Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-generatedprecheck 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 H be a nonempty set of subgroups of G (Subgroup). Then the intersection

K  =  ⋂H∈HH  =  { x∈G  :  x∈H for every H∈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 H of subgroups of G, and K the intersection of the members of H.

[L1]

Each H∈H contains e, is closed under the operation, and is closed under inverses (Subgroup).

[L2]

One-step test: a nonempty subset S⊆G with gh−1∈S for all g,h∈S is a subgroup (One-step subgroup test: a nonempty H⊆G is a subgroup iff gh−1∈H for all g,h∈H; the identity and the inverses of H are then those of G).

Proof

technique · direct
1.1

K⊆G, since every member of H is a subset of G and H is nonempty.

givenL1
1.2

e∈K, since e∈H for every H∈H; in particular K is nonempty.

L1
1.3

Let g,h∈K and let H∈H be arbitrary. Then g,h∈H, so h−1∈H by closure under inverses and gh−1∈H by closure under the operation.

L1given
2.1

Since H was arbitrary in step 1.3, gh−1 lies in every member of H, that is gh−1∈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 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 ⟨S⟩ generated by a subset, the cyclic subgroup ⟨g⟩, 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 · two levels

10 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