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 $|Y|\leq n-4$. 



Size of a certain union of the product of stablizers in  alternating groups
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 $n!/4$.


