A simple group of order 168 doesn’t have subgroups of order 14

Question

Let $G$ be a simple group of order 168. Prove that $G$ doesn't have subgroups of order 14.

I know there are eight 7-Sylows in $G$ and that the normalizer of a 7-Sylow has order 21, it was a previous exercise.

My idea: Assume there is a subgroup $H<G$ of order 14 and then find a nontrivial normal subgroup. But I don't know how to proceed. Can someone give me any hints?

Answer 1

Let $H \lt G$ with $|H|=14$ and let $P \in Syl_7(H)$, then $P$, having index $2$, is normal in $H$, hence $H \subseteq N_G(P)$. Observe that $P \in Syl_7(G)$. But $|G:N_G(P)|=8$, hence $|N_G(P)|=21$ implying $14 \mid 21$, a contradiction to Lagrange's Theorem.