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

Question

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
$\psi$:$(12)\mapsto(15)(23)(46)$, $(13)\mapsto(14)(26)(35)$, $(14)\mapsto(13)(24)(56)$, $(15)\mapsto(12)(36)(45)$, $(16)\mapsto(16)(25)(34)$, $\psi\in \operatorname{Aut}(S_6)\setminus \operatorname{Inn}(S_6)$. — Andrews, Nov 07 '19 at 11:13
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

score 9 · Accepted Answer · answered Oct 27 '19 at 19:43

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.

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

1 Answers1