Question about $\operatorname{Aut}(S_6)$ and $\operatorname{Aut}(A_6)$

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!

• 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. – Matt Samuel Oct 28 '19 at 17:29
• @MattSamuel Is there a easy way to show center of $\operatorname{Aut}(S_6)$ is trivial? I made a mistake before. – Andrews Oct 28 '19 at 17:36
• Not that I know of, but with the previous edit it seemed like something you were assuming. – Matt Samuel Oct 28 '19 at 17:44
• 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 $\psi f(\sigma)\psi^{-1}\ne f(\sigma)$ – Robert Chamberlain Oct 28 '19 at 18:39
• $\operatorname{Aut}(S_6)/\operatorname{Inn}(S_6)\cong\mathbb Z_2$ 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. – 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

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

• 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). – Thomas Browning Nov 4 '19 at 3:36
• @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. – Andrews Nov 4 '19 at 5:44