Proving a group homomorphism: $\phi: G\rightarrow \text{Aut}(G), g\mapsto g^*$

Question

Given $\phi: G\rightarrow Aut(G), g\mapsto g^*$

and $g^*: G\rightarrow G, x\mapsto gxg^{-1}$

where $g,g^{-1},x\in G$

I need to prove that $\phi$ is a homomorphism ($\phi(gh)=\phi(g)\phi(h)$).

So, I started like this:

$\phi(gh)=ghx(gh)^{-1}=ghxg^{-1}h^{-1}$ <-here's where I'm stuck.

I considered this step: $=gh(gg^{-1})xg^{-1}h^{-1}=(ghg^{-1})(gxg^{-1})h^{-1}=g^*g^*h^{-1}$,

but somehow that looks wrong.

How do I go from here?

Answer 1

I see two misunderstandings here:

1. As addressed in the comments, for $g, h$ in any group $G$ we have $(gh)^{-1} = h^{-1}g^{-1}$, not $g^{-1}h^{-1}$.

2. The operation in $\operatorname{Aut}(G)$ is function composition.

Therefore for every $g,h,x \in G$ we have $$ \phi(gh)(x) = (gh)x(gh)^{-1} = g(hxh^{-1})g^{-1} = g\big(h^*(x)\big)g^{-1} = g^*\big(h^*(x)\big) = (g^* \circ h^*)(x) $$ or, in other words, $\phi(gh) = \phi(g) \circ \phi(h)$.