**Proof plan.** The forward direction is immediate from the subgroup axioms. For the converse, we recover the identity (by setting $a = b$), then inverses (by setting $a = e$), and finally closure (by combining inverses with the hypothesis). Each step uses only the single condition $ab^{-1} \in H$. **Step 1 (Identity).** Since $H$ is non-empty, there exists $g \in H$. Setting $a = g$ and $b = g$ gives $e = gg^{-1} \in H$. **Step 2 (Inverses).** For any $a \in H$, setting $b = a$ in the hypothesis gives $a^{-1} = ea^{-1} \in H$. **Step 3 (Closure).** For $a, b \in H$, Step 2 gives $b^{-1} \in H$, so the hypothesis applied to $a$ and $b^{-1}$ yields $ab = a(b^{-1})^{-1} \in H$.