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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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    Thank you so much joriki & Clive -- this helps me out a lot. I'm not that familiar with typesets yet.2012-08-11

1 Answers 1

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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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    yes, those 4s sho$u$ld have been ns. I will fix them now.2012-08-11