# Showing that every automorphism of $S_3$ is a conjugation without using the orders of the elements

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.

• What is your injective map $f \colon \mathrm{Aut}(S_3) \to S_3$?
– Nick
Nov 21 '20 at 19:31
• 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 $Aut(S_3)$. Nov 21 '20 at 19:56
• @Nick math.stackexchange.com/q/3913926/850849 just notice that $Bij(X) \cong S_3$ Nov 22 '20 at 7:32
• But how do you define your map ${\rm Aut}(S_3) \to {\rm Bij}(X)$ without using orders of elements? Nov 24 '20 at 9:40
• 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. Nov 24 '20 at 9:46

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.