One-Step Subgroup Test

Theorem
Let $$G$$ be a group and $$H$$ be a nonempty subset of $$G$$. If $$\forall a,b \in H$$, $$ab^{-1} \in H$$, then $$H$$ is a subgroup of $$G$$.

Proof
(Associativity) Since the operation on $$H$$ is the same as that on $$G$$ and since $$G$$ is a group, the operation is associative.

(Identity) Since $$H$$ is nonempty, $$\exists x \in H$$. If we take $$a=x$$ and $$b=x$$, then $$ab^{-1}=xx^{-1}=e \in H$$, where $$e$$ is the identity element.

(Inverses) If we take $$a=e$$ and $$b=x$$, then $$ab^{-1}=ex^{-1}=x^{-1} \in H$$. Thus, $$H$$ is closed under taking inverses.

(Closure) Let $$x,y \in H$$. Then $$y^{-1} \in H$$, so we may take $$a=x$$ and $$b=y^{-1}$$. So, $$ab^{-1}=x(y^{-1})^{-1}=xy \in H$$. Thus, H has closure.

Therefore, H is a subgroup of G.

QED