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.
group-theory finite-groups
asked Nov 16 at 10:10
Ashish
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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
asked Nov 16 at 10:10
Ashish
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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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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.
group-theory finite-groups
asked Nov 16 at 10:10
Ashish
121
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
asked Nov 16 at 10:10
Ashish
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asked Nov 16 at 10:10
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asked Nov 16 at 10:10
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asked Nov 16 at 10:10
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asked Nov 16 at 10:10
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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 $G/Z(G)$.
answered Nov 16 at 10:40
Robert Chamberlain
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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 $G/Z(G)$.
answered Nov 16 at 10:40
Robert Chamberlain
4,0351521
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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 $G/Z(G)$.
answered Nov 16 at 10:40
Robert Chamberlain
4,0351521
add a comment |
up vote
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down vote
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 $G/Z(G)$.
answered Nov 16 at 10:40
Robert Chamberlain
4,0351521
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 $G/Z(G)$.
answered Nov 16 at 10:40
Robert Chamberlain
4,0351521
answered Nov 16 at 10:40
Robert Chamberlain
4,0351521
answered Nov 16 at 10:40
Robert Chamberlain
4,0351521
answered Nov 16 at 10:40
Robert Chamberlain
4,0351521
4,0351521
add a comment |
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