LemmaStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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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.

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The intersection of a nonempty family of subgroups of $G$ is a subgroup of $G$

Statement

Let $G$ be a group (Group and abelian group) and let $\mathcal{H}$ be a nonempty set of subgroups of $G$ (Subgroup). Then the intersection

$K \;=\; \bigcap_{H \in \mathcal{H}} H \;=\; \{\, x \in G \;:\; x \in H \text{ for every } H \in \mathcal{H} \,\}$

is a subgroup of $G$ . In particular the intersection of two subgroups is a subgroup.

Facts & Assumptions

Given: A group $G$ with identity $e$ , a nonempty set $\mathcal{H}$ of subgroups of $G$ , and $K$ the intersection of the members of $\mathcal{H}$ .

[L1]

Each $H \in \mathcal{H}$ contains $e$ , is closed under the operation, and is closed under inverses (Subgroup).

[L2]

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

Proof

technique · direct

1.1

$K \subseteq G$ , since every member of $\mathcal{H}$ is a subset of $G$ and $\mathcal{H}$ is nonempty.

givenL1

1.2

$e \in K$ , since $e \in H$ for every $H \in \mathcal{H}$ ; in particular $K$ is nonempty.

1.3

Let $g, h \in K$ and let $H \in \mathcal{H}$ be arbitrary. Then $g, h \in H$ , so $h^{-1} \in H$ by closure under inverses and $g h^{-1} \in H$ by closure under the operation.

L1given

2.1

Since $H$ was arbitrary in step 1.3, $g h^{-1}$ lies in every member of $\mathcal{H}$ , that is $g h^{-1} \in K$ .

step 1.3

3.1

$K$ is a nonempty subset of $G$ satisfying the one-step test, hence a subgroup of $G$ .

step 1.1step 1.2step 2.1L2∎

Remarks

The hypothesis that $\mathcal{H}$ is nonempty is load bearing. The intersection of the empty family of subsets of $G$ is not a subset of $G$ by any convention used here; the statement is made for a nonempty family so that step 1.1 is available.
This lemma is what makes The subgroup $\langle S \rangle$ generated by a subset, the cyclic subgroup $\langle g \rangle$ , and cyclic groups legitimate: the family of subgroups containing a given subset $S$ is nonempty, since $G$ itself belongs to it, so its intersection is a subgroup, and it is by construction the smallest subgroup containing $S$ .

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 14 results over 12 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

Subgroup (Wikipedia) (standard reference, not scraped)

The intersection of a nonempty family of subgroups of GGG is a subgroup of GGG