I'd like to demonstrate that every automorphism of $S_3$ is a conjugation but without looking at the order of the elements. My attempt is the following :

I've already proved that there is an injective homomorphism $f : Aut(S_3) \hookrightarrow S_3$ and also that the conjugation map $Ad$ is an automorphism.

Now let's consider $g : S_3 \longrightarrow Aut(S_3)$ such as $g(\sigma) = Ad(\sigma)$. It is easy to prove that $g$ is an injective homomorphism.

Let $\psi$ be in $Aut(S_3)$ and $\sigma$ be in $S_3$. Thus we have :

$$g \circ f : Aut(S3) \hookrightarrow S_3 \hookrightarrow Aut(S_3)$$ $$ (g \circ f)(\psi) = g(\sigma) = Ad(\sigma)$$

Because of the two injections, we can deduce that $S_3 \cong Aut(S_3)$ and thus $g \circ f$ is an automorphism. I think it also proves that there is as much elements in $Aut(S_3)$ than conjugations by the elements of $S_3$. Therefore, as the set of conjugations by the elements of $S_3$ is a subset of $Aut(S_3)$ we can deduce that every automorphism of $S_3$ is a conjugation.

I'm not quite sure about what I said. Could you please help me ?

Thanks a lot.

  • 1
    $\begingroup$ What is your injective map $f \colon \mathrm{Aut}(S_3) \to S_3$? $\endgroup$
    – Nick
    Nov 21, 2020 at 19:31
  • $\begingroup$ It is probably expecting you to say that $S_3$ is generated by the $3$ transpositions. Since you like arrows, the center is trivial thus $S_3$ embeds into $Aut(S_3)$. $\endgroup$
    – reuns
    Nov 21, 2020 at 19:56
  • $\begingroup$ @Nick math.stackexchange.com/q/3913926/850849 just notice that $Bij(X) \cong S_3$ $\endgroup$
    – Laury
    Nov 22, 2020 at 7:32
  • $\begingroup$ But how do you define your map ${\rm Aut}(S_3) \to {\rm Bij}(X)$ without using orders of elements? $\endgroup$
    – Derek Holt
    Nov 24, 2020 at 9:40
  • $\begingroup$ I am not convinced that this question makes much sense. It would be useful to prove that an automorphism of ${\rm Sym}(3)$ induces a permutation of the set $X = \{(1,2),(1,3),(2,3)\}$. OK, suppose not, and suppose for example we had an automorphism $\phi$ with $\phi((1,2)) = (1,2,3)$. Then (with $e$ equals identity) $e = \phi(e) = \phi((1,2)(1,2)) = \phi(1,2)\phi(1,2)=$ $ (1,2,3)(1,2,3) = (1,3,2)$, contradiction. But we have surreptitiously been using the fact that $(1,2)$ has order $2$ and $(1,2,3)$ does not, so it is dubious to claim that we have not looked at orders of elements. $\endgroup$
    – Derek Holt
    Nov 24, 2020 at 9:46

1 Answer 1


I'd attempt the following geometric argument.

Consider 2-simplex (an equilateral triangle on the plane). Its symmetry group is precisely $S_3.$ Moreover, $S_3$ acts simply transitively on the "flags" (sequences of faces ordered by inclusion) of faces of the triangle. In other words: it acts simply transitively on the poset of faces $\mathcal{P}(S_3)$ of $\Delta^2.$ Thus we can identify the poset of faces with the group $S_3$ as sets.

Now consider an arbitrary automorphism $\varphi:S_3\to S_3.$ It induces an automorphism of $\mathcal{P}(S_3).$ Geometric realization of this poset=barycentric subdivision of $\Delta^2.$ Any symmetry of $\Delta^2$ clearly induces an automorhism of the barycentric subdivision and vice versa, hence automorphisms of $S_3$ is the group $S_3$ itself.


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