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Suppose the symmetric group $S_n$ acts transitively on a set $X$, i.e. for every $x, y \in X$, $\exists g \in S_n$ such that $gx = y$.

Show that either $|X| \le 2$ or $|X| \ge n$.

Small steps towards the solution:

  • As $S_n$ acts transitively on a set $X$, the whole of $X$ is one single orbit under the action of $S_n$.

  • By the Orbit-Stabilizer Theorem, then, $|X|$ = $|S_n : \text{Stabilizer of }x|$ for any $x \in X$.

  • We also know that the Stabilizer of any $x \in X$ is a subgroup of $S_n$.

  • When $|X| = 2$, the Stabilizer of $x$ is the alternating group $A_n$.

I'm halfway but can't get the final result. Any help would be much appreciated, as always. Thank you.

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Following joriki's comment, I've $\LaTeX$'d the post; click 'edit' to see what I did to get the subscripts, $\le$ signs, etc. –  Clive Newstead Aug 11 '12 at 10:36
    
Thank you so much joriki & Clive -- this helps me out a lot. I'm not that familiar with typesets yet. –  Conan Wong Aug 11 '12 at 14:13
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2 Answers

up vote 6 down vote accepted

For $n < 4$ the result is clear. For $n = 4$ the result is false - we have a surjection $S_4 \to S_3$ by killing the unique normal subgroup of order 4, given by the products of transpositions (thanks to cocopuffs for fixing an error in this before).

For $n > 4$, the map $S_n \to \text{Aut}_\text{Set}(X) = S_{\# X}$ induced by your action must either be injective or have kernel $A_n$ or $S_n$. In the first case we must have $\#X \ge n$ and in the others we have $\#X = 2$ or $1$ respectively.

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But the coset space has $6$ elements in the case $n=4$ –  Cocopuffs Aug 11 '12 at 9:57
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Yes. Maybe you mean let $S_{4}$ act by conjugation on the $3$ non-identity elements of that normal Klein $4$-subgroup. That is a transitive action of $S_{4}$ on a set $X$ of size $3.$ –  Geoff Robinson Aug 11 '12 at 10:00
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@Geoff Robinson that's better (more concerete) than what i meant, i meant that the quotient by that group is $S_3$ so we get a transitive action on a 3 element set. –  user29743 Aug 11 '12 at 10:03
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@Cocopuffs the question is equivalent to the existence of an arbitrary subgroup of index between $2$ and $n$, not just a normal one. –  user29743 Aug 11 '12 at 10:04
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@Cocopuffs: Your first comment about the index was right. Your second comment isn't correct (you dohave the right nomal subgroups, of course). $S_{4}$ does have a subgroup of index $3$, a Sylow $2$-subgroup. Therefore, there is a homomorphism from $S_{4}$ to $S_{3}$ (in fact surjective). The kernel is the normal Klein $4$-subgroup of course. –  Geoff Robinson Aug 11 '12 at 10:06
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Let $S_n\to G$ is ephimorhism and $|G|<n$. Let us prove that $|G|<3$. We have $\mathbb{C}$ representation of $S_n$, $(V,\pi)$ and $\dim V<n$. We can assume that $(V,\pi)$ is irreducible. I will use theory of representation of symmetric group. Ireeducible representation of $S_n$ is diagramm of Young(http://en.wikipedia.org/wiki/Young_diagram#Diagrams). And If demission of this representation(see http://en.wikipedia.org/wiki/Representation_theory_of_the_symmetric_group)

Is it answer?

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Thank you user37274 for your Rep theory approach as well, but there are some gaps in your description that I can't quite follow. Thanks anyway. –  Conan Wong Aug 11 '12 at 14:20
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