Prove that $\operatorname{Aut}(S_n)$ isomorphic to $S_n$ when $n\geq 3$, and $n \neq 6$.

I can see that the automorphisms of $S_n$ have the same structure as $S_n$. But I am having trouble finding an isomorphism.

  • 1
    $\begingroup$ It is easy to find the isomorphism. For any group $G$ there is a natural homomorphism $G \to \operatorname{Aut}(G)$. The hard part is to prove that it is actually surjective in the case of $S_n$ for $n \geq 3, n \neq 6$. $\endgroup$ – Dune Aug 15 '14 at 12:54
  • $\begingroup$ Who is the natural homomorphism?? $\endgroup$ – Jam Aug 15 '14 at 13:07
  • $\begingroup$ Think about conjugation. $\endgroup$ – Dune Aug 15 '14 at 13:09
  • $\begingroup$ $g--->(x--->gx^(-1)g)$ $\endgroup$ – Jam Aug 15 '14 at 13:17
  • 2
    $\begingroup$ @ManolisLyviakis: right, you want to show that $S_n\cong$ Inn($S_n$) and that Inn($S_n$)=Aut($S_n$). I don't know what textbook you are using, but Dummit and Foote break the problem into manageable chunks. $\endgroup$ – mwmjp Aug 15 '14 at 13:35

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