Can one recover $G$ from $G/H$? I was thinking about a discussion that took place in the comments of some question on this site about quotient groups and isomorphic copies of these.
One of the persons mentioned that in $G/H$ you have more information than in $K$($\simeq G/H$) because in $G/H$ you still have information about $G,H$. 
Indeed, you can recover $H$ as the unit of the group $G/H$ and taking the union, one has $G=\bigcup G/H$ . 
So set-theoretically, "$(G,H) \to G/H$" is injective (I put quotemarks there because $G/H$ is not only a function of the sets $G,H$ but also of multiplication on $G$). 
My question is the following : can you recover the group $G$ knowing the group $G/H$ ? (That is,also recover the multiplication, not only the underlying set)
I guess a precise way to ask the question is : let $C$ be the category of couples $(G,H)$ where $G$ is a group (from here on, this means $G = (E, \cdot)$ where $\cdot$ satisfies the group axioms) and $H$ is a normal subgroup (here $H$ is only the underlying set though) and of group morphisms that preserve the normal subgroup ($f : (G,H) \to (K,N)$ has $f(H)\subset N$) and let $F$ be the functor $C\to \bf{Grp}$ that sends $(G,H)$ to $G/H$ and $f : (G,H) \to (K,N)$ to the canonically associated $\overline{f} : G/H \to K/N$. Is $F$ injective on objects ? 
(Though the functorial analysis here isn't necessary, it seemed like a good way of phrasing it)
This says precisely this : given $G/H$ I know $G, H$, but do I know the multiplication ?
EDIT: As pointed out in the comments, and in Hagen Von Eitzen's answer, the "trivial case" $G/G$ shows that the question as it now stands is trivial. But a more interesting (I hope) question comes out of it: for which normal subgroups can we recover $G$, or in a more global approach, for what subcategories $D$ of $C$ is the restriction of $F$ to $D$ injective on objects ? 
For instance, the (full) subcategory of couples $(G,\{e\})$ has this property (as knowing what $\{a\}\{b\} = a\{e\} b\{e\}$ is in $G/{e}$ tells you what $ab$ is). 
 A: Note that the answer that follows resulted from me misinterpreting the question. I understood "recover $G$" to mean recover $G$ up to an isomorphism that preserves the given group structure on $G/H$. But the intention was to recover the group operation on $G$ exactly. Since, whenever $|H|>1$ there is no way of identifying the identity element of $G$, it is only possible to recover $G$ exactly when $|H|=1$.
For any group structure on $H$, is possible that $G = G/H \times H$. So we can only recover $G$ from $G/H$ if there is a unique group structure on $H$, and if the direct product is the only extension.
There is a unique group structure on $H$ if and only if $H$ is finite and has order equal to either $1$ or a prime. If $|H|=1$ then $G$ is determined by $G/H$, so assume that $|G/H|=p$ is prime.
For the extension to be unique, we require first that there is no nontrivial action of $G/H$ on $H$ or equivalently that the commutator quotient $H_1(G/H)$ of $G/H$ is finite and has order coprime to $p-1$. If that holds, then $H \le Z(G)$.
Finally, we require that there is no nonsplit central extension of $H$ by $G/H$, which is equivalent to $H_1(G/H)$ having order coprime to $p$, and the Schur Multiplier $H_2(G/H)$ of $G/H$ being finite and having order coprime to $p$.
Summing up, the conditions for $G$ to be uniquely determined by $G/H$ are that either (i) $|H|=1$; or (ii) $|H|=p$ is finite and prime, $H_1(G/H)$ is finite of order coprime to $p(p-1)$, and $H_2(G/H)$ is finite of order coprime to $p$.
A: Let $G$ be a group of order two with underlying set $\{a,b\}$ (but I won't tell you which of these is the neutral element).
Then $G/G$ is a group with underlying set $\{\{a,b\}\}$ and trivial operation, $\{a,b\}*\{a,b\}=\{a,b\}$.
Clearly, we can recover the underlying set $\{a,b\}$ from this. But all we know about $G/G$ is symmetric with respect to swapping $a\leftrightarrow b$. 
Consequently, with this knowledge alone we cannot determine which of $a,b$ is the neutral element of $G$.
