Self-duality of the subgroup lattice of $G\times H$

Question

Let $G$ and $H$ be finite groups and $\gcd(|G|,|H|)=1$.

Suppose the lattice of all subgroups of $G\times H$ is self-dual. Is the lattice of all subgroups of $G$ self-dual?

Answer 1

The answer is Yes. This is seen as follows:

First, let us say that a group $G$ has a dual $\overline{G}$ (which is another group), if there is an order-reversing bijection $\delta\colon L(G) \to L(\overline{G})$ from $L(G)$, the subgroup lattice of $G$, onto the subgroup lattice $L(\overline{G})$ of $\overline{G}$. Groups with a dual have been studied by Suzuki and Zacher, and finite groups (even locally finite groups) with a dual are essentially classified. It follows from this classification that a (locally) finite group with a dual is even self-dual, that is, the lattice of subgroups is self-dual. (See Corollary 8.2.5 in Roland Schmidt's book Subgroup Lattices of Groups.)

Thus it suffices to see that $G$ has a dual when the subgroup lattice of $G\times H$ is self-dual. Let $\varepsilon\colon L(G\times H) \to L(G\times H)$ be the duality of the subgroup lattice. Then $\varepsilon$ induces an order-reversing bijection between the subgroups of $G$ and the subgroups of $G\times H$ containing $G^{\varepsilon}$. (I view $G$ and $H$ as subgroups of $G\times H$, so that I can apply $\varepsilon$ to $G$.) I claim that $G^{\varepsilon}$ is normal in $G\times H$. It will then follow that the factor group $(G\times H)/G^{\varepsilon}$ is a dual of $G$.

To see that $G^{\varepsilon}$ is normal, first observe that $G\times H = \langle G^{\varepsilon}, H^{\varepsilon} \rangle$, since $G\cap H = 1$ and $\varepsilon$ is a duality of the subgroup lattice. Thus we only need to show that $H^{\varepsilon}$ normalizes $G^{\varepsilon}$. Let $x\in H^{\varepsilon}$, and let $\varphi$ be the map sending a subgroup $U\leq G\times H$ to the subgroup $( ( U^{\varepsilon} )^x )^{ \varepsilon^{-1} }$. Then $\varphi$ is a lattice isomorphism of $L(G\times H)$ onto itself, and $H^{\varphi}=H$. As $G$ and $H$ have coprime orders, $G$ is the unique complement of $H$ in $G\times H$. Thus $G^{\varphi} = G$, too. It follows that $(G^{\varepsilon})^x = G^{\varepsilon}$. As $x\in H^{\varepsilon}$ was arbitrary, this means that $H^{\varepsilon}$ normalizes $G^{\varepsilon}$, and so $G^{\varepsilon}$ is normal in $G\times H$, as claimed.
(Remark: one can even show that $G\times H$ is the direct product of $G^{\varepsilon}$ and $H^{\varepsilon}$ and that these subgroups have coprime order, too. See Lemma 8.1.8 in Schmidt's book, from which I took the argument in the last paragraph.)