Powers $g^n$: natural exponents in a monoid and integer exponents in a group, with $g^0=e$

Natural exponents, in a monoid. Let $(M,\cdot ,e)$ be a monoid (Semigroup and monoid) and $g\in M$. By the recursion theorem (The recursion theorem), applied with the set $M$, the element $e$ and the function $x\mapsto x\cdot g$ from $M$ to $M$, there is exactly one function $\mathbb{N}\to M$, written $n\mapsto g^n$, with

$g^0=e,g^{\sigma (n)}=g^n\cdot g(n\in \mathbb{N})\mathrm{.}$

In particular $g^0=e$ for every $g$, including $g=e$, and $g^1=g^{\sigma (0)}=e\cdot g=g$. Since $\mathbb{N}$ contains $0$ (The natural numbers $\mathbb{N}$ (von Neumann)), the exponent $0$ is a genuine value of the definition and not a separate convention.

Integer exponents, in a group. Let $G$ be a group (Group and abelian group) and $g\in G$. Write $\iota :\mathbb{N}\to \mathbb{Z}$ for the embedding $\iota (k)=[(k,0)]$ of The naturals embed in the integers, which is injective, preserves addition, multiplication and order, and has as image exactly the nonnegative integers. For $x\in \mathbb{Z}$ define

• $g^x:=g^k$, the natural power, when $0\le x$ and $x=\iota (k)$;
• $g^x:=(g^k{)}^{-1}$ when $x<0$ and $-x=\iota (k)$.

Why this is well defined. The order on $\mathbb{Z}$ is total and antisymmetric (The integers form a totally ordered ring, Order on the integers), so exactly one of $0\le x$ and $x<0$ holds and the two clauses never both apply. In the first clause $x$ is nonnegative, so $x=\iota (k)$ for some $k\in \mathbb{N}$, and $k$ is unique because $\iota$ is injective. In the second clause $x<0$ gives $0=x+(-x)<0+(-x)=-x$ by compatibility of the order with addition (The integers form a totally ordered ring, Arithmetic on the integers), so $-x$ is a positive integer and again $-x=\iota (k)$ for a unique $k$. The inverse $(g^k{)}^{-1}$ is a single determined element by In a monoid, a left inverse and a right inverse of the same element are equal; hence an invertible element has exactly one inverse, and it is two-sided and Left inverse, right inverse, and invertible element of a monoid. Finally the two readings of $g^{\iota (k)}$, as a natural power and as an integer power, agree by construction, so no ambiguity is introduced.

Abbreviation. In an exponent we write $k$ for the integer $\iota (k)$ when a natural number $k$ is used where an integer is expected; this is unambiguous because $\iota$ is injective and preserves the arithmetic and the order, and because the two readings of $g^k$ agree as just noted.

Additive notation. When the group is written additively the same object is written $ng$ or $n\cdot g$ rather than $g^n$, with $0g=0$ and $\sigma (n)g=ng+g$; the definitions are identical, only the symbols differ.

• The case split is on the integer, not on a representative. An integer is an equivalence class $[(a,b)]$ of pairs of naturals (The integers as equivalence classes of pairs of naturals), so a definition of $g^x$ by a formula in $a$ and $b$ would carry a well-definedness obligation. The definition above avoids that entirely: it splits on the sign of $x$, which is a property of the integer itself, and then uses the unique natural supplied by The naturals embed in the integers.

• Negative exponents need a group, not merely a monoid. The second clause uses an inverse, which a monoid need not have. Natural powers are available in any monoid, and every statement below says which of the two settings it is in.

• The exponent is an index, an element of $\mathbb{N}$ or of $\mathbb{Z}$, and never an element of the group. No canonical-natural machinery (The canonical natural $\iota (n)=n\cdot 1_F$ of a field) is involved: that item is about the image of $\mathbb{N}$ inside a field, a different map with the same traditional name.

• The laws $g^{m+n}=g^m{g}^n$, $(g^m{)}^n=g^{mn}$ and, when $g$ and $h$ commute, $(gh{)}^n=g^n{h}^n$ are proved in Exponent laws in a group: $g^{m+n}=g^m{g}^n$ and $(g^m{)}^n=g^{mn}$ for all $m,n\in \mathbb{Z}$, and $(gh{)}^n=g^n{h}^n$ when $g$ and $h$ commute. The last one is false without the commuting hypothesis.