Show there are are no simple groups of order 1638

Question

I'm trying to use Sylow Theory to show there are no simple groups of order 1638. I got so far as to factor $1638 = 2*3^2*7*13$ and compute that we must have $n_2|819$ $n_3 \in \{1,7,3\}$ $n_7 \in \{1,78\}$ and $n_{13} \in \{1,14\}$. But from there none of the usual tricks seem to work. How can I solve this?

Answer 1

Suppose that $n_{13}=14$. Then $G$ embeds as a transitive subgroup of $S_{14}$. The point stabiliser $H$ has order $3^213$ and has a normal subgroup $K$ of order $13$. But the normalizer of $K$ in $S_{13}$ has order $12\times 13$ which is not a multiple of $3^213$.