The strategy is a direct two-way argument using the subgroup properties of $H$. **Step 1: Forward ($aH = bH \implies a^{-1}b \in H$).** Since $b = be \in bH = aH$, there exists $h \in H$ with $b = ah$. Then $a^{-1}b = h \in H$. **Step 2: Backward ($a^{-1}b \in H \implies aH = bH$).** Suppose $a^{-1}b = k \in H$, so $b = ak$. Then $b \in aH$, and also $b \in bH$. Since $b$ belongs to both $aH$ and $bH$, these cosets share an element. By the [Cosets Partition the Group](/theorems/781) theorem (two cosets that share an element must be equal), $aH = bH$.