# Why isn't $\text{Out}(S_6) \cong \{ e \}$?

I am learning about automorphisms, and came across the fact that:

$$\text{Out} (S_n) = \{e \} \hspace{15pt} \text{when} \hspace{5pt} n \neq 6$$

I see why this is true for n in general, since $$\text{Aut} (S_n) \cong S_n$$ and I can convince myself that $$\text{Inn}(S_n) \cong S_n$$, meaning that:

$$\text{Out} (S_n) \cong \text{Aut}(S_n) / \text{Inn}(S_n) \cong \{e \}$$

What fails when n = 6?

• For the reason see for example here. For $n=6$ there exists an effective action of $S_n$ on a set of $n$ elements, other than the natural one. – Dietrich Burde Oct 27 '19 at 19:20
• Possible duplicate of Outer Automorphisms of $S_n$. – Dietrich Burde Oct 27 '19 at 19:24
• $\psi$:$(12)\mapsto(15)(23)(46)$, $(13)\mapsto(14)(26)(35)$, $(14)\mapsto(13)(24)(56)$, $(15)\mapsto(12)(36)(45)$, $(16)\mapsto(16)(25)(34)$, $\psi\in \operatorname{Aut}(S_6)\setminus \operatorname{Inn}(S_6)$. – Andrews Nov 7 '19 at 11:13
• See this question for further reading. – Andrews Nov 7 '19 at 11:15
• For the sake of completeness here we explain why an automorphism mapping transpositions to transpositions is automatically inner. As Derek Holt said, it is not difficult at all. – Jyrki Lahtonen Nov 18 '19 at 16:00

For all $$n \ne 6$$, single transpositions have larger centralizers in $$S_n$$ than elements of order $$2$$ in any other conjugacy classes. So any automorphism of $$S_n$$ must map transpositions to transpositions, and from that it is not too difficult to deduce that the automorphism must be inner.
But in $$S_6$$, the elements $$(1,2)$$ and $$(1,2)(3,4)(5,6)$$ both have centralizers of order $$48$$, and it turns out that there is an automorphism of $$S_6$$ that maps one to the other.