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)S6?




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!


Answer



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).


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