Consider the map $\psi : G \rightarrow Aut(G)$ given by $\psi(g) = \phi_{g}.$ Prove that $\psi$ is a homomorphism

Question

In a previous problem, I already proved that $\phi_{g}$ is an isomorphism, for $\phi_{g}(x)=gxg^{-1}$ so knowing that $\phi_{g} = gxg^{-1}$ is an isomorphism will certainly help here. Anyway, this is what I get:

Let $a,b \in G$ $\psi(ab) = \phi(ab)$ If $\phi_{g} = gxg^{-1},$ then $\phi_{gh} = ghx(gh)^{-1} = ghxh^{-1}g^{-1}$ $\psi(ab)=\phi_{ab}=abxb^{-1}a^{-1}$

Then I get stuck. I don't know where to go from here. I know that I need to show $\psi(ab)=\phi_{ab}=\phi_{a}\phi_{b}=\psi(a)\psi(b)$ but we don't know that G is abelian so I can't rearrange everything to my liking

Answer 1

If we want to prove that any two maps are same then we should show that both maps agree on all arguments where they are defined.

So to prove:

$\psi(gh)=\psi(g)°\psi(h)$

i.e, $\phi_{gh}=\phi_g ° \phi_h$

We have to show that $\phi_{gh}(x)=\phi_g ° \phi_h(x) \ \forall x \in G$

Take any $x\in G$ ,

$\phi_{gh}(x)=(gh)x(gh)^{-1}=ghxh^{-1}g^{-1} \tag{1}$

And

$\phi_g ° \phi_h(x)=\phi_g(\phi_h(x))=\phi_g(hxh^{-1})=ghxh^{-1}g^{-1} \tag{2}$

From (1) and (2), Conclusion follows .