Suppose that $G$ is a permutation group of degree $11$ and order $7920=11 \cdot 10 \cdot 9 \cdot 8$. Prove that $G$ is simple.
A hint to the problem is to show that $|N_G(P)|=55$ where $P \in{\rm Syl}_{11}(G)$. I've shown this. But I don't know how to link that with simplicity of $G$.
Let $N$ be a nontrivial normal subgroup of $G$. Then, since $G$ has prime degree, it is primitive, and so $N$ is transitive. So $|N|$ is divisible by $11$ and we can assume that $P \le N$.
By the Frattini Argument, $G = NN_G(P)$. If $N_G(P) \not\le N$, then, since $|N_G(P)|=55$, we must have $P = N_N(P)$. But then $N$ has a normal $p$-complement $K$ by Burnside's Transfer Theorem, and $K$ is characteristic in $N$ and hence normal in $G$, and since $K$ cannot be transitive, we must have $K=1$. But then $N = P$ is normal in $G$, which we know is false.
So $N_G(P) \le N$ and $N=G$, and we have proved that $G$ is simple.