Is a group uniquely determined by the sets {ab,ba} for each pair of elements a and b?

Question

This question has been cross-posted here on mathoverflow.

For a given group $G=(S,\cdot)$ with underlying set $S$, consider the function $$ F_G:S\times S\to\mathcal P(S)\\ F_G(a,b):=\{a\cdot b,~b\cdot a\} $$ from $S\times S$ to the power set of $S$.

I'd like to figure out how much information from $G$ is encoded by $F_G$. In particular, does $F_G$ determine the group $G$ up to isomorphism?

Given different groups $G_1$ and $G_2$ with underlying sets $S_1$ and $S_2$, suppose that a function $\varphi:S_1\to S_2$ has the property $\varphi\bigl(F_{G_1}(a,b)\bigr)=F_{G_2}\bigl(\varphi(a),\varphi(b)\bigr)$ for all $a$ and $b$ in $G_1$; if $\varphi^{-1}$ exists and has this property as well, let's say that $F_{G_1}\cong F_{G_2}$. For one thing, if $G_1\cong G_2$ then $F_{G_1}\cong F_{G_2}$. Going the other way, if $F_{G_1}\cong F_{G_2}$ then $|G_1|=|G_2|$ and $Z(G_1)\cong Z(G_2)$.

Given $F_G$, it's easy to find out which pairs of elements in $G$ commute, which subsets of $G$ constitute subgroups of $G$, and which subsets of $G$ are generating sets of $G$. Moreover, if $F_{G_1}\cong F_{G_2}$ then $G_1$ and $G_2$ must have the same cycle graph. This means that if $F_{G_1}\cong F_{G_2}$ and the order of these groups is less than 16, then $G_1\cong G_2$. Indeed, according to Wikipedia's page on cycle graphs, "For groups with fewer than 16 elements, the cycle graph determines the group (up to isomorphism)."

The question that's been plaguing me is whether $F_{G_1}\cong F_{G_2}\implies G_1\cong G_2$ in the general finite case. I think that a counterexample would need to involve groups of order 16 or larger. Any ideas?

EDIT:

This question has been resolved on mathoverflow. The answer is yes, $F_G$ does determine $G$ up to isomorphism.

Answer 1

So far I can tell we get the following list of things. When I say we "know a subgroup," I mean we know its underlying set as a subset of $G$, not its algebraic structure. For $X,Y\subseteq G$ we can extend the definition of $F$ so that $F(X,Y)=\bigcup_{x\in X,y\in Y}F(x,y)$.

• Identity: the element $e\in G$ is the unique $x\in G$ such that $F(x,x)=\{x\}$.
• Inverses: given $g\in G$, its inverse $g^{-1}\in G$ is the unique $x\in G$ such that $F(x,g)=\{e\}$.
• Powers of elements: since we have inverses and $\{g^n\}=F(g,g^{n-1})$, we can recursively compute any integer power of an element.
• Orders of elements cyclic submonoids, cycle graph: via powers of elements.
• Torsion and $p$-torsion: via orders of elements.
• Conjugacy classes: since $F(b,F(a,b^{-1}))=\{bab^{-1},a,b^{-1}ab\}$, the conjugacy class of a given element $a\in G$ is given by $\bigcup_{b\in G}F(b,F(a,b^{-1}))$.
• Commuting pairs of elements: $a,b\in G$ commute iff $F(a,b)$ is a singleton.
• Centralizers of subsets and center: follows from commuting pairs of elements.
• Cyclic orbits under inner automorphisms: given $b\in G$, the orbit of $G$ under the map $x\mapsto bxb^{-1}$ is given by applying the formula in conjugacy classes recursively.
• Normalizers of subsets and normal subsets: given a subset $X\subseteq G$, its normalizer is the set of all elements $g\in G$ for which $gXg^{-1}=X$. This is equivalent to $X$ being a union of orbits under the inner automorphism $x\mapsto gxg^{-1}$, which we know. Normal subsets follow.
• Subgroups generated by subsets: Given a subset $S$, the subgroup $\langle S\rangle$ is the intersection of all subsets of $G$ containing $S$, closed under inverses, and such that $x,y\in S\Rightarrow F(x,y)\subseteq S$.
• Lattice of subgroups and normal subgroups: we can identify which subsets are subgroups using subgroup generated by property ($S$ is a subgroup iff $S=\langle S\rangle$), then order them according to inclusion. By testing for the subgroup property and normal subset property both, we can identify normal subgroups too.
• Sylow subgroups: subgroups maximal with respect to containing $p$-torsion elements.
• Maximal/minimal (normal) subgroups: these can be identified by looking at the lattice of subgroups or lattice of normal subgroups.
• Relative normality. If $H\le K\le G$ then $H\triangleleft K$ can be tested by restricting $F_G$ to $K$ to get the function $F_K$, and then we can identify it as a normal subgroup.
• Nilpotence: nilpotent iff all maximal subgroups are normal.
• Frattini subgroup: intersect all maximal subgroups.
• Chief factors and chief series: use lattice of normal subgroups.
• Fitting subgroup: to compute the centralizer of a chief factor $H/K$, first partition $H$ into cosets of $K$ via $aK=F_H(a,K)$ (since $K$ is normal in $H$), then find all $g\in G$ such that each coset of $K$ is stable under conjugation by $g$, i.e. is a union of cyclic orbits. Then intersect all centralizers of chief factors for the fitting subgroup.
• Socle: generate a subgroup from the union of minimal normal subgroups.
• Set of commutators. Commutators are of the form $(ab)(ba)^{-1}$. So write $F(a,b)=\{x,y\}$ and then compute $H(a,b)=F(x,y^{-1})\cup F(x^{-1},y)$. The set of commutators is $\bigcup_{a,b\in G}H(a,b)$.
• Derived subgroup and derived series: used subgroup generated by and set of commutators to get $G'$. Since $F$ restricted to $G'$ is $F_{G'}$, simply iterated to get the derived series.
• Solvability: follows from derived series.

It's been suggested that we take $G$ a nonabelian group and $H,K$ two nonisomorphic groups of the same order and consider $G\times H$ versus $G\times K$. In this case, since the cycle data of $H$ and $K$ will be different, the cycle data of $G\times H$ and $G\times K$ will differ, so their $F$'s will be different if $G,H,K$ are finite, and so this does not create any finite counterexamples.

Answer 2

I don't know the answer off the top of my head. But if I had to tackle this problem, this is how I'd go about it: As I recall, the first non-Abelian simple group is of order 60 and it's isumorphic to A5. This fact is important in this case since your claim is obviously true if the groups are Abelian since Z(G) is equal to G in this case. (Also,just to be careful here, if they're isomorphic, then clearly BOTH groups must be Abelian.Prove it,it's not hard.) Also, clearly all the subgroups of Abelian groups are normal. Therefore, it seems to me the cases you want to check are the non-Abelian simple groups since these are the finite groups there may be some question as to the validity of your implication.

What you really want to know is where does exactly does the implication break down if at all. I'd begin with the alternating group A5 and some Abelian group of the same order (60) and test your hypothesis. Clearly is the other group is Abelian, it can't be isomorphic to A5. Also, the centers cannot be isomorphic either.

I'd have to work on this for awhile,but that should get you started Good luck!