How to count the conjugates of an exotic $S_5$?

Question

It can be read off the The Elliott configuration - a $5$-coloring of $K6$ - that $S_6$ has an exotic $S_5$ subgroup (it's not a point stabilizer) which I will call $X_5 = \langle (1\;3\;6\;4\;5), (1\;2)(3\;6)(4\;5) \rangle$.

How can I prove that there are exactly 6 conjugates of $X_5$?

I thought that I could use orbit-stabilizer theorem with $G = S_6$ acting on itself by conjugation, then the number of conjugates of $X_5$ is $|G|/|N_{G}(X_5)|$.. but I don't know how to compute the denominator.

• Source: https://cp4space.wordpress.com/2012/11/24/outer-automorphism-of-s6/

Answer 1

Denote $c = [G : N_G(X_5)]$. By Lagrange's theorem $c = 1, 2, 3$ or $6$. The only non-trivial proper normal subgroup of $S_6$ is $A_6$, so $c$ cannot be $1$. For the same reason $c$ cannot be $3$, because otherwise $N_G(X_5)$ would contain the kernel of a homomorphism $G \rightarrow S_3$. Also, $c$ cannot be $2$ because then $N_G(X_5) = A_6$, but $S_5$ does not embed into $A_6$. The only possibility is $c = 6$.

Answer 2

my notes on the answer

• $|G:N_G(X_5)| = 1$ would imply $X_5$ normal in $S_6$ which is absurd because only $A_6$ is normal in $S_6$.
• $|G:N_G(X_5)| = 2$ would imply (again because $A_6$ is the only normal subgroup) that $N_G(X_5) = A_6$ but there's no embedding $S_n \to A_{n+1}$
• $|G:N_G(X_5)| = 3$ would imply there are 3 conjugates of $X_5$ so we have a homomorphism $S_6 \to S_3$ and we know $X_5^g = X_5$ for every $g \in X_5$ so it can't have trivial kernel but it must have otherwise we get a normal subgroup in $S_6$.

I don't understand why we can't use an argument of type 3 for the first two though?

I don't know why there's no embedding (injective homomorphism) $S_n \to A_{n+1}$, it's easy to show this for even $n$ by the orders but not clear how for odd $n$.