0
$\begingroup$

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?

|cite|improve this question
$\endgroup$
  • 5
    $\begingroup$ $ghx(gh)^{-1} \neq ghxg^{-1}h^{-1}$, rather $ghx(gh)^{-1} = ghxh^{-1}g^{-1}$. $\endgroup$ – Slade Nov 13 '15 at 9:50
  • $\begingroup$ @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? $\endgroup$ – de_dust Nov 13 '15 at 10:03
2
$\begingroup$

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

|cite|improve this answer
$\endgroup$
  • $\begingroup$ Am I right if I say, that the kernel of $\phi$ is only the identity element of $G$? $\endgroup$ – de_dust Nov 13 '15 at 10:47
  • $\begingroup$ 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? $\endgroup$ – A.P. Nov 13 '15 at 11:03
  • $\begingroup$ 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? $\endgroup$ – de_dust Nov 13 '15 at 11:09
  • $\begingroup$ Now I understand. Thank you! $\endgroup$ – de_dust Nov 13 '15 at 11:18
  • $\begingroup$ 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 $g^* = \text{Id}_G \iff g \in Z(G)$. $\endgroup$ – A.P. Nov 13 '15 at 11:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.