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Motivation

I think of the "structure" of a topological space $X$ as being the limit operator on functions $I\to X$ where $I$ could be the natural numbers or another topological space -- in this sense, a topological homomorphism (continuous function) $f$ is a function that commutes with the limit operation $f(\lim x)=\lim f(x)$, similar to how a group homomorphism commutes with group multiplication $f(\mathrm{mult}(x,y))=\mathrm{mult}(f(x),f(y))$ and a linear transformation commutes with linear combination.

Nonetheless, it can be shown that this structure can be determined uniquely by the set of open sets on $X$. One may also understand these open sets to be the "sub-(topological spaces)" of $X$ as the topology of $X$ is inherited by them exactly (well, the closed sets are also a "dual" kind of sub-topological spaces).

Similarly, given a set $V$ and a list of subsets that we call "subspaces" (which would have to satisfy some properties), one can determine the vector space up to isomorphism (i.e. we can find its dimension).

I wonder if something like this can be done with groups. Given a set $G$ and a list of subsets we call its "subgroups", can we determine the group up to isomorphism? At least for finite sets?

Example given the set $\{0, 1, 2, 3\}$, we'd be given the following "subgroup structure" on it: $\{\{0\},\{0,2\},\{0,1,2,3\}\}$, and the group being described is $C_4$. The positions of 1 and 3 aren't determined, but the group is still determined to isomorphism.

Martin Sleziak
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Abhimanyu Pallavi Sudhir
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    Exactly what information are you getting? Do you get the operation, or just the underlying sets? If you get the operation, so that you are actually getting the subgroups, do you exclude the whole group? Do you get how the underlying sets related to one another (the lattice of subgroups) or just their orders? Both? The lattice by itself does not determine the group, and neither does a list of orders of subgroups (even with multiplicity). – Arturo Magidin Aug 15 '19 at 05:58
  • Each group is its own subgroup. Same for vector spaces and subspaces: The unique space of maximal dimension (in the finite dimensional case) will be the space itself.

    Could you specify what exactly you are looking for? E.g. do you want a certain set of subgroups that always define the overlying group? If yes, did you try some sets out, e.g. normal divisors? One set that might be interesting would be the cyclic groups, i.e. the question if a finite group is uniquely determined up to iso. by the order of its elements (might be wrong, no idea, but looks interesting at least).

     – Dirk Aug 15 '19 at 05:59
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    @Dirk: No, groups are not uniquely determined by the multiset of the orders of its elements. There is an abelian group of order $p^3$ in which every non-identity element has order $p$, and there is a nonabelian group of order $p^3$ in which every non-identity element has order $p$ for any odd prime $p$: the abelian group is $C_p\times C_p\times C_p$, and the nonabelian one is the Heisenberg group over $\mathbb{F}_p$. – Arturo Magidin Aug 15 '19 at 06:01
  • Though I don’t have a counterexample off the top of my head, if I had to guess it would be that the answer is “no” and that a counterexample can be found within $p$-groups. They are notorious for providing counterexamples for almost any finite collection of invariants one might try. – Arturo Magidin Aug 15 '19 at 06:06
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    @ArturoMagidin You're getting just the underlying sets, listing all their elements. The question is if we can determine the operation. See the added example. – Abhimanyu Pallavi Sudhir Aug 15 '19 at 06:07
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    A more precise statement of the question is: suppose $G$ and $H$ are groups and $f:G\to H$ is a bijection such that for all $S\subseteq G$, $S$ is a subgroup of $G$ iff $f(S)$ is a subgroup of $H$. Then must $G$ and $H$ be isomorphic as groups? – Eric Wofsey Aug 15 '19 at 08:04
  • You could also ask for the stronger conclusion that $f$ itself is an isomorphism of groups, but that has very easy counterexamples (arbitrarily permute the non-identity elements of a cyclic group of prime order, or take the inverse map on any non-abelian group). – Eric Wofsey Aug 15 '19 at 08:09
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    Have a look at https://math.stackexchange.com/questions/14588/does-the-order-lattice-of-subgroups-and-lattice-of-factor-groups-uniquely-det and see whether it answers your question. – Gerry Myerson Aug 15 '19 at 08:23
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    There is a recent preprint of James Wilson which constructs (lots of) non-isomorphic finite groups with isomorphic lattices of subgroups, and lots of other properties: https://arxiv.org/abs/1612.01444 If Jack Schmidt's answer in Gerry's link isn't what your after then try looking in this article. – user1729 Aug 15 '19 at 10:26
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    I think Jack Schmidt's answer provides the desired counterexample (with two groups of order $3^5$; $p$-groups, as I guessed). The function $f$ is a bijection of underlying sets that respects "is a subgroup of" and induces an isomorphism of subgroup lattices. It even induces an isomorphism between corresponding proper subgroups (thus, is a stronger identification than the one asked in the current question; and respects normality and isomorphisms of proper quotients), but the two groups are not isomorphic. – Arturo Magidin Aug 15 '19 at 14:52
  • Had a chance to look into these comments, Abhi? – Gerry Myerson Aug 19 '19 at 12:51
  • Are you still here, Abhi? – Gerry Myerson Aug 20 '19 at 13:02
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    @Gerry Eric Wofsey's answer here answers my question -- I'll have a look at your link for the finite counter-example. – Abhimanyu Pallavi Sudhir Aug 20 '19 at 16:33

1 Answers1

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Here is a counterexample for infinite groups. Consider $G=\mathbb{Z}[1/p]$ and $H=\mathbb{Z}[1/q]$ for distinct primes $p$ and $q$. In both of these groups, every finitely generated subgroup is cyclic, and thus a subset is a subgroup iff it is either a cyclic subgroup or a nested union of cyclic subgroups. Now consider the bijection $f:G\to H$ given by $f(ap^nq^m)=ap^mq^n$ where $n\in\mathbb{Z}$, $m\in\mathbb{N}$, and $a$ is an integer not divisible by $p$ or $q$ (or $a=0$). Then $f$ and $f^{-1}$ both preserve the divisibility relation, and thus map cyclic subgroups to cyclic subgroups, and thus map all subgroups to subgroups. Thus $G$ and $H$ have isomorphic subgroup structures, but are not isomorphic as groups.

I don't know about the finite case but here is an observation. By induction on the order of the group, we can assume we already know the isomorphism class of all the proper subgroups of our group. So for instance, if it is true that a finite group is determined up to isomorphism by its order and the number of proper subgroups of each isomorphism type that it has, we could conclude that a finite group is determined up to isomorphism by its subgroup structure. I don't know whether that statement is true though, and I wouldn't be surprised if it's false.

Eric Wofsey
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