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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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    You can use $\TeX$ on this site. To do so, enclose it in dollar signs; single dollar signs for inline formulas, double dollar signs for displayed equations. You can also see the $\TeX$ code for any math formatting you see on this site by right-clicking on it and selecting "Show Math As:TeX Commands".2012-08-11
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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.2012-08-11
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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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    But the coset space has $6$ elements in the case $n=4$2012-08-11
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    oh, you're right!2012-08-11
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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.$2012-08-11
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    For $n=4$ the result holds as well since ${1}$, $V_4$, $A_4$ and $S_4$ are the only normal subgroups of $S_4$, all of whose index are $\le 2$ or $\ge 4$2012-08-11
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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.2012-08-11
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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.2012-08-11
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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.2012-08-11
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    @GeoffRobinson Ooh you're right. My argument reducing it to normal subgroups doesn't work for $n=4$. Sorry!2012-08-11
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    @countinghaus I have posted an answer above. Can you check it please?2012-08-11
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    @countinghaus Thank you for your time and help. In the 2nd paragraph of your answer (after the edits thus far), do you mean "kernel An or Sn"? Thought it might be a typo. Thanks.2012-08-11
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    @Cocopuffs and GeoffRobinson -- Thank you very much for your scrutiny and comments. Much appreciated too.2012-08-11
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    yes, those 4s should have been ns. I will fix them now.2012-08-11