Let \(H\le G\). Prove that \[ xH=yH\iff x^{-1}y\in H\iff y^{-1}x\in H. \]
Solution
Suppose first that \(xH=yH\). Since \(y\in yH=xH\), there is some \(h\in H\) with \[ y=xh. \] Hence \[ x^{-1}y=h\in H. \] Conversely, if \(x^{-1}y=h\in H\), then \(y=xh\), and therefore \[ yH=xhH=xH. \] Thus \[ xH=yH\iff x^{-1}y\in H. \]
Finally, \[ x^{-1}y\in H\iff (x^{-1}y)^{-1}=y^{-1}x\in H, \] because subgroups are closed under inverses. Hence all three conditions are equivalent.