# Prove function is a homomorphism

#### mathjam0990

Let G be a group. Fix g âˆˆ G. Define a map Ï† : G â†’ G by Ï†(x) = gxg^âˆ’1

Prove: Ï† is an isomorphism

What I Know: I already showed it is bijective. Now, I need help proving the homomorphism part. I know by definition for all a,b in G, f(ab)=f(a)f(b)

Question: How do I show this? For some reason I am getting confused and I don't think it's that difficult but I can't grasp it.

What I Have Done: Let a=gag^-1 and b=gbg^-1. Then f(ab)=f(gag^-1 * gbg^-1) Is this even correct or did I start off totally wrong?

Thanks!!

#### mehoul

What I Have Done: Let a=gag^-1 (...)
you need to prove that f(ab)=f(a)f(b) for all a,b not just for those satisfying the equation a=gag^{-1}

If a,b are any elements of G, f(ab) = g(ab)g^{-1} = ?

#### Country Boy

Math Team
For any a and b, $$\displaystyle f(ab)= gabg^{-1}$$, $$\displaystyle f(a)= gag^{-1}$$, $$\displaystyle f(b)= gbg^{-1}$$. Now, what is f(a)f(b)?

Last edited by a moderator:
Similar threads