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

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.

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 $3 = 1 \cdot 2 + 1$ .

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

37 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)
Union (set theory) (Wikipedia) (standard reference, not scraped)