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 $G$

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.

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 $x x^{-1} = e \in H$.

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 $ff = fe$ gives $f = e$.

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

The second claim, inverses. Let $x \in H$ with inverse $x^{\ast} \in H$ for the restricted operation, so $x^{\ast} x = f = e$ by step 1.3. Since $x$ is invertible in $G$, uniqueness of inverses gives $x^{\ast} = x^{-1}$; in particular $x^{-1} \in H$.

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 $gh \in H$.

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.

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.

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 $G$.