Let $G$ be a group of order $525$ and suppose $G$ is not simple. Then $G$ must have $15$ Sylow $7$-subgroups (of order $7$). Let $H$ be such a subgroup. Then $[G:N(H)] = 15$, where $N(H)$ is the normalizer of $H$ in $G$. Thus $N(H)$ has order $35$, whence $N(H)$ is cyclic.
MY QUESTION:
Let $K$ be a subgroup of $H$ of order $5$. Prove that $N(K)\ge N(H)$.