The subgroup criterion

Forward. If $H$ is a subgroup, it is closed under products and inverses, so $ab^{-1}\in H$.

Converse. Suppose $ab^{-1}\in H$ whenever $a,b\in H$.
- *Identity.* Since $H\neq\emptyset$, take $a\in H$. Then $aa^{-1}=e\in H$.
- *Inverse.* For $b\in H$, $eb^{-1}=b^{-1}\in H$.
- *Closure.* For $a,b\in H$, $b^{-1}\in H$, so $a(b^{-1})^{-1}=ab\in H$.

Conclusion. $H$ is a subgroup.