## Compute a bound on the index of Inn(Sym(6)) in Aut(Sym(6))

This exercise shows that $[\mathsf{Aut}(S_6) : \mathsf{Inn}(S_6)] \leq 2$.

1. Let $K$ be the conjugacy class of transpositions in $K_6$ and let $L$ be any other conjugacy class of elements of order 2. Prove that $|K| \neq |L|$ unless $L$ is the class consisting of products of three disjoint 2-cycles. Deduce that $\mathsf{Aut}(S_6)$ has a subgroup of index at most 2 that maps transpositions to transpositions.
2. Prove that $[\mathsf{Aut}(S_6) : \mathsf{Inn}(S_6)] \leq 2$. [Hint: Show that any automorphism that sends transpositions to transpositions is inner.]

1. Every element of order 2 in $S_6$ is a product of 2-cycles, and there are three such cycle shapes: $2$, $2,2$, and $2,2,2$. Let $K_1$, $K_2$, and $K_3$ denote the conjugacy classes of these elements. By this previous exercise, $|K_1| = 15$, $|K_2| = 45$, and $|K_3| = 15$.

In the previous exercise we saw that $\mathsf{Aut}(S_6)$ acts on the conjugacy classes of $S_6$ by application. Now if $\varphi \in \mathsf{Aut}(S_6)$, then $\varphi[K_1]$ is a conjugacy class of elements of order 2 and having cardinality $|K_1|$, so that $\varphi[K_1] \in \{K_1,K_3\}$. By the Orbit-Stabilizer Theorem, $[\mathsf{Aut}(S_6) : \mathsf{stab}_{\mathsf{Aut}(S_6)}(K_1)] = |\mathsf{Aut}(S_6) \cdot K_1| \leq 2$.

2. Suppose $\varphi \in \mathsf{Aut}(S_6)$ maps transpositions to transpositions. By a lemma to the previous exercise, we have $(1\ 2) \mapsto (a\ b_2)$, …, $(1\ 6) \mapsto (a\ b_6)$ for some distinct elements $a,b_2,\ldots,b_6 \in \{1,2,\ldots,6\}$. Consider now $S_6 \leq S_7$ as a subgroup; we can extend $\varphi$ to an automorphism of $S_7$ by defining $(1\ 7) \mapsto (a\ 7)$. Since every automorphism of $S_7$ is inner, we have $\varphi(\sigma) = \tau \sigma \tau^{-1}$ for some $\tau \in S_7$. In particular, note that $(\tau(1)\ \tau(7)) = (a\ 7)$, so that $\tau(7) = 7$. Thus $\tau$ is in (the canonical isomorphic copy of) $S_6$, so that $\varphi$ is in fact an inner automorphism of $S_6$.

Thus $\mathsf{stab}(K_1) \leq \mathsf{Inn}(S_6) \leq \mathsf{Aut}(S_6)$, and we have $[\mathsf{Aut}(S_6) : \mathsf{Inn}(S_6)] \leq 2$.

I think that $\mathsf{stab}_{S_6}(K_1)$ should be $\mathsf{stab}_{\mathsf{Aut_{S_6}}}(K_1)$.