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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 $K$.

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

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  • $\begingroup$ 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? $\endgroup$
    – ulilaka
    Apr 23 '14 at 4:18
  • $\begingroup$ 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. $\endgroup$
    – Ted
    Apr 23 '14 at 5:22

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