On this document an outer automorphism of $S_6$ is constructed. I would like to use this construction to prove that $\mathrm{Aut}(S_6)\cong S_6\rtimes_\varphi\mathbb{Z}_2$. The idea would be to find an outer automorphism of order 2. If $F$ is the outer automorphism constructed in the document above then I can prove that the order of $F$ is even. Any outer automorphism is of the form $F\gamma_\sigma$ for some inner automorphism $\gamma_\sigma$. I can also prove that $F^2$ is inner so $F^2=\gamma_\tau$. I think one needs to use $\tau$ to construct $\sigma$ so that $F\gamma_\sigma$ has order 2. This is basically the approach used in this paper. But it seems to me the the construction of the outer automorphism done in the paper is a bit different than the construction I am interested in. I feel like there should be a way to modify Rotman's argument to make it work for the first construction. Any ideas?
The automorphism group of $S_6$ is isomorphic to a semidirect prodct
abstract-algebraautomorphism-groupfinite-groupsgroup-theorysymmetric-groups
Related Solutions
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
$\text{Aut}(A_6)=\text{Aut}(S_6)$.
Any such group $\widetilde{G}$ would have $G \trianglelefteq \widetilde{G}$ such that the map $\widetilde{G} \rightarrow \operatorname{Aut}(G)$ is surjective, giving an isomorphism $\widetilde{G} / C_{\widetilde{G}}(G) \rightarrow \operatorname{Aut}(G)$.
Take for example $G = D_8$, dihedral of order $8$.
You could take $\widetilde{G} = D_{16}$, dihedral of order $16$. This contains two normal subgroups isomorphic to $G$, and for both the map $\widetilde{G} \rightarrow \operatorname{Aut}(G)$ is surjective.
You could also take $\widetilde{G} = \operatorname{SD}_{16}$ (semidihedral group of order $16$), which contains a normal subgroup isomorphic to $G$, such that $\widetilde{G} \rightarrow \operatorname{Aut}(G)$ is surjective.
So it seems to me the $\widetilde{G}$ you are looking for is not unique.
Best Answer
I don't see how Rotman's construction can easily be used to find an outer automorphism of order $2$. That method uses the action on the cosets of a transitive subgroup of $S_6$ that is isomorphic to $S_5$. But the resulting outer automorphism depends on the numbering of the six cosets, and there are $720$ possible numberings. So you are effectively just looking for the innter automorphism $\gamma_\sigma$ that you describe.
I think the easiest way to do this is just to write down an outer automorphism of order $2$. To define the automorphism it suffices to specify the images of the transpositions $(1,2),(2,3),(3,4),(4,5),(5,6)$, each of which should be the product of three disjoint transpositions.
From the Coxeter presentation of $S_6$, it follows that such an assignment will define an automorphism provided that the images of two generators $x,y$ commute if and only if $x$ and $y$ do. This makes it possible to find the sought after automorphism of order 2 without too much difficulty.
I did this and found the automorphism of of order 2 below. This was not so hard. I knew that the iamges of $(1,2)$, $(3,4)$ and $(5,6)$ must all commute, so I wrote those down first, then I just found the images of $(2,3)$ and $(4,5)$ by trial and error.
$(1,2) \mapsto (1,2)(3,4)(5,6)$;
$(2,3) \mapsto (1,6)(2,3)(4,5)$;
$(3,4) \mapsto (1,2)(3,5)(4,6)$;
$(4,5) \mapsto (1,4)(2,3)(5,6)$;
$(5,6) \mapsto (1,2)(3,6)(4,5)$.