# Does $[H_i , G_i]$ distributive imply $[H_1 \times H_2, G_1 \times G_2]$ modular?

Let $$L(G)$$ be the subgroup lattice of $$G$$ and $$[H, G]$$ an interval in $$L(G)$$.

A lattice $$(L, \wedge, \vee)$$ is distributive if $$a∨(b∧c) = (a∨b) ∧ (a∨c)$$, $$\forall a,b,c \in L$$, and is modular if we just have $$a ≤ c \Rightarrow a ∨ (b ∧ c) = (a ∨ b) ∧ c.$$ Obviously, distributive implies modular.

Theorem (Ore, 1938): A group $$G$$ is locally cyclic iff $$L(G)$$ is distributive.
Theorem (Lukacs-Palfy 1986): A group $$G$$ is abelian iff $$L(G \times G)$$ is modular.

Corollary: If $$L(G)$$ is distributive then $$L(G \times G)$$ is modular.

We wonder whether the above corollary can be extended to intervals:
Question: Does $$[H_i , G_i]$$ distributive imply $$[H_1 \times H_2, G_1 \times G_2]$$ modular?

Examples: If $$H_i$$ is a maximal subgroup of $$G_i$$ then $$[H_i, G_i]$$ is (obviously) distributive, and $$[H_1 \times H_2, G_1 \times G_2]$$ is modular (see the corollary here).

Question 2: Does $$[H_i \times H_i , G_i \times G_i]$$ modular imply $$[H_1 \times H_2, G_1 \times G_2]$$ modular?

Remark : $$[H_i, G_i]$$ modular does not imply $$[H_1 \times H_2, G_1 \times G_2]$$ modular, in general. For example, $$L(Q_8)$$ is modular but $$L(Q_8 \times Q_8)$$ is not, because the quaternion group $$Q_8$$ is not abelian.

Generalization of Q1: Does $$[H_i, G_i]$$ distributive imply $$[\prod_{i \in I} H_i, \prod_{i \in I} G_i]$$ modular?

Generalization of Q2: Does $$[H_i \times H_i , G_i \times G_i]$$ modular imply $$[\prod_{i \in I} H_i , \prod_{i \in I} G_i]$$ modular?
(we can assume $$G_i$$ and $$I$$ finite if necessary)

Optional question : Is $$[\prod_{i} H_i , \prod_{i} G_i] \simeq \prod_{i} [H_i , G_i]$$ if $$|G_i:H_i|$$ are pairwise relatively prime?

Proof of the theorem of Ore: see the transcript here.

Lemma (see here): Let $$G$$ be a group, then its normal subgroup lattice is modular.

Proof of the theorem of Lukacs-Palfy (coming from this paper): If $$G$$ is abelian, then $$G \times G$$ is abelian, and every subgroup is normal, so by the lemma above, $$L(G \times G)$$ is modular.

Now suppose that $$L(G \times G)$$ is modular.
First we show that each subgroup $$H$$ of $$G$$ is normal. Consider the following subgroups of $$G \times G$$ : $$X=H[H,G] \times 1$$, $$Y = \{(g,g) : g \in G \}$$, $$Z = H \times 1$$. Then $$X \ge Z$$, hence by modularity $$X \wedge (Y \vee Z) = (X \wedge Y) \vee Z$$. Now $$X \le Y \vee Z$$ and $$X \wedge Y = 1$$, so we have $$X=Z$$, thus $$[H,G] \le H$$, i.e. $$H \triangleleft G$$.
Now it follows that $$[x,y] \in \langle x \rangle \wedge \langle y \rangle$$ for any $$x,y \in G$$. Let $$[x,y] = x^k$$ then $$y^{-1}xy = x^{k+1}$$ and $$x^k = y^{-1}x^k y = x^{(k+1)k}$$ so $$x^{k^2} = 1$$. If $$x$$ has infinite order then $$k=0$$ and $$[x,y] = 1$$. Since elements of finite relatively prime orders obviously commute, it remains to prove that if $$x$$ and $$y$$ have $$p$$-power orders ($$p$$ prime) then $$[x,y] = 1$$. We may assume that $$\vert x \vert \ge \vert y \vert$$ and let $$\phi : \langle y \rangle \to \langle x \rangle$$ be a one-to-one homomorphism which is the identity on $$\langle z \rangle = \langle y \rangle \wedge \langle x \rangle$$. Now consider the following subgroups of $$G \times G$$ : $$X = \langle (x,x),(1,[x,y]) \rangle$$, $$Y = \langle (\phi(y),y) \rangle$$ and $$Z = \langle (x,x) \rangle$$. Then $$X \ge Z$$, hence by modularity $$X \wedge (Y \vee Z) = (X \wedge Y) \vee Z$$. Now $$X \le Y \vee Z$$ and $$X \wedge Y \le$$ $$(\langle x \rangle \times \langle x \rangle) \wedge Y \le \langle (z,z) \rangle \le Z$$, so we have $$X = Z$$, thus $$[x,y] = 1$$ in this case as well. $$\square$$

Corollary: Let $$G$$ be a finite abelian group, $$H$$ a group. If the subgroup lattices of $$G \times G$$ and $$H \times H$$ are isomorphic then $$G \simeq H$$.
proof: The subgroup lattices of the square are modular, hence $$H$$ is abelian by our theorem. The direct decomposition of $$G \times G$$ into a product of cyclic subgroups of prime power orders corresponds to a direct decomposition of $$H \times H$$. It is easy to see that the corresponding cyclic factor have equal orders, hence $$G \times G \simeq H \times H$$, so $$G \simeq H$$. $$\square$$.

• Obviously, if Q1 and G2 are true, then G1 (and Q2) is also true. – Sebastien Palcoux Feb 18 '14 at 12:13
• I think you can get a positive answer to the optional question using the ideas about projections and intersections from my answer to a previous question of yours, at least when the index set $I$ is finite. Consider first the case $|I|=2$. As the two Projection/Intersection quotients in any intermediate subgroup are isomorphic but have coprime orders, they are both trivial. Now use induction on $|I|$. – John Shareshian Feb 19 '14 at 4:41