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?

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    $\begingroup$ The condition on the orders of $G$ and $H$ imply that every subgroup of $G\times H$ is of the form $A\times B$ with $A\leq G$ and $B\leq H$. Hence, the lattice of subgroups of $G\times H$ is equal to the product of the lattices of the subgroups of $G$ and of $H$. $\endgroup$ – Arturo Magidin Aug 10 '14 at 22:54
  • $\begingroup$ Yes. But it proves the inverse of the question. btw probably there is a counterexample for the question above. $\endgroup$ – Minimus Heximus Aug 11 '14 at 11:01
  • $\begingroup$ I realize the observation above doesn't answer the question; but it now looks more like: if the product of two lattices is self-dual, is one of the lattices self-dual? And that does indeed look suspect. $\endgroup$ – Arturo Magidin Aug 11 '14 at 15:04

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.)

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  • $\begingroup$ Does the inverse of your "order-reversing bijection" an order-reversing bijection too? $\endgroup$ – Minimus Heximus Jun 8 '15 at 1:01
  • $\begingroup$ and what is $G^{\varepsilon}$? $\endgroup$ – Minimus Heximus Jun 8 '15 at 1:05
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    $\begingroup$ @user47958: I use exponential notation for maps, so $G^{\varepsilon}$ is the image of $G$ under the map $\varepsilon$, denoted $\varepsilon(G)$ in a different notation. Here I implicitely identify $G$ with a subgroup of $G\times H$. "Order-reversing bijection" means that $X \leq Y \iff \varepsilon(X) \geq \varepsilon(Y)$ for subgroups $X$, $Y\leq G\times H$. With $X= \varepsilon^{-1}(U)$ and $Y=\varepsilon^{-1}(V)$, this yields that $\varepsilon^{-1}$ is order-reversing too. $\endgroup$ – Frieder Ladisch Jun 8 '15 at 12:26
  • $\begingroup$ Thanks. It seems correct. Still I have to check some details. But I accept it. $\endgroup$ – Minimus Heximus Jun 8 '15 at 15:18

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