Show isomorphism for element g in group G

[bennyska]

Homework Statement

let G be a group and let g be one fixed element of G. Show that the map i_g, such that i_g(x) = gxg' for x in G, is an isomorphism of G with itself.

Homework Equations

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?

 

[pbandjay]

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

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

 

[bennyska]

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 i_g, such that i_g(x) = gxg' for x in G, is an isomorphism of G with itself.
g' is the inverse of g, i think.

 

[VeeEight]

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, i_g(xy) = i_g(x)i_g(y)

 

[bennyska]

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

 

[bennyska]

and for the bijection:

injection
let i_g(x) = i_g(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.