From (1), (2), (3), [Aut(S6):Inn(S6)]=2.
My question:
1. How to prove Aut(S6)≅S6⋊φZ2?
2. How to prove Aut(S6)≇S6×Z2?
3. How to prove Aut(A6)≅Aut(S6)?
My effort:
1. For 1, it remains to show there exists σ∈Aut(S6)∖Inn(S6) s.t. σ2=id.
2. For 2, Z(S6×Z2)=Z2, it's sufficient to show Z(Aut(S6))≠Z2.
3. For 3, I proved Aut(Sn)⩽Aut(An) (Is this correct?) and [Aut(A6):Inn(S6)]⩽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 ψ∈Aut(S6)∖Inn(S6) of order 2, and I wonder if there's a way to avoid doing so, i.e. find an element of order 2 in Aut(S6)∖Inn(S6) without writing it out explicitly.
2. I used the specific element ψ to show Z2≅⟨ψ⟩ is not normal in Aut(S6), I wonder if we can analysis the center of Aut(S6) instead. And what is center of Aut(S6)?
3. Is there a better way to prove Aut(A6)≅Aut(S6)?
Thanks for your time and effort!
Answer
For 1, there exists ψ∈Aut(S6)∖Inn(S6) s.t. ψ2=id.
ψ:(12)↦(15)(23)(46),(13)↦(14)(26)(35),(14)↦(13)(24)(56),(15)↦(12)(36)(45),(16)↦(16)(25)(34).
Therefore Aut(S6)≅S6⋊Z2.
For 2, we have short exact sequence for groups: 1→S6f→Aut(S6)π→Z2→1, Z2={±1,×}.
This sequence right splits, so there exists homomorphism g:Z2→Aut(S6) s.t. π∘g=id.
Let g(−1)=ψ∉Inn(S6), then g(1)=ψ2=id.
f:S6→Inn(S6), g:Z2→⟨ψ⟩.
Claim: ⟨ψ⟩ is not normal subgroup of Aut(S6), so Aut(S6)≇S6×Z2.
For σ∈S6, define γσ∈Inn(S6) to be action by conjugation of σ.
It's sufficient to prove γσψγ−1σ≠ψ, i.e.γσψ≠ψγσ for some σ∈S6.
Let σ=(12), γσψ((12))=γσ((15)(23)(46))=(12)(15)(23)(46)(12)=(13)(25)(46).
ψγσ(12)=ψ((12))=(15)(23)(46). γσψ≠ψγσ for σ=(12).
Thus Aut(S6)≅S6⋊Z2 and Aut(S6)≇S6×Z2.
For 3, fix 1≠α∈An, cα∈Inn(An) is action by conjugation of α.
Define φ:Aut(Sn)→Aut(An), φ(β)=βcαβ−1 for β∈Aut(Sn).
Easy to check φ is monomorphism, so Aut(Sn)⩽Aut(An)
Together with [Aut(A6):Inn(Sn)]⩽2 and [Aut(S6):Inn(Sn)]=2, we have
Aut(A6)=Aut(S6).
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