Session-authored (Fable 5 assisted)

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

✓ 11 results · all verified · 9 also independently AI-judged

Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs. The 2 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Monoids, Groups and Subgroups: Examples and Counterexamples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

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

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.

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

$(Z, +)$ is an abelian group, $(Z, \cdot)$ is a commutative monoid that is not a group, and its group of units is ${1, - 1}$

Example

Let $Z$ be the integers with the operations of Arithmetic on the integers. Then

$(Z, +, 0)$ is an abelian group (Group and abelian group);
$(Z, \cdot, 1)$ is a commutative monoid (Semigroup and monoid) which is not a group, because $0$ has no multiplicative inverse;
its group of units (The invertible elements of a monoid form a group under the restricted operation) is $Z^{\times} = {1, - 1}$ , and $1 \neq - 1$ .

Facts & Assumptions

Given: The integers $Z$ with $+$ , $\cdot$ , $0$ and $1$ (The integers as equivalence classes of pairs of naturals, Arithmetic on the integers), and the embedding $ι : N \to Z$ of The naturals embed in the integers.

[L1]

In $Z$ : addition is associative and commutative, $x + 0 = x$ , and every $x = [(a, b)]$ has the additive inverse $- x = [(b, a)]$ with $x + (- x) = 0$ ; multiplication is associative and commutative with $x \cdot 1 = x$ , and it distributes over addition. These are the ring axioms, verified one by one in the proof of The integers form a commutative ring (The integers form a commutative ring, Arithmetic on the integers).

[L2]

The order on $Z$ is total, antisymmetric and transitive, is compatible with addition, and positives are closed under multiplication (The integers form a totally ordered ring, Order on the integers).

[L3]

$ι$ is injective, preserves addition, multiplication and order, and its image is exactly the nonnegative integers; $ι (0) = 0$ and $ι (1) = 1$ (The naturals embed in the integers).

[L4]

On $N$ : every $k \neq 0$ is a successor $σ (j) = 1 + j$ , so $k \neq 0$ implies $1 \leq k$ (Every nonzero natural number is a successor, Addition is commutative, Order on the natural numbers, The natural numbers $N$ (von Neumann)).

[L5]

A group is a monoid in which every element is invertible; the units of a monoid form a group (Group and abelian group, Semigroup and monoid, Left inverse, right inverse, and invertible element of a monoid, Left identity, right identity, and two-sided identity for a binary operation, The invertible elements of a monoid form a group under the restricted operation).

Verification

technique · direct

1.1

Addition on $Z$ is associative and commutative, $x + 0 = x$ and hence $0 + x = x$ , and every $x$ has a two-sided additive inverse $- x$ ; so $(Z, +, 0)$ is an abelian group.

L1L5

1.2

Multiplication on $Z$ is associative and commutative and $x \cdot 1 = x = 1 \cdot x$ ; so $(Z, \cdot, 1)$ is a commutative monoid.

L1L5

1.3

For every $y \in Z$ , $0 \cdot y = 0$ : by distributivity $0 \cdot y = (0 + 0) \cdot y = 0 \cdot y + 0 \cdot y$ , and adding $- (0 \cdot y)$ gives $0 = 0 \cdot y$ .

1.4

$1 \neq 0$ in $Z$ : $ι$ is injective with $ι (1) = 1$ and $ι (0) = 0$ , and $1 \neq 0$ in $N$ since $1 = σ (0)$ contains $0$ as an element while $0 = \emptyset$ has none.

L3L4

1.5

Discreteness in $Z$ : if $0 < x$ then $1 \leq x$ . Indeed $x \geq 0$ gives $x = ι (k)$ with $k \in N$ , and $k \neq 0$ since $ι (0) = 0 \neq x$ ; so $1 \leq k$ in $N$ and, $ι$ preserving the order, $1 = ι (1) \leq ι (k) = x$ .

L3L4

2.1

$0$ is not invertible in $(Z, \cdot, 1)$ : $0 \cdot y = 0 \neq 1$ for every $y$ . Hence $(Z, \cdot, 1)$ is not a group.

step 1.3step 1.4L5

2.2

$1$ and $- 1$ are units: $1 \cdot 1 = 1$ , and $(- 1) (- 1) = 1$ because $(- 1) (- 1) + (- 1) = (- 1) ((- 1) + 1) = (- 1) \cdot 0 = 0$ , so $(- 1) (- 1)$ is the additive inverse of $- 1$ , which is $1$ .

step 1.3L1

2.3

If $0 < u$ and $1 \leq v$ then $u \leq u v$ : from $1 \leq v$ we get $0 \leq v - 1$ , so $u (v - 1) \geq 0$ , either factor being possibly zero, and $u v - u = u (v - 1) \geq 0$ gives $u \leq u v$ .

step 1.5L1L2

2.4

Let $x y = 1$ . Then $x \neq 0$ and $y \neq 0$ by step 1.3 and step 1.4. If $x > 0$ and $y < 0$ then $0 < x (- y) = - (x y) = - 1$ , so $1 < 0$ , contradicting $0 < 1$ , which holds by step 1.4 and totality since $1 \geq 0$ ; the case $x < 0$ , $y > 0$ is the same with the names interchanged. So $x$ and $y$ are both positive or both negative.

step 1.3step 1.4L1L2L3

3.1

Both positive: $1 \leq x$ and $1 \leq y$ by step 1.5, so $x \leq x y = 1$ by step 2.3, and with $1 \leq x$ antisymmetry gives $x = 1$ .

step 1.5step 2.3L2

4.1

Both negative: then $0 < - x$ and $0 < - y$ and $(- x) (- y) = x y = 1$ by ring arithmetic, so $- x = 1$ by step 3.1, that is $x = - 1$ .

step 3.1L1L2

5.1

By steps 2.2, 2.4, 3.1 and 4.1 the units of $(Z, \cdot, 1)$ are exactly $1$ and $- 1$ ; and $1 \neq - 1$ , since $1 = - 1$ would give $1 + 1 = 0$ , while $0 < 1$ gives $1 < 1 + 1$ and hence $0 < 1 + 1$ . So $Z^{\times} = {1, - 1}$ , a group under multiplication with two elements.

step 2.2step 2.4step 3.1step 4.1L1L2L5∎

Remarks

The Statement of The integers form a commutative ring is quoted here by its content, not by its name. That theorem says $Z$ is "a commutative ring with multiplicative identity", a phrase not defined at this point in the reading order; what is used above is the list of equations its proof verifies one at a time. Nothing here presupposes a definition of a ring.
Cancellation without invertibility. $(Z, \cdot)$ satisfies cancellation by nonzero elements yet is not a group, and $(N, +)$ is a sharper example still, being cancellative outright (A commutative monoid in which cancellation holds need not be a group: $(N, +)$ ).
$Z^{\times} = {1, - 1}$ is the first finite group in the library that is not trivial. It is cyclic of order $2$ , generated by $- 1$ .

CounterexampleConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27Open item page →

A commutative monoid in which cancellation holds need not be a group: $(N, +)$

Statement refuted

False claim: every commutative monoid (Semigroup and monoid) in which the cancellation law holds, that is in which $x + z = y + z$ implies $x = y$ , is a group (Group and abelian group).

The natural numbers under addition refute it: $(N, +, 0)$ is a commutative monoid, cancellation holds in it, and it is not a group, because $1$ has no additive inverse.

Facts & Assumptions

Given: $N$ with addition defined by $m + 0 = m$ and $m + σ (n) = σ (m + n)$ (Addition of natural numbers), and $0 = \emptyset$ , $σ (n) = n \cup {n}$ , $1 = σ (0)$ (The natural numbers $N$ (von Neumann)).

[L1]

Addition is a binary operation $N \times N \to N$ (Addition of natural numbers, Binary operation on a set; associativity, commutativity, and a subset closed under the operation).

[L2]

Addition is associative (Addition is associative) and commutative (Addition is commutative).

[L3]

$0 + n = n$ for every $n$ (Left identity for addition), and $n + 0 = n$ by the defining recursion (Addition of natural numbers).

[L4]

Cancellation: $m + k = n + k$ implies $m = n$ (Addition is cancellative).

[L5]

A monoid is an associative binary operation with a two-sided identity; a group is a monoid in which every element has a two-sided inverse (Semigroup and monoid, Group and abelian group, Left inverse, right inverse, and invertible element of a monoid, Left identity, right identity, and two-sided identity for a binary operation).

[L6]

The refuted claim: every commutative cancellative monoid is a group.

Counterexample

technique · direct

1.1

Addition is a binary operation on $N$ , associative and commutative.

L1L2

1.2

$0$ is a two-sided identity for addition: $n + 0 = n$ by the recursion and $0 + n = n$ by [L3]. Hence $(N, +, 0)$ is a commutative monoid.

L3L5

1.3

Cancellation holds: $m + k = n + k$ implies $m = n$ , and by commutativity $k + m = k + n$ implies $m = n$ as well.

L2L4

1.4

For every $k \in N$ , $σ (k) \neq 0$ : the set $σ (k) = k \cup {k}$ has $k$ as an element, whereas $0 = \emptyset$ has no elements.

given

2.1

$1$ has no additive inverse in $N$ : for any $k \in N$ , $1 + k = k + 1 = k + σ (0) = σ (k + 0) = σ (k) \neq 0$ , so no $k$ satisfies $1 + k = 0$ .

step 1.4L2given

3.1

Hence $(N, +, 0)$ is not a group, since a group requires every element to be invertible and $1 \in N$ is not.

step 2.1L5

4.1

By steps 1.2, 1.3 and 3.1 the monoid $(N, +, 0)$ is commutative and cancellative but not a group, so the claim of [L6] is false.

step 1.2step 1.3step 3.1L6∎

Remarks

This is the sharpest available refutation, not merely a refutation. The weaker observation that $(N, +)$ is not a group leaves open the possibility that cancellation is what is missing; the point here is that cancellation is present and still does not suffice. In a group cancellation is a theorem (Cancellation in a group: $g x = g y$ or $x g = y g$ forces $x = y$ ; equivalently left and right translation by $g$ are bijections of $G$ , so $g x = h$ and $x g = h$ each have exactly one solution), so the implication runs one way only.
What is missing is exactly invertibility, and the standard remedy is to adjoin it: the construction of $Z$ from $N$ (The integers as equivalence classes of pairs of naturals) is precisely the passage from this cancellative monoid to a group containing it, which is why the pairs $(a, b)$ there stand for formal differences.
The element $1$ is not special: no $n \neq 0$ has an additive inverse in $N$ , by the same computation with $n$ in place of $1$ once $n$ is written as a successor. One witness suffices to refute the claim.

ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passverified 2026-08-02 (claude-opus-5)Open item page →

For $n \geq 1$ the congruence classes modulo $n$ form an abelian group $(Z / n, +)$ of order $n$ , generated by the class of $1$

Example

Fix a natural number $n \geq 1$ and write $\overset{ˉ}{n} : = ι (n)$ for the corresponding positive integer, $ι : N \to Z$ being the embedding of The naturals embed in the integers. For $a, b \in Z$ define

$a \equiv b (m o d n) : ⟺ a - b = q \overset{ˉ}{n} for some q \in Z,$

that is, $\overset{ˉ}{n} ∣ (a - b)$ (Division with remainder in $Z$ : for $a \in Z$ and $b > 0$ there are unique $q, r \in Z$ with $a = q b + r$ and $0 \leq r < b$ ). Then:

$\equiv$ is an equivalence relation on $Z$ (Equivalence relation, equivalence class, and the quotient set $A / \sim$ ); its classes are the congruence classes modulo $n$ , and the quotient set is written $Z / n$ ;
$[a] + [b] : = [a + b]$ is a well-defined binary operation on $Z / n$ , and $(Z / n, +, [0])$ is an abelian group (Group and abelian group);
$Z / n$ is finite of order $n$ : $∣ Z / n ∣ = n$ (The order $∣ G ∣$ of a finite group and the order $ord (g)$ of an element, with $ord (g) = \infty$ when no positive power of $g$ is the identity);
$Z / n = ⟨ [1] ⟩$ , so it is cyclic, generated by the class of $1$ (The subgroup $⟨ S ⟩$ generated by a subset, the cyclic subgroup $⟨ g ⟩$ , and cyclic groups).

The hypothesis $n \geq 1$ is needed for claim 3, and for claim 3 only: at $n = 0$ the relation is equality, so $Z / 0$ has one class for each integer and is infinite, and there is no natural number $n$ with $∣ Z / 0 ∣ = n$ . Claims 1, 2 and 4 do hold at $n = 0$ , where $Z / 0$ is an infinite cyclic group generated by $[1]$ .

Facts & Assumptions

Given: A natural number $n \geq 1$ , the integer $\overset{ˉ}{n} = ι (n)$ , and the relation $a \equiv b$ meaning $a - b = q \overset{ˉ}{n}$ for some $q \in Z$ (The integers as equivalence classes of pairs of naturals, Arithmetic on the integers).

[L1]

$Z$ is a commutative ring, with $u - v : = u + (- v)$ (The integers form a commutative ring, Arithmetic on the integers); its order is total and antisymmetric and compatible with addition, and positives are closed under multiplication (The integers form a totally ordered ring, Order on the integers).

[L2]

$ι$ is injective, preserves addition, multiplication and order, and its image is exactly the nonnegative integers; $ι (0) = 0$ , $ι (1) = 1$ ; hence $\overset{ˉ}{n} > 0$ because $n \geq 1$ (The naturals embed in the integers, Order on the natural numbers, The natural numbers $N$ (von Neumann)). Moreover $n = {s \in N : s < n}$ : a natural number is exactly the set of the naturals below it (On $N$ the order is membership: $m < n ⟺ m \in n$ ).

[L3]

An equivalence relation is a reflexive, symmetric and transitive relation; $[a] = {b : a \sim b}$ and the quotient set is the set of classes (Equivalence relation, equivalence class, and the quotient set $A / \sim$ ); and $a \sim b$ if and only if $[a] = [b]$ (The equivalence classes of an equivalence relation are nonempty, cover $A$ , and are pairwise equal or disjoint; conversely every such cover arises from exactly one equivalence relation).

[L4]

Division with remainder: for $a \in Z$ and $b > 0$ there are unique $q, r \in Z$ with $a = q b + r$ and $0 \leq r < b$ (Division with remainder in $Z$ : for $a \in Z$ and $b > 0$ there are unique $q, r \in Z$ with $a = q b + r$ and $0 \leq r < b$ ).

[L5]

A group is a monoid all of whose elements are invertible; abelian means commutative; the order $∣ G ∣$ of a finite group is the unique natural $m$ with $G \approx m$ ; $⟨ g ⟩$ is the smallest subgroup containing $g$ and equals the set of integer powers of $g$ (Group and abelian group, 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, The order $∣ G ∣$ of a finite group and the order $ord (g)$ of an element, with $ord (g) = \infty$ when no positive power of $g$ is the identity, The subgroup $⟨ S ⟩$ generated by a subset, the cyclic subgroup $⟨ g ⟩$ , and cyclic groups, $⟨ g ⟩ = {g^{n} : n \in Z}$ , and every cyclic group is abelian, Finite, countably infinite, countable, uncountable, Equinumerous sets, $A \approx B$ and $A ⪯ B$ , Injection, surjection, bijection).

[L6]

Powers in a group written additively: $g^{0}$ is the identity, $g^{σ (m)} = g^{m} + g$ , and $g^{x} = - g^{k}$ when $x < 0$ and $- x = ι (k)$ (Powers $g^{n}$ : natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$ ).

[L7]

Induction on $N$ (The principle of mathematical induction); on $N$ exactly one of $s < t$ , $s = t$ , $t < s$ holds (Trichotomy of the order on $N$ ).

Verification

technique · direct

1.1

$\equiv$ is reflexive, since $a - a = 0 = 0 \cdot \overset{ˉ}{n}$ ; symmetric, since $a - b = q \overset{ˉ}{n}$ gives $b - a = (- q) \overset{ˉ}{n}$ ; and transitive, since $a - b = q \overset{ˉ}{n}$ and $b - c = q^{'} \overset{ˉ}{n}$ give $a - c = (a - b) + (b - c) = (q + q^{'}) \overset{ˉ}{n}$ . So it is an equivalence relation, and $Z / n$ is its quotient set.

L1L3

1.2

The operation is well defined: if $a \equiv a^{'}$ and $b \equiv b^{'}$ , say $a - a^{'} = q \overset{ˉ}{n}$ and $b - b^{'} = q^{'} \overset{ˉ}{n}$ , then $(a + b) - (a^{'} + b^{'}) = (a - a^{'}) + (b - b^{'}) = (q + q^{'}) \overset{ˉ}{n}$ , so $a + b \equiv a^{'} + b^{'}$ and $[a + b] = [a^{'} + b^{'}]$ . Hence $[a] + [b] : = [a + b]$ depends only on the classes.

L1L3

1.3

$ι$ reflects the order: if $ι (s) < ι (t)$ then $s < t$ , since otherwise $t \leq s$ would give $ι (t) \leq ι (s)$ , contradicting antisymmetry.

L1L2L7

1.4

Distinct such representatives give distinct classes: let $s, t \in N$ with $s < n$ , $t < n$ and $[ι (s)] = [ι (t)]$ , so $ι (s) - ι (t) = q \overset{ˉ}{n}$ for some $q$ , that is $ι (s) = q \overset{ˉ}{n} + ι (t)$ . Both $ι (s)$ and $ι (t)$ are nonnegative and, $ι$ preserving the order, both are $< ι (n) = \overset{ˉ}{n}$ . So $ι (s)$ is written as $q \overset{ˉ}{n} + ι (t)$ with $0 \leq ι (t) < \overset{ˉ}{n}$ and also as $0 \cdot \overset{ˉ}{n} + ι (s)$ with $0 \leq ι (s) < \overset{ˉ}{n}$ ; the uniqueness clause of division with remainder forces $q = 0$ and $ι (t) = ι (s)$ , whence $s = t$ by injectivity of $ι$ .

L1L2L3L4

1.5

In the group $Z / n$ the $k$ -th power of $[1]$ in additive notation is $[ι (k)]$ for every $k \in N$ : the set of $k$ for which this holds contains $0$ , since the identity is $[0] = [ι (0)]$ , and is closed under $σ$ , since the power at $σ (k)$ is the power at $k$ plus $[1]$ , that is $[ι (k)] + [1] = [ι (k) + 1] = [ι (σ (k))]$ .

L1L2L6L7

2.1

$(Z / n, +, [0])$ is an abelian group: associativity, commutativity, the identity law $[a] + [0] = [a + 0] = [a]$ and the inverse law $[a] + [- a] = [0]$ all follow from the corresponding identities in $Z$ applied to representatives, which is legitimate by step 1.2.

step 1.2L1L5

2.2

Every class has a representative $r$ with $0 \leq r < \overset{ˉ}{n}$ : given $a$ , divide $a = q \overset{ˉ}{n} + r$ with $0 \leq r < \overset{ˉ}{n}$ , so $a - r = q \overset{ˉ}{n}$ and $[a] = [r]$ . Moreover $r \geq 0$ gives $r = ι (s)$ for a unique $s \in N$ , and $ι (s) < ι (n)$ gives $s < n$ .

step 1.3L2L3L4

3.1

The map $φ : n \to Z / n$ with $φ (s) = [ι (s)]$ is well defined, the elements of the natural number $n$ being exactly the naturals $s < n$ ; it is surjective by step 2.2 and injective by step 1.4, hence a bijection. So $Z / n \approx n$ and $∣ Z / n ∣ = n$ .

step 2.2step 1.4L2L5

3.2

For a negative integer $x$ with $- x = ι (k)$ , the $x$ -th power of $[1]$ is the inverse of $[ι (k)]$ , namely $[- ι (k)] = [x]$ ; with step 1.5 this gives that the set of integer powers of $[1]$ is ${[a] : a \in Z} = Z / n$ .

step 2.1step 1.5L1L2L6

4.1

Since $⟨ [1] ⟩$ is exactly the set of integer powers of $[1]$ , step 3.2 gives $⟨ [1] ⟩ = Z / n$ , so $Z / n$ is cyclic, generated by $[1]$ .

step 3.2L5

5.1

Claims 1 to 4 are established in steps 1.1, 2.1, 3.1 and 4.1.

step 1.1step 2.1step 3.1step 4.1∎

Remarks

The hypothesis $n \geq 1$ is carried by the title and by the statement, not left implicit. It is used twice and in an essential way: $\overset{ˉ}{n} > 0$ is what makes Division with remainder in $Z$ : for $a \in Z$ and $b > 0$ there are unique $q, r \in Z$ with $a = q b + r$ and $0 \leq r < b$ applicable in step 2.2, and it is what makes the count in step 3.1 come out as $n$ . At $n = 0$ the relation $\equiv$ is equality on $Z$ , the quotient set is in bijection with $Z$ itself, and the group is infinite.
No greatest common divisor is used anywhere above, and none is available at this point in the reading order. Only division with remainder is needed. The multiplicative structure of $Z / n$ , where the units are the classes coprime to $n$ , does need gcd theory and belongs to a later page.
The elements are congruence classes, that is subsets of $Z$ , and $[a]$ is a set. Nothing above ever names an element of $Z / n$ except through a representative, which is why step 1.2 has to be checked before the operation may be written down at all.

