Going through examples of applications of the Sylow theorems in Fraleigh's book, when proving that no group of order 36 is simple, after concluding that $| H \cap K | = 3$ for two $3$-Sylows $H$,$K$, I can understand that $| N(H \cap K )|$ must be a multiple of $9$ by the first Sylow theorem, as in this question. What I can't understand is why he automatically rules out $9$, stating it has to be a $> 1$ multiple.
The same happened in the case of a order $48$ group, in which he says that for any two $2$-Sylows $H$, $K$, $| N(H \cap K )|$ must be a >1 multiple of $16$ since $| H \cap K | = 8$, but I could convince myself of that by counting, since in this particular case, $H \cap K$ is normal in both $H$ and $K$.
$H \cap K$ is normal in both $H$ and $K$ in the 36 case also, because in a group of order 9, every subgroup of order 3 is normal. So $N(H \cap K)$ contains both $H$ and $K$. Since $H$ and $K$ are distinct subgroups of order 9, we must have $N(H \cap K) > 9$.