LemmaStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-28

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$(\mathbb{Z}, \cdot, 1)$ is a commutative monoid whose group of units is $\{1, -1\}$ ; equivalently $u \mid 1$ holds exactly for $u = 1$ and $u = -1$

Statement

$(\mathbb{Z}, \cdot, 1)$ , with the multiplication of Arithmetic on the integers, is a commutative monoid (Semigroup and monoid). Its group of units (Left inverse, right inverse, and invertible element of a monoid, The invertible elements of a monoid form a group under the restricted operation) is

$\mathbb{Z}^{\times} \;=\; \{\, 1, \, -1 \,\},$

and these are two distinct elements. Equivalently, for $u \in \mathbb{Z}$ the condition $u \mid 1$ (Divisibility in $\mathbb{Z}$ : $d \mid a$ when $a = dq$ for some integer $q$ ) holds exactly when $u = 1$ or $u = -1$ .

Facts & Assumptions

Given: $\mathbb{Z}$ with the operations of Arithmetic on the integers, and the embedding $\iota : \mathbb{N} \to \mathbb{Z}$ , $\iota(k) = [(k,0)]$ , of The naturals embed in the integers.

[L1]

$\mathbb{Z}$ is a commutative ring: multiplication is a function $\mathbb{Z} \times \mathbb{Z} \to \mathbb{Z}$ and is associative and commutative, $x \cdot 1 = x$ , and every $x$ has an additive inverse $-x$ , with $-(-x) = x$ and $(-x)(-y) = xy$ (The integers form a commutative ring, Arithmetic on the integers).

[L2]

The order on $\mathbb{Z}$ is total, antisymmetric and transitive, and positives are closed under multiplication; $x < y$ means $x \le y$ together with $x \ne y$ (The integers form a totally ordered ring, Order on the integers).

[L3]

A binary operation on a set $S$ is a function $S \times S \to S$ ; a monoid is a set with an associative binary operation and a two-sided identity, and it is commutative when the operation is (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, Semigroup and monoid).

[L4]

In a monoid $M$ , $u$ is a unit when it has a two-sided inverse, and $M^{\times}$ denotes the set of units; $M^{\times}$ is a group under the restricted operation (Left inverse, right inverse, and invertible element of a monoid, The invertible elements of a monoid form a group under the restricted operation, Group and abelian group).

[L5]

$u \mid a$ means $a = uv$ for some $v \in \mathbb{Z}$ (Divisibility in $\mathbb{Z}$ : $d \mid a$ when $a = dq$ for some integer $q$ ).

[L6]

If $d \mid a$ and $a \ne 0$ then $d \ne 0$ and $|d| \le |a|$ (If $d \mid a$ and $a \ne 0$ then $d \ne 0$ and $|d| \le |a|$ ; hence the set of divisors of a nonzero integer is bounded above by $|a|$ ).

[L7]

$|x| = x$ when $x \ge 0$ and $|x| = -x$ when $x \le 0$ (The absolute value $|a|$ of an integer); $|x| \ge 0$ and $|x| = 0$ exactly when $x = 0$ (Absolute value in $\mathbb{Z}$ : $|a| \ge 0$ ; $|a| = 0$ exactly when $a = 0$ ; $|-a| = |a|$ ; $|ab| = |a|\,|b|$ ; $-|a| \le a \le |a|$ ; and $|a| \le c$ exactly when $-c \le a \le c$ ).

[L8]

$\iota$ is injective, preserves addition, multiplication and order, and its image is exactly the set of nonnegative integers; $\iota(0) = 0$ and $\iota(1) = 1$ (The naturals embed in the integers, The integers as equivalence classes of pairs of naturals).

[L9]

On $\mathbb{N}$ : $m < n$ if and only if $\sigma(m) \le n$ (Discreteness: $\sigma(n)$ is the immediate successor); $0 = \varnothing$ , $1 = \sigma(0) = \{0\}$ and $2 = \sigma(1) = \{0,1\}$ , so $1 \ne 0$ and $2 \ne 0$ (The natural numbers $\mathbb{N}$ (von Neumann)); and $0 \le k$ for every $k$ (Order on the natural numbers).

Proof

technique · direct

1.1

Multiplication on $\mathbb{Z}$ is a function $\mathbb{Z} \times \mathbb{Z} \to \mathbb{Z}$ , hence a binary operation, and it is associative and commutative; $1$ is a two-sided identity, since $x \cdot 1 = x$ and, by commutativity, $1 \cdot x = x$ . So $(\mathbb{Z}, \cdot, 1)$ is a commutative monoid.