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

$Sym ({1, 2, 3})$ has exactly six elements, is non-abelian, and its elements have orders $1$ , $2$ and $3$

Example

Let $X = {1, 2, 3}$ , the three-element set of natural numbers $1 = σ (0)$ , $2 = σ (1)$ , $3 = σ (2)$ , and let $G = Sym (X)$ be its symmetric group (The symmetric group $Sym (X)$ : the bijections of a set $X$ under composition), a group under composition ( $Sym (X)$ is a group under composition, and it is non-abelian whenever $X$ has at least three distinct elements). Then:

$G$ has exactly six elements, namely

$i d, (1 2), (1 3), (2 3), (1 2 3), (1 3 2),$

so $G \approx 6$ and $∣ G ∣ = 6$ (The order $∣ G ∣$ of a finite group and the order $ord (g)$ of an element, with $ord (g) = \infty$ when no positive power of $g$ is the identity);
$G$ is not abelian: $(1 2) \circ (2 3) = (1 2 3)$ while $(2 3) \circ (1 2) = (1 3 2)$ ;
the orders of its elements are $ord (i d) = 1$ , $ord = 2$ for each of the three transpositions, and $ord = 3$ for each of the two $3$ -cycles.

Facts & Assumptions

Given: $X = {1, 2, 3}$ and $G = Sym (X)$ with the operation $f \circ g$ , $(f \circ g) (x) = f (g (x))$ , and identity $i d$ (The symmetric group $Sym (X)$ : the bijections of a set $X$ under composition).

[L1]

$1$ , $2$ and $3$ are pairwise distinct natural numbers: $2 = σ (1) = 1 \cup {1}$ and $3 = σ (2) = 2 \cup {2} = {0, 1, 2}$ , so $1 \in 2$ , $2 \in 3$ and $1 \in 3$ , while no natural number is a member of itself (The natural numbers $N$ (von Neumann), Every natural number is a transitive set and is not a member of itself); so $X$ has exactly three elements.

[L2]

$Sym (X)$ is a group under composition; the cycle symbols $(a b)$ and $(a b c)$ denote the permutations described in The symmetric group $Sym (X)$ : the bijections of a set $X$ under composition ( $Sym (X)$ is a group under composition, and it is non-abelian whenever $X$ has at least three distinct elements, Group and abelian group).

[L3]

Two functions $X \to X$ are equal exactly when they agree at every point; a function on $X$ is determined by the triple $(f (1), f (2), f (3))$ , and $f$ is a bijection of $X$ exactly when those three values are pairwise distinct, since three distinct values in a three-element set exhaust it (Injection, surjection, bijection).

[L4]

Powers: $f^{0} = i d$ , $f^{1} = f$ and $f^{σ (k)} = f^{k} \circ f$ (Powers $g^{n}$ : natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$ ); $f^{j + k} = f^{j} \circ f^{k}$ and $(f^{j})^{k} = f^{j k}$ (Exponent laws in a group: $g^{m + n} = g^{m} g^{n}$ and $(g^{m})^{n} = g^{m n}$ for all $m, n \in Z$ , and $(g h)^{n} = g^{n} h^{n}$ when $g$ and $h$ commute); $ord (f)$ is the least $k \geq 1$ with $f^{k} = i d$ (The order $∣ G ∣$ of a finite group and the order $ord (g)$ of an element, with $ord (g) = \infty$ when no positive power of $g$ is the identity).

[L5]

Finiteness and counting: $A \approx m$ means a bijection $A \to m$ exists, and $∣ A ∣ = m$ is then the unique such natural (Finite, countably infinite, countable, uncountable, Equinumerous sets, $A \approx B$ and $A ⪯ B$ , The order $∣ G ∣$ of a finite group and the order $ord (g)$ of an element, with $ord (g) = \infty$ when no positive power of $g$ is the identity).

[L6]

If $ord (f) = k$ then $f^{j} \neq i d$ for $1 \leq j < k$ , and conversely a $k$ with $f^{k} = i d$ and $f^{j} \neq i d$ for $1 \leq j < k$ is the order (The order $∣ G ∣$ of a finite group and the order $ord (g)$ of an element, with $ord (g) = \infty$ when no positive power of $g$ is the identity, If $ord (g) = n$ then $g^{k} = e$ iff $k$ is an integer multiple of $n$ , the powers $g^{0}, \dots, g^{n - 1}$ are distinct, and $⟨ g ⟩$ has exactly $n$ elements; if $g$ has infinite order then $g^{j} = g^{k}$ only for $j = k$ ).

Verification

technique · direct

1.1

Writing each element of $G$ as the triple of its values at $1, 2, 3$ , the six listed permutations are $i d = (1, 2, 3)$ , $(2 3) = (1, 3, 2)$ , $(1 2) = (2, 1, 3)$ , $(1 2 3) = (2, 3, 1)$ , $(1 3 2) = (3, 1, 2)$ and $(1 3) = (3, 2, 1)$ . These six triples are pairwise different, so the six permutations are pairwise different.

L1L2L3

1.2

Each transposition $τ \in {(1 2), (1 3), (2 3)}$ satisfies $τ \circ τ = i d$ , since it exchanges two points and fixes the third, so applying it twice returns every point to itself; and $τ \neq i d$ , since it moves two points.

L2L3

1.3

$ord (i d) = 1$ , since ${i d}^{1} = i d$ and $1$ is the least natural that is $\geq 1$ .

L4L6

2.1

Every element of $G$ is one of the six. Let $f \in G$ ; its triple has pairwise distinct entries by [L3]. There are three possible values for $f (1)$ ; for each, two remaining values for $f (2)$ ; and then $f (3)$ is forced to be the one element of $X$ left over. Running through those six combinations produces exactly the six triples listed in step 1.1.

step 1.1L1L3

2.2

Composites, computed pointwise. $(1 2) \circ (2 3)$ sends $1 \mapsto 1 \mapsto 2$ , $2 \mapsto 3 \mapsto 3$ , $3 \mapsto 2 \mapsto 1$ , giving the triple $(2, 3, 1)$ , which is $(1 2 3)$ . And $(2 3) \circ (1 2)$ sends $1 \mapsto 2 \mapsto 3$ , $2 \mapsto 1 \mapsto 1$ , $3 \mapsto 3 \mapsto 2$ , giving $(3, 1, 2)$ , which is $(1 3 2)$ .

step 1.1L2L3

2.3

Let $c = (1 2 3)$ , so $c$ sends $1 \mapsto 2 \mapsto 3 \mapsto 1$ . Then $c \circ c$ sends $1 \mapsto 3$ , $2 \mapsto 1$ , $3 \mapsto 2$ , which is the triple $(3, 1, 2)$ , that is $(1 3 2)$ ; and $c \circ c \circ c$ sends $1 \mapsto c (3) = 1$ , $2 \mapsto c (1) = 2$ , $3 \mapsto c (2) = 3$ , so it is $i d$ .

step 1.1L2L3

2.4

Each transposition $τ$ has $ord (τ) = 2$ : $τ^{2} = τ \circ τ = i d$ by step 1.2, while $τ^{1} = τ \neq i d$ , so $2$ is the least $k \geq 1$ with $τ^{k} = i d$ .

step 1.2L4L6

3.1

By steps 1.1 and 2.1 the set $G$ consists of exactly the six listed elements, so the map $6 \to G$ sending $0, 1, 2, 3, 4, 5$ to $i d, (1 2), (1 3), (2 3), (1 2 3), (1 3 2)$ respectively is a bijection; hence $G \approx 6$ and $∣ G ∣ = 6$ . This is claim 1.

step 1.1step 2.1L5

3.2

By step 2.2 the two composites differ, since the triples $(2, 3, 1)$ and $(3, 1, 2)$ differ at $1$ ; so $G$ is not abelian, which is claim 2.

step 2.2L3

3.3

Each of $c = (1 2 3)$ and $c^{2} = (1 3 2)$ has order $3$ : by step 2.3, $c^{3} = i d$ , $c^{1} = c \neq i d$ and $c^{2} = (1 3 2) \neq i d$ , so $ord (c) = 3$ ; and $(c^{2})^{3} = c^{6} = i d$ with $c^{2} \neq i d$ and $(c^{2})^{2} = c^{4} = c \neq i d$ , so $ord (c^{2}) = 3$ as well.

step 2.3L4L6

4.1

Claims 1, 2 and 3 are steps 3.1, 3.2 and steps 1.3, 2.4 and 3.3 taken together; the orders occurring are exactly $1$ , $2$ and $3$ .

step 3.1step 3.2step 1.3step 2.4step 3.3∎

Remarks

This is the first explicitly non-abelian group in the library. Claim 2 instantiates the general statement of $Sym (X)$ is a group under composition, and it is non-abelian whenever $X$ has at least three distinct elements at the three distinct points $1, 2, 3$ ; the pair of transpositions used there is the pair used here.
No element of $G$ has order $6$ , by claim 3, so by If $ord (g) = n$ then $g^{k} = e$ iff $k$ is an integer multiple of $n$ , the powers $g^{0}, \dots, g^{n - 1}$ are distinct, and $⟨ g ⟩$ has exactly $n$ elements; if $g$ has infinite order then $g^{j} = g^{k}$ only for $j = k$ every cyclic subgroup $⟨ f ⟩$ has at most three elements and none of them is all of $G$ : the group is not cyclic. That also follows from claim 2, every cyclic group being abelian ( $⟨ g ⟩ = {g^{n} : n \in Z}$ , and every cyclic group is abelian).
The elements $1$ , $2$ , $3$ here are the natural numbers of The natural numbers $N$ (von Neumann), hence sets; nothing in the computation uses anything about them beyond their being three distinct objects.

CounterexampleConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27Open item page →

$(g h)^{n} = g^{n} h^{n}$ fails without commutativity: two transpositions in $Sym ({1, 2, 3})$ with $(g h)^{2} \neq g^{2} h^{2}$

Statement refuted

False claim: in every group $G$ , $(g h)^{n} = g^{n} h^{n}$ for all $g, h \in G$ and all $n$ (Powers $g^{n}$ : natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$ ).

The claim is true when $g$ and $h$ commute, which is exactly what Exponent laws in a group: $g^{m + n} = g^{m} g^{n}$ and $(g^{m})^{n} = g^{m n}$ for all $m, n \in Z$ , and $(g h)^{n} = g^{n} h^{n}$ when $g$ and $h$ commute asserts and exactly why that lemma carries the hypothesis in its statement and in its title. Without the hypothesis it fails, already at $n = 2$ : in $Sym ({1, 2, 3})$ take the transpositions $g = (1 2)$ and $h = (2 3)$ . Then $g^{2} h^{2} = i d$ while $(g h)^{2} \neq i d$ .

Facts & Assumptions

Given: $X = {1, 2, 3}$ with $1$ , $2$ , $3$ pairwise distinct; $Sym (X)$ with composition $(f \circ k) (x) = f (k (x))$ and identity $i d$ (The symmetric group $Sym (X)$ : the bijections of a set $X$ under composition); and $g = (1 2)$ , $h = (2 3)$ , so $g$ exchanges $1$ and $2$ and fixes $3$ , while $h$ exchanges $2$ and $3$ and fixes $1$ .

[L1]

$Sym (X)$ is a group under composition ( $Sym (X)$ is a group under composition, and it is non-abelian whenever $X$ has at least three distinct elements, Group and abelian group, Left identity, right identity, and two-sided identity for a binary operation).

[L2]

$f^{2} = f^{σ (1)} = f^{1} \circ f = f \circ f$ , and $f^{0} = i d$ (Powers $g^{n}$ : natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$ ).

[L3]

Two permutations are equal exactly when they agree at every point of $X$ (Injection, surjection, bijection, The symmetric group $Sym (X)$ : the bijections of a set $X$ under composition).

[L4]

The refuted claim: $(g h)^{n} = g^{n} h^{n}$ in every group, for all elements and all $n$ .

Counterexample

technique · direct

1.1

$g \circ g = i d$ : $g$ sends $1 \mapsto 2 \mapsto 1$ and $2 \mapsto 1 \mapsto 2$ and fixes $3$ twice. Hence $g^{2} = i d$ .

L2L3given

1.2

$h \circ h = i d$ for the same reason, with the pair $2, 3$ in place of $1, 2$ . Hence $h^{2} = i d$ .

L2L3given

1.3

The composite $g \circ h$ sends $1 \mapsto h (1) = 1 \mapsto g (1) = 2$ , sends $2 \mapsto h (2) = 3 \mapsto g (3) = 3$ , and sends $3 \mapsto h (3) = 2 \mapsto g (2) = 1$ .

L3given

2.1

$g^{2} h^{2} = i d \circ i d = i d$ .

step 1.1step 1.2L1

2.2

$(g h)^{2} = (g \circ h) \circ (g \circ h)$ sends $1 \mapsto 2 \mapsto 3$ by step 1.3, so it does not fix $1$ , since $3 \neq 1$ .

step 1.3L2L3given

3.1

Therefore $(g h)^{2} \neq i d = g^{2} h^{2}$ , and the claim of [L4] is false; the elements $g$ and $h$ do not commute, since $g \circ h$ sends $2$ to $3$ while $h \circ g$ sends $2 \mapsto g (2) = 1 \mapsto h (1) = 1$ .

step 2.1step 2.2L3L4∎

Remarks

What survives without commutativity is the expansion $(g h)^{2} = g h g h$ , which is all that associativity gives. Turning it into $g^{2} h^{2}$ requires exchanging the middle $h$ and $g$ , and that exchange is the commuting hypothesis in Exponent laws in a group: $g^{m + n} = g^{m} g^{n}$ and $(g^{m})^{n} = g^{m n}$ for all $m, n \in Z$ , and $(g h)^{n} = g^{n} h^{n}$ when $g$ and $h$ commute, nothing weaker.
Why this matters for the library. The published Laws of integer exponents states $(a b)^{n} = a^{n} b^{n}$ with no commuting hypothesis and is correct, because it is a statement about a field, where multiplication is commutative by definition. Carrying that form over to groups would be a false generalisation, and this item is the witness.
The same pair $g, h$ witnesses that $Sym ({1, 2, 3})$ is not abelian ( $Sym ({1, 2, 3})$ has exactly six elements, is non-abelian, and its elements have orders $1$ , $2$ and $3$ ), which is no accident. At $n = 2$ the equation $(g h)^{2} = g^{2} h^{2}$ reads $g h g h = g g h h$ , and cancelling $g$ on the left and $h$ on the right (Cancellation in a group: $g x = g y$ or $x g = y g$ forces $x = y$ ; equivalently left and right translation by $g$ are bijections of $G$ , so $g x = h$ and $x g = h$ each have exactly one solution) turns it into $h g = g h$ ; so at this exponent the law holds for a pair exactly when the pair commutes, and every non-abelian group supplies a counterexample.

ExampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27Open item page →

The Klein four-group as the subgroup ${i d, (12) (34), (13) (24), (14) (23)}$ of $Sym ({1, 2, 3, 4})$ : abelian of order $4$ , non-cyclic, every non-identity element of order $2$

Example

Let $X = {1, 2, 3, 4}$ , four pairwise distinct natural numbers, and work in $Sym (X)$ (The symmetric group $Sym (X)$ : the bijections of a set $X$ under composition, $Sym (X)$ is a group under composition, and it is non-abelian whenever $X$ has at least three distinct elements). Put

$a = (1 2) (3 4), b = (1 3) (2 4), c = (1 4) (2 3),$

each being the composite of the two disjoint transpositions shown, and set $V = {i d, a, b, c}$ . Then:

$V$ is a subgroup of $Sym (X)$ (Subgroup) with four distinct elements, so $∣ V ∣ = 4$ (The order $∣ G ∣$ of a finite group and the order $ord (g)$ of an element, with $ord (g) = \infty$ when no positive power of $g$ is the identity);
$V$ is abelian, with multiplication table generated by $a^{2} = b^{2} = c^{2} = i d$ , $a b = b a = c$ , $a c = c a = b$ and $b c = c b = a$ ;
every element of $V$ other than $i d$ has order $2$ ;
$V$ is not cyclic (The subgroup $⟨ S ⟩$ generated by a subset, the cyclic subgroup $⟨ g ⟩$ , and cyclic groups).

$V$ is called the Klein four-group.

Facts & Assumptions

Given: $X = {1, 2, 3, 4}$ and the permutations $a, b, c$ of $X$ acting as follows: $a$ sends $1 \mapsto 2$ , $2 \mapsto 1$ , $3 \mapsto 4$ , $4 \mapsto 3$ ; $b$ sends $1 \mapsto 3$ , $3 \mapsto 1$ , $2 \mapsto 4$ , $4 \mapsto 2$ ; $c$ sends $1 \mapsto 4$ , $4 \mapsto 1$ , $2 \mapsto 3$ , $3 \mapsto 2$ (The symmetric group $Sym (X)$ : the bijections of a set $X$ under composition).

[L1]

$1$ , $2$ , $3$ , $4$ are pairwise distinct natural numbers, since each is a member of every later one and no natural number is a member of itself (The natural numbers $N$ (von Neumann), Every natural number is a transitive set and is not a member of itself).

[L2]

$Sym (X)$ is a group under composition, with identity $i d$ ( $Sym (X)$ is a group under composition, and it is non-abelian whenever $X$ has at least three distinct elements, Group and abelian group).

[L3]

Two permutations agree exactly when they agree at every point of $X$ (Injection, surjection, bijection).

[L4]

One-step test: a nonempty subset $H$ of a group with $x y^{- 1} \in H$ for all $x, y \in H$ is a subgroup (One-step subgroup test: a nonempty $H \subseteq G$ is a subgroup iff $g h^{- 1} \in H$ for all $g, h \in H$ ; the identity and the inverses of $H$ are then those of $G$ , Subgroup).

[L5]

Powers and order: $f^{0} = i d$ , $f^{1} = f$ , $f^{2} = f \circ f$ , and $ord (f)$ is the least $k \geq 1$ with $f^{k} = i d$ (Powers $g^{n}$ : natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$ , The order $∣ G ∣$ of a finite group and the order $ord (g)$ of an element, with $ord (g) = \infty$ when no positive power of $g$ is the identity).

[L6]

If $ord (g) = m$ is finite then $∣ ⟨ g ⟩ ∣ = m$ (If $ord (g) = n$ then $g^{k} = e$ iff $k$ is an integer multiple of $n$ , the powers $g^{0}, \dots, g^{n - 1}$ are distinct, and $⟨ g ⟩$ has exactly $n$ elements; if $g$ has infinite order then $g^{j} = g^{k}$ only for $j = k$ ); $⟨ g ⟩$ is the smallest subgroup containing $g$ (The subgroup $⟨ S ⟩$ generated by a subset, the cyclic subgroup $⟨ g ⟩$ , and cyclic groups); $∣ A ∣ = m$ means $A \approx m$ , and that natural is unique (Finite, countably infinite, countable, uncountable, Equinumerous sets, $A \approx B$ and $A ⪯ B$ , The order $∣ G ∣$ of a finite group and the order $ord (g)$ of an element, with $ord (g) = \infty$ when no positive power of $g$ is the identity).

Verification

technique · direct

1.1

The four elements are pairwise distinct: at the point $1$ they take the values $i d (1) = 1$ , $a (1) = 2$ , $b (1) = 3$ and $c (1) = 4$ , which are pairwise distinct.

L1L3given

1.2

Each of $a$ , $b$ , $c$ is its own inverse: $a \circ a$ sends $1 \mapsto 2 \mapsto 1$ , $2 \mapsto 1 \mapsto 2$ , $3 \mapsto 4 \mapsto 3$ and $4 \mapsto 3 \mapsto 4$ , so $a \circ a = i d$ ; the same computation with the corresponding pairs gives $b \circ b = i d$ and $c \circ c = i d$ .

