Given that $G/H=\{xH:x \in G\}$ is group under operation $(xH)(yH)=(xyH)$ . Then $H$ is normal subgroup of $G$

Question

let $G$ is any group and $H$ is its subgroup , such that $G/H=\{xH:x \in G\}$ is group under operation $(xH)(yH)=(xyH)$ . Then show that $H$ is normal subgroup of $G$

if $H$ is not normal ,then there exist some $x\in G$ such that $xH \neq Hx$, clearly $x\notin H$ now how can i get a contradiction . I know then operation on $G/H$ is not well defined. but how to prove that ?

any hint. Thanks in advanced

Answer 1

We clearly have $xhH=xH$ for $x\in G, \ h\in H$. Then, by being well defined, we arrive to $$(xhx^{-1})H=xhH\cdot x^{-1}H=xH\cdot x^{-1}H=(xx^{-1})H=eH=H$$ Thus, $xhx^{-1}\in H$.