# The only group of order $255$ is $\mathbb Z_{255}$ ( Using Sylow and the $N/C$ Theorem)

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 $|G|$.

• $|N(H)/C(H)|$ is equal to the index of $C(H)$ in $N(H)$ hence divides $|N(H)|=|G|$. – Quang Hoang Aug 24 '14 at 18:57
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 $H$).