L2L3given

1.3

$a \circ b = c$ : it sends $1 \mapsto b (1) = 3 \mapsto a (3) = 4$ , $2 \mapsto 4 \mapsto 3$ , $3 \mapsto 1 \mapsto 2$ , $4 \mapsto 2 \mapsto 1$ , and $c$ sends $1 \mapsto 4$ , $2 \mapsto 3$ , $3 \mapsto 2$ , $4 \mapsto 1$ . And $b \circ a = c$ : it sends $1 \mapsto a (1) = 2 \mapsto b (2) = 4$ , $2 \mapsto 1 \mapsto 3$ , $3 \mapsto 4 \mapsto 2$ , $4 \mapsto 3 \mapsto 1$ .

L3given

1.4

$a \circ c = b$ and $c \circ a = b$ : the first sends $1 \mapsto c (1) = 4 \mapsto a (4) = 3$ , $2 \mapsto 3 \mapsto 4$ , $3 \mapsto 2 \mapsto 1$ , $4 \mapsto 1 \mapsto 2$ ; the second sends $1 \mapsto a (1) = 2 \mapsto c (2) = 3$ , $2 \mapsto 1 \mapsto 4$ , $3 \mapsto 4 \mapsto 1$ , $4 \mapsto 3 \mapsto 2$ ; and $b$ sends $1 \mapsto 3$ , $2 \mapsto 4$ , $3 \mapsto 1$ , $4 \mapsto 2$ .

L3given

1.5

$b \circ c = a$ and $c \circ b = a$ : the first sends $1 \mapsto c (1) = 4 \mapsto b (4) = 2$ , $2 \mapsto 3 \mapsto 1$ , $3 \mapsto 2 \mapsto 4$ , $4 \mapsto 1 \mapsto 3$ ; the second sends $1 \mapsto b (1) = 3 \mapsto c (3) = 2$ , $2 \mapsto 4 \mapsto 1$ , $3 \mapsto 1 \mapsto 4$ , $4 \mapsto 2 \mapsto 3$ ; and $a$ sends $1 \mapsto 2$ , $2 \mapsto 1$ , $3 \mapsto 4$ , $4 \mapsto 3$ .

L3given

2.1

$V$ is nonempty and closed under $x \mapsto x^{- 1}$ : ${i d}^{- 1} = i d$ and each of $a, b, c$ is its own inverse by step 1.2.

step 1.2L2

2.2

$V$ is closed under composition: composing $i d$ with anything returns that element, each of $a, b, c$ composed with itself gives $i d$ by step 1.2, and the six mixed products are computed in steps 1.3, 1.4 and 1.5, each landing in $V$ .

step 1.2step 1.3step 1.4step 1.5L2

2.3

$V$ is abelian: the products computed in steps 1.3, 1.4 and 1.5 agree in either order, $i d$ commutes with everything, and each element commutes with itself. With step 1.2 this is the table of claim 2.

step 1.2step 1.3step 1.4step 1.5L2

2.4

Each of $a, b, c$ has order $2$ : it is not $i d$ by step 1.1, so $ord \neq 1$ , and its square is $i d$ by step 1.2; hence $2$ is the least $k \geq 1$ with the $k$ -th power equal to $i d$ . This is claim 3, $ord (i d) = 1$ being immediate.

step 1.1step 1.2L5

3.1

Hence for $x, y \in V$ one has $y^{- 1} \in V$ and $x y^{- 1} \in V$ , so $V$ is a subgroup of $Sym (X)$ by the one-step test.

step 2.1step 2.2L4

4.1

$V$ has exactly four elements: the map $4 \to V$ sending $0, 1, 2, 3$ to $i d, a, b, c$ is a bijection by step 1.1, so $V \approx 4$ and $∣ V ∣ = 4$ . This with step 3.1 is claim 1.

step 1.1step 3.1L6

5.1

$V$ is not cyclic: if $V = ⟨ g ⟩$ for some $g \in V$ , then $g$ has finite order and $∣ ⟨ g ⟩ ∣ = ord (g)$ , so $ord (g) = ∣ V ∣ = 4$ ; but every element of $V$ has order $1$ or $2$ by step 2.4, and $4 \neq 1$ and $4 \neq 2$ . This is claim 4.

step 4.1step 2.4L6∎

Remarks

An abelian group need not be cyclic. $V$ is the smallest witness of that, and it separates the two conditions that $⟨ g ⟩ = {g^{n} : n \in Z}$ , and every cyclic group is abelian relates in one direction only: cyclic implies abelian, never the converse.
The four-group is realised here inside a symmetric group rather than as a product. The external direct product of two groups is introduced on a later page, so $V$ is exhibited as a set of four explicit permutations and checked by hand; nothing above uses any construction the library has not built.
Every non-identity element having order $2$ is exactly what rules out a generator: a cyclic group of order $4$ must contain an element of order $4$ (If $ord (g) = n$ then $g^{k} = e$ iff $k$ is an integer multiple of $n$ , the powers $g^{0}, \dots, g^{n - 1}$ are distinct, and $⟨ g ⟩$ has exactly $n$ elements; if $g$ has infinite order then $g^{j} = g^{k}$ only for $j = k$ ), and $(Z / 4, +)$ is such a group (For $n \geq 1$ the congruence classes modulo $n$ form an abelian group $(Z / n, +)$ of order $n$ , generated by the class of $1$ ). So there are at least two groups of order $4$ that are not the same group. Classifying the groups of a given order needs machinery from a later page and is not attempted here.

ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passverified 2026-08-02 (claude-opus-5)Open item page →

The eight vertex permutations of a square form a non-abelian subgroup of $Sym ({1, 2, 3, 4})$ of order $8$ , generated by a $4$ -cycle and one diagonal swap

Example

Let $X = {1, 2, 3, 4}$ and regard its four elements as the vertices of a square read in cyclic order, so that the edges are the four pairs

$E = {{1, 2}, {2, 3}, {3, 4}, {4, 1}}$

and the two remaining pairs ${1, 3}$ , ${2, 4}$ are the diagonals. Call a permutation $f$ of $X$ a vertex symmetry of the square when, for all $x \neq y$ in $X$ , ${x, y} \in E$ if and only if ${f (x), f (y)} \in E$ .

Put $r = (1 2 3 4)$ and $s = (1 3)$ in $Sym (X)$ (The symmetric group $Sym (X)$ : the bijections of a set $X$ under composition) and

$D = {i d, r, r^{2}, r^{3}, s, r s, r^{2} s, r^{3} s},$

where juxtaposition is composition and powers are those of Powers $g^{n}$ : natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$ . Then:

$r^{4} = i d$ , $s^{2} = i d$ and $s r = r^{- 1} s$ ;
$D$ is a subgroup of $Sym (X)$ (Subgroup) whose eight listed elements are pairwise distinct, so $∣ D ∣ = 8$ (The order $∣ G ∣$ of a finite group and the order $ord (g)$ of an element, with $ord (g) = \infty$ when no positive power of $g$ is the identity), and $D = ⟨ {r, s} ⟩$ (The subgroup $⟨ S ⟩$ generated by a subset, the cyclic subgroup $⟨ g ⟩$ , and cyclic groups);
$D$ is not abelian: $r s \neq s r$ ;
$D$ is exactly the set of vertex symmetries of the square.

Facts & Assumptions

Given: $X = {1, 2, 3, 4}$ ; $r$ the permutation sending $1 \mapsto 2 \mapsto 3 \mapsto 4 \mapsto 1$ ; $s$ the permutation exchanging $1$ and $3$ and fixing $2$ and $4$ ; $E$ the four edge pairs listed above (The symmetric group $Sym (X)$ : the bijections of a set $X$ under composition).

[L1]

$1, 2, 3, 4$ are pairwise distinct natural numbers (The natural numbers $N$ (von Neumann), Every natural number is a transitive set and is not a member of itself).

[L2]

$Sym (X)$ is a group under composition with identity $i d$ ( $Sym (X)$ is a group under composition, and it is non-abelian whenever $X$ has at least three distinct elements, Group and abelian group); two permutations are equal exactly when they agree at every point (Injection, surjection, bijection).

[L3]

Exponent laws in a group: $f^{j + k} = f^{j} f^{k}$ , $f^{- k} = (f^{k})^{- 1}$ and $(f^{j})^{k} = f^{j k}$ (Exponent laws in a group: $g^{m + n} = g^{m} g^{n}$ and $(g^{m})^{n} = g^{m n}$ for all $m, n \in Z$ , and $(g h)^{n} = g^{n} h^{n}$ when $g$ and $h$ commute, Powers $g^{n}$ : natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$ ).

[L4]

Cancellation in a group (Cancellation in a group: $g x = g y$ or $x g = y g$ forces $x = y$ ; equivalently left and right translation by $g$ are bijections of $G$ , so $g x = h$ and $x g = h$ each have exactly one solution).

[L5]

One-step test for subgroups (One-step subgroup test: a nonempty $H \subseteq G$ is a subgroup iff $g h^{- 1} \in H$ for all $g, h \in H$ ; the identity and the inverses of $H$ are then those of $G$ , Subgroup); $⟨ S ⟩$ is the smallest subgroup containing $S$ (The subgroup $⟨ S ⟩$ generated by a subset, the cyclic subgroup $⟨ g ⟩$ , and cyclic groups).

[L6]

If $ord (f) = m$ then $f^{0}, \dots, f^{m - 1}$ are pairwise distinct and $⟨ f ⟩$ has exactly $m$ elements (If $ord (g) = n$ then $g^{k} = e$ iff $k$ is an integer multiple of $n$ , the powers $g^{0}, \dots, g^{n - 1}$ are distinct, and $⟨ g ⟩$ has exactly $n$ elements; if $g$ has infinite order then $g^{j} = g^{k}$ only for $j = k$ , The order $∣ G ∣$ of a finite group and the order $ord (g)$ of an element, with $ord (g) = \infty$ when no positive power of $g$ is the identity); $∣ A ∣ = m$ means $A \approx m$ for the unique such natural (Finite, countably infinite, countable, uncountable, Equinumerous sets, $A \approx B$ and $A ⪯ B$ ).

Verification

technique · direct

1.1

Powers of $r$ , computed pointwise: $r^{2}$ sends $1 \mapsto 3$ , $2 \mapsto 4$ , $3 \mapsto 1$ , $4 \mapsto 2$ ; $r^{3}$ sends $1 \mapsto 4$ , $2 \mapsto 1$ , $3 \mapsto 2$ , $4 \mapsto 3$ ; and $r^{4}$ sends every point back to itself, so $r^{4} = i d$ . None of $r$ , $r^{2}$ , $r^{3}$ is $i d$ , each moving $1$ . Hence $ord (r) = 4$ and $r^{- 1} = r^{3}$ .

