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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 $G \simeq H$. $\square$.

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  • $\begingroup$ Obviously, if Q1 and G2 are true, then G1 (and Q2) is also true. $\endgroup$ – Sebastien Palcoux Feb 18 '14 at 12:13
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    $\begingroup$ 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 $|I|$. $\endgroup$ – John Shareshian Feb 19 '14 at 4:41

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