Proof of Additive and Multiplicative Groups of Numbers

Routine facts about nonempty classes and restrictions of binary operations are used without comment (Toolkit for AL1).

We begin with the additive groups of (i). The integers $\mathbb{Z}$ form a commutative ring by The Integers Form an Ordered Ring Containing the Natural Numbers as Its Positive Elements (i). The rationals $\mathbb{Q}$ and the reals $\mathbb{R}$ are ordered fields, in particular fields, by The Rational Numbers Form an Archimedean Ordered Field Containing the Integers (i) and The Real Numbers Form an Ordered Field in Which Every Nonempty Set Bounded Above Has a Supremum (i), and $\mathbb{C}$ is a field by The Complex Numbers Form a Field Containing the Real Numbers: i Squared Is -1 and Not Real, the Canonical Form a + bi and Its Arithmetic, and Numerals (i); a field is a commutative ring by Fields (i). In each of the four, the additive laws say that addition is an associative operation with identity $0$, that is, a monoid, in which

$$x+(-x)=0=(-x)+x,$$

so that every element is a unit. Hence each is a group, abelian because addition is commutative (Groups and Abelian Groups (ii)).

Next let $F$ be one of the fields $\mathbb{Q}$, $\mathbb{R}$, $\mathbb{C}$. The set $F\setminus\{0\}$ contains $1\neq0$ and is closed under products by Rules of Arithmetic in a Commutative Ring: Zero, Signs and Squares, and No Zero Divisors in a Field (iv), and the multiplicative laws make it a monoid with identity $1$. Each $x\neq0$ has the inverse $1/x$, which is again nonzero since $x\cdot(1/x)=1\neq0$ (Rules of Arithmetic in a Commutative Ring: Zero, Signs and Squares, and No Zero Divisors in a Field (i)). So $F\setminus\{0\}$ is an abelian group, exactly as for the additive groups.

The ray $(0,\infty)$ is nonempty, as $1>0$. For $z,w>0$ we have $1/w>0$ by Rules of Arithmetic and Order in an Ordered Field (viii), so $zw^{-1}=z\cdot(1/w)>0$ as a product of positive reals. Hence $(0,\infty)$ is a subgroup of $\mathbb{R}\setminus\{0\}$ by Subgroup Criterion and Basic Examples (i), and so a group with the restricted operation by Subgroup Criterion and Basic Examples (ii); it is abelian since $\mathbb{R}\setminus\{0\}$ is. This completes (ii).

For (iii), let $S=\{z\in\mathbb{C}:|z|=1\}$. Its elements are nonzero by Conjugation and Modulus of Complex Numbers: Field Automorphism, Real and Imaginary Parts, Multiplicativity, Inverses, Component Bounds, Triangle Inequalities and Finite Sums (iii), and $1\in S$: indeed $|1|=|1\cdot1|=|1|^{2}$ by Conjugation and Modulus of Complex Numbers: Field Automorphism, Real and Imaginary Parts, Multiplicativity, Inverses, Component Bounds, Triangle Inequalities and Finite Sums (iv), and $|1|\neq0$ by Conjugation and Modulus of Complex Numbers: Field Automorphism, Real and Imaginary Parts, Multiplicativity, Inverses, Component Bounds, Triangle Inequalities and Finite Sums (iii), so $|1|=1$. If $z,w\in S$, then by Conjugation and Modulus of Complex Numbers: Field Automorphism, Real and Imaginary Parts, Multiplicativity, Inverses, Component Bounds, Triangle Inequalities and Finite Sums (iv) and Conjugation and Modulus of Complex Numbers: Field Automorphism, Real and Imaginary Parts, Multiplicativity, Inverses, Component Bounds, Triangle Inequalities and Finite Sums (v)

$$|zw^{-1}|=|z|\,|w^{-1}|=|z|/|w|=1,$$

so $zw^{-1}\in S$, and $S$ is a subgroup of $\mathbb{C}\setminus\{0\}$ by Subgroup Criterion and Basic Examples (i).

For the real exponential in (iv), the function takes values in $(0,\infty)$ by Basic Properties of the Exponential Function (iii) and preserves the operation by Basic Properties of the Exponential Function (ii), so it is a homomorphism (Group Homomorphisms and Isomorphisms (i)); it is bijective onto $(0,\infty)$ by Basic Properties of the Exponential Function (xi), and therefore an isomorphism by Group Homomorphisms and Isomorphisms (ii). (Here $\exp_{\mathbb{R}}$ is the function of Complex Analysis: Standing Notation (iv).)

For the complex exponential, $\exp z\neq0$ for every $z\in\mathbb{C}$ by The Complex Exponential: the Functional Equation, Conjugates, Real Arguments and the Modulus (iii), so the function takes values in $\mathbb{C}\setminus\{0\}$, and it preserves the operation by The Complex Exponential: the Functional Equation, Conjugates, Real Arguments and the Modulus (ii); so it is a homomorphism by Group Homomorphisms and Isomorphisms (i).