No simple group of order 2016

Let $|G| = 2016 = 2^{5}\cdot 3^{2} \cdot 7$.

By Sylow theorems let $n_{7}$ be the number of Sylow $7$-subgroups.

Then by Sylow theorems, if $n_{7} \neq 1$.

We proceed through cases:

Case 1: Suppose $n_{7} = 8$.

We have $[G: N_{G}(P)] = n_{7} = 8$, where $P$ is a sylow $7$-subgroup. Then $\left|N_{G}(P)\right| = 4\cdot 9 \cdot 7$.

Now, let $G$ act on the set of left cosets of $N_{G}(P)$, then this action affords a homomorphism: $\phi: G \rightarrow S_{8}$.

Now, suppose $G$ is simple, then $\phi$ is injection into $S_{8}$. If $G$ has a subgroup of index $2$, then it would not be simple and so $G$ is a subgroup of $S_{8}$ which does not have a subgroup of index $2$ it follows from the Second Isomorphism Theorem that $G$ must be in $A_{8}$.

Fact 1: if $P$ is a Sylow $7$-subgroup in $S_{8}$, then $\left|N_{A_{8}}(P)\right| = \frac{1}{2}\left|N_{S_{8}}(P) \right|$.

Fact 2: $\left|N_{S_{p +1}}(P) \right| = p(p-1)$, it follows that $|N_{S_{8}}(P)| = 6 \cdot 7$.

It follows from Fact 1 that $\left|N_{A_{8}}(P) \right| = 21$. But $N_{G}(P) \leq N_{A_{8}}(P)$ and so we should have $\left |N_{G}(P) \right|$ dividing $21$, which is absurd and so $G$ must not be simple.