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

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

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 $n g$ or $n \cdot g$ rather than $g^{n}$, with $0 g = 0$ and $\sigma(n) g = n g + g$; 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.