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I'm doing an exercise were I had to first prove that all automorphisms of $S_3$ induce a permutation in $X= \{ \alpha \in S_3 \mid{\rm order}(\alpha) = 2\}$, which was easy enough.

Now I have to prove that ${\rm Aut}(S_3) \cong S_3$.

I noticed that evaluating an automorphism in an element of $X$ and taking the product of that with any other cycle doesn't work, since there's no way you're getting all of $S_3$ out of that. Evaluation in general doesn't seem too useful in general.

I tried googling for a bit and found a pdf that says the following:

Any automorphism of $S_3$ must send elements of order $2$ to elements of order $2$; in this case the only elements of order $2$ are the transpositions, so an element of ${\rm Aut}(S_3)$ permutes the transpositions: that is, we obtain a map $\varphi:{\rm Aut}(S_3) \rightarrow S_{ \{(1\,2),(2\,3),(1\,3)\}}$ . Since the transpositions generate $S_3$, this map is injective. On the other hand $\#{\rm Aut}(S_3) \geq 6$, since ${\rm Aut}(S_3)$ contains the inner automorphisms and $Z(S_3)$ is trivial. Hence $\varphi$ is an isomorphism.

I understand that since the $2$-cycles generate $S_3$ any automorphism is determined by its values in them but I'm not sure what $\varphi$ would look like nor why it has to exist exactly.

Any help would be greatly appreciated.

Edit: I think I've got it. I've got an isomorphism between ${\rm Aut}(S_3)$ and $S_{\{(12),(13),(23)\}}$ by $\varphi (f) = (f(1\,2)\,f(1\,3)\,f(2\,3))$, and then I have an isomorphism between $S_{\{(12),(13),(23)\}}$ and $S_3$ by relabeling the elements, i.e., $(12) \rightarrow 1 $, $(23) \rightarrow 2$ and $(23) \rightarrow 3$.

Shaun
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Martin Williams
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  • Are you sure $(23)\rightarrow 3$, not $(13)\rightarrow 3$ ? – KON3 Sep 14 '15 at 07:16
  • There was a typo, one of the $(23)$ in "$(23) \rightarrow2$ and $(23) \rightarrow 3$" should be a $(13)$. I'm just looking at the problem again for the first time in a while, but I don't see why it would matter if $(23) \rightarrow3$ or $(13) \rightarrow3$. – John Williams Sep 15 '15 at 14:06

2 Answers2

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$S_3$ is generated by its elements of order $2$, and the elements of $S_3$ of order $2$ are exactly the single transpositions. So all you need to show is that if you send transpositions to transpositions (through the automorphism on $S_3$), then this is equivalent to just permuting the 3 underlying elements. And then you need that permuting the $3$ elements gives an automorphism on $S_3$ always, but this is trivial.

user2566092
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  • I've managed to proved that any automorphism of $S_3$ is determined by it's values in the single transpositions. That way I have an isomorphism between Aut($S_3$) and the permutations of the single transpositions, which I think is similar to what you meant in the first part. But now I think I need an isomorphisms from the permutations on the 2-cycles to $S_3$ itself, don't I? – Martin Williams Apr 28 '14 at 19:03
  • I think I've got it. I've got an isomorphism between Aut($S_3$) and $S_{{(12),(13),(23)}}$ by $\phi (f) = (f(1,2),f(1,3),f(2,3))$, and then I have an isomorphism between $S_{{(12),(13),(23)}}$ and $S_3$ by relabeling the elements ie $(12) \rightarrow 1 $ , $(23) \rightarrow 2$ and $(23) \rightarrow 3$. – Martin Williams Apr 28 '14 at 19:29
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Automorphisms send conjugacy classes to conjugacy classes, while preserving elements' order. Therefore, in $S_3$, they are class-preserving, and hence cycle type-preserving, and finally inner. So: $$\operatorname{Aut}(S_3)=\operatorname{Inn}(S_3)\stackrel{Z(S_3)=\{()\}}{\cong} S_3$$

citadel
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