Let $G$ be a group. Consider the set $P(G)$ of all pairs $(H,N)$ of subgroups of $G$ such that $N$ is a normal subgroup of $H$. Consider the relation $\leq_G$ on $P(G)$ defined as follows: $(H,N)\leq_G(H',N')$ if $N'\subset N$ and $H\subset NH'=\{nh \, | \, n\in N,h\in H'\}$. It is immediate to check that $\leq_G$ is a partial order on $P(G)$.

**Question**: Does this poset have a name? Does it have been studied before?

**Motivation**: Assume that $G$ acts on some space $S$, and there are two well-behaved subsets $A\subset B$ of $S$. We expect that the *external symmetry group* of $B$ is a subgroup of the *external symmetry group* of $A$ and the *internal symmetry group* of $A$ is a subgroup of the *internal symmetry group* of $B$. Basically, the **symmetry group** of $A$ is the subgroup of $G$ which contains that elements preserving $A$ (globally), the **external symmetry group** of $A$ is the subgroup of $G$ consisting of the elements that also fix the structure inside $A$, e.g., fix all points of $A$. The **internal symmetry group** of $A$ is the symmetry group of $A$ on its own; it is obtained by taking quotient of the symmetry group of $A$ by its external symmetry group.

So in above definition, $H$ is the symmetry group of some object $A$, $N$ is its external symmetry group and it should be a normal subgroup of $H$. The internal symmetry group of $A$ is then the quotient group $H/N$. The relation $(H,N)\leq_G(H',N')$ means that the object that has symmetry groups $(H,N)$ is contained in the object that has symmetry groups $(H',N')$. So we expect that $N'\subset N$ and $H/N\subset H'/N'$, which means $H\subset NH'$.

For example, let $S$ be a set and $P_S$ be its permutation group. A subset $A$ of $S$ has symmetry group $P_A \times P_{S/A}$ and external symmetry group $P_{S/A}$. It can be checked that for two subsets $A,B$ of $S$, $A\subset B$ if and only if $P_{S/B}\subset P_{S/A}$ and $P_A\times P_{S/A}\subset P_{S/A}(P_B\times P_{S/B})=P_B\times P_{S/A}$.