# Existence of inclusions of finite groups with a particular lattice property

Definition : Let $\sim$ be the equivalence relation on inclusions of finite groups, generated by :
$(H \subset G) \sim (\phi(H) \subset \phi(G))$, with $\phi: G \to L$ a finite group morphism and with $ker(\phi) \subset H$.
(See the appendix below)

Let $(H \subset G)$ be a maximal inclusion of finite groups:

Question : Is there an inclusion of finite groups $(A \subset B)$ having one and only one non-trivial intermediate subgroup $(A \subset P \subset B)$, and with $(A \subset P) \sim (P \subset B) \sim (H \subset G)$ ?
(Unicity of $(A \subset B)$ up to $\sim$ ?)

Remark: obviously, if $H$ is a normal subgroup of $G$ then $(H \subset G) \sim (\{ 1 \} \subset G/H)$ and by maximality, $G/H \simeq \mathbb{Z}_p$ with $p$ prime, so, we can choose $(A \subset B) = (\{ 1 \} \subset \mathbb{Z}_{p^2})$.
Then, we can restrict to $(H \subset G)$ such that $H$ does not contain any non-trivial normal subgroup of $G$.

Examples of $(A \subset B)$ as above, with $(H \subset G)\not\sim (\{ 1 \} \subset \mathbb{Z}_{p})$, are also welcome (if they exist).

Application: the theory of group-subgroup subfactors (see here).

Let $(H \subset G)$ be a maximal inclusion of finite groups, and let $n$ be a positive integer.

Generalization : Is there an inclusion of finite groups $(A \subset B)$ whose lattice of intermediate subgroups is a single chain $A = P_0 \subset P_1 \subset ... \subset P_n = B$ with $(P_i \subset P_{i+1}) \sim (H \subset G)$? (Unicity of $(A \subset B)$ up to $\sim$ ?)

Remark: We can write a remark as above, with $\mathbb{Z}_{p^n}$

Appendix on the equivalence relation $\sim$ :

Let $\sim_1$ generated by $(H \subset G) \sim_1 (H/K \subset G/K)$ with $K \subset H$ a normal subgroup of $G$.

Let $\sim_2$ generated by $(H \subset G) \sim_2 (\phi(H) \subset L)$ with $\phi: G \to L$ isomorphism.

Let $\sim_3$ gen. by $(H_1 \subset G_1) \sim_3 (H_2 \subset G_2)$, $\phi \in Aut(G_1 \times G_2)$, $\phi(H_1 \times G_2) = G_1 \times H_2$.

Obviously:
($\sim_1$ et $\sim_2$) $\Leftrightarrow$ $\sim$
$\sim_3$ $\Rightarrow$ $\sim$
$\sim$ $\not\Rightarrow$ $\sim_3$.

• I don't understand your first remark (in fact I don't believe it). How do you prove it? Feb 15, 2014 at 9:07
• I also had a question about your first remark, that I hoped you might be able to help me with. If $G$ is a perfect group and $H$ a normal subgroup how is $G/H =\mathbb{Z}/p\mathbb{Z}$? Feb 19, 2014 at 21:15
• @NeilHoffman : By assumption, $(H \subset G)$ is a maximal inclusion. But, by definition, $H$ is a maximal subgroup of $G$ if there is no non-trivial intermediate subgroup i.e. $H \subset K \subset G$ implies $K \in \{H, G \}$. Now if in addition $H$ is normal, then $G/H$ is a maximal group, i.e. it admits no non-trivial subgroup. Now let $M$ be a maximal group and $e \neq g \in M$ then $N=\langle g \rangle$ is a subgroup of $M$, so $N=M$ by maximality. Then $M$ is cyclic, so isomorphic to $\mathbb{Z}$ or $\mathbb{Z}_n$ and by maximality isomorphic to $\mathbb{Z}_p$ with $p$ prime. Feb 19, 2014 at 21:42
• See here for a reformulation of $\sim$ using actions on cosets and permutation isomorphism. Feb 19, 2014 at 22:50
• $(\sim \not\Rightarrow \sim_3)$ because if $K \subset H$ is a normal subgroup of $G$ then $(H \subset G) \sim (H/K \subset G/K)$ but in general $H \times G/K \not\simeq G \times H/K$ (for example take $G=\mathbb{Z}_4$ and $H=K=\mathbb{Z}_2$). Feb 19, 2014 at 22:57