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I'd like to have a better understanding of this old valuable answer (I've tried to comment the post, but with no feedback).

I'm reading (correctly?):

  • the first two Lemmata as: $\operatorname{Aut}(S_n)/\operatorname{Inn}(S_n)\cong\phi(\operatorname{Aut}(S_n))$, where $\phi$ is the action of $\operatorname{Aut}(S_n)$ on the set of the conjugacy classes of $S_n$;
  • the third Lemma as: for $n\ne 6$, there's no class of involutions with the same size of the class of transpositions.

But how does from this follow that every automorphism of $S_{n\ne 6}$ is class-preserving (namely that for $n\ne 6$ the action is trivial)? For instance, $S_7$ has two conjugacy classes of involutions which have also the same size ($105$): how, from the three Lemmata, can one rule out the existence of some (outer) automorphism swapping such two classes? I feel like I'm misunderstanding (or not properly valorizing) something, so that I don't see how the Theorem follows.

As per the comments, now I see that, w.r.t. the action of $\operatorname{Aut}(S_n)$ on the set of conjugacy classes of $S_n$:

  • from the second lemma, follows that $\operatorname{Stab}(\operatorname{cl}((ij)))=\operatorname{Inn}(S_n)$;
  • from the third lemma, follows that, for $n\ne 6$, $\operatorname{Stab}(\operatorname{cl}((ij)))=\operatorname{Aut}(S_n)$.

Therefore, for $n\ne 6$, $\operatorname{Aut}(S_n)=\operatorname{Inn}(S_n)$. Thanks to @runway44 for their comments.

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    I don't see any comment from you under my old answer, though maybe you deleted it (I haven't used my anon account in quite awhile). According to the first lemma, if an automorphism stabilizes the class of transpositions (which it must, by the second lemma), then it's inner. If the automorphism is inner (which, remember, is the conclusion we want) then it follows it preserves all classes, not just the class of transpositions. Also, I'm curious what word translates to the English verb "valorizing." – anon Mar 31 '21 at 07:08
  • evaluating (or perhaps validating)? – Derek Holt Mar 31 '21 at 07:19
  • @runway44, yes I deleted the comment while posting this question. As for "valorizing" (of course I'm not English native...), I meant "making the most of" or "taking advantage of". As for the rest, I'll think of it some more. Thanks. –  Mar 31 '21 at 07:37
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    Cool. Keep in mind the fact that automorphisms are class-preserving is not used in the proof of the theorem, rather it is a consequence of the theorem. The proof the the theorem only uses the fact automorphisms preserve the one class of transpositions. – anon Mar 31 '21 at 08:07
  • @runway44 Maybe I got the point: Lemma 2 states: $\operatorname{Stab}(\operatorname{cl}((ij)))=\operatorname{Inn}(S_n)$. Lemma 3 states: for $n\ne 6$, $\operatorname{Stab}(\operatorname{cl}((ij)))=\operatorname{Aut}(S_n)$. Ergo, for $n\ne 6$, $\operatorname{Aut}(S_n)=\operatorname{Inn}(S_n)$. Does it fit? –  Mar 31 '21 at 11:02
  • Lemma 3 doesn't "state" that ${\rm Stab}({\rm cl}((ij)))={\rm Aut}(S_n)$, but it does imply it. – anon Mar 31 '21 at 11:07
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    @runway44, yeah, because it implies that the conjugacy class of the transpositions is fixed by any automorphism. Strictly speaking, also the Lemma 2 doesn't "state" that, but implies that because $\operatorname{Inn}(S_n)\le\operatorname{Stab}(\operatorname{cl}((ij)))$ trivially holds. –  Mar 31 '21 at 11:14

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