ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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For any field $F$ , $(F, +)$ and $(F ∖ {0}, \cdot)$ are abelian groups; in particular $(Q, +)$ , $(Q ∖ {0}, \cdot)$ , $(R, +)$ and $(R ∖ {0}, \cdot)$

Example

Let $F$ be a field (Field). Then

$(F, +, 0)$ is an abelian group (Group and abelian group), the additive group of $F$ ;
$(F ∖ {0}, \cdot, 1)$ is an abelian group, the multiplicative group of $F$ , often written $F^{\times}$ .

Since the rationals (The rationals form a field) and the reals (The reals form a field) are fields, this gives four groups at once: $(Q, +)$ , $(Q ∖ {0}, \cdot)$ , $(R, +)$ and $(R ∖ {0}, \cdot)$ . All four are infinite and abelian.

Facts & Assumptions

Given: A field $F$ with operations $+$ and $\cdot$ and distinguished elements $0 \neq 1$ (Field).

[L1]

Axiom (A): addition is associative and commutative, $x + 0 = x$ for all $x \in F$ , and every $x$ has an additive inverse $- x$ with $x + (- x) = 0$ (Field).

[L2]

Axiom (M): multiplication is associative and commutative, $x \cdot 1 = x$ for all $x \in F$ , and every $x \neq 0$ has a multiplicative inverse $x^{- 1}$ with $x \cdot x^{- 1} = 1$ (Field).

[L3]

Axiom (D): $x (y + z) = x y + x z$ ; and $0 \neq 1$ (Field).

[L4]

A group is a monoid in which every element is invertible; a monoid is an associative binary operation on a set together with a two-sided identity; abelian means the operation is commutative; a subset closed under an operation inherits it (Group and abelian group, Semigroup and monoid, Binary operation on a set; associativity, commutativity, and a subset closed under the operation, Left identity, right identity, and two-sided identity for a binary operation, Left inverse, right inverse, and invertible element of a monoid).

[L5]

$Q$ is a field (The rationals form a field) and $R$ is a field (The reals form a field).

Verification

technique · direct

1.1

Addition is a binary operation on $F$ , associative and commutative by (A).

1.2

$0$ is a two-sided identity for $+$ : $x + 0 = x$ by (A), and $0 + x = x + 0 = x$ by commutativity. So $(F, +, 0)$ is a commutative monoid.

L1L4

1.3

Every $x \in F$ is invertible for $+$ : $x + (- x) = 0$ by (A) and $(- x) + x = 0$ by commutativity, so $- x$ is a two-sided inverse. Hence $(F, +, 0)$ is an abelian group.

L1L4

1.4

For every $z \in F$ , $z \cdot 0 = 0$ : by (D), $z \cdot 0 = z (0 + 0) = z \cdot 0 + z \cdot 0$ , and adding $- (z \cdot 0)$ to both sides gives $0 = z \cdot 0$ .

L1L3

2.1

$F ∖ {0}$ is closed under multiplication. Let $x, y \neq 0$ and suppose $x y = 0$ . Multiplying by $x^{- 1}$ and regrouping, $y = 1 \cdot y = (x^{- 1} x) y = x^{- 1} (x y) = x^{- 1} \cdot 0 = 0$ , contradicting $y \neq 0$ . So $x y \neq 0$ , and multiplication restricts to a binary operation on $F ∖ {0}$ .

step 1.4L2L4

3.1

$1 \in F ∖ {0}$ because $0 \neq 1$ , and $x \cdot 1 = x = 1 \cdot x$ for every $x$ by (M) and commutativity; the restricted operation is associative and commutative because it is so on $F$ . Hence $(F ∖ {0}, \cdot, 1)$ is a commutative monoid.

step 2.1L2L3L4

4.1

Every $x \in F ∖ {0}$ is invertible there: (M) supplies $x^{- 1} \in F$ with $x x^{- 1} = 1$ , and $x^{- 1} \neq 0$ since otherwise $1 = x \cdot 0 = 0$ , contradicting $0 \neq 1$ ; commutativity gives $x^{- 1} x = 1$ as well, so $x^{- 1}$ is a two-sided inverse lying in $F ∖ {0}$ .

step 1.4step 3.1L2L3

5.1

By steps 1.3, 3.1 and 4.1, $(F, +, 0)$ and $(F ∖ {0}, \cdot, 1)$ are abelian groups, and applying this to the fields $Q$ and $R$ gives the four groups listed.

step 1.3step 3.1step 4.1L4L5∎

Remarks

This is the cheapest bridge from the analysis half of the library to the algebra half. The field axioms as stated in Field already say " $(F, +)$ is an abelian group" and " $(F ∖ {0}, \cdot)$ is an abelian group"; what the verification above adds is that the group axioms in the sense of Group and abelian group really do all follow from the equations that axiom spells out, including the one thing the spelling out leaves implicit, namely that $F ∖ {0}$ is closed under multiplication.
Step 2.1 is where the absence of zero divisors is used, and it is a consequence of the field axioms rather than an extra hypothesis. In a general commutative ring the corresponding statement is false, and the nonzero elements need not be closed under multiplication.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

21 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Field (mathematics) (Wikipedia) (standard reference, not scraped)
Abelian group (Wikipedia) (standard reference, not scraped)

For any field FF, (F,+)(F,+) and (F∖{0},⋅)(F∖{0},⋅) are abelian groups; in particular (Q,+)(Q,+), (Q∖{0},⋅)(Q∖{0},⋅), (R,+)(R,+) and (R∖{0},⋅)(R∖{0},⋅)