Let $G$ be a group such that $G =255 = 3.5.17$. Let $H$ be a sylow $17$ sub group of $G$.
Then, by Sylow's theorem : the number of sylow $17$ sub groups can be $n=1,18,35,…$ . Since, $n$ should divide $15$ also, $n=1$ is the only possibility.
Hence, if $H$ is a subgroup of order $17$, then it's normal as well. Hence, the normalizer of H $ = N(H)=G$.
Since, by the $N/C$ theorem : $N(H)/C(H)$ is isomorphic to a sub group of $Aut(H)$.
Hence, $|N(H)/C(H)|$ divides $|Aut(H)|=|Aut(\mathbb Z_{17)}|=|U(17)|=16$
i.e. $|N(H)/C(H)|$ divides $16$
I got till this point. But one hint to prove this question says that $|N(H)/C(H)|$ divides $255$ as well. I am not able to understand why it divides $|G|$.
Could $N(H)/C(H)$ be a subgroup of $G$ to be able to divide $|G|$.
Thank you for your help.
Best Answer
As you pointed out $G = N(H)$ since there is only one Sylow-17 subgroup, so $255 = |G| = |N(H)| = |C(H)| \cdot |N(H)/C(H)|$ i.e. $|N(H)/C(H)|$ divides $255$.
Thus, $|N(H)/C(H)|$ divides both 255 and 16, so must be equal to $1$ i.e. $G = N(H) = C(H)$. This tells us that everything in $G$ commutes with $H$ (in particular, elements of the Sylow-3 and Sylow-5 subgroups commute with the elements of $H$).