Let $G$ be a group of order $\ge3$

If $Z(\operatorname{Aut}(G))$ is trivial then so is $Z(G)$?

By assumption, we can choose any $g$ and $g'$ in $G$ s.t. $g\neq g'$ and $g,g'\neq e$

Since any inner automorphisms are not in $Z(\operatorname{Aut}(G))$, there exists $x$ s.t. $c_g c_g'(x)\neq c_g' c_g(x)$ i.e. $gg'xg'^{-1}g^{-1}\neq g'gxg^{-1}g'^{-1}$

So, $gg'\neq g'g$ for any $g,g'$, hence $g\notin Z(G)$.

It is my solution. Is it right?

  • $\begingroup$ Your proof is not correct. You pick $g \in G$ distinct from the identity, and you want to show that $g \notin Z(G)$. In order to show this, you pick $g'$ with the property that $c_g$ and $c_{g'}$ do not commute: but there is no justification on existence of $g'$. $\endgroup$ – Crostul Apr 2 '16 at 10:21
  • 1
    $\begingroup$ $Q_8$ is a counterexample. ${\rm Aut}(G) \cong S_4$. $\endgroup$ – Derek Holt Apr 2 '16 at 11:46

Not completely. We have the homomorphism $G\to\operatorname{Aut}(G)$, $g\mapsto c_g$ and $Z(G)$ is its kernel. So in order to use that $c_g\notin Z(\operatorname{Aut}(G))$ you seem to assume already that $g\notin Z(G)$.

| cite | improve this answer | |

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.