Does the hypergraph of subgroups determine a group?

Question

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?

Answer 1

In the comments to the question, I notice something which might be an error, or at least is an incomplete response. It is pointed out in the comments that there exist nonisomorphic groups with isomorphic subgroup lattices. While true, that fact doesn't answer this question, since it is possible to have isomorphic subgroup lattices and nonisomorphic subgroup hypergraphs. Even if you assume that the groups have the same order and isomorphic subgroup lattices, it does not immediately follow that they have isomorphic subgroup hypergraphs. What is needed is a subgroup preserving and reflecting bijection between the groups.

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I am sure the negative answer to this question can be found in Roland Schmidt's book. But I would like to point to a theorem I coauthored after Schmidt's book was published, which applies to this question. Namely:

Thm. For any finite $N$, there is a finite set $X=X_N$ and $N$ binary operations $\circ_i$ defined on $X$ such that

(1) $G_i = (X,\circ_i)$ is a group for all $i$,
(2) $G_i\not\cong G_j$ when $i\neq j$, and
(3) for all $i, j$, the groups $G_i^{\kappa}$ and $G_j^{\kappa}$ have exactly the same subgroups (as sets) for all cardinals $\kappa$.

The last item means that, for any fixed $\kappa$, the subgroup hypergraphs of $G_i^{\kappa}$ and $G_j^{\kappa}$ are equal for any $i$ and $j$.

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The paper is

Keith A. Kearnes and Agnes Szendrei,
Groups with identical subgroup lattices in all powers.
J. Group Theory 7 (2004), no. 3, 385--402.

Answer 2

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.

Answer 3

As @Keith Kearnes says, the negative answer ought to be somewhere in Roland Schmidt's book. Unless I'm mistaken, it suffices to find two non isomorphic groups with isomorphic coset lattices. Indeed, the elements of $G$ correspond directly with the cosets of the trivial subgroup. By applying the regular group action, you can take any such element (or 1-coset) to be the identity element. The subgroups are the cosets containing the identity. Thus you can determine from the coset lattice the sets of elements in each subgroup (as in the question).

Non-isomorphic groups with isomorphic coset lattices exist, as is discussed in Chapter 9.4 of Schmidt. Something that I like about this discussion is that it is relatively easy to find concrete examples.

I believe that the groups with identities [625, 9] and [625, 10] in the GAP small groups library are explicit examples with isomorphic coset lattices, hence with isomorphic subgroup hypergraphs. These groups can be found by following the ideas of Example 9.4.14(b) in Schmidt's book. I haven't checked all the details in GAP, but these two groups at least have the same StructureDescription of (C25 x C5) : C5, and same number of subgroups at each level.

Schmidt constructs these groups by taking certain semidirect products $H \rtimes \mathbb{Z}_p$, where $H$ is the nonabelian group of order $p^3$ and exponent $p$. The construction doesn't seem to work for $p=3$, and indeed the result from Huppert that Schmidt cites in the example holds only for $p>3$.