Proof. This is a straightforward proof. We have that aH = bH. Multiplying both sides with b^{-1}, we get b^{-1}(aH) = b^{-1}(bH), which implies that b^{-1}aH = H. This means that b^{-1}a is an element of H, because subgroups contain the identity element. We could write this in summary:
\begin{align*} aH = bH &\iff b^{-1}(aH) = b^{-1}(bH) \\ &\iff b^{-1}aH = H \\ &\iff b^{-1}a \in H. \end{align*}