Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27
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.

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

Statement

Let G be a group (Group and abelian group) with identity e and let H⊆G be nonempty. Then H is a subgroup of G (Subgroup) if and only if

gh−1∈Hfor all g,h∈H.

Moreover, if H⊆G is nonempty, closed under the operation of G, and is a group under that restricted operation with some identity element f and some inverse x∗ for each x∈H, then f=e and x∗=x−1 for every x∈H; so H is a subgroup in the sense of Subgroup, and "subgroup" and "subset that is a group under the restricted operation" agree.

Facts & Assumptions

Given: A group G with identity e, and a nonempty subset H⊆G.

[L1]

A subgroup is a subset containing e and closed under the operation and under inverses; it is then a group under the restricted operation, with identity e and with the inverses of G (Subgroup).

[L2]

The group laws in G: associativity, the two-sided identity e, and two-sided inverses (Group and abelian group).

[L3]

(g−1)−1=g (In a group e−1=e, (g−1)−1=g and (gh)−1=h−1g−1, the order of the last product being essential).

[L4]

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

[L5]

Uniqueness of inverses in a monoid, in the sharp form: if x is invertible and yx=e or xy=e, then y=x−1 (In a monoid, a left inverse and a right inverse of the same element are equal; hence an invertible element has exactly one inverse, and it is two-sided).

Proof

technique · direct
1.1L1

Necessity. Suppose H is a subgroup and let g,h∈H. Then h−1∈H by closure under inverses, and gh−1∈H by closure under the operation.

1.2givenL2choose

Sufficiency, the identity. Suppose gh−1∈H for all g,h∈H. Since H is nonempty, choose x∈H; taking g=h=x gives xx−1=e∈H.

1.3givenL2L4

The second claim, the identity. Let H be nonempty, closed under the operation, and a group under the restricted operation with identity f∈H. Then ff=f in H, hence in G; also fe=f in G. Cancelling f on the left in ff=fe gives f=e.

2.1step 1.2givenL2

Sufficiency, inverses. Let h∈H. Taking g=e, which lies in H by step 1.2, gives eh−1=h−1∈H.

2.2step 1.3L5

The second claim, inverses. Let x∈H with inverse x∗∈H for the restricted operation, so x∗x=f=e by step 1.3. Since x is invertible in G, uniqueness of inverses gives x∗=x−1; in particular x−1∈H.

3.1step 2.1givenL3

Sufficiency, products. Let g,h∈H. By step 2.1, h−1∈H, so applying the hypothesis to the pair g and h−1 gives g(h−1)−1∈H, and (h−1)−1=h, so gh∈H.

3.2step 1.3step 2.2L1

Hence such an H contains e by step 1.3, is closed under the operation by assumption, and is closed under inverses by step 2.2: it is a subgroup in the sense of Subgroup.

4.1step 1.1step 1.2step 2.1step 3.1L1

Steps 1.2, 2.1 and 3.1 verify (S1), (S3) and (S2), so H is a subgroup; with step 1.1 this proves the equivalence.

5.1step 4.1step 3.2∎

The one-step test characterises subgroups among nonempty subsets, and a nonempty subset that is a group under the restricted operation is a subgroup with the same identity and the same inverses as G.

Remarks

  • Why the second claim is needed at all. Nothing in the phrase "is a group under the restricted operation" forces the identity of that group to be the identity of G; the hypothesis only says some element acts as an identity within H. Cancellation in G is what collapses the two, and it is available because G is a group. In a monoid the corresponding statement is false: a subset closed under the operation can be a monoid whose identity is not the identity of the ambient monoid, as {0} inside (Z,⋅,1) shows, where 0 is an idempotent acting as an identity on that subset.

  • Nonemptiness cannot be dropped from the one-step test, since the empty set satisfies the condition vacuously and is not a subgroup: it does not contain e.

  • Closure under the operation alone is not enough, even for a nonempty subset: the nonnegative integers inside (Z,+) are closed under addition and contain 0, but are not a subgroup, as recorded on the companion page.

Depends on

Dependency tree · 2 levels

Sources