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Let $n>15$ and $A=A_n$ be the alternating group of degree $n$ on the set $\{1,\dots,n\}$. For any subset $X$ of $\{1,\dots,n\}$, $Stab_A(X)$ denote as usual the set of all permutations $\sigma$ of $A$ such that $x^\sigma=x$ for all $x\in X$ (the usual stablizer of $X$ in $A$ under the usual action of $A$ on $\{1,\dots,n\}$.

Is the following set has the size greater that $n!/4$?

$$\bigcup Stab_A(X) \times Stab_A(Y),$$ where the pair $(X,Y)$ runs over all pair of subsets $X$ and $Y$ of $\{1,\dots,n\}$ such that $X \cap Y=\varnothing$, $X\cup Y={1,\dots,n}$ and $|X|\leq n-4$ and $|Y|\leq n-4$.

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Consider the cycle breakdown of a permutation in $A_n$. The only reason that a permutation could not be in this set is if it includes a cycle of size greater than $n-4$.

Proof: Otherwise it is always possible to choose $X$ and $Y$ such that each cycle is contained completely in $X$ or $Y$. Then one can write the permutation as a product of two, one in the stabilizer of $X$ and one in the stabilizer of $Y$.

It is not too hard to count the number of permutations with large cycles.

$\frac{n!}{n-3}+\frac{n!}{n-2}+\frac{n!}{n-1}+\frac{n!}{n}< .3 n! $

is an easy upper bound. Since half of these are not alternating, the number of alternating permutations not of this type is at least $n!/4$.

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  • $\begingroup$ @Will. Thanks. you mean in the first line ``could NOT be" $\endgroup$
    – Alireza Abdollahi
    Commented Dec 23, 2011 at 20:25
  • $\begingroup$ @will. I think here in the proof something is wrong or I am in wrong!: Suppose $n=20$ and take the permutation $t=(1,2,3,4,5,6,7,8)(9,10,11,12,13,14,15,16,17,18,19,20)$ Certainly $t\in A_{20}$, but what are choices for $X$ and $Y$ such that $(1,2,3,4,5,6,7,8) \in Stab_{A_n}(X)$ and $(9,10,11,12,13,14,15,16,17,18,19,20) \in Stab_{A_n}(Y)$? One can easily find choices for $X$ and $Y$ such that these cycles lie in $Stab_{S_n}(X)$ and $Stab_{S_n}(Y)$. So I think there is a gap in the proof! Am I right? $\endgroup$
    – Alireza Abdollahi
    Commented Dec 24, 2011 at 15:06
  • $\begingroup$ You are correct. My proof is in error. I can't yet think up a nice way to enumerate all additional bad permutations. $\endgroup$
    – Will Sawin
    Commented Dec 24, 2011 at 17:59

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