# Show isomorphism for element g in group G

1. Feb 16, 2010

### bennyska

1. The problem statement, all variables and given/known data
let G be a group and let g be one fixed element of G. Show that the map ig, such that ig(x) = gxg' for x in G, is an isomorphism of G with itself.

2. Relevant equations

3. The attempt at a solution
not even really understanding the question. can someone break it down for me, and explain what the question is asking? am i trying to find a function, or rather show that i(x) preserves the structure?

Last edited: Feb 16, 2010
2. Feb 16, 2010

### pbandjay

Re: isomorphism

Does it specify what is F? Or is g an element of G?

Also, does it specify what exactly is g' in ig(x) = gxg'?

3. Feb 16, 2010

### bennyska

Re: isomorphism

i'm sorry, it's supposed to be:
let G be a group and let g be one fixed element of G. Show that the map ig, such that ig(x) = gxg' for x in G, is an isomorphism of G with itself.
g' is the inverse of g, i think.

4. Feb 16, 2010

### VeeEight

Re: isomorphism

I am assuming that g is an element of G and that g' is the inverse of g.

The problem is giving you a function and asking you to show that it is an isomorphism. To show that it is an isomorphism, you must show it is a bijection and that for all x, y in G, ig(xy) = ig(x)ig(y)

5. Feb 16, 2010

### bennyska

Re: isomorphism

so for the latter part:
ig(xy)=gxyg'
=gxg'gyg'
=ig(x)*ig(y)
is that right?

Last edited: Feb 16, 2010
6. Feb 16, 2010

### bennyska

Re: isomorphism

and for the bijection:

injection
let ig(x) = ig(y) for x,y in G. then
gxg'=gyg'
g'gxg'=g'gy'g inverses
xg'=yg'
x=y by right cancellation

surjection
let x,y in G. then
y=gxg'
yg=gxg'g
yg=gx
g'yg=g'gx
g'yg=x

is this right? i kind of feel like i'm using normal multiplication to do this, ignoring commutativity.