operatorname{ord}(g) \ge 1$ whenever it is finite, and $\operatorname{ord}(g) = 1$ exactly when $g = e$, since $g^{1} = g$.

Every element of a finite group has finite order. If $G$ is finite then $S_g \ne \varnothing$ for every $g \in G$, by In a finite group, every element $g$ satisfies $g^{n} = e$ for some natural $n \ge 1$, so $\operatorname{ord}(g)$ is a natural number.

• Two well-definedness obligations, both discharged above and neither parenthetical. For $|G|$ it is that a finite set is equinumerous with exactly one natural number, which is claim 3 of The pigeonhole principle on $\mathbb{N}$. For $\operatorname{ord}(g)$ it is that the set $S_g$ has a least element when it is nonempty, which is The well-ordering principle, and that it is nonempty at all in the finite case, which is In a finite group, every element $g$ satisfies $g^{n} = e$ for some natural $n \ge 1$. Neither quantity is definable before those three items.

• $|G|$ and $\operatorname{ord}(g)$ are the same kind of thing, and that is not an accident. If $\operatorname{ord}(g) = n$ then $g^{k} = e$ iff $k$ is an integer multiple of $n$, the powers $g^{0}, \dots, g^{n-1}$ are distinct, and $\langle g \rangle$ has exactly $n$ elements; if $g$ has infinite order then $g^{j} = g^{k}$ only for $j = k$ proves $|\langle g \rangle| = \operatorname{ord}(g)$: the order of an element is the order of the group it generates. The shared word and the shared notation are justified by that identity, not by convention.

• $\infty$ is not a value in $\mathbb{N}$, so a statement such as "$\operatorname{ord}(g) \mid k$" is asserted only when the order is finite; the infinite case is always stated separately. The order on $\mathbb{N}$ used above is the additive one of Order on the natural numbers, and $\mathbb{N}$ contains $0$ (The natural numbers $\mathbb{N}$ (von Neumann)), which is why $S_g$ carries the condition $k \ge 1$: $g^{0} = e$ holds for every $g$ and says nothing.