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

[moo5003]

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

 

[Dick]

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.

 

[moo5003]

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?

 

[Dick]

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.