Is the group (G,dG) isomorphic to the original group G?

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Homework Help Overview

The discussion revolves around the properties of a group \( G \) and a derived operation \( dG \) defined on it. The original poster is exploring whether the group structure given by \( (G, dG) \) can uniquely determine the original group \( G \) with its operation \( * \). The focus is primarily on the implications of this relationship and the nature of the groups involved.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster notes that the operation \( dG \) leads to every element being of order 2, suggesting a structure similar to the Klein group. They express uncertainty about the next steps. Other participants suggest considering the implications of knowing \( (G, dG) \) and whether it allows for the reconstruction of \( (G, *) \). There is also a question about the meaning of \( G = H \) in the context of isomorphism versus equality of groups.

Discussion Status

The discussion is active, with participants exploring the relationship between the groups and the implications of their operations. Some guidance has been provided regarding the nature of the relationship between \( G \) and \( H \), specifically that they can only be shown to be the same up to isomorphism.

Contextual Notes

Participants are navigating the definitions and implications of group operations and isomorphisms, with a focus on the uniqueness of group structures derived from different operations. There is an acknowledgment of the complexity involved in proving such relationships.

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Homework Statement


Exercise 1.2:2.
(i) If G is a group
Define an operation dG on |G| by dG(x, y) = x*y^-1.
Does the group given by (G,dG) determine the original group G with *
(I.e., if G1 and G2 yield the same pair, (G1,dG1) = (G2,dG2) , must G1 = G2?)

There is a part II, but I would rather focus on I first.

The Attempt at a Solution



So, I started by noting that G,dG forces every element to be of order 2 since:
x dG x = x*x^-1 = e = x dG x^-1 thus x^-1 = x

Thus G,dG is a klein group. I'm not sure how to proceed, any hint would be appreciated
 
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If you know (G,*) then you know (G,dG). Now ask yourself, if you know (G,dG) then can you figure out what (G,*) is? Hint: dG(x,y^(-1))=x*y.
 
Starting from (G,dG)

dG(x,y) = x*y^-1

thus

x*y = dG(x,y^-1)

Since (G,dG) = (H,dH)

x*y = dG(x,y^-1) = dH(x,y^-1) = x*y (* in terms of H)

Thus G = H

I guess I'm a little confused what it means for G = H. Am I trying to show that they have the same universe and operation or that they are the same upto isomorphism?
 
You can only show that they are the same up to isomorphism. That's the strongest sense of 'same' you can ever hope to prove.
 

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