In the definition of a quotient group, does the subgroup have to be normal? Let $G$ be a group and let $H \leq G$. Does $H$ need to be a normal subgroup to have the quotient group $G/H$?

Progress
I think yes. By definition, a subgroup $H$ is normal if and only if it is the kernel of some homomorphism. Moreover, $G/H$ makes sense if and only if $H$ is the kernel of some homomorphism. Putting these together, we obtain the answer. Does this seem right?
 A: Yes, normality is required. This answer starts by assuming that $H$ is not normal but that $G/H$ forms a group, and finds a contradiction.
So, suppose $H$ is a subgroup of $G$ but not normal in $G$, and suppose that the multiplication of $G$ inherited by $G/H$ gives a group: $G/H$ with multiplication $(gH)\cdot (hH)=(gh)H$ forms a group.
Firstly, note that the element $H(=1H)$ is the identity of the "group" $G/H$, as $gH\cdot H=gH$ for all $g\in G$ (and in a group, if you have a right identity then it is also a left identity).
Now, as $H$ is not normal in $G$ there exists some $g\in G$ such that $gHg^{-1}\neq H$. Consider $(gH)\cdot(g^{-1}H)$. As the multiplication in $G$ is inherited by $G/H$, this has to be equal to the identity of the "group" $G/H$. Hence, $gHg^{-1}H=H$. This implies that$^{\dagger}$ $gHg^{-1}=H$, a contradiction.

$^{\dagger}$I'll leave the implication as an exercise. You have that the map $H\times H\rightarrow H$, $(h_1, h_2)\mapsto gh_1g^{-1}h_2$ is a bijection. Use this to show that $H\rightarrow H$, $h_1\mapsto gh_1g^{-1}$ is also a bijection.
