4
$\begingroup$

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!

$\endgroup$
  • 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
1
$\begingroup$

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

$\text{Aut}(A_6)=\text{Aut}(S_6)$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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

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.