Let G be a finite group and S be the set of Sylow p-subgroups of G for a
prime p dividing the order of G. Assume that |S|>1.


Let U and V be two disjoint non-empty subsets of S such that,


$$\bigcup\_{P\in U}P=\bigcup\_{P\in V}P=\bigcup\_{P\in S}P$$ and


$$\bigcap\_{P\in U}P=\bigcap\_{P\in V}P=\bigcap\_{P\in S}P$$ .


Is there any group with this property ?



About the set of Sylow-$p$ subgroups of $G$
Yes I have found an example where this occurs.


The group $G$ is an extension $3^3:2^2$, of an elementary abelian group of order $27$ by one of order $4$, where the generators of the group of order $4$ act as the diagonmal matrices $(-1,-1,1)$ and $(1,-1,-1)$ on the group $3^3$. Then $G$ embeds into $S\_9$ with generators
$$(2,3)(5,6), (5,6)(8,9), (1,2,3), (4,5,6), (7,8,9).$$


Let $S$ be the set of Sylow $2$-subgroups of $G$. Then $|S|=27$ and $|\cup S| = 28$, and I had no difficulty at all, by making random choices, in finding $U$ and $V$ of sizes $13$ and $14$ with the specified properties.


