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. $\square$