Given that $G/H=\{xH:x \in G\}$ is group under operation $(xH)(yH)=(xyH)$ . Then $H$ is normal subgroup of $G$ 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 
 A: 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, $xhx^{-1}\in H$. 
A: The set $G/H$ is well-defined even if $H$ is not normal, because we can just define $G/H$ as the set of all left cosets on $H$. The problem is with the fact that, if $H$ is not normal, this set does not satisfy the group axioms under the operation $(xH)(yH)=(xy)H$. Now, there is probably a way to do this through a proof by contradiction using the group axioms, like you are trying to do. However, I feel like it might be easier to do a more indirect proof using homomorphisms.
Let's assume $G/H$ is a group under the operation $(xH)(yH)=(xy)H$. Then, let us define the homomorphism $\phi : G \rightarrow G/H$ by $\phi(g)=gH$. This is a homomorphism because:
$$\phi(gH)\phi(g_2H)=(gH)(g_2)H=(gg_2)H=\phi(gg_2)$$
Now, what is the kernel of this homomorphism? Well, the identity in $G/H$ is $H$, so we need to solve $\phi(g)=gH=H$. This happens if and only if $g \in H$, so the kernel of $\phi$ is $H$. However, the kernel of any homomorphism is a normal subgroup of the domain, so $H$ is a normal subgroup of $G$.
