Sunday, October 13, 2019

abstract algebra - Question about operatornameAut(S6) and operatornameAut(A6)



From (1), (2), (3), [Aut(S6):Inn(S6)]=2.



My question:



1. How to prove Aut(S6)S6φZ2?




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 σAut(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 ψAut(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 Z2ψ 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 ψAut(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)S6Z2.






For 2, we have short exact sequence for groups: 1S6fAut(S6)πZ21, Z2={±1,×}.



This sequence right splits, so there exists homomorphism g:Z2Aut(S6) s.t. πg=id.



Let g(1)=ψInn(S6), then g(1)=ψ2=id.

f:S6Inn(S6), g:Z2ψ.



Claim: ψ is not normal subgroup of Aut(S6), so Aut(S6)S6×Z2.



For σS6, define γσInn(S6) to be action by conjugation of σ.



It's sufficient to prove γσψγ1σψ, i.e.γσψψγσ for some σS6.



Let σ=(12), γσψ((12))=γσ((15)(23)(46))=(12)(15)(23)(46)(12)=(13)(25)(46).




ψγσ(12)=ψ((12))=(15)(23)(46). γσψψγσ for σ=(12).



Thus Aut(S6)S6Z2 and Aut(S6)S6×Z2.






For 3, fix 1αAn, cαInn(An) is action by conjugation of α.



Define φ:Aut(Sn)Aut(An), φ(β)=βcαβ1 for βAut(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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