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 ?
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.
