# Ruling out orders when applying Sylow's theorems

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 $N(H \cap K) > 9$.

• Ok, I figured that out but I was wondering if there was anyway to scape from knowlege that $p^2$-order groups are abelian. Apparently not, it was fundamental actually, right? Apr 23 '14 at 4:18
• You can also use the fact that in a group of order $p^n$, every subgroup of order $p^{n-1}$ is normal. That would cover both of your examples.
– Ted
Apr 23 '14 at 5:22