on such group whose inner automorphisms group isomorphic to S3. [duplicate]











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  • Is there a way to describe all finite groups $G$ such that $operatorname{Aut}(G) cong S_3$?

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Let $frac{G}{Z(G)}≅S_3$, such that $S_3$ is permutation group on 3 letters and $Z(G)$ is non trivial central subgroup of $G$. What are the possibility of group $G$ and does there exist always an non-inner automorphism group $G$ ?
or if inner automorphism group is given then what we can say about $G$ ?
Thanks.






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marked as duplicate by Derek Holt, Nicky Hekster group-theory
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 Nov 16 at 11:24


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    • Is there a way to describe all finite groups $G$ such that $operatorname{Aut}(G) cong S_3$?

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    Let $frac{G}{Z(G)}≅S_3$, such that $S_3$ is permutation group on 3 letters and $Z(G)$ is non trivial central subgroup of $G$. What are the possibility of group $G$ and does there exist always an non-inner automorphism group $G$ ?
    or if inner automorphism group is given then what we can say about $G$ ?
    Thanks.






    group-theory finite-groups







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     Nov 16 at 11:24


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      • Is there a way to describe all finite groups $G$ such that $operatorname{Aut}(G) cong S_3$?

        2 answers




      Let $frac{G}{Z(G)}≅S_3$, such that $S_3$ is permutation group on 3 letters and $Z(G)$ is non trivial central subgroup of $G$. What are the possibility of group $G$ and does there exist always an non-inner automorphism group $G$ ?
      or if inner automorphism group is given then what we can say about $G$ ?
      Thanks.






      group-theory finite-groups







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      This question already has an answer here:




      • Is there a way to describe all finite groups $G$ such that $operatorname{Aut}(G) cong S_3$?

        2 answers




      Let $frac{G}{Z(G)}≅S_3$, such that $S_3$ is permutation group on 3 letters and $Z(G)$ is non trivial central subgroup of $G$. What are the possibility of group $G$ and does there exist always an non-inner automorphism group $G$ ?
      or if inner automorphism group is given then what we can say about $G$ ?
      Thanks.





      This question already has an answer here:




      • Is there a way to describe all finite groups $G$ such that $operatorname{Aut}(G) cong S_3$?

        2 answers







      group-theory finite-groups




      group-theory finite-groups






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      asked Nov 16 at 10:10









      Ashish

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          It's not too hard using Sylow subgroups to show $S_3$ has no non-trivial non-split central extensions, so the only possible $G$ are those of the form $Gcong S_3times A$ with $A$ abelian.



          If $Z(G)=1$ then $Gcong S_3$ has no non-inner automorphisms.



          In general, given $G/Z(G)$ there's not a lot we can say about $G$. The best I can think of is if $G$ is perfect then it is a quotient of the Schur cover of $G/Z(G)$.






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            It's not too hard using Sylow subgroups to show $S_3$ has no non-trivial non-split central extensions, so the only possible $G$ are those of the form $Gcong S_3times A$ with $A$ abelian.



            If $Z(G)=1$ then $Gcong S_3$ has no non-inner automorphisms.



            In general, given $G/Z(G)$ there's not a lot we can say about $G$. The best I can think of is if $G$ is perfect then it is a quotient of the Schur cover of $G/Z(G)$.






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              It's not too hard using Sylow subgroups to show $S_3$ has no non-trivial non-split central extensions, so the only possible $G$ are those of the form $Gcong S_3times A$ with $A$ abelian.



              If $Z(G)=1$ then $Gcong S_3$ has no non-inner automorphisms.



              In general, given $G/Z(G)$ there's not a lot we can say about $G$. The best I can think of is if $G$ is perfect then it is a quotient of the Schur cover of $G/Z(G)$.






              share|cite|improve this answer























                up vote
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                up vote
                0
                down vote









                It's not too hard using Sylow subgroups to show $S_3$ has no non-trivial non-split central extensions, so the only possible $G$ are those of the form $Gcong S_3times A$ with $A$ abelian.



                If $Z(G)=1$ then $Gcong S_3$ has no non-inner automorphisms.



                In general, given $G/Z(G)$ there's not a lot we can say about $G$. The best I can think of is if $G$ is perfect then it is a quotient of the Schur cover of $G/Z(G)$.






                share|cite|improve this answer












                It's not too hard using Sylow subgroups to show $S_3$ has no non-trivial non-split central extensions, so the only possible $G$ are those of the form $Gcong S_3times A$ with $A$ abelian.



                If $Z(G)=1$ then $Gcong S_3$ has no non-inner automorphisms.



                In general, given $G/Z(G)$ there's not a lot we can say about $G$. The best I can think of is if $G$ is perfect then it is a quotient of the Schur cover of $G/Z(G)$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 16 at 10:40









                Robert Chamberlain

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