# Poset defined on pairs of subgroups

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 $$P_A\times P_{S/A}\subset P_{S/A}(P_B\times P_{S/B})=P_B\times P_{S/A}$$.