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Theorem: For the symmetric group $S_n$, all subgroups of index $n$ are point stabilizers (for the natural action of $S_n$ on $\{1,\ldots,n\}$) when $n\neq 6$.

I am looking for a proof of this fact. Although I am seeking a proof at the level of an undergraduate student, I cannot find any kind of proofs. If someone could provide me a proof of this fact, I would be grateful.

Matcha Latte
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S.Lara
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2 Answers2

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There is a proof in B. Huppert, "Endliche Gruppen I" , Springer-Verlag, New York, 1967. It is Satz 5.5 of chapter II on page 175/176. But it is in German. The theorem in this book also includes case $n = 6$. For $n = 6$ see also https://math.stackexchange.com/q/1385135.

psl2Z
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Here is an outline of a proof which, with more details included, need not go beyond the level of an advanced undergraduate course.

Lets assume that $n \ge 5$ and that the simplicity of $A_n$ is known, from which you can easily deduce that $A_n$ is the only proper nontrivial normal subgroup of $S_n$.

Suppose $H < S_n$ with $|S_n:H|=n$ where $H$ is not a point stabilizer. Then the permutation action of $S_n$ on the cosets of $H$ must be faithful, and so its image is $S_n$. So this action is a homomorphism $S_n \to S_n$ that maps a point stabilizer in $S_n$ to $H$, and hence it is an outer automorphism of $S_n$.

So now we have reduced the problem to the well-known result (mentioned in a comment by Ethan Bolker) that $S_n$ has no outer automorphisms for $n \ne 6$.

I expect proofs of that have appeared here before, but here is an outline. Let $\phi:S_n \to S_n$ be an automorphism. The first step is to prove that, when $n \ne 6$, the map $\phi$ must map transpositions to transpositions. You could do that either by

(i) proving that if $\tau,\rho \in S_n$ with $\tau$ a transposition and $\rho$ a non-transposition of order $2$ then $|C_{S_n}(\tau)| > |C_{S_n}(\rho)|$ when $n \ne 6$, or

(ii) showing that the order of the product $\tau\tau^g$ for $g \in S_n$ can only have order $1,2$ or $3$, whereas $\rho\rho^g$ can have other orders, again when $n \ne 6$.

I prefer (ii) because it involves more group theory rather than number theory.

Having shown that $\phi$ maps transpositions to transpositions, we can compose with an inner automorphism and assume that $\phi: (1,2) \mapsto (1,2)$. Then the sets of transpositions $A_1 := \{(1,i) : i>3\}$ and $A_2 := \{(2,i) : i > 3\}$ are maximal with the property that the product of any two of them or any of them with $(1,2)$ has order $3$, and so $\phi$ must map $\{A_1,A_2\}$ to itself, and by composing with the inner automorphism induced by $(1,2)$ if necessary, we can assume $\phi:A_1 \mapsto A_1, A_2 \mapsto A_2$. and then by composing again with an inner automorphism we get $\phi:(1,i) \mapsto (1,i)$ for all $i \ge 2$, and hence $\phi$ is the identity, and the original $\phi$ was inner.

Derek Holt
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  • Thanks!! But I don't understand why i or ii implies that $\phi$ maps transpositions to transpositions, could you clarify that claim? – S.Lara Nov 06 '24 at 10:15
  • For (i), a group isomorphism $\phi:G \to H$ maps $C_G(g)$ to $C_H(\phi(g))$ for all $g \in G$, so for the automorphism $\phi$ of $S_n$ we have $|C_{S_n}(\tau) |= |C_{S_n}(\phi(\tau))|$. A similar argument works for (ii). – Derek Holt Nov 06 '24 at 16:45