Let $G$ be a group and let $H$ be a subgroup. If $H$ is normal in $G$, then $G/H$ has a group structure. But in general, can there be a groupoid structure on $G/H$(left cosets or right cosets) that generalizes the normal case?



Groupoid structure on G/H?
One answer to your question is that there is always the notion of an "action groupoid", although this does not reproduce the *group* structure on $G/H$ when $H$ is normal.


Let $G$ be a group acting on a set $X$. (There are generalizations when both $X,G$ are groupoids.) Then the **action groupoid** $X//G$ is the groupoid with objects $X$, and morphisms $X\times G$. More precisely, if $x,y \in X$, then $\hom(x,y) = \{ g\in G \text{ s.t. } gx = y\}$. The groupoid axioms are essentially obvious.


For example, let $X = G/H$ be the set of left $H$-cosets. Then the action groupoid is very simple: it is connected (any object is isomorphic to any other), with $\text{aut}(e) = H$, where $e = eH$ is the trivial left $H$-coset. And $\hom(e,g) = gH$, where $g = gH$ is a coset.
 So as a discrete groupoid, this action groupoid is equivalent to $\{\text{pt}\}//H$, also sometimes called the "classifying groupoid" $\mathcal B H$, because the geometric realization of the nerve of this groupoid is the usual classifying space of $H$.


In short-hand, we have the following equation:
$$ (G/H) //G \cong 1//H$$
where $\cong$ denotes equivalence of groupoids. (Actually, since I'm talking about left actions, I should probably write $G \backslash \backslash X$ for the action groupoid, and so the equation really should be $G \backslash \backslash (G/H) = H \backslash \backslash 1$, but typing "/" is much faster than typing "\backslash", so I won't use the better notation.)


But you are probably asking a different question. Recall that when $H$ is normal, then $G/H$ has a group structure, which is to say there is a groupoid with one object and whose morphisms are elements of $G/H$. Of course, as you know, if $H$ is not normal, then $G/H$ does not have a natural group structure, because in general $g\_1Hg\_2H$ is not a left $H$-coset.


You can try to do the following. Any set is naturally a groupoid with only trivial morphisms, and then the set $G/H$ is equivalent to the groupoid $G//H$, where $H$ acts on the set $G$ by translation = right multiplication. (This is because $H$ acts on $G$ freely.) But $G$ is actually more than a set: it is a group. So let's think about it as a "groupal groupoid" or "2-group", i.e. a 2-groupoid with only one object; in this case, it will also only have identity 2-morphisms.


Then I guess you should try to form the "action 2-group" or something, by adding 2-morphisms for the translations by $H$. But I think that if you do, you no longer have a groupal groupoid: I think that if $H$ is not normal, then the group multiplication is not a functor from the action groupoid $G//H$.


The other only thing I can think of is to define $K = \text{Norm}\_GH$, the normalizer, and then $K/H$ is a group that embeds in $G/H$, so let the objects of your groupoid be cosets of $K$ and the morphisms given by $K/H$?


So, long story short: in the way that I think you are hoping, no, $G/H$ is not naturally a groupoid.


