MHB Prove function is a homomorphism

[mathjam0990]

Let G be a group. Fix g ∈ G. Define a map φ : G → G by φ(x) = gxg^−1

Prove: φ 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!

 

[Euge]

Hi mathjam0990,

The trick here is to use the fact that for all $x\in G$, $x^{-1}x = e$. Let $a,b\in G$. Then

$$\varphi(a)\varphi(b) = (gag^{-1})(gbg^{-1}) = ga(g^{-1}g)bg^{-1} = gaebg^{-1} = gabg^{-1} = \varphi(ab).$$

Therefore, $\varphi$ is a homomorphism.

 

[mathjam0990]

Euge, thank you so much! I never would have guessed that. Much appreciated!