TITLE: Does the hypergraph of subgroups determine a group?
QUESTION [20 upvotes]: A hypergraph is a pair $H=(V,E)$ where $V\neq \emptyset$ is a set and $E\subseteq{\cal P}(V)$ is a collection of subsets of $V$. We say two hypergraphs $H_i=(V_i, E_i)$ for $i=1,2$ are isomorphic if there is a bijection $f:V_1\to V_2$ such that $f(e_1) \in E_2$ for all $e_1\in E_1$, and $f^{-1}(e_2) \in E_1$ for all $e_2\in E_2$.
If $G$ is a group, denote by $\text{Sub}(G)$ the collection of the subgroups of $G$.
Are there non-isomorphic groups $G,H$ such that the hypergraphs $(G, \text{Sub}(G))$ and $(H, \text{Sub}(H))$ are isomorphic?

REPLY [11 votes]: For large prime $p$, there are uncountably many non-isomorphic Tarski monsters of exponent $p$. 
For these groups $G$, the subgroups lattice consists of basically a partition of $G\smallsetminus\{1\}$ into countably many subsets of cardinal $p-1$ (so the subgroups are the whole group, $\{1\}$ and the union of $\{1\}$ with any component of the partition. These hypergraphs are obviouly isomorphic.