Let $G$ be a simple group of order $1092 =2^2\cdot 3\cdot7\cdot13.$ Prove that $G$ has a single conjugacy class of subgroups of index $14,$ but no subgroup of index $13.$
Extending from this post: the subgroup of index 14 in the simple group of order 1092, I want to prove that there is no subgroups of index $13$. I have managed to show that $G$ has a single conjugacy class of subgroups of index $14$ using the hints provided in the previous post. However, I can't seem to show that there is no subgroup of index $13.$
So far I tried to use the technique mentioned on page 204-205 of Dummit and Foote.
"Let $G$ be a simple group with a proper subgroup of index $k$. If $P\in{\rm Syl}_p(G)$ and if $P$ is also a Sylow $p$-subgroup of $A_k$, then $|N_G(P)|$ must divide $|N_{A_k}(P)|$"
Let $P$ be a $13$-Sylow subgroup of $G$. Then, from my calculations, $|N_G(P)|=|N_{A_k}(P)|=2\cdot 3\cdot 13$ which is sadly not sufficient to get the desired contradiction.
Is there any other ways to improve on my technique? If not, can somebody provide some hints on how to proceed.
Best Answer
A subgroup $H$ of index $13$ would have order $84$, and so by Sylow's Theorem would have a normal Sylow $7$-subgroup $P$.
So $H \le N_G(P)$. If $H = N_G(P)$, then $G$ has $13$ Sylow $7$-subgroups, contradicting Sylow's Theorem. So by Lagrange we must have $N_G(P) = G$, contradicting simplicity.