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

This is a cross-posted question, originally active here on math.stackexchange.

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?

The answer is yes: the map $F_G$ determines the group. Mansfield makes good use of this in his elementary proof of the fact that the group determinant determines the group (v. Lemma 4 in his paper).
Suppose that the set $S$ is known to both of us (say with an agreed order) and that I have a specific (but unknown to you) $n \times n$ matrix $A$ giving the group operation of $G$. I give you only the function $F_G.$ Then, it turns out, you can recover the the set $\{{A,A^t\}}.$ Either one will determine $G$ uniquely, but you will not know which it was that I used.
• So in other words, multiplication is determined up to opposites (the opposite group being the group with the same underlying set, but multiplication done in reverse order). Conversely, it's clear that $F_G = F_{G'}$ if $G'$ is the opposite of $G$, so this is the best we can hope for. – Colin Reid Oct 28 '14 at 0:55