# Proving that Aut($S_3$) is isomorphic to $S_3$

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

Now I have to prove that $Aut(S_3) \cong S_3$. I noticed that evaluating an automorphism in an element of X and taking the product of that with any other cycle doesnt 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 Aut($S_3$) permutes the transpositions: that is, we obtain a map $\phi$ : 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 #Aut($S_3) \geq 6$, since Aut($S_3$) contains the inner automorphisms and Z($S_3$) is trivial. Hence φ is an isomorphism.

I understand that since the 2-cycles generate $S_3$ any automorphism is determined by it's values in them but I'm not sure what $\phi$ 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 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 $(23) \rightarrow 3$.

• Are you sure $(23)\rightarrow 3$, not $(13)\rightarrow 3$ ? – Anjan3 Sep 14 '15 at 7:16
• 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

$S_3$ 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.
• 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