Let $H \le G$ be a solvable subgroup of the finite group $G$ such that for each prime $p$ and each Sylow $p$-subgroup $S$ of $G$ we have $$ S \cap H \in \mbox{Syl}_p(H). $$ Then $H$ is subnormal in $G$, i.e. there exists a sequence of subgroups $H \unlhd A_1 \unlhd A_2 \unlhd \ldots \unlhd A_n \unlhd G$.
I tried to suppose that $H$ is not subnormal and we have $H < G$ with $N_G(H) = H$ (by going over to factor groups this poses no real restriction), but I have no idea how to proceed. As $H < G$ I guess there must be some $S \in \mbox{Syl}_p(G)$ such that $H \cap S$ is properly contained in $S$, then I know that $H \cap S < N_G(H\cap S) \le S$, maybe that helps. Also if $p$ divides the order of $H$, then each $S \in \mbox{Syl}_p(G)$ has non-trivial intersection with $H$ by hypothesis.
Also note that in each finite group $G$ with subgroup $H$ we have for $P \in \mbox{Syl}_p(H)$ some $S \in \mbox{Syl}_p(G)$ such that $P = H \cap S$, so the above is the converse assumption to this statement.
So any ideas how to solve this?