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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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 $G$

Statement

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

$g h^{- 1} \in H for all g, h \in H .$

Moreover, if $H \subseteq G$ 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 $x \in H$ , then $f = e$ and $x^{*} = x^{- 1}$ for every $x \in H$ ; 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 $H \subseteq G$ .

[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 $(g h)^{- 1} = h^{- 1} g^{- 1}$ , the order of the last product being essential).

[L4]

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

[L5]

Uniqueness of inverses in a monoid, in the sharp form: if $x$ is invertible and $y x = e$ or $x y = e$ , 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

1.1

Necessity. Suppose $H$ is a subgroup and let $g, h \in H$ . Then $h^{- 1} \in H$ by closure under inverses, and $g h^{- 1} \in H$ by closure under the operation.

1.2

Sufficiency, the identity. Suppose $g h^{- 1} \in H$ for all $g, h \in H$ . Since $H$ is nonempty, choose $x \in H$ ; taking $g = h = x$ gives $x x^{- 1} = e \in H$ .

givenL2choose

1.3

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

givenL2L4

2.1

Sufficiency, inverses. Let $h \in H$ . Taking $g = e$ , which lies in $H$ by step 1.2, gives $e h^{- 1} = h^{- 1} \in H$ .

step 1.2givenL2

2.2

The second claim, inverses. Let $x \in H$ with inverse $x^{*} \in H$ 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 $x^{- 1} \in H$ .

step 1.3L5

3.1

Sufficiency, products. Let $g, h \in H$ . By step 2.1, $h^{- 1} \in H$ , so applying the hypothesis to the pair $g$ and $h^{- 1}$ gives $g (h^{- 1})^{- 1} \in H$ , and $(h^{- 1})^{- 1} = h$ , so $g h \in H$ .

step 2.1givenL3

3.2

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.

step 1.3step 2.2L1

4.1

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.

step 1.1step 1.2step 2.1step 3.1L1

5.1

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 $G$ .

step 4.1step 3.2∎

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 $(Z, \cdot, 1)$ 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

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

Subgroup test (Wikipedia) (standard reference, not scraped)
Subgroup (Wikipedia) (standard reference, not scraped)

One-step subgroup test: a nonempty H⊆GH⊆G is a subgroup iff gh−1∈Hgh−1∈H for all g,h∈Hg,h∈H; the identity and the inverses of HH are then those of GG