Does the hypergraph of subgroups determine a group?

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?

• So there is the important vertex which is characterized by being on every hyperedge. – AHusain Oct 6 '18 at 16:43
• Possible duplicate of Does subgroup structure of a finite group characterize isomorphism type? – verret Oct 6 '18 at 19:44
• This question reminds me a bit of a conjecture of Thompson -- he found two non-isomorphic maximal subgroups of $M_{23}$ which both contained the same number of elements of the same order, for every order. This led him to conjecture that if two groups $G$ and $H$ both have the same number of elements of the same order, for every order, and $G$ is solvable, then so is $H$. (This is now proved I believe.) I wonder if one could ask something similar here: Suppose $G$ and $H$ have the same subgroup hypergraph and $G$ is solvable. Then $H$ is solvable... – Nick Gill Oct 8 '18 at 9:37
• Nice comment @NickGill - I think your question is worth asking! – Dominic van der Zypen Oct 8 '18 at 9:44
• In answer to @NickGill's question: the answer is "yes". Indeed, if G and H have the same lattice of subgroups, then they are already both (super)solvable or both not (super)solvable. There are effective ways to detect solvability from the subgroup lattice: Roland Schmidt gave a characterization in terms of chains of modular elements with a flavor entirely similar to that of the definition. My own favorite characterization is the topological/combinatorial characterization of John Shareshian. – Russ Woodroofe Oct 8 '18 at 12:21

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.

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 $$j$$.

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.

• Let me mention that your construction yields groups with square-free order (which is quite opposite, and of independent interest, to the case of $p$-groups provided in Russ' answer). – YCor Oct 8 '18 at 11:02

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.

• By the way, the question was previously asked as a comment here mathoverflow.net/questions/309096/… and I gave this answer there (although I ignored the word "hypergraph" which is the relevant notion). – YCor Oct 7 '18 at 21:55

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 $$p>3$$.

• As far as subgroup of order 5, these groups have 1 normal subgroup, 1 conjugacy class of 5 sgs, and 1 conjugacy class of 25 sgs. (Added this comment when I fixed a 5 that should have been p, after @YCor pointed out a problem.) – Russ Woodroofe Oct 8 '18 at 8:04