No simple group for order $5103 = 3^6 \cdot 7$
using the small index argument and assuming the group is simple:
consider $n_{3}$ this must divide the rest of the order of the group and be congruent to $1$ modulo $3$ so $n_3 = 7$
Further there exists $\varphi : G \to S_{7}$ since $\mid G : N_{G}(P_{3}) \mid \, = 7$ and $ker(\varphi) \le N_{G}(P_{3})$ assuming this kernel is the identity we get $G \simeq S_{P} \le S_7$ hence $\mid G \mid$ $\displaystyle \Bigg \vert \ \mid S_7\mid$
but the solution says $|G| \nmid 8!$ and I see that $n_3 \in \{1,7\}$ and $1+7 = 8$ so I see how they got $8$ but why $8$? aren't I supposed to be looking at $S_7$ and not $S_8$
EDIT: another group of order from the same question is for order $4125 = 5^3 \cdot 3 \cdot 11$, the solution uses $n_5 = 11$ from $n_5 \in \{ 1,3,11 \}$ but the uses $14!$ so now I am more confused at which $S_k$ how to decide what $K = ??$