The order $\mathrm{\mid }G\mathrm{\mid }$ of a finite group and the order $\mathrm{ord}(g)$ of an element, with $\mathrm{ord}(g)=\mathrm{\infty }$ 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 $A⪯B$). That natural number is unique: if $G\approx n$ and $G\approx n^{\mathrm{\prime }}$ then $n\approx n^{\mathrm{\prime }}$, since $\approx$ is symmetric and transitive, and then $n=n^{\mathrm{\prime }}$ by claim 3 of The pigeonhole principle on $\mathbb{N}$. The order of $G$ is that unique natural number, written $\mathrm{\mid }G\mathrm{\mid }$. A group is infinite when its underlying set is not finite, and $\mathrm{\mid }G\mathrm{\mid }$ 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

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

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

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

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

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

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

• Two well-definedness obligations, both discharged above and neither parenthetical. For $\mathrm{\mid }G\mathrm{\mid }$ 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 $\mathrm{ord}(g)$ 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.

• $\mathrm{\mid }G\mathrm{\mid }$ and $\mathrm{ord}(g)$ are the same kind of thing, and that is not an accident. If $\mathrm{ord}(g)=n$ then $g^k=e$ iff $k$ is an integer multiple of $n$, the powers $g^0,\dots ,g^{n-1}$ are distinct, and $⟨g⟩$ has exactly $n$ elements; if $g$ has infinite order then $g^j=g^k$ only for $j=k$ proves $\mathrm{\mid }⟨g⟩\mathrm{\mid }=\mathrm{ord}(g)$: 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.

• $\mathrm{\infty }$ is not a value in $\mathbb{N}$, so a statement such as "$\mathrm{ord}(g)\mid k$" 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.