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I'm reading Artin's Algebra, Edition 1. In Chapter 5 there's proposition (8.4):

Let $c_g$ denote conjugation by $g$, the map $c_g(x) = gxg^{-1}$. The map $f: S_3 \rightarrow Aut(S_3)$ from the symmetric group to its group of automorphisms which is defined by the rule $g\mapsto c_g$ is bijective.

Artin provides the proof as:

Let $A$ denote the group of automorphisms of $S_3$. We know from Chapter 2 (3.4) that $c_g$ is an automorphism. Also, $c_{gh}=c_gc_h$ because $c_{gh}(x) = (gh)x(gh)^{-1}= ghxh^{-1}g^{-1}= c_g(c_h(x))$ for all $x$. This shows that $f$ is a homomorphism. Now conjugation by $g$ is the identity if and only if $g$ is in the center of the group. The center of $S_3$ is trivial, so $f$ is injective.

It is to prove surjectivity of $f$ that we look at a permutation representation of $A$. The group $A$ operates on the set $S_3$ in the obvious way; namely, if $\alpha$ is an automorphism and $s \in S_3$, then $\alpha s = \alpha (s)$. Elements of $S_3$ of different orders will be in distinct orbits for this operation. So $A$ operates on the subset of $S_3$ of elements of order $2$. This set contains the three elements $\{y, xy, x^2y\}$. If an automorphism a fixes both $xy$ and $y$, then it also fixes their product $xyy = x$. Since $x$ and $y$ generate $S_3$, the only such automorphism is the identity. This shows that the operation of $A$ on $\{y, xy, x^2y\}$ is faithful and that the associated permutation representation $A \rightarrow Perm\{y, xy, > x^2y\}$ is injective. So the order of $A$ is at most $6$. Since $f$ is injective and the order of $S_3$ is $6$, it follows that $f$ is bijective.

I don't understand the bolded line: why "Elements of $S_3$ of different orders will be in distinct orbits for this operation"?

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This is because a permutation in $S_n$ (in a finite group $G$) and its image by an automorphism of $S_n$ ($G$) have the same order.

Indeed, if $\alpha$ is an automorphism, $\alpha(s)^k=e\iff s^k=e$.

There results that permutations in the orbit of $s$ under $\operatorname{Aut}(S_n)$ have the same order as $s$. By contrapositive, permutations with different orders can't be in the same orbit.

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  • $\begingroup$ I still don't get it. What you said is , for $s\in S_n$ and $\alpha \in Aut(S_n)$, let $s' = \alpha s$, then $s$ and $s'$ has the same order -- but why? Secondly, what Artin says is, if $s\in S_n$ and $t\in S_n$ has different order, then $s$ and $t$ will be in different orbit -- how could I reach this from your point? $\endgroup$ – athos Apr 28 '16 at 11:16
  • $\begingroup$ I've added some details. Is it clearer now? $\endgroup$ – Bernard Apr 28 '16 at 11:22
  • $\begingroup$ Thank you so much! Now I got it. $\endgroup$ – athos Apr 29 '16 at 4:07

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