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

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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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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.