Automorphisms of a group G

Dummit and Foote Section 4.4 Automorphisms Exercise 1 reads as follows:

Let $\displaystyle \sigma \in Aut(G)$ and $\displaystyle \phi_g$ is conjugation by g, prove that $\displaystyle \sigma \phi_g \sigma^{-1}$ = $\displaystyle \phi_{\sigma (g)}$.

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A start to the proof is as follows:

$\displaystyle \sigma \phi_g \sigma^{-1}$

= $\displaystyle \sigma (\phi_g (\sigma^{-1} (x)))$

= $\displaystyle \sigma (g. \sigma^{-1} (x). g^{-1} )$

= $\displaystyle \sigma (g). x . \sigma (g^{-1} )$

Now we have completed the proof if $\displaystyle \sigma (g^{-1}) = ( {\sigma (g) )}^{-1}$

But why is $\displaystyle \sigma (g^{-1}) = ( {\sigma (g) )}^{-1}$ in this case??

Can anyone please help clarify this situation?

Peter

Re: Automorphisms of a group G

I have now thought about this and suspect that the answer to my own question is as follows:

For an automorphism we have $\displaystyle \phi (e) = e $

Thus $\displaystyle \phi (e) = \phi (g g^{-1}) = \phi (g) \phi( g^{-1} ) = e $

and from this it follows that $\displaystyle \phi (g^{-1}) = {[\phi (g)]}^{-1} $

Am I correct?

Peter

Re: Automorphisms of a group G

Quote:

Originally Posted by

**Bernhard** I have now thought about this and suspect that the answer to my own question is as follows:

For an automorphism we have $\displaystyle \phi (e) = e $

Thus $\displaystyle \phi (e) = \phi (g g^{-1}) = \phi (g) \phi( g^{-1} ) = e $

and from this it follows that $\displaystyle \phi (g^{-1}) = {[\phi (g)]}^{-1} $

Am I correct?

Peter

Yes, that is correct.