From (1), (2), (3), $[\operatorname{Aut}(S_6):\operatorname{Inn}(S_6)]=2$.
My question:
$1$. How to prove $\operatorname{Aut}(S_6)\cong S_6\rtimes_\varphi \mathbb Z_2$?
$2$. How to prove $\operatorname{Aut}(S_6)\not\cong S_6\times \mathbb Z_2$?
$3$. How to prove $\operatorname{Aut}(A_6)\cong \operatorname{Aut}(S_6)$?
My effort:
$1$. For 1, it remains to show there exists $\sigma\in \operatorname{Aut}(S_6)\setminus \operatorname{Inn}(S_6)$ s.t. $\sigma^2=\text{id}$.
$2$. For 2, $Z(S_6\times\mathbb Z_2)=\mathbb Z_2$, it's sufficient to show $Z(\operatorname{Aut}(S_6))\neq\mathbb Z_2$.
$3$. For 3, I proved $\operatorname{Aut}(S_n)\leqslant\operatorname{Aut}(A_n)$ (Is this correct?) and $[\operatorname{Aut}(A_6):\operatorname{Inn}(S_6)]\leqslant 2$.
Update:
I wrote my answer below, but there still remain three questions:
$1$. I copied the result from a book to give an explicit element $\psi\in\operatorname{Aut}(S_6)\setminus \operatorname{Inn}(S_6)$ of order $2$, and I wonder if there's a way to avoid doing so, i.e. find an element of order $2$ in $\operatorname{Aut}(S_6)\setminus \operatorname{Inn}(S_6)$ without writing it out explicitly.
$2$. I used the specific element $\psi$ to show $\mathbb Z_2\cong \langle \psi\rangle$ is not normal in $\operatorname{Aut}(S_6)$, I wonder if we can analysis the center of $\operatorname{Aut}(S_6)$ instead. And what is center of $\operatorname{Aut}(S_6)$?
$3$. Is there a better way to prove $\operatorname{Aut}(A_6)\cong \operatorname{Aut}(S_6)$?
Thanks for your time and effort!
Best Answer
For 1, there exists $\psi\in \operatorname{Aut}(S_6)\setminus \operatorname{Inn}(S_6)$ s.t. $\psi^2=\text{id}$.
$\quad\psi:(12)\mapsto(15)(23)(46), (13)\mapsto(14)(26)(35), (14)\mapsto(13)(24)(56),\\\qquad (15)\mapsto(12)(36)(45), (16)\mapsto(16)(25)(34).$
Therefore $\operatorname{Aut}(S_6)\cong S_6\rtimes\mathbb Z_2$.
For 2, we have short exact sequence for groups: $1\to S_6\overset{f}{\to}\operatorname{Aut}(S_6)\overset{\pi}{\to} \mathbb Z_2\to 1 $, $\mathbb Z_2=\{\pm1,\times\}$.
This sequence right splits, so there exists homomorphism $g:\mathbb Z_2 \to \operatorname{Aut}(S_6)$ s.t. $\pi\circ g=\text{id}.$
Let $g(-1)=\psi\not\in \operatorname{Inn}(S_6)$, then $g(1)=\psi^2=\text{id}$. $f:S_6\to \operatorname{Inn}(S_6)$, $g:\mathbb Z_2 \to \langle\psi\rangle$.
Claim: $\langle\psi\rangle$ is not normal subgroup of $\operatorname{Aut}(S_6)$, so $\operatorname{Aut}(S_6)\not \cong S_6\times\mathbb Z_2$.
For $\sigma\in S_6$, define $\gamma_\sigma \in \operatorname{Inn}(S_6)$ to be action by conjugation of $\sigma$.
It's sufficient to prove $\gamma_\sigma\psi\gamma_\sigma^{-1}\neq\psi$, i.e.$\gamma_\sigma\psi\neq\psi\gamma_\sigma$ for some $\sigma\in S_6$.
Let $\sigma=(12)$, $\gamma_\sigma\psi((12))=\gamma_\sigma((15)(23)(46))=(12)(15)(23)(46)(12)=(13)(25)(46)$.
$\psi\gamma_\sigma(12)=\psi((12))=(15)(23)(46)$. $\gamma_\sigma\psi\neq\psi\gamma_\sigma$ for $\sigma=(12)$.
Thus $\operatorname{Aut}(S_6)\cong S_6\rtimes\mathbb Z_2$ and $\operatorname{Aut}(S_6)\not \cong S_6\times\mathbb Z_2$.
For 3, fix $1\neq\alpha\in A_n$, $c_\alpha\in\text{Inn}(A_n)$ is action by conjugation of $\alpha$.
Define $\varphi:\text{Aut}(S_n)\to\text{Aut}(A_n)$, $\varphi(\beta)=\beta c_\alpha \beta^{-1}$ for $\beta\in \text{Aut}(S_n)$.
Easy to check $\varphi$ is monomorphism, so $\text{Aut}(S_n)\leqslant\text{Aut}(A_n)$
Together with $[\text{Aut}(A_6):\text{Inn}(S_n)]\leqslant2$ and $[\text{Aut}(S_6):\text{Inn}(S_n)]=2$, we have
$\text{Aut}(A_6)=\text{Aut}(S_6)$.