Proof of Exponent Laws for Integer Powers in a Group

Throughout, $G$ is a monoid in which every element is a unit (Groups and Abelian Groups (i)), brackets in products are dropped by associativity, and powers with exponents in $\mathbb{N}_{0}$ are those of the monoid $G$ (Integer Powers in a Group (i)); routine facts about natural powers are used without comment (Toolkit for AL1).

We begin with two facts about natural exponents. For $x\in G$ and $a\in\mathbb{N}_{0}$, the element $x$ is a unit, so Exponent Laws for Natural Powers in a Monoid (vi) gives

$$(x^{a})^{-1}=(x^{-1})^{a}.\tag{A}$$

Moreover,

$$g^{-a}=(g^{-1})^{a}\qquad(a\in\mathbb{N}_{0}).\tag{B}$$

For $a\ge1$ this is the definition of negative powers in Integer Powers in a Group (i), and for $a=0$ both sides equal $e$.

Next, (i) and (ii). By Exponent Laws for Natural Powers in a Monoid (i), $g^{1}=g^{0}g=eg=g$. For $n\ge0$, $e^{n}=e$ by Exponent Laws for Natural Powers in a Monoid (iv). For $n<0$,

$$e^{n}=(e^{-1})^{-n}=e^{-n}=e,$$

by Integer Powers in a Group (i), Cancellation Laws and Basic Inverse Identities in a Group (ii), and the case $n\ge0$ applied to $-n$.

We now prove (v). If $n\ge0$, both $g^{-n}$ and $(g^{n})^{-1}$ equal $(g^{-1})^{n}$, by (B) and by (A) with $x=g$. If $n<0$, put $a=-n\in\mathbb{N}$, so that $g^{-n}=g^{a}$. By Integer Powers in a Group (i), (A) with $x=g^{-1}$, and Cancellation Laws and Basic Inverse Identities in a Group (iii),

$$(g^{n})^{-1}=((g^{-1})^{a})^{-1}=((g^{-1})^{-1})^{a}=g^{a},$$

and by Integer Powers in a Group (i) and again Cancellation Laws and Basic Inverse Identities in a Group (iii), $(g^{-1})^{n}=((g^{-1})^{-1})^{a}=g^{a}$.

For (iii) we first show that

$$g^{k+1}=g^{k}g\qquad\text{for every }k\in\mathbb{Z}.$$

For $k\ge0$ this is Exponent Laws for Natural Powers in a Monoid (i). For $k\le-1$ put $a=-k-1\in\mathbb{N}_{0}$; by (B), Exponent Laws for Natural Powers in a Monoid (i) for $g^{-1}$, and $g^{-1}g=e$ (Uniqueness of the Identity Element and of Inverses in a Group (ii)),

$$g^{k}g=(g^{-1})^{a+1}g=(g^{-1})^{a}g^{-1}g=(g^{-1})^{a}=g^{k+1}.$$

Multiplying $g^{k}=g^{k-1}g$ on the right by $g^{-1}$ gives $g^{k-1}=g^{k}g^{-1}$ for every $k\in\mathbb{Z}$. Now fix $m\in\mathbb{Z}$. Since $(g^{-1})^{a}=g^{-a}$ by (B), it suffices to show, by induction on $a\in\mathbb{N}_{0}$ (Induction on ℕ₀ from 0 for a Condition Expressed by a Predicative Formula (ii)), that

$$g^{m+a}=g^{m}g^{a}\qquad\text{and}\qquad g^{m-a}=g^{m}(g^{-1})^{a}.$$

For $a=0$ both read $g^{m}=g^{m}e$. If both hold for $a$, then the two rules just proved and Exponent Laws for Natural Powers in a Monoid (i) give

$$g^{m+a+1}=g^{m+a}g=g^{m}g^{a}g=g^{m}g^{a+1},\qquad g^{m-a-1}=g^{m-a}g^{-1}=g^{m}(g^{-1})^{a}g^{-1}=g^{m}(g^{-1})^{a+1}.$$

Since every $n\in\mathbb{Z}$ is $a$ or $-a$ for some $a\in\mathbb{N}_{0}$ (The Integers Form an Ordered Ring Containing the Natural Numbers as Its Positive Elements (vi)), this proves (iii).

For (iv), recall that $n=a$ or $n=-a$ for some $a\in\mathbb{N}_{0}$ (The Integers Form an Ordered Ring Containing the Natural Numbers as Its Positive Elements (vi)). Consider first $n=a$ and argue by induction on $a$. For $a=0$, $(g^{m})^{0}=e=g^{0}=g^{m\cdot0}$. If $(g^{m})^{a}=g^{ma}$, then by Exponent Laws for Natural Powers in a Monoid (i) and (iii)

$$(g^{m})^{a+1}=(g^{m})^{a}g^{m}=g^{ma}g^{m}=g^{ma+m}=g^{m(a+1)}.$$

If $n=-a$ with $a\in\mathbb{N}$, then two applications of (v) and the first case give

$$(g^{m})^{-a}=((g^{m})^{a})^{-1}=(g^{ma})^{-1}=g^{-ma}.$$

For (vi), the case $n\ge0$ is Exponent Laws for Natural Powers in a Monoid (v). From $gh=hg$ and Cancellation Laws and Basic Inverse Identities in a Group (iv) we get

$$h^{-1}g^{-1}=(gh)^{-1}=(hg)^{-1}=g^{-1}h^{-1},$$

so $g^{-1}$ and $h^{-1}$ commute as well. For $n<0$ put $a=-n$. By Integer Powers in a Group (i), the identity just displayed, Exponent Laws for Natural Powers in a Monoid (v) for the commuting pair $g^{-1},h^{-1}$, and Integer Powers in a Group (i) again,

$$(gh)^{n}=((gh)^{-1})^{a}=(g^{-1}h^{-1})^{a}=(g^{-1})^{a}(h^{-1})^{a}=g^{n}h^{n}.$$

Finally, (vii). We first show $f(x^{a})=f(x)^{a}$ for $x\in G$ and $a\in\mathbb{N}_{0}$, by induction on $a$. For $a=0$ it says $f(e)=e_{H}$, which is Group Homomorphisms Preserve the Identity Element and Inverses (i). If it holds for $a$, then by Exponent Laws for Natural Powers in a Monoid (i), in $G$ and in $H$, and Group Homomorphisms and Isomorphisms (i),

$$f(x^{a+1})=f(x^{a}x)=f(x^{a})f(x)=f(x)^{a}f(x)=f(x)^{a+1}.$$

This is the case $n\ge0$. For $n<0$ put $a=-n$; by Integer Powers in a Group (i) in $G$ and in $H$, the case just proved, and Group Homomorphisms Preserve the Identity Element and Inverses (ii),

$$f(g^{n})=f((g^{-1})^{a})=f(g^{-1})^{a}=(f(g)^{-1})^{a}=f(g)^{n}.$$