Is it possible to find an isomorphism between two groups with a new operation?

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Homework Help Overview

The discussion revolves around the properties of groups and the exploration of isomorphisms between a group (G,·) and a newly defined group (G,*) using a different operation. The original poster seeks to demonstrate that the new group is isomorphic to the original group.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to define an isomorphism and seeks hints on how to establish it. Some participants question the use of conjugate actions, while others propose specific functions as potential isomorphisms.

Discussion Status

Participants are actively exploring different approaches to finding an isomorphism, with one participant claiming to have found a suitable function. There is a mix of ideas being shared, but no explicit consensus has been reached regarding the best method.

Contextual Notes

One participant notes that the new group is referred to as the opposite group, which may influence the understanding of its properties and applications in group theory.

TorKjellsson
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Homework Statement


Let [itex](G,\cdot)[/itex] be a group. Defining the new operation * such that [itex]a*b = b \cdot a[/itex] it is pretty easy to show that [itex](G,*)[/itex] is a group. Show that this new group is isomorphic to the old one.


Homework Equations





The Attempt at a Solution



I have been experimenting with the possibility to define an isomorphism [itex]\phi (a*b) = b a[/itex] but can't really seem to get it right. Can anyone give me a hint on how to find the isomorphism?


Tor
 
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Will the conjugate action work?
 
Hm, actually after doing some more exercises I figured out a function that works. I am not quite sure of what you mean by conjugate action but I post my solution here:

Let [itex]\phi: (G,*) \rightarrow (G, \cdot)[/itex] such that [itex]\phi (a) = a^{-1}[/itex]. Then [itex]\phi(a) \phi(b) = a^{-1} b^{-1}[/itex]. Consider now [itex]\phi(a*b) = (ba)^{-1} = a^{-1} b^{-1}[/itex] and we see that [itex]\phi(a) \phi(b) = \phi(a*b)[/itex].
 
Ooo nice. I was thinking of something like a --> ga(g-1). But that doesn't work.
 
This group is called the opposite group. It is important for studying the relation between left- and right- group actions. Just thought I'd throw that fun fact in there.
 

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