# Products of maximal inclusions of finite groups with a non-obvious intermediate

Let $(H_1 \subset G_1)$ and $(H_2 \subset G_2)$ be core-free maximal inclusions of finite groups.

Their product, the inclusion $(H_1 \times H_2 \subset G_1 \times G_2)$, admits four obvious intermediate subgroups : $H_1 \times H_2$, $G_1 \times G_2$, $H_1 \times G_2$ and $G_1 \times H_2$.

Question : If $(H_1 \times H_2 \subset G_1 \times G_2)$ admits a non-obvious intermediate subgroup, what are the properties of $(H_1 \subset G_1)$ and $(H_2 \subset G_2)$ ?

Here are examples of "prospective" properties, ordered from the weakest to the strongest :

• $[G_1 : H_1]=[G_2 : H_2]$ ?
• $(H_1 \subset G_1) \sim (H_2 \subset G_2)$ ?
• $(H_1 \subset G_1) \sim (H_2 \subset G_2) \sim (\{e\} \subset \mathbb{Z}_p)$ ?

If I understand correctly, each $H_i$ is maximal in $G_i$ and core-free. In particular, unless $G_i$ has prime order $p$, $H_i$ is not normal in $G_i$.

Let $X$ be a subgroup of $G_1 \times G_2$. For $i=1,2$ let $X^i$ be the projection of $X$ onto $G_i$ and let $X_i$ be the intersection of $X$ with $G_i$. Then $X_i$ is a normal subgroup of $X^i$, and $X^1/X_1 \cong X^2/X_2$. If $X$ contains $H_1 \times H_2$ as in your situation, then each $X_i$ contains $H_i$. As $H_i$ is maximal in $G_i$, $X_i \in \{H_i,G_i\}$. If I have understood your assumptions correctly, unless $G_i$ has prime order $p$, it follows that $X_i=X^i$ (in which case also $X_{3-i}=X^{3-i}$ and $X=X^1 \times X^2$). So, unless each $G_i$ has prime order $p$, the four groups you list are the only intermediate subgroups.

Here is a development of the argument of John Shareshian :

Theorem: Let $(H_i \subset G_i)$ be core-free maximal inclusions of groups, then $(H_1 \times H_2 \subset G_1 \times G_2)$ admits a non-obvious intermediate subgroup iff $G_1 \simeq G_2 \simeq \mathbb{Z}_p$.

Proof: $H_i$ is core-free in $G_i$ (i.e. if $K \subset H_i$ is a normal subgroup of $G_i$ then $K=\{ e \}$), so if $H_i$ is a normal subgroup, then $H_i=\{ e \}$, and $G_i \simeq \mathbb{Z}_p$ ($p$ prime) by maximality assumption.
So $G_i \not\simeq \mathbb{Z}_p$ (for a $p$ prime) iff $H_i$ is not normal in $G_i$.

Let $X$ be a subgroup of $G_1 \times G_2$.

Let $X^1 = \{g_1 \in G_1 : \exists g_2 \in G_2 \text{ with } (g_1,g_2) \in X \}$ the projection of $X$ onto $G_1$.
Let $X_1 = \{h \in G_1 : (h,e) \in X \}$ the intersection of $X$ with $G_1 \times \{e \}$, projected onto $G_1$.

$X_1$ is a normal subgroup of $X^1$ because if $g_1 \in X^1$ and $h \in X_1$, then $g_1^{-1}hg_1 \in X_1$ because $(g_1^{-1}hg_1,e)=(g_1,g_2)^{-1}(h,e)(g_1,g_2) \in X$

By the same way, we define $X^2$ and a normal subgroup $X_2$.
Let $\phi : X^1/X_1 \to X^2/X_2$ with $\phi(g_1X_1)= g_2X_2$ such that $(g_1,g_2) \in X$
It is well-defined because if $(g_1^{-1}g'_1,g_2^{-1}g'_2) \in X_1 \times X_2$, then $(g'_1,g_2), (g_1,g'_2), (g'_1,g'_2) \in X$
It's obviously surjective, and injective because if $\phi(g_1X_1)= X_2$ then $(g_1,e) \in X$ and so $g_1 \in X_1$
Conclusion : $X^1/X_1 \simeq X^2/X_2$

Now if $X$ is an intermediate subgroup of $(H_1 \times H_2 \subset G_1 \times G_2)$ then $H_i \subset X_i$,
and the maximality assumption forces $X_i, X^{i} \in \{ H_i,G_i \}$.

Next if $G_1 \not\simeq \mathbb{Z}_p$ then $H_1$ is not normal in $G_1$, but $X_1$ is normal in $X^1$, so $X_1=X^1=H_1$ or $G_1$.
But, $X^1/X_1 \simeq X^2/X_2$, so $X_2=X^2=H_2$ or $G_2$ as well.
It follows that if $(g_1,g_2) \in X$ then $(g_1,e), (e,g_2) \in X$, so that $X = X_1 \times X_2$
Conclusion $X = H_1 \times H_2$, $G_1 \times H_2$, $H_1 \times G_2$ or $G_1 \times G_2$, i.e. there is not non-obvious intermediate.

Idem if $G_2 \not\simeq \mathbb{Z}_q$, ...

Finally , if $G_1 \simeq \mathbb{Z}_p$ and $G_2 \simeq \mathbb{Z}_q$, there is a non-obvious intermediate subgroup iff $p=q$. $\square$

Definitions: A lattice $(L, \wedge, \vee)$ is :
- Distributive if $a∨(b∧c) = (a∨b) ∧ (a∨c)$
- Modular if $a ≤ c \Rightarrow a ∨ (b ∧ c) = (a ∨ b) ∧ c$
$(\forall a,b,c \in L)$
Obviously, distributive $\Rightarrow$ modular.

Theorems : A finite group $G$ is
- Cyclic iff $\mathcal{L}(G)$ is distributive (Ore 1938)
- Abelian iff $\mathcal{L}(G \times G)$ is modular (Lukacs-Palfy 1986)
(see here thm2.3 p431 and thm6.5 p449)

Corollary: If $(H_i \subset G_i)$ are maximal inclusions of groups, then the lattice of intermediate subgroups $\mathcal{L}(H_1 \times H_2 \subset G_1 \times G_2)$ is modular.

Proof : If $(H_1 \times H_2 \subset G_1 \times G_2)$ admits no non-obvious intermediate subgroup, then the lattice is isomorphic to $\mathcal{L}(\mathbb{Z}_6)$ which is distributive (because $\mathbb{Z}_6$ cyclic), so modular.
Else, the lattice is $\mathcal{L}(\mathbb{Z}_p \times \mathbb{Z}_p)$ and is modular because $\mathbb{Z}_p$ is abelian. $\square$