If $G/H$ is a group, then does $H$ have to to be normal? If $H$ is normal then $G/H$ is a group. But is the converse true? That is, if $G/H$ is a group, does this mean that $H$ is normal?
Or are there any counterexamples?
 A: The notation $G/H$ is ambiguous unless $H$ is normal; thus, $G/H$ is well-defined if and only if $H$ is normal.
Proof:
When $H$ is not normal, there's an issue of whether we use left or right cosets in $G$. Lets say we pick left cosets:
$$
\{gH\colon g\in G\}.
$$
If we want the multiplication operation on this set to be compatible with that of $G$, we are forced to define:
$$
aH\cdot bH\equiv (ab)H.
$$
For this to be well-defined, we need that for every $h_1,h_2\in H$, there is some $h_3\in H$ such that
$$
ah_1bh_2=abh_3.
$$
Let's choose $h_2=e$. Then $ah_1b=abh_3$, and cancelling the $a$ yields $h_1b=bh_3$. Thus the cosets $Hb$ and $bH$ are equal. Since $b$ was arbitrary, it follows that $H$ is normal.
A: I think the question is this: Does there exist a non-normal subgroup $H \subset G$, such that $(xH)(yH) := (xy)H$ makes $G/H$ (which is nowhere ambiguous, because it is just the set of cosets) into a well-defined group?
The answer is no:
Let $H$ be not normal, hence there is a $a \in G$ with $aH \neq Ha$. Take $b \in aH \setminus Ha$.
On the one hand, we have $(bH)(a^{-1}H) = (aH)(a^{-1}H) = (aa^{-1})H=H$.
On the other hand, we have $(bH)(a^{-1}H) = (ba^{-1})H \neq H$, since $ba^{-1} \notin H$ by the assumption $b\notin Ha$.
Thus the induced group structure is not well-defined.
Of course you could define any other group-structure on $G/H$, which is not induced by the one on $G$.
A: There are some good answers here.
One thing to think about is that it isn't just $G/H$ being a group we care about; in general, for any subgroup $H \subset G$ we can define the set of cosets $G/H$. This is a set, and so we can define some group structure on it however we like. So in a certain sense, $G/H$ can always be a group regardless of the normality of $H$.
However, what we want is more than that: we want a compatibility between the group operations; that is, we want the group structure on $G$ and on $G/H$ to be related. Specifically, we want that the natural set-theoretic function
$$
\phi : G \to G/H
$$
should be a homomorphism of groups. This is only true if $H$ is normal.
A: If $G/H$ is a group, then the natural map $\phi: G \longrightarrow G/H$ has $H$ as its kernel. 
Claim: any subgroup appearing as the kernel of a group homomorphism is a normal subgroup.
Proof: Call $H$ our kernel subgroup. Then we want to show that $ghg^{-1} \in H$. But then $\phi(ghg^{-1}) = \phi(g) \phi(h) \phi(g)^{-1} = \phi(g)\phi(g)^{-1} = e$ the identity element, and thus $ghg^{-1}$ is also in the kernel. Thus the kernel of a homomorphism is a normal subgroup. $\diamondsuit$
This necessarily gives that $H$ is normal.
By the way, this is also iff. So if $H$ is a normal subgroup, then it appears as the kernel of the natural descent map. But implicit within this statement is that $G/H$ is a group.
