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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 $X_5 = \langle (1\;3\;6\;4\;5), (1\;2)(3\;6)(4\;5) \rangle$.

enter image description here

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.

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Your generators for X5 only generate a group of order 12, specifically, the symmetries of the hexagon. – Josh B. Mar 7 '13 at 20:09
@JoshB., thanks very much for catching that! I don't know why I had that on my page.. – user58512 Mar 7 '13 at 20:17
Think of what the possibilities are for the normalizer. If it has index 2 then it must be what group? What are the subgroups of index 3 in the ambient group? – Tobias Kildetoft Mar 7 '13 at 20:23
up vote 1 down vote accepted

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 $c = 6$.

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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 $n$.

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You can use the simplicity of $A_n$ for $n \geq 5$. When $n \geq 5$, there is no embedding $S_n \rightarrow A_{n+1}$ because $A_{n+1}$ does not have a proper subgroup of index $< n+1$. – Mikko Korhonen Mar 10 '13 at 19:59
@m.k., thank you very much! – user58512 Mar 10 '13 at 20:00
You can't use the same argument in the first two cases as in the third because $S_6$ does have a subgroup of index $1$ and also of index $2$. Just the existence of a subgroup $H$ of index $1$ or $2$ is not enough to find a contradiction. You have to use some property of $H$ (if $H$ has index $1$ use the fact that it's the normalizer of $X_5$, if it has index $2$ you use the fact that it contains $S_5$, so $H$ cannot be $A_6$). – Mikko Korhonen Mar 10 '13 at 20:29
@m.k., I'm a bit confused aren't we using the fact that it's a normal subgroup of $S_6$ (since it came from the kernel of a homomorphism)? – user58512 Mar 10 '13 at 20:31

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