# Group Isomorphism

• Aug 13th 2007, 10:55 PM
r7iris
Group Isomorphism
It is very urgent!!!! need help!!!!

Let G be any group and let g be an element of G. Define the function f:G-->G by f(a) = g^(-1)*a*g (a is in G)(thus f takes every element to its conjuate by g). Show that f is an isomorphism from G to itself. Show, by example, that f need not be the identity function.
• Aug 14th 2007, 05:26 AM
ThePerfectHacker
Quote:

Originally Posted by r7iris
It is very urgent!!!! need help!!!!

Let G be any group and let g be an element of G. Define the function f:G-->G by f(a) = g^(-1)*a*g (a is in G)(thus f takes every element to its conjuate by g). Show that f is an isomorphism from G to itself. Show, by example, that f need not be the identity function.

This is a very special automorphism called "conjugation automorphism".

It is a homomorphism,
$\phi(x)\phi(y) = gx^{-1}g^{-1}gyg^{-1}=gxyg^{-1}=\phi(xy)$

Is is one-to-one,
$\phi (x) = \phi(y) \implies gxg^{-1} = gyg^{-1} \implies x=y$

Is is onto,
$\forall y\in G \mbox{ choose }x=gyg^{-1}\implies \phi(x) = y$