Prove that aH = bH iff b^-1a in H

Let H be a subgroup of G and a,b \in G. Then aH = bH if and only if b^{-1}a \in H.

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*}

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