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"?


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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.