Operation defined on a group uniquely determines that group

Question

This is an exercise (2.2:3) in An Invitation to General Algebra and Universal Constructions by George M. Bergman:

If $G$ is a group, let us define an operation $\delta_{G}$ on $|G|$ by $\delta_{G}(x, y) = xy^{-1}x$. Does the pair $G' = (|G|, \delta_{G})$ determine the group $(|G|, {}\cdot{} , ^{-1}, e)$? (I.e., if $G_1$ and $G_2$ yield the same pair, $G'_1 = G'_2$, must $G_1 = G_2$?)

($|G|$ means underlying set of a group).

If $G'_1 = G'_2$ then they are equal as pairs and that means that $|G_1| = |G_2|$ and $\delta_{G_1} = \delta_{G_2}$. Let $\circ$, $^{-1}$, $e_1$ (resp. $\star$ , $^{'}$, $e_2$) be binary operation, operation of taking inverse and identity operation in group $G_1$ (resp. $G_2$). By virtue of $\delta_{G_1} = \delta_{G_2}$ we get $$ x \circ y^{-1} \circ x = x \star y^{'} \star x \tag{1}$$ for every $x, y$ of $|G_1|$. Plugging various $x$ and $y$ in $(1)$ we can get following relations of identities: $$ e_1 \star e_1 = {e_2}^{-1} \tag{2}$$ $$ e_2 \circ e_2 = {e_1}^{'} \tag{3}$$ $$ {e_2}^{-1} \star e_1 = {e_2}^{-1} \circ {e_2}^{-1} \tag{4}$$ And I'm stuck with that: can't get any reasonable relation which would allow to get equalitity of identities or equality of inverses or just distributivity between two group operations (to prove that groups are equal). On the other hand, to disprove that one must find a counterexample e.g. in groups of small order (i.e. two different group structures on the same set giving the same derived operations), but it is tedious.

Answer 1

Since the question asks for equality of operations rather than isomorphism: take $G$ to be any elementary abelian $2$-group. Then $xy^{-1} x = y$ so the operation $\delta_G$ gives you no information and, for example, there are already two such structures on the two-element set (given by choosing one or the other of the two elements to be the identity).

Answer 2

This question is was answered a year and a half ago, but let me add a comment. In the answer given, it is argued that if $G$ is a nontrivial elementary abelian $2$ group, then $(|G|,\delta_G)$ does not contain enough information to determine which element of $G$ is the identity element. In fact, if $G$ is ANY nontrivial group, then $(|G|,\delta_G)$ does not contain enough information to determine which element is the identity element. This is because $\delta_G$ is an idempotent operation ($\delta_G(x,x)=x$) and the idempotent subclone of a group cannot identify the identity element.

More concretely, let $G$ be any nontrivial group and choose any $a\in G-\{e\}$. The operation $x\otimes y:=xa^{-1}y$ is a group multiplication on the set $|G|$. This multiplication has identity element $a$ (which was chosen essentially arbitrarily) and inverse operation $\textrm{inv}_{\otimes}(x)=ax^{-1}a$. Finally, $\delta_{G_\otimes}(x,y)=x\otimes \textrm{inv}_{\otimes}(y)\otimes x = xa^{-1}ay^{-1}aa^{-1}x=xy^{-1}x=\delta_G(x,y)$. This shows that $(|G|, \otimes, \textrm{inv}_{\otimes}, a)$ and $(|G|,\cdot, {}^{-1},e)$ have the same $\delta$-function but different identity elements. These groups also have different multiplication tables. They will also have different inverse operation tables unless $a$ is a central involution.