In joint work with Martha Kilpack of BYU, we've been looking at the lattice of closure operators on subgroup lattices.

Recall that if $P$ is a partially ordered set, then a function $\phi\colon P\to P$ is a closure operator on $P$ if and only if (i) $\phi$ is increasing: $a\leq \phi(a)$ for all $a\in P$; (ii) $\phi$ is isotone: if $a\leq b$, then $\phi(a)\leq \phi(b)$; and (iii) $\phi$ is idempotent: $\phi(\phi(a))=\phi(a)$ for all $a\in \phi$.

We can place a partial order on the closure operators on $P$, by saying $\phi\leq \psi$ if and only if $\phi(a)\leq \psi(a)$ for all $a\in P$.

If $P$ is a lattice (any two elements $a$ and $b$ have a least upper bound $a\vee b$ and a greatest lower bound $a\wedge b$), then the closure operators on $P$ form a lattice themselves, setting $(\phi\wedge\psi)(a) = \phi(a)\wedge\psi(a)$, and $(\phi\vee\psi)(a) = \wedge \theta(a)$, where $\theta$ ranges over all closure operators $\theta\geq \phi$ and $\theta\geq \psi$.

It is known that every finite lattice is isomorphic to a sublattice of a lattice of subgroups of a (possibly infinite) group, $\mathrm{sub}(G)$; it is also known that every finite lattice is isomorphic to an \textit{interval} $[H,K]$ in a lattice of subgroups of some (possibly infinite) group $G$ (where $[H,K]$ is the sublattice of all subgroups $M$ with $H\leq M\leq K$). It is an important open question whether every finite lattice is an interval in the lattice of a subgroups of a finite group $G$.

We asked: for which lattices $L$ is the lattice of closures operators on $L$, $\mathrm{co}(L)$, isomorphic to the lattice of subgroups of a group, $\mathrm{sub}(G)$?

We began by looking at finite lattices $L$ that were themselves of the form $\mathrm{sub}(H)$ for some finite group $H$. We proved:

Theorem. The lattice of closure operators on a finite subgroup lattice, $\mathrm{co}(\mathrm{sub}(H))$ is isomorphic to a subgroup lattice if and only if $H$ is cyclic of prime power order.

Note that $\mathrm{sub}(H)$, when $H$ is cyclic of prime power order, is a finite chain. We then extended this result to the case of the lattice of closure operators on any finite lattice:

Theorem. Let $L$ be a finite lattice. Then $\mathrm{co}(L)$ is isomorphic to a subgroup lattice if and only if $L$ is a finite chain.

For infinite lattices things get more complicated, since $\mathrm{sub}(G)$ is always an algebraic lattice: every subset has a least upper bound and greatest lower bound, and every element is the least upper bound for a (possibly infinite) set of compact elements; an element $c$ is compact if, whenever it is bounded above by a least upper bound of a set $X$, there is a finite subset $X_0$ of $X$ such that $c\leq \mathrm{lub}(X_0)$. For lattices of the form $\mathrm{sub}(G)$, a closure operator is algebraic if and only if $\phi(H)$ is the join of the $\phi(K)$, where $K$ ranges over all finitely generated subgroups of $H$. My co-author had proven that if $L$ is an algebraic lattice, then the collection of all algebraic closure operators on $L$, $\mathrm{aco}(L)$, is also an algebraic lattice. So the question then becomes: for which groups $G$ is $\mathrm{aco}(\mathrm{sub}(G))$ a subgroup lattice?

We've proven that if $G$ is infinite and torsion, then $\mathrm{aco}(\mathrm{sub}(G))$ is a subgroup lattice if and only if $G$ is the Prufer $p$-group; and that if $G$ has both nontrivial torsion and torsionfree elements, then $\mathrm{aco}(\mathrm{sub}(G))$ is never a subgroup lattice. We are still working on the infinite torsionfree case.