# Operation defined on a group uniquely determines that group

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.

• $|G|$ usually denotes the cardinality of the set $G$. Jan 6, 2021 at 12:03
• Yes, but in this book it means underlying set of a group. That's exactly what author writes. Jan 6, 2021 at 15:46
• Perhaps it is helpful to read the solution for Example $2.2.2$ with $\delta_{G}(x, y) = xy^{-1}$. Jan 6, 2021 at 16:17
• I solved 2.2:2, but can't handle this one. In 2.2:2 just plug $x$ instead of $y$ in $x \circ y^{-1} = x \star y^{'}$ and get $e_1 = e_2$. Equality of inverses and binary operations follows. Jan 6, 2021 at 16:48
• @DietrichBurde: Bergman distinguishes between the algebraic structure (which involves both the underlying set and the operations and/or relations defined on it) and the underlying set. So he uses that notation throughout (and being very explicit about it, and about the clash with its common use) for underlying set, and uses $\mathrm{card}(X)$ for the cardinality of a set $X$. The book was originally typeset in troff, where there were fewer options vis a vis available fonts, and the notation originated in that version. Jan 6, 2021 at 22:27

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).
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.