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

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?

• $ghx(gh)^{-1} \neq ghxg^{-1}h^{-1}$, rather $ghx(gh)^{-1} = ghxh^{-1}g^{-1}$. – Slade Nov 13 '15 at 9:50
• @Slade: But now I get: $\phi(gh)=ghx(gh)^{-1}=ghxh^{-1}g^{-1}=gh^*g^{-1}=g^*=\phi(g)$. Shouldn't it be $\phi(gh)=\phi(g)\phi(h)$ for $\phi$ to be a homomorphism? – de_dust Nov 13 '15 at 10:03

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, $\phi(gh) = \phi(g) \circ \phi(h)$.

• Am I right if I say, that the kernel of $\phi$ is only the identity element of $G$? – de_dust Nov 13 '15 at 10:47
• No, @de_dust. The kernel of $G$ is the centre $Z(G)$, i.e. the set of elements of $G$ that commute with every other element. Do you understand why? – A.P. Nov 13 '15 at 11:03
• No, can you explain? I thought: $Ker(\phi)=\{g\in G|g^*=Id_G\}$ So, $g^*=Id_G\iff gxg^{-1}=x\iff g=e$ Therefore: $Ker(\phi)=\{e\}$ Did I make a mistake? – de_dust Nov 13 '15 at 11:09
• Now I understand. Thank you! – de_dust Nov 13 '15 at 11:18
• Yes, the mistake is in the second "iff". Indeed, multiplying (on the right) both sides of $gxg^{-1} = x$ by $g$ we get $gx = xg$. This means that $g^* = \text{Id}_G \iff g \in Z(G)$. – A.P. Nov 13 '15 at 11:20