Struggling to understand proof, there is no simple group of order $525$

Prove there is no simple group of order $525.$

$N(H)$ is the normalizer of a set $H$.

Proof:Let $L_7$ be a Sylow $7$ subgroup of $G$. It follows that $|N(L_7)|=35$ by the sylow theorems. Let $L$ be a subgroup of $N(L_7)$ of order $5$. Since $N(L_7)$ is cyclic $N(L) geq N(L_7)$( I do not understand this part of the proof), so that $35$ divides $|N(L)|$. But L is contained in a sylow $5$ subgroup which is abelian, thus $25$ divides $|N(L)|$ as well. It follows that $175$ divides $|N(L)|$, now the index theorem gives a contradiction.

I am having a lot of confusion proving nonabelian groups of specific orders are not simple, because all of the new facts appearing about normalizers, which I have never learnt before. How should I better learn how to derive facts about normalizers.

In particular in this proof, I know why $|N(L_7)|=35$, I do not need help on that, but why does this line hold?

Let $L$ be a subgroup of order $5$. Since $N(L_7)$ is cyclic $N(L) geq N(L_7)$ so that $N(L_7)$ is a subgroup of $N(L)$?