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 $[\operatorname{Aut}(A_6):\operatorname{Inn}(S_6)]\leqslant 2$.


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!

  • 2
    $\begingroup$ If you know the center is trivial, you don't need to do all that work. $S_6\times\mathbb Z_2$ clearly has nontrivial center. $\endgroup$ – Matt Samuel Oct 28 '19 at 17:29
  • $\begingroup$ @MattSamuel Is there a easy way to show center of $\operatorname{Aut}(S_6)$ is trivial? I made a mistake before. $\endgroup$ – Andrews Oct 28 '19 at 17:36
  • $\begingroup$ Not that I know of, but with the previous edit it seemed like something you were assuming. $\endgroup$ – Matt Samuel Oct 28 '19 at 17:44
  • 1
    $\begingroup$ It's not clear what your contradiction is. I think the easiest way to show ${\rm Aut}(S_6)\not\cong S_6\times\mathbb{Z}_2$ is to let $\psi\in{\rm Aut}(S_6)\setminus{\rm Inn}(S_6)$, let $\sigma\in S_6$ with $\psi(\sigma)\ne\sigma$ and show $\psi f(\sigma)\psi^{-1}\ne f(\sigma)$ $\endgroup$ – Robert Chamberlain Oct 28 '19 at 18:39
  • 1
    $\begingroup$ $\operatorname{Aut}(S_6)/\operatorname{Inn}(S_6)\cong\mathbb Z_2$ does not imply that $\operatorname{Aut}(S_6)\cong S_6\rtimes_\varphi \mathbb Z_2$. (The extension may not split.) Although in this case it is true. $\endgroup$ – verret Oct 28 '19 at 20:19

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


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  • $\begingroup$ Your answer to part 1 only shows that there exists a non-inner automorphism of order 2 (not that $Aut(S_6)$ is a semidirect product). $\endgroup$ – Thomas Browning Nov 4 '19 at 3:36
  • $\begingroup$ @ThomasBrowning I showed $\mathbb Z_2=\langle \psi\rangle$ can be seen as a subgroup of $\operatorname{Aut}(S_6)$, together with $ \psi\in\operatorname{Aut}(S_6)\setminus\operatorname{Inn}(S_6)$ and $\operatorname{Inn}(S_6)$ is of index 2, $\operatorname{Aut}(S_6)$ is a semidirect product by definition. $\endgroup$ – Andrews Nov 4 '19 at 5:44

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