If $H$ be a subgroup of a group $G$ such that $Ha \not=Hb$ implies that $aH\not=bH$.Then how can I show that $gHg^{-1}\subset H$ $\forall$ $g\in G$?
I do not think I have made any progress.However, this is what I have done: For any $h\in H$,$(ghg^{-1})(gh^{-1}g^{-1})=e$ where $e$ is the identity element of the group $G$.
And for $h,k\in H$, $(ghg^{-1})(gkg^{-1})=g(hk)g^{-1}$ which is in $gHg^{-1}$ as $hk\in H$. Thus $gHg^{-1}$ is a group.
I do not think I am getting anywhere.