Subgroup lattice isomorphic to the power set lattice If $G$ is a group, we denote by $\text{Sub}(G)$ the lattice of all subgroups of $G$, ordered by $\subseteq$. Given a cardinal $\kappa$, is there a group $G$ with $\text{Sub}(G) \cong {\cal P}(\kappa)$ where ${\cal P}(\kappa)$ denotes the power set lattice of all subsets of $\kappa$?
 A: The only such groups are of the form $\bigoplus_{p\in I}C_p$ for some set $I$ of primes. So the only possible $\kappa$ are the countable ones.
Indeed, given such $G$, one can identify $\kappa$ to the set of minimal subgroups, and every nontrivial subgroup contains a minimal one (hence $G$ is torsion). Given two minimal subgroups $C\neq C'$ (hence both of prime order), it follows from the assumption that the subgroup $H$ generated by $C,C'$ has only 4 subgroups: $H,C,C'$ and $1$. First, this forces $H$ to have finitely many subgroups and hence it's finite. Moreover it readily follows that $H=C\oplus C'$ and $C,C'$ have distinct prime order (if $|C|\neq |C'|$, then one identifies $C$ and $C'$ as Sylow subgroups and ok; if $|C|=|C'|=p$, if $H$ has order $\ge p^3$ one has more subgroups, and for $|H|=p^2$ one gets a contradiction since the number of subgroups is either $3$ or $p+3\neq 4$). Finally, $G$ has no cyclic subgroup of non-prime $p$-power order, since one would get a finite initial segment in the lattice not of the right form. Considering $x\in G$, it has finite order and hence is a product of elements of prime order. Since the minimal subgroup pairwise commute, $G$ is abelian and hence is a direct sum as described.

Edit: here's a more general result. Suppose that the lattice of subgroups satisfies (a) for all $a<b<c$, the interval $[a,c]=\{d:a\le d\le c\}$ is not a chain. (b) for all $a\le b,a\le c$ such that both $[a,b]$ and $[a,c]$ are finite, the interval $[a,a\vee c]$ is finite. (c) for all $a\le b,a\le c$ with $b\neq c$ such that $[a,b]=\{a,b\}$ and $[a,c]=\{a,c\}$, we have $[a,b\vee c]=\{a,b,c,b\vee c\}$. Then the group has the above form.
(This applies when the lattice of subgroups is a Boolean algebra, even without assuming it to have atoms.)
Indeed, first (b) implies that for no prime $p$, $\mathbf{Z}/p^2\mathbf{Z}$ is not a isomorphic to a subquotient of $G$. In particular, $G$ is torsion, and all elements have square-free order. This implies that $G$ is generated by its subgroups of prime order. Next, (b) implies that any finite union of cyclic subgroups of prime order is contained in a finite subgroup (since a group with finitely many subgroups has to be finite.) Finally, (c) implies (as already done in the previous paragraph) that any two cyclic subgroups of prime order have distinct order and centralize each other. Whence the conclusion.
