Quotienting kills the subgroups lattice below an element; what kills upwards? Consider a group $G$, and a normal subgroup $H$. Now, the fourth isomorphism theorem tells us that there is a bijection between:


*

*$\{ \text{subgroups $H'$ such that $H \subseteq H'  $} \} \leftrightarrow \{ \text{subgroups of $G/H$} \}$
What is the corresponding operation for:


*

*$\{ \text{subgroups $H'$ such that $H' \subseteq H $} \} \leftrightarrow \{ \text{subgroups  $G \star H$} \}$
What is this operator $\star: Grp \times Grp \rightarrow Grp$? Is there such an operation for any algebraic structure? I can't think of such an operation off-hand, other than the naive operation of $G \star H \equiv H$ (that is, restrict your attention to $H$)
However, perhaps there are some legitimate algebraic structures $H$, where one can find an isomorphic copy of $H$ in $G$ which is not equal to $H$, but I don't know of any such examples off-hand.
 A: You are interested in an operation $\star$ which truncates the lattice of subgroups in a manner dual to the quotienting operation.
That is, we know that there is a bijective correspondence
$$ 
\{ \text{subgroups of } G \text{ containing N} \} 
\longleftrightarrow
\{ \text{subgroups of } G/N \}
$$
Indeed, this bijective correspondence is an isomorphism of lattices when we view the collection of subgroups as a lattice under inclusion.
Is there some operation $\star$ so that we have an isomorphism of lattices
$$
\{ \text{subgroups of } G \text{ contained in H} \}
\longleftrightarrow
\{ \text{subgroups of } G \star H \}?
$$
In the comments I claimed that the only such operation is $G \star H = H$. I'll give three quick arguments for why this should be the case, and I'll also outline a specific group in which such a $\star$ operation exists (though, as we will see, the construction of $\star$ depends heavily on the specifics of the group we construct it for).

First, and simplest justification:
If we take $G \star H = H$, everything works. And moreover, everything works in a canonical way. It is easy to see that for any algebraic structure $\mathscr{A}$ with a substructure $\mathscr{B}$, we will have an analogous isomorphism 
$$
\{ \text{substructures of } \mathscr{A} \text{ contained in } \mathscr{B} \}
\longleftrightarrow
\{ \text{substructures of } \mathscr{B} \}
$$
The fact that this works in this level of generality means we are on to something. 

Second, somewhat more formal justification, but still fairly heuristic:
By "everything works in a canonical way" in the above section, I mean there is a natural isomorphism between the lattices. This can be formalized in the language of category theory, and indeed category theory provides more reasons $G \star H = H$ should be the "correct" dual notion to quotienting. 
The "subgroup inclusion" operation is dual to the "quotient projection" operation. They are equalizers and coequalizers respectively (though for this notion of "dual" to be made perfectly precise, we need to work with abelian groups, since you cannot quotient by non-normal subgroups). If there is another operation $\star$ which also works, it is not as categorically obvious. 
Category theory tends to not unnecessarily duplicate universal-type properties.
This is good heuristic evidence that such a $\star$ is either quite complicated, or quite noncanonical, if it exists at all.

The third, and final, nail in the coffin is to work out what such a $\star$ operation would need to do. I admit that my "counterexample" of symmetric groups in the comments was slightly hasty, so I've given a better one here.
We should be able to apply it to any group in order to get an isomorphism of lattices between $\{ \text{subobjects of } G \text{ contained in } H \}$ and $\text{Sub}(G \star H)$. Of course, we already know that the first thing is really $\text{Sub}(H)$. 
So $\text{Sub}(H)$ and $\text{Sub}(G \star H)$ have to be isomorphic as lattices. 
But it is known that the Klein 4 Group $K$ is the only group having $M_3$ as its lattice of subgroups. So for any group containing $K$ as a subgroup, $G \star K = K$ is forced. Constructions in universal algebra and category theory tend to be "polymorphic" in that they do the same thing regardless of what $G$ and $H$ are. The fact that $H=K$ forces $G \star H = H$ means this equation is probably true for every $H$, since otherwise you need to somehow "case" on what $H$ is, which is substantially less elegant. I suppose this still doesn't fully rule out the existence of such a function $\star$, but it certainly puts its existence on thin ice (especially if you want to find an analogue for any old algebraic theory, as opposed to just groups). 

Now for one special case where such a $\star$ operation does exist, albeit for somewhat silly reasons.
Consider $G = \bigoplus_\mathbb{Z} \mathbb{Z}/2$. That is, it is integer many copies of $\mathbb{Z}/2$. Then any subgroup $H$ of $G$ is isomorphic to each of the $\mathbb{Z}$ possible "shifts" of the subgroup $H$, where the $k$th shift is
$$(x_i~|~i \in \mathbb{Z}) \in H \mapsto (x_{i-k}~|~i \in \mathbb{Z}).$$ 
You can pick any $k$ you like, and define $G \star H$ to be this shift will satisfy what you want. Of course, this structure is highly dependent on the construction of $G$. In particular, nothing like this could work for finite groups.

I hope this helps ^_^