L1L2L3L6given

1.2

$s^{2} = i d$ , since $s$ exchanges $1$ and $3$ and fixes $2$ and $4$ , so applying it twice fixes every point; hence $s^{- 1} = s$ and $s \neq i d$ .

L2L3given

2.1

$s r = r^{3} s$ . Both sides are computed pointwise: $s \circ r$ sends $1 \mapsto r (1) = 2 \mapsto s (2) = 2$ , $2 \mapsto 3 \mapsto 1$ , $3 \mapsto 4 \mapsto 4$ , $4 \mapsto 1 \mapsto 3$ ; while $r^{3} \circ s$ sends $1 \mapsto s (1) = 3 \mapsto r^{3} (3) = 2$ , $2 \mapsto 2 \mapsto 1$ , $3 \mapsto 1 \mapsto 4$ , $4 \mapsto 4 \mapsto 3$ . The two agree at every point. With step 1.1 this gives $s r = r^{- 1} s$ , which with claim 1's other two equations completes claim 1.

step 1.1L2given

2.2

$s \notin {i d, r, r^{2}, r^{3}}$ : $s$ fixes $2$ , while $r$ , $r^{2}$ and $r^{3}$ send $2$ to $3$ , $4$ and $1$ respectively, and $s \neq i d$ by step 1.2.

step 1.1step 1.2L1L2

3.1

By induction from step 2.1, $s r^{k} = r^{- k} s$ for every $k \in Z$ : the case $k = 0$ is trivial, and $s r^{k + 1} = (s r^{k}) r = r^{- k} s r = r^{- k} r^{- 1} s = r^{- (k + 1)} s$ , while applying $s$ on both sides of $s r^{k} = r^{- k} s$ and using $s^{2} = i d$ gives the statement for $- k$ .

step 1.2step 2.1L3

3.2

The eight listed elements are pairwise distinct. The four powers $r^{0}, r^{1}, r^{2}, r^{3}$ are pairwise distinct because $ord (r) = 4$ . If $r^{i} s = r^{j} s$ then $r^{i} = r^{j}$ by cancelling $s$ on the right, so $i = j$ for $i, j$ among $0, 1, 2, 3$ . And $r^{i} s = r^{j}$ would give $s = r^{j - i}$ , contradicting step 2.2.

step 1.1step 2.2L4L6

4.1

$D$ is closed under composition. A product of two listed elements has the form $(r^{i} s^{ε}) (r^{j} s^{δ})$ with $ε, δ \in {0, 1}$ . If $ε = 0$ it equals $r^{i + j} s^{δ}$ ; if $ε = 1$ then, moving $s$ past $r^{j}$ by step 3.1, it equals $r^{i} r^{- j} s s^{δ} = r^{i - j} s^{1 + δ}$ . In either case, reducing the exponent of $r$ using $r^{4} = i d$ and the exponent of $s$ using $s^{2} = i d$ gives one of the eight listed elements.

step 1.1step 1.2step 3.1L3

4.2

$D$ is closed under inverses: $(r^{i})^{- 1} = r^{- i}$ , again one of the four powers after reduction; and $(r^{i} s)^{- 1} = s^{- 1} (r^{i})^{- 1} = s r^{- i} = r^{i} s$ by step 3.1 and $s^{- 1} = s$ , so each of the four elements $r^{i} s$ is its own inverse.

step 1.1step 1.2step 3.1L3

4.3

$D$ is not abelian: $s r = r^{3} s$ by step 2.1, while $r s = r^{1} s$ ; if these were equal then $r^{3} = r$ by cancelling $s$ on the right, contradicting the distinctness of the powers of $r$ . This is claim 3.

step 1.1step 2.1step 3.2L4

5.1

$D$ is a subgroup: it contains $i d$ , is closed under composition by step 4.1 and under inverses by step 4.2, so $x y^{- 1} \in D$ for $x, y \in D$ and the one-step test applies. Its eight elements are distinct by step 3.2, so the map $8 \to D$ listing them is a bijection and $∣ D ∣ = 8$ .

step 3.2step 4.1step 4.2L5L6

6.1

$D = ⟨ {r, s} ⟩$ : $D$ is a subgroup containing $r$ and $s$ , so $⟨ {r, s} ⟩ \subseteq D$ ; conversely any subgroup containing $r$ and $s$ contains every $r^{i} s^{ε}$ , hence contains $D$ , so $D \subseteq ⟨ {r, s} ⟩$ . This with step 5.1 is claim 2.

step 5.1L5

7.1

Every element of $D$ is a vertex symmetry. The map $r$ carries the four edges ${1, 2}, {2, 3}, {3, 4}, {4, 1}$ to ${2, 3}, {3, 4}, {4, 1}, {1, 2}$ , so it maps $E$ onto $E$ ; being a bijection of $X$ , it therefore also carries each of the two non-edges ${1, 3}$ , ${2, 4}$ to a non-edge, and the "if and only if" holds. The map $s$ carries those four edges to ${3, 2}, {2, 1}, {1, 4}, {4, 3}$ , again onto $E$ , so the same applies. The vertex symmetries form a subgroup, since the defining condition is preserved by composition and, being an equivalence, by inverses; hence it contains $⟨ {r, s} ⟩ = D$ .

step 6.1L2L5given

8.1

Conversely let $f$ be a vertex symmetry. The neighbours of a point $x$ , meaning the $y$ with ${x, y} \in E$ , are exactly $r (x)$ and $r^{3} (x)$ , and these two are distinct because $r^{2} (x) \neq x$ by step 1.1. There is a unique $i \in {0, 1, 2, 3}$ with $r^{i} (1) = f (1)$ , since those four values are $1, 2, 3, 4$ ; put $f_{1} = r^{- i} f$ , again a vertex symmetry, with $f_{1} (1) = 1$ .

step 1.1step 5.1step 7.1L2given

9.1

Since ${1, 2} \in E$ , the pair ${f_{1} (1), f_{1} (2)} = {1, f_{1} (2)}$ is an edge, so $f_{1} (2)$ is a neighbour of $1$ , that is $f_{1} (2) \in {2, 4}$ . Since ${2, 3} \in E$ , the point $f_{1} (3)$ is a neighbour of $f_{1} (2)$ , and it differs from $f_{1} (1) = 1$ because $f_{1}$ is injective; the neighbours of $2$ are $1$ and $3$ and the neighbours of $4$ are $3$ and $1$ , so in both cases $f_{1} (3) = 3$ . Then $f_{1} (4)$ is the one element of $X$ not already taken.

step 8.1L2given

10.1

So either $f_{1}$ is $1 \mapsto 1$ , $2 \mapsto 2$ , $3 \mapsto 3$ , $4 \mapsto 4$ , that is $f_{1} = i d$ and $f = r^{i}$ ; or $f_{1}$ is $1 \mapsto 1$ , $2 \mapsto 4$ , $3 \mapsto 3$ , $4 \mapsto 2$ . In the second case $f_{1} = r^{2} s$ , since $r^{2} \circ s$ sends $1 \mapsto s (1) = 3 \mapsto r^{2} (3) = 1$ , $2 \mapsto 2 \mapsto 4$ , $3 \mapsto 1 \mapsto 3$ and $4 \mapsto 4 \mapsto 2$ ; hence $f = r^{i} r^{2} s = r^{i + 2} s \in D$ .

step 1.1step 9.1L2L3

11.1

By steps 7.1, 9.1 and 10.1 the vertex symmetries of the square are exactly the elements of $D$ , which is claim 4; claims 1, 2 and 3 are steps 2.1, 6.1 and 4.3.

step 2.1step 6.1step 4.3step 7.1step 9.1step 10.1∎

Remarks

The square is a combinatorial object here, not a geometric one. The identification of these eight permutations with the rigid motions of a square in the Euclidean plane is not available at this point in the reading order: $R^{2}$ with its metric comes much later. What is used instead is the edge relation $E$ , and claim 4 says the group is exactly the symmetry group of that relation, which is what "vertex permutations of a square" means here.
Every element is a rotation or a reflection, in the sense that $D$ splits as the four powers of $r$ and the four elements $r^{i} s$ ; step 4.2 shows each of the latter is its own inverse, matching the geometric picture in which a reflection applied twice is the identity.
The relation $s r = r^{- 1} s$ of claim 1 is the whole reason $D$ is closed: it is what lets any word in $r$ and $s$ be pushed into the normal form $r^{i} s^{ε}$ , which is step 4.1. Without it the eight elements would not obviously be all of $⟨ {r, s} ⟩$ .

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

$n Z$ is a subgroup of $(Z, +)$ for every $n \in Z$ , and every subgroup of $(Z, +)$ has this form

Example

Work in the abelian group $(Z, +, 0)$ . For $n \in Z$ put

$n Z : = {n k : k \in Z} .$

Then:

$n Z$ is a subgroup of $(Z, +)$ for every $n \in Z$ (Subgroup), and $n Z = ⟨ n ⟩$ (The subgroup $⟨ S ⟩$ generated by a subset, the cyclic subgroup $⟨ g ⟩$ , and cyclic groups);
conversely, every subgroup $H \leq (Z, +)$ equals $n Z$ for some $n \geq 0$ ; and $n$ may be taken to be $0$ if $H = {0}$ and otherwise the least positive element of $H$ .

In particular every subgroup of $(Z, +)$ is cyclic.

Facts & Assumptions

Given: The abelian group $(Z, +, 0)$ (The integers as equivalence classes of pairs of naturals, Arithmetic on the integers, Group and abelian group), and the embedding $ι : N \to Z$ of The naturals embed in the integers.

[L1]

$Z$ is a commutative ring, with $u - v : = u + (- v)$ (The integers form a commutative ring, Arithmetic on the integers); its order is total, antisymmetric, transitive and compatible with addition, and positives are closed under multiplication (The integers form a totally ordered ring, Order on the integers).

[L2]

$ι$ is injective, preserves addition, multiplication and order, and its image is exactly the nonnegative integers; $ι (0) = 0$ , $ι (1) = 1$ (The naturals embed in the integers).

