1
$\begingroup$

Let $G$ is non-abelian and simple group. Let $I ={\rm Inn}(G) \cong G$, $A = {\rm Aut}(G)$ and $B = {\rm Aut}(A)$. Since $Z(A)=1$, we have $A \cong {\rm Inn}(A)$, so we can identify $A$ with the normal subgroup ${\rm Inn}(A)$ of $B$.

Now $C:=C_B(I)$ is also a normal subgroup of $B$ and so $[C,A] \le C \cap A $.

Now my question is why $C \cap A=1$???

Again why $C:=C_B(I)$ is a normal subgroup of $B$??

$\endgroup$
  • 1
    $\begingroup$ This title isn't very informative, consider editing it $\endgroup$ – Belgi Oct 8 '14 at 10:15
  • $\begingroup$ You suggest one... $\endgroup$ – user152715 Oct 8 '14 at 10:18
  • 1
    $\begingroup$ $C \cap A = 1$ because $A$ is the automorphism group of $G$, and no nontrivial automorphism can fix every element of $G$. To get $[C,A] \le C$ you don't need $C$ to be normal in $B$, you only need $A$ to normalize $C$, which is true because $A$ normalizes $I$. $\endgroup$ – Derek Holt Oct 8 '14 at 10:34
  • $\begingroup$ So I think I improved the title, a little, but I'd be happy to see someone do better. $\endgroup$ – Gerry Myerson Oct 8 '14 at 10:56

Your Answer

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

Browse other questions tagged or ask your own question.