# A group isomorphism

1. Jul 21, 2011

### TorKjellsson

1. The problem statement, all variables and given/known data
Let $(G,\cdot)$ be a group. Defining the new operation * such that $a*b = b \cdot a$ it is pretty easy to show that $(G,*)$ is a group. Show that this new group is isomorphic to the old one.

2. Relevant equations

3. The attempt at a solution

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

Best,
Tor

2. Jul 21, 2011

### Oster

Will the conjugate action work?

3. Jul 21, 2011

### TorKjellsson

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 $\phi: (G,*) \rightarrow (G, \cdot)$ such that $\phi (a) = a^{-1}$. Then $\phi(a) \phi(b) = a^{-1} b^{-1}$. Consider now $\phi(a*b) = (ba)^{-1} = a^{-1} b^{-1}$ and we see that $\phi(a) \phi(b) = \phi(a*b)$.

4. Jul 21, 2011

### Oster

Ooo nice. I was thinking of something like a --> ga(g-1). But that doesnt work.

5. Jul 21, 2011

### Kreizhn

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.