[L3]

One-step test for subgroups, written additively: a nonempty $H \subseteq Z$ with $x - y \in H$ for all $x, y \in H$ is a subgroup (One-step subgroup test: a nonempty $H \subseteq G$ is a subgroup iff $g h^{- 1} \in H$ for all $g, h \in H$ ; the identity and the inverses of $H$ are then those of $G$ , Subgroup).

[L4]

$⟨ g ⟩$ is the smallest subgroup containing $g$ , and equals the set of integer powers of $g$ , which in additive notation are the multiples of $g$ (The subgroup $⟨ S ⟩$ generated by a subset, the cyclic subgroup $⟨ g ⟩$ , and cyclic groups, $⟨ g ⟩ = {g^{n} : n \in Z}$ , and every cyclic group is abelian, Powers $g^{n}$ : natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$ ).

[L5]

Division with remainder: for $a \in Z$ and $b > 0$ there are $q, r \in Z$ with $a = q b + r$ and $0 \leq r < b$ (Division with remainder in $Z$ : for $a \in Z$ and $b > 0$ there are unique $q, r \in Z$ with $a = q b + r$ and $0 \leq r < b$ ).

[L6]

Every nonempty subset of $N$ has a least element (The well-ordering principle).

[L7]

On $N$ : exactly one of $s < t$ , $s = t$ , $t < s$ holds (Trichotomy of the order on $N$ ); every $k \neq 0$ is a successor $σ (j) = 1 + j$ , so $k \neq 0$ gives $1 \leq k$ (Every nonzero natural number is a successor, Addition is commutative, Order on the natural numbers, The natural numbers $N$ (von Neumann)).

[L8]

Induction on $N$ (The principle of mathematical induction).

Verification

technique · direct

1.1

$n Z$ is nonempty, containing $0 = n \cdot 0$ ; and for $x = n k$ , $y = n k^{'}$ in it, $x - y = n k - n k^{'} = n (k - k^{'}) \in n Z$ by distributivity. By the one-step test $n Z$ is a subgroup of $(Z, +)$ .

L1L3

1.2

In the additive group $(Z, +)$ the $k$ -th power of $n$ is $n k$ . For $k \in N$ this holds by induction: the $0$ -th power is the identity $0 = n \cdot 0$ , and the power at $σ (k)$ is the power at $k$ plus $n$ , which is $n ι (k) + n = n (ι (k) + 1) = n ι (σ (k))$ . For a negative integer $x$ with $- x = ι (k)$ , the $x$ -th power is the additive inverse of $n ι (k)$ , namely $n (- ι (k)) = n x$ .

L1L2L4L8

1.3

Let $H \leq (Z, +)$ with $H \neq {0}$ . Choose $x \in H$ with $x \neq 0$ ; then $- x \in H$ as well, and by totality one of $x$ , $- x$ is positive, so $H$ contains a positive integer.

L1L3choose

2.1

Hence $⟨ n ⟩$ , the set of integer powers of $n$ , is exactly ${n k : k \in Z} = n Z$ . With step 1.1 this is claim 1.

step 1.1step 1.2L4

2.2

Every positive integer is $ι (k)$ for a unique $k \in N$ with $k \neq 0$ , hence with $k \geq 1$ . So $S : = {k \in N : k \geq 1 and ι (k) \in H}$ is nonempty by step 1.3; let $m$ be its least element and put $n : = ι (m)$ , a positive element of $H$ .

step 1.3L2L6L7

3.1

$n Z \subseteq H$ : by step 2.1, $n Z = ⟨ n ⟩$ , and $⟨ n ⟩$ is contained in every subgroup containing $n$ , in particular in $H$ .

step 2.1step 2.2L4

4.1

$H \subseteq n Z$ : let $h \in H$ and divide $h = q n + r$ with $0 \leq r < n$ , legitimate since $n > 0$ . Then $q n \in n Z \subseteq H$ , so $r = h - q n \in H$ because $H$ is closed under subtraction. If $r > 0$ then $r = ι (t)$ with $t \geq 1$ , and $ι (t) < ι (m)$ gives $t < m$ , since otherwise $m \leq t$ would give $ι (m) \leq ι (t)$ , contradicting antisymmetry; that puts $t$ in $S$ below its least element, which is impossible. Hence $r = 0$ and $h = q n \in n Z$ .

step 2.2step 3.1L1L2L3L5L7

5.1

So $H = n Z$ with $n$ the least positive element of $H$ ; and if $H = {0}$ then $H = 0 Z$ , since $0 \cdot k = 0$ for every $k$ . This is claim 2, and with claim 1 it shows every subgroup of $(Z, +)$ is $⟨ n ⟩$ for some $n \geq 0$ , hence cyclic.

step 2.1step 2.2step 3.1step 4.1L1L4∎

Remarks

Both halves use the division algorithm, but only the second one visibly. The first half is pure closure arithmetic; the second is the standard argument that a subgroup containing a least positive element $n$ can contain nothing strictly between the multiples of $n$ , and it is exactly Division with remainder in $Z$ : for $a \in Z$ and $b > 0$ there are unique $q, r \in Z$ with $a = q b + r$ and $0 \leq r < b$ that produces the offending remainder.
$n$ and $- n$ generate the same subgroup, since $(- n) k = n (- k)$ , so the $n$ in claim 2 is unique only after normalising it to be nonnegative. The normalisation is what the phrase "the least positive element" achieves.
Inclusion among these subgroups is divisibility: $m Z \subseteq n Z$ holds exactly when $n ∣ m$ . Indeed the inclusion applied to $k = 1$ gives $m \in n Z$ , that is $m = n c$ for some $c$ ; and conversely $m = n c$ gives $m k = n (c k) \in n Z$ for every $k$ . The systematic study of the divisibility relation belongs to a later page.

CounterexampleConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27Open item page →

A nonempty subset of a group closed under the operation need not be a subgroup: the nonnegative integers inside $(Z, +)$

Statement refuted

False claim: if $G$ is a group and $H \subseteq G$ is nonempty, contains the identity, and is closed under the operation of $G$ , then $H$ is a subgroup of $G$ (Subgroup).

The set of nonnegative integers inside the additive group of $Z$ refutes it: $H = {x \in Z : 0 \leq x}$ contains $0$ , is closed under addition, and is not a subgroup, because $1 \in H$ while $- 1 \notin H$ .

Facts & Assumptions

Given: The abelian group $(Z, +, 0)$ (The integers as equivalence classes of pairs of naturals, Arithmetic on the integers, Group and abelian group) and the subset $H = {x \in Z : 0 \leq x}$ , which by The naturals embed in the integers is exactly the image of $ι : N \to Z$ .

[L1]

$Z$ is a commutative ring; its order is total, antisymmetric and transitive and is compatible with addition (The integers form a commutative ring, The integers form a totally ordered ring, Order on the integers, Arithmetic on the integers).

[L2]

$ι$ is injective with image exactly the nonnegative integers, and $ι (0) = 0$ , $ι (1) = 1$ (The naturals embed in the integers).

[L3]

A subgroup must contain the identity and be closed under the operation and under inverses (Subgroup); equivalently, a nonempty $H$ is a subgroup exactly when $x - y \in H$ for all $x, y \in H$ (One-step subgroup test: a nonempty $H \subseteq G$ is a subgroup iff $g h^{- 1} \in H$ for all $g, h \in H$ ; the identity and the inverses of $H$ are then those of $G$ ).

[L4]

A subset closed under an operation inherits it as a binary operation (Binary operation on a set; associativity, commutativity, and a subset closed under the operation).

[L5]

The refuted claim: a nonempty subset of a group containing the identity and closed under the operation is a subgroup.

Counterexample

technique · direct

1.1

$0 \in H$ , since $0 \leq 0$ ; so $H$ is nonempty and contains the identity of $(Z, +)$ .

L1given

1.2

$H$ is closed under addition: if $0 \leq x$ and $0 \leq y$ then $x \leq x + y$ by compatibility of the order with addition, and $0 \leq x$ , so $0 \leq x + y$ by transitivity. Hence $+$ restricts to a binary operation on $H$ , and $(H, +, 0)$ is a commutative monoid.

L1L4given

1.3

$0 < 1$ : the integer $1 = ι (1)$ lies in the image of $ι$ , so $0 \leq 1$ , and $1 \neq 0$ because $ι$ is injective and $1 \neq 0$ in $N$ . Hence $1 \in H$ .

L2given

2.1

$- 1 \notin H$ : adding $- 1$ to both sides of $0 < 1$ gives $- 1 < 0$ , so $0 \leq - 1$ fails by antisymmetry.

step 1.3L1

3.1

Therefore $H$ is not closed under inverses, since $1 \in H$ and its additive inverse $- 1$ is not in $H$ ; so $H$ is not a subgroup of $(Z, +)$ .

step 1.3step 2.1L3

4.1

By steps 1.1, 1.2 and 3.1 the set $H$ is a nonempty subset containing the identity and closed under the operation which is not a subgroup, so the claim of [L5] is false.

step 1.1step 1.2step 3.1L5∎

Remarks

Closure under inverses is an independent condition, and this is why the economical criterion of One-step subgroup test: a nonempty $H \subseteq G$ is a subgroup iff $g h^{- 1} \in H$ for all $g, h \in H$ ; the identity and the inverses of $H$ are then those of $G$ tests $x y^{- 1}$ rather than $x y$ : the single expression carries both closure requirements at once.
For a finite subset the claim would be true, since a nonempty finite subset of a group closed under the operation is a subgroup; the witness above is necessarily infinite. That finiteness result is not proved in the library at this point in the reading order, and nothing here rests on it.
$H$ is a perfectly good commutative monoid, by step 1.2, and $ι$ is a bijection from $N$ onto $H$ carrying addition to addition (The naturals embed in the integers). So the example is the same phenomenon as A commutative monoid in which cancellation holds need not be a group: $(N, +)$ , seen from inside a group.

False statementConstruction: Literature-sourcedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)verified 2026-07-29 (claude-sonnet-5)Open item page →

FALSE: The union of two subgroups is a subgroup

Statement

False claim: if $H$ and $K$ are subgroups of a group $G$ (Subgroup), then $H \cup K$ is a subgroup of $G$ .

