**Theorem** ([Øystein Ore, 1938](https://projecteuclid.org/journals/duke-mathematical-journal/volume-4/issue-2/Structures-and-group-theory-II/10.1215/S0012-7094-38-00419-3.short)): A finite group $G$ is cyclic iff its lattice of subgroups $\mathcal{L}(G)$ is [distributive](https://en.wikipedia.org/wiki/Distributive_lattice).  

*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](https://www.gap-system.org/).  

The converse is false: $(S\_2 \subset S\_4)$ is a cyclic inclusion with a non-distributive lattice.




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*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](https://en.wikipedia.org/wiki/Subfactor) by using the coproduct of minimal projections on the $2$-boxes space (see [here](https://mathoverflow.net/q/179179/34538) and [there](https://mathoverflow.net/q/179188/34538)).




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*Remark*: We give here the proof of the stronger version of the theorem, for the l[ocally cyclic groups](https://en.wikipedia.org/wiki/Locally_cyclic_group) (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](https://books.google.fr/books/about/Subgroup_Lattices_of_Groups.html?id=EuVadOnix5MC&redir_esc=y) 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](https://en.wikipedia.org/wiki/Least_common_multiple) and [gcd](https://en.wikipedia.org/wiki/Greatest_common_divisor) are distributive.



Generalization of a theorem of Øystein Ore in group theory
The answer of the main question is **yes**. Let $G$ be a finite group and $H$ a subgroup. 


*Definition*: The group $G$ is called $H$-cyclic if $\exists g \in G$ such that $\langle H,g \rangle = G$.  

Note that: $\langle H,g \rangle = G \Leftrightarrow \langle Hg \rangle = G$.  

**Ore's theorem for intervals** (1938): If the interval $[H,G]$ is distributive, then $G$ is $H$-cyclic.  

*proof*: see [here](http://projecteuclid.org/download/pdf_1/euclid.dmj/1077490633), Theorem 7 p269. 


The following is a new proof.


*Lemma 0*: The top interval of a distributive lattice, is boolean.  

*proof*: see [here](https://mathoverflow.net/q/212340/34538).  

*Lemma 1:* If $H$ is a maximal subgroup then $G$ is $H$-cyclic.  

*proof*: immediate.  

*Lemma 2:* Let $[K,G]$ be the top interval of $[H,G]$. If $G$ is $K$-cyclic then it is $H$-cyclic.  

*proof*: straightforward. 


So the proof of Ore's theorem for intervals reduces to the following Theorem. 


*Theorem*: If the interval $[H,G]$ is boolean, then $G$ is $H$-cyclic.  

*proof*: Let $M$ be a coatom of $[H,G]$, and $M^{\complement}$ its complement. By induction on the rank of the lattice (and Lemma 1), we can assume $M$ and $M^{\complement}$ both $H$-cyclic, i.e. there are $a, b \in G$ such that $\langle H,a \rangle = M$ and $\langle H,b \rangle = M^{\complement}$. Let $g=a b$ then $a=g b^{-1}$ and $b=a^{-1}g$, so $\langle H,a,g \rangle = \langle H,g,b \rangle = \langle H,a,b \rangle = M \vee M^{\complement} = G$.  

Now, $\langle H,g \rangle = \langle H,g \rangle \vee H = \langle H,g \rangle \vee (M \wedge M^{\complement})$, but by distributivity $\langle H,g \rangle \vee (M \wedge M^{\complement}) = (\langle H,g \rangle \vee M \rangle) \wedge (\langle H,g \rangle \vee M^{\complement} \rangle)$.  

 So $ \langle H,g \rangle = \langle H,a,g \rangle \wedge \langle H,g,b \rangle = G$. The result follows. $\square$ 


The converse of Ore's theorem is false because $\langle S\_2,(1234) \rangle = S\_4$ but $[S\_2, S\_4]$ is not distributive. 


