**Theorem** (Øystein Ore, 1938): A finite group $G$ is cyclic iff its lattice of subgroups $\mathcal{L}(G)$ is distributive.

*Proof*: see below.

Let $(H \subset G)$ be an inclusion of finite groups and $\mathcal{L}(H \subset G)$ its lattice of intermediate subgroups.

I would like to generalize the above theorem of Øystein Ore to the inclusions of finite groups, i.e. find an equivalent formulation of the following property $(D)$, in term of "cyclic" notions. $$(D) \ \ \ \ \mathcal{L}(H \subset G) \text{ is distributive } $$ *Definition*: An inclusion of finite groups $(H \subset G)$ is **cyclic** if it checks $(C_0)$.

$$(C_0) \ \ \ \exists g \in G \text{ such that } \langle H,g \rangle = G$$ *Examples*: The maximal inclusions are cyclic, and $(\{e\} \subset G)$ is cyclic iff $G$ is cyclic.

**Question**: $(D)$ $\Rightarrow$ $(C_0)$ ?

*Remark*: I've checked that it's true for index $[G:H]<32$ and $\vert G \vert \le 5.10^5$ with GAP.

The converse is false: $(S_2 \subset S_4)$ is a cyclic inclusion with a non-distributive lattice.

*Bonus question*: How complete $(C_0)$ for having an equivalence with $(D)$?

I've tried all the following completions of strictly increasing strongness:

- $(C_1) \ $ $\forall K$, $H \le K \le G$, $\exists g \in G$ such that $\langle H,g \rangle = K$
- $(C_2) \ $ $\forall K$, $H \le K \le G$, $\exists g \in G$, $\exists n \ge 0$ such that $\langle H,g \rangle = G$ and $\langle H,g^n \rangle = K$
- $(C_3) \ $ $\exists g \in G$, $\forall K$, $H \le K \le G$, $\exists n \ge 0$ such that $\langle H,g^n \rangle = K$

*Conjecture*: $(D)$ is strictly between $(C_1)$ and $(C_3)$.

I've checked it for index $[G:H] < 32$ and $\vert G \vert \le 10^4$. Moreover $(D)$ is *not orderable* with $(C_2)$.

*Remark*: All these notions and questions can be extended to the theory of subfactors by using the coproduct of minimal projections on the $2$-boxes space (see here and there).

*Remark*: We give here the proof of the stronger version of the theorem, for the locally cyclic groups (i.e. every finitely generated subgroup is cyclic), because it's more relevant for an attempt of generalization.

**Proof** of the theorem of Øystein Ore (coming from this book p12-13):

Suppose first that $\mathcal{L}(G)$ is distributive and let $a,b \in G$. We have to show that $\langle a,b \rangle$ is cyclic. Because $\langle a \rangle \wedge \langle b \rangle$ is centralized by $a$ and $b$, $(\langle a \rangle \wedge \langle b \rangle) \leq Z(\langle a,b \rangle)$. Also,$\langle ab \rangle \vee \langle a \rangle = \langle a,b \rangle = \langle ab \rangle \vee \langle b \rangle$, and then by distributivity $$ \langle ab \rangle \vee (\langle a \rangle \wedge \langle b \rangle) = (\langle ab \rangle \vee \langle a \rangle) \wedge (\langle ab \rangle \vee \langle b \rangle) = \langle a,b \rangle $$ So, $\langle a,b \rangle / (\langle a \rangle \wedge \langle b \rangle) \simeq \langle ab \rangle /(\langle ab \rangle \wedge (\langle a \rangle \wedge \langle b \rangle) )$ is cyclic, and then $\langle a,b \rangle$ is abelian, as cyclic extension of a central subgroup. By the structure of finitely generated abelian groups, there are $c,d \in G$ such that $\langle a,b \rangle = \langle c \rangle \times \langle d \rangle$. As we've already shown, $\langle c,d \rangle / \langle c \rangle \wedge \langle d \rangle$ is cyclic. Because $\langle c \rangle \wedge \langle d \rangle = 1$, $\langle a,b \rangle = \langle c,d \rangle $ is cyclic.

Now suppose that $G$ is locally cyclic and let $A,B,C \in \mathcal{L}(G)$. Because $G$ is abelian,

we just need to verify that $A(B \wedge C) = AB \wedge AC$. Clearly, $A(B \wedge C) \leq AB \wedge AC$.

Let $x \in AB \wedge AC$, then $x=ab=a'c$ with $a,a' \in A$, $b \in B$ and $c \in C$.

Because $G$ is locally cyclic, $\exists g \in G$ such that $\langle a,a',b,c \rangle = \langle g \rangle$. Next, $ab=a'c$ implies that $\langle g \rangle = (A \wedge \langle g \rangle)(B \wedge \langle g \rangle)=(A \wedge \langle g \rangle)(C \wedge \langle g \rangle)$. If one of the three subgroups $A \wedge \langle g \rangle$, $B \wedge \langle g \rangle$, $C \wedge \langle g \rangle$ is trivial, then $x=b=c \in B \wedge C$ or $x \in A$. In each case, $x \in A(B \wedge C)$.

So suppose that all these subgroups are non-trivial and let $n,r,s$ the respective indices of $A \wedge \langle g \rangle$, $B \wedge \langle g \rangle$, $C \wedge \langle g \rangle$ in $\langle g \rangle$. So $(n,r) = 1 = (n,s)$, and then $(n,rs) = 1$ and so $\langle g \rangle = \langle g^n \rangle \langle g^{rs} \rangle = (A \wedge \langle g \rangle) (B \wedge C \wedge \langle g \rangle) \leq A(B \wedge C)$.

Again, it follows that $x \in A(B \wedge C)$. So $A(B \wedge C) = AB \wedge AC$ as expected. $\square$

*Remark*: For a finite cyclic group $G$, the second part of the proof can be reduced to the facts that $G$ has exactly one subgroup of order $d$ for every divisor $d$ of $ord(G)$, and that lcm and gcd are distributive.