The corresponding statement for intersections is true and is The intersection of a nonempty family of subgroups of $G$ is a subgroup of $G$ . For unions it fails, and the smallest natural witness is inside $(Z, +)$ : take $H = 2 Z = {2 k : k \in Z}$ and $K = 3 Z = {3 k : k \in Z}$ , both subgroups by the one-step test (One-step subgroup test: a nonempty $H \subseteq G$ is a subgroup iff $g h^{- 1} \in H$ for all $g, h \in H$ ; the identity and the inverses of $H$ are then those of $G$ ). Then $2 \in H$ and $3 \in K$ , so both lie in $H \cup K$ , while their sum $5$ lies in neither.

Throughout, $2 : = 1 + 1$ , $3 : = 2 + 1$ , $4 : = 3 + 1$ and $5 : = 4 + 1$ in $Z$ .

Facts & Assumptions

Given: The abelian group $(Z, +, 0)$ (The integers as equivalence classes of pairs of naturals, Arithmetic on the integers, Group and abelian group), the sets $H = 2 Z = {2 k : k \in Z}$ and $K = 3 Z = {3 k : k \in Z}$ , and the numerals $2 = 1 + 1$ , $3 = 2 + 1$ , $4 = 3 + 1$ , $5 = 4 + 1$ .

[L1]

$Z$ is a commutative ring (The integers form a commutative ring, Arithmetic on the integers); its order is total, antisymmetric and transitive and is compatible with addition (The integers form a totally ordered ring, Order on the integers).

[L2]

$0 < 1$ in $Z$ : $1 = ι (1)$ lies in the image of $ι$ , hence $0 \leq 1$ , and $1 \neq 0$ because $ι$ is injective and $1 \neq 0$ in $N$ (The naturals embed in the integers).

[L3]

One-step test, written additively: a nonempty $S \subseteq Z$ with $x - y \in S$ for all $x, y \in S$ is a subgroup of $(Z, +)$ ; a subgroup contains the identity and is closed under the operation and under inverses (One-step subgroup test: a nonempty $H \subseteq G$ is a subgroup iff $g h^{- 1} \in H$ for all $g, h \in H$ ; the identity and the inverses of $H$ are then those of $G$ , Subgroup).

[L4]

Division with remainder: for $a \in Z$ and $b > 0$ there is exactly one pair $(q, r)$ of integers with $a = q b + r$ and $0 \leq r < b$ (Division with remainder in $Z$ : for $a \in Z$ and $b > 0$ there are unique $q, r \in Z$ with $a = q b + r$ and $0 \leq r < b$ ).

[L5]

A subgroup is closed under the operation (Subgroup, One-step subgroup test: a nonempty $H \subseteq G$ is a subgroup iff $g h^{- 1} \in H$ for all $g, h \in H$ ; the identity and the inverses of $H$ are then those of $G$ ).

[L6]

The refuted claim: for all subgroups $H, K$ of a group $G$ , the union $H \cup K$ is a subgroup of $G$ .

Refutation

technique · direct

1.1

$0 < 1 < 2 < 3$ : adding $1$ to $0 < 1$ gives $1 < 2$ , and adding $1$ again gives $2 < 3$ ; transitivity chains them. In particular $0 \leq 1 < 2$ and $0 \leq 2 < 3$ .

L1L2

1.2

$2 \cdot 2 = 4$ and $3 \cdot 1 = 3$ : by distributivity $2 \cdot 2 = 2 (1 + 1) = 2 + 2 = (2 + 1) + 1 = 3 + 1 = 4$ , and $3 \cdot 1 = 3$ by the multiplicative identity law.

L1given

1.3

For every $n \in Z$ the set $n Z = {n k : k \in Z}$ is a subgroup of $(Z, +)$ : it contains $0 = n \cdot 0$ and is therefore nonempty, and for $x = n k$ and $y = n k^{'}$ in it, $x - y = n k - n k^{'} = n (k - k^{'}) \in n Z$ by distributivity, so the one-step test applies. In particular $H = 2 Z$ and $K = 3 Z$ are subgroups of $(Z, +)$ .

L1L3

2.1

$5 = 2 \cdot 2 + 1$ and $5 = 3 \cdot 1 + 2$ : the first is $4 + 1 = 5$ by step 1.2, and the second is $3 + 2 = (3 + 1) + 1 = 4 + 1 = 5$ by associativity.

step 1.2L1given

3.1

$5 \notin 2 Z$ : if $5 = 2 k$ then $5 = k \cdot 2 + 0$ with $0 \leq 0 < 2$ , while step 2.1 gives $5 = 2 \cdot 2 + 1$ with $0 \leq 1 < 2$ ; uniqueness of the pair $(q, r)$ forces $0 = 1$ , contradicting $0 < 1$ .

step 1.1step 2.1L1L2L4

3.2

$5 \notin 3 Z$ : if $5 = 3 k$ then $5 = k \cdot 3 + 0$ with $0 \leq 0 < 3$ , while step 2.1 gives $5 = 1 \cdot 3 + 2$ with $0 \leq 2 < 3$ ; uniqueness forces $0 = 2$ , contradicting $0 < 1 < 2$ .

step 1.1step 2.1L1L2L4

4.1

$2 = 2 \cdot 1 \in 2 Z = H$ and $3 = 3 \cdot 1 \in 3 Z = K$ , so both lie in $H \cup K$ ; but $2 + 3 = 5$ lies in neither by steps 3.1 and 3.2, hence not in $H \cup K$ .

step 1.3step 2.1step 3.1step 3.2L1given

5.1

So $H \cup K$ is not closed under the operation of $(Z, +)$ and is therefore not a subgroup, although $H$ and $K$ both are by step 1.3; the claim of [L6] is false.

step 1.3step 4.1L5L6∎

Remarks

Intersections behave, unions do not. An intersection of subgroups is always a subgroup (The intersection of a nonempty family of subgroups of $G$ is a subgroup of $G$ ), which is what makes $⟨ S ⟩$ definable as the smallest subgroup containing $S$ (The subgroup $⟨ S ⟩$ generated by a subset, the cyclic subgroup $⟨ g ⟩$ , and cyclic groups). There is no corresponding "largest subgroup contained in $H \cup K$ " construction on the union side, and the failure above is why.
The exact condition. For subgroups $H, K$ of a group $G$ , the union $H \cup K$ is a subgroup if and only if $H \subseteq K$ or $K \subseteq H$ . One direction is immediate, the union then being the larger of the two. For the other, suppose neither inclusion holds and choose $h \in H ∖ K$ and $k \in K ∖ H$ . If $h k$ were in $H$ then $k = h^{- 1} (h k) \in H$ , and if $h k$ were in $K$ then $h = (h k) k^{- 1} \in K$ ; both contradict the choice, so $h k \notin H \cup K$ and the union is not closed.
The witness above is an instance of that criterion: neither $2 Z$ nor $3 Z$ contains the other, since $2 \notin 3 Z$ and $3 \notin 2 Z$ , both by the uniqueness argument of steps 3.1 and 3.2 applied to $2 = 0 \cdot 3 + 2$ and $3 = 1 \cdot 2 + 1$ .

Sources

Standard references

Recommended treatments; not extraction sources.

Monoids, Groups and Subgroups: Examples and Counterexamples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

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},⋅)

Example

Facts & Assumptions

Verification

Remarks

(Z,+)(Z,+) is an abelian group, (Z,⋅)(Z,⋅) is a commutative monoid that is not a group, and its group of units is {1,−1}{1,−1}

Example

Facts & Assumptions

Verification

Remarks

A commutative monoid in which cancellation holds need not be a group: (N,+)(N,+)

Statement refuted

Facts & Assumptions

Counterexample

Remarks

For n≥1n≥1 the congruence classes modulo nn form an abelian group (Z/n,+)(Z/n,+) of order nn, generated by the class of 11

Example

Facts & Assumptions

Verification

Remarks

Sym⁡({1,2,3})Sym({1,2,3}) has exactly six elements, is non-abelian, and its elements have orders 11, 22 and 33

Example

Facts & Assumptions

Verification

Remarks

(gh)n=gnhn(gh)n=gnhn fails without commutativity: two transpositions in Sym⁡({1,2,3})Sym({1,2,3}) with (gh)2≠g2h2(gh)2=g2h2

Statement refuted

Facts & Assumptions

Counterexample

Remarks

The Klein four-group as the subgroup {id,(12)(34),(13)(24),(14)(23)}{id,(12)(34),(13)(24),(14)(23)} of Sym⁡({1,2,3,4})Sym({1,2,3,4}): abelian of order 44, non-cyclic, every non-identity element of order 22

Example

Facts & Assumptions

Verification

Remarks

The eight vertex permutations of a square form a non-abelian subgroup of Sym⁡({1,2,3,4})Sym({1,2,3,4}) of order 88, generated by a 44-cycle and one diagonal swap

Example

Facts & Assumptions

Verification

Remarks

nZnZ is a subgroup of (Z,+)(Z,+) for every n∈Zn∈Z, and every subgroup of (Z,+)(Z,+) has this form

Example

Facts & Assumptions

Verification

Remarks

A nonempty subset of a group closed under the operation need not be a subgroup: the nonnegative integers inside (Z,+)(Z,+)

Statement refuted

Facts & Assumptions

Counterexample

Remarks

FALSE: The union of two subgroups is a subgroup

Statement

Facts & Assumptions

Refutation

Remarks

Sources

Standard references

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

$(Z, +)$ is an abelian group, $(Z, \cdot)$ is a commutative monoid that is not a group, and its group of units is ${1, - 1}$

A commutative monoid in which cancellation holds need not be a group: $(N, +)$

For $n \geq 1$ the congruence classes modulo $n$ form an abelian group $(Z / n, +)$ of order $n$ , generated by the class of $1$

$Sym ({1, 2, 3})$ has exactly six elements, is non-abelian, and its elements have orders $1$ , $2$ and $3$

$(g h)^{n} = g^{n} h^{n}$ fails without commutativity: two transpositions in $Sym ({1, 2, 3})$ with $(g h)^{2} \neq g^{2} h^{2}$

The Klein four-group as the subgroup ${i d, (12) (34), (13) (24), (14) (23)}$ of $Sym ({1, 2, 3, 4})$ : abelian of order $4$ , non-cyclic, every non-identity element of order $2$

The eight vertex permutations of a square form a non-abelian subgroup of $Sym ({1, 2, 3, 4})$ of order $8$ , generated by a $4$ -cycle and one diagonal swap

$n Z$ is a subgroup of $(Z, +)$ for every $n \in Z$ , and every subgroup of $(Z, +)$ has this form

A nonempty subset of a group closed under the operation need not be a subgroup: the nonnegative integers inside $(Z, +)$