Take a group $G$, and consider the set of all its subgroups. Is there a natural way to define multiplication of subgroups, in such a manner that the set forms a group? If so, how is the operation constricted, and what is this group called?
The reason I came up with the question and why it might seem natural is this. A consequence of the isomorphism theorems is that $HH'/H\cong H'$ whenever $H,H'$ are disjoint subgroups and $H$ is a normal subgroup of $G$, and furthermore, if $H,H'$ both happen to be normal subgroups in $G$ with $H\leq H'\leq G$, then $(G/H)/(H'/H)\cong G/H'$. So in some twisted sense, it seems like for at least a certain situations and for certain types of subgroups of $G$, there may be a natural method of defining multiplication that allows us to actually "do arithmetic" over the subgroups of $G$? This is in the sense that, over this proposed group, we can actually just "cancel $H$" in $(G/H)/(H'/H)\cong G/H'$ as a consequence of how multiplication is defined. It seems very interesting (to me, at least) what the consequences of defining such a group might be.
Sorry if the question is vague, but here's an idea of what I'm not looking for: Take any group, consider all its $n$ subgroups, and define multiplication in some arbitrary way, such as the multiplication over $\mathbb Z/n\mathbb Z$. If we do this, we completely lose the fact that these elements started out as subgroups of a group, and the interesting algebraic structure is lost! Instead, I'm interested in whether there is such a way of defining this group of subgroups that preserves the algebraic properties of the original group $G$.
A naïve construction I tried that doesn't work: for two subgroups $H,H'$ of an abelian group $G$, define $HH'=\{hh':h\in H,h'\in H'\}$. It's quite easy to verify this always gives a subgroup of $G$, so we have closure, and associativity and such are easily checked too. The problem is that most subgroups of $G$ have no inverse in this "group" of subgroups for a general group $G$, because for $HH'=1$, we require $hh'=1$ for every $h\in H$, $h'\in H'$ which is not possible in all but the most trivial cases. As noted in the comments, if we removed the hypothesis that $G$ is abelian, then its subgroups need not be normal, and this construction fails even more badly since we don't even have closure anymore. So in general, this construction would not go anywhere close to working.
Any thoughts, or links to something that has already been done, are greatly appreciated!