CounterexampleConstruction: 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.
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.

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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.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

34 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

Subgroup (Wikipedia) (standard reference, not scraped)
Submonoid (Wikipedia) (standard reference, not scraped)
Cancellation property (Wikipedia) (standard reference, not scraped)

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