L1L3

1.2

For $u \in \mathbb{Z}$ , being a unit of this monoid means $uv = 1$ for some $v$ , the two equations $uv = 1$ and $vu = 1$ being the same by commutativity; and $uv = 1$ for some $v$ is precisely $u \mid 1$ . So $\mathbb{Z}^{\times} = \{\, u : u \mid 1 \,\}$ .

L1L4L5

1.3

Both $1$ and $-1$ lie in $\mathbb{Z}^{\times}$ : $1 \cdot 1 = 1$ and $(-1)(-1) = 1 \cdot 1 = 1$ .

L1L4

1.4

$0 \le 1$ , since $1 = \iota(1)$ lies in the image of $\iota$ , which is the set of nonnegative integers; hence $|1| = 1$ . Also $1 \ne 0$ , since $\iota$ is injective and $1 \ne 0$ in $\mathbb{N}$ .

L7L8

1.5

Discreteness of $\mathbb{Z}$ : if $0 < x$ then $1 \le x$ . Indeed $x \ge 0$ , so $x = \iota(k)$ for some $k \in \mathbb{N}$ ; $k \ne 0$ because $\iota(0) = 0 \ne x$ ; hence $0 < k$ in $\mathbb{N}$ , so $1 = \sigma(0) \le k$ , and applying $\iota$ , which preserves the order, gives $1 = \iota(1) \le \iota(k) = x$ .

L2L8L9

1.6

$1 \ne -1$ : otherwise $1 + 1 = 1 + (-1) = 0$ , whereas $1 + 1 = \iota(1) + \iota(1) = \iota(1+1) = \iota(2)$ and $\iota(2) \ne \iota(0) = 0$ because $\iota$ is injective and $2 \ne 0$ in $\mathbb{N}$ .

L1L8L9

2.1

Let $u \mid 1$ . Since $1 \ne 0$ , [L6] gives $u \ne 0$ and $|u| \le |1| = 1$ .

step 1.4L6

3.1

Also $|u| \ge 0$ and $|u| \ne 0$ , so $0 < |u|$ , whence $1 \le |u|$ ; with step 2.1 and antisymmetry this gives $|u| = 1$ .

step 1.5step 2.1L2L7

4.1

From $|u| = 1$ : if $u \ge 0$ then $u = |u| = 1$ , and if $u \le 0$ then $-u = |u| = 1$ , so $u = -(-u) = -1$ . By totality one of the two holds, so $u = 1$ or $u = -1$ .

step 3.1L1L2L7

5.1

Combining, $\mathbb{Z}^{\times} = \{\, u : u \mid 1 \,\} = \{1, -1\}$ , a two-element set, and by [L4] it is a group under multiplication, with identity $1$ and with each of its elements its own inverse.

step 1.2step 1.3step 4.1step 1.6L4∎

Remarks

Why this is proved here and not cited. The published example $(\mathbb{Z}, +)$ is an abelian group, $(\mathbb{Z}, \cdot)$ is a commutative monoid that is not a group, and its group of units is $\{1, -1\}$ records the same fact, $\mathbb{Z}^{\times} = \{1,-1\}$ , but it lives on an examples page, and pages of that kind are leaves in the library's reading order: nothing later may depend on them. The statement is therefore re-established here, on a spine, so that later pages have a citable home for it. The two agree; neither rests on the other.
The proof is an application of the divisor bound, not a computation. What makes $\{1,-1\}$ the whole answer is that $u \mid 1$ forces $|u| \le 1$ (If $d \mid a$ and $a \ne 0$ then $d \ne 0$ and $|d| \le |a|$ ; hence the set of divisors of a nonzero integer is bounded above by $|a|$ ) while $u \ne 0$ forces $|u| \ge 1$ (discreteness), and antisymmetry closes the gap. Nothing about the decimal shape of an integer is used.

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 67 results over 25 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Unit (ring theory) (Wikipedia) (standard reference, not scraped)

(Z,⋅,1)(\mathbb{Z}, \cdot, 1)(Z,⋅,1) is a commutative monoid whose group of units is {1,−1}\{1, -1\}{1,−1}; equivalently u∣1u \mid 1u∣1 holds exactly for u=1u = 1u=1 and u=−1u = -1u=−1