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.

  • $\begingroup$ See the dual version of this post here. $\endgroup$ Feb 1, 2015 at 9:21

2 Answers 2


This is not a proper answer, but is a bit too long for a comment. You should look at the following paper:

Groups and Lattices by P.P. Palfy

First, he states a stronger version of Ore's result:

Theorem: Let $G$ be any group. The subgroup lattice $\mathcal{L}(G)$ is distributive if and only if the group $G$ is locally cyclic. (i.e. every finitely generated subgroup of $G$ is cyclic.)

Second he states a theorem (proved by various people at various times, sometimes erroneously):

Theorem If $D$ is a finite distributive lattice, then $D$ is isomorphic (as a lattice) to the lattice of normal subgroups of some finite group $G$.

Finally the result that may be of most interest to you:

Corollary 5.3 For every finite distributive lattice $D$, there exists a finite group $G$ such that, for $H=\{(g,g) \mid g\in G\}$, then the lattice of subgroup inclusions $\mathcal{L}(H\subset (G\times G))$ is isomorphic to $D$.

Perhaps you know all this already...

  • 1
    $\begingroup$ There's also Roland Schmidt's nice book "Subgroup lattices of groups", which goes into a bit more depth on most of these and many related topics. $\endgroup$ Aug 30, 2014 at 1:49
  • $\begingroup$ @RussWoodroofe: yes I've transcribed the proof of the Ore's theorem from this book. Unfortunately, neither this book nor Palfy's paper answer my question. $\endgroup$ Aug 30, 2014 at 9:10

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, Theorem 7 p269.

The following is a new proof.

Lemma 0: The top interval of a distributive lattice, is boolean.
proof: see here.
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.


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