There was previous task was same but with $N = 185$. And I prove it by showing that number of Sylow subgroups is 1 for every prime $p\mid N$. But there I have some options $N_5 \in \{1, 51\}$, $N_{17} = 1$, $N_3 \in \{1, 85\}$.
I've tried to get contradiction from $N_5 = 51$ or $N_3=85$, but I didn't manage to do it
I understand that it's impossible to have $N_5 = 51$ and $N_3=85$ at the same time.
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$\begingroup$The key to $N=255$ is the observation that it is squarefree and coprime with its totient function.
Let $n_q$ denote the number of Sylow $q$-subgroups.
So, let us begin by the observation than any group $G$ of order $15$ is cyclic. Indeed, let us merely note $n_3\mid 5$, $n_3\equiv 1\text{ mod }3$ and so $n_3=1$. Similarly, $n_5\mid 3$ and $n_5\equiv 1\text{ mod }5$ which gives that $n_3=1$. Thus, we see that both the Sylow subgroups $P,Q$ of $G$ are both normal. Since evidently $P\cap Q=\{e\}$, and $PQ=G$ (since $\displaystyle |PQ|=\frac{|P||Q|}{|P\cap Q|}=\frac{15}{1}=15=|G|$) we may conclude that $G\cong P\times Q\cong\mathbb{Z}/3\times\mathbb{Z}/5\cong\mathbb{Z}/15$.
Now, to the case of $|G|=255$. Let $P$ be a Sylow $17$-subgroup. Note then that $n_{17}\mid 15$ and $n_{17}\cong 1\text{ mod }17$ gives $n_{17}=1$. Thus, we see that $P\unlhd G$. Now, here comes the step that is different than the standard approach. Since $P$ is normal we have that $G$ acts on $P$ by conjugation, giving us a group map $G\to\text{Aut}(P)$ whose kernel is $C_G(P)$. Now, note that since $P$ is just $C_{17}$ we have that $|\text{Aut}(P)|=16$. Note now that since $(|G|,16)=1$ you can conclude from the first isomorphism theorem that the image of the group map $G\to\text{Aut}(P)$ is trivial, and so the kernel is everything. This says that $C_G(P)=G$, and so $P\subseteq Z(G)$. Note then that $G/P$ is a group of order $15$ which, by what we have said in the previous paragraph, is cyclic. Now, it is a common theorem that if you mod out a group by a subgroup of its center and you get something cyclic, your group must have been abelian the whole time. So, from all of this we can conclude that $G$ is abelian in which case, by any method that tickles your fancy, you get that $G$ must just be $\mathbb{Z}/255$.
While the above may be ad hoc, it was actually just a simple application of the following (EXTREMELY USEFUL!) theorems:
Theorem: Let $G$ be a finite group and $N\unlhd G$. Then, if $(|G|,|\text{Aut}(N)|)=1$ then $N\leqslant Z(G)$.
and
Theorem: If $G$ is a group and $N\leqslant Z(G)$ such that $G/N$ is cyclic of finite order, then $G$ is abelian.
These allow us to prove the more general theorem:
Theorem: Let $G$ be a finite group such that $(|G|,\varphi(|G|))=1$ (where $\varphi$ is the totient function), then $G$ is cyclic.
As a side note, a cool fact is that the integer $n$ has the property that the ONLY subgroup of order $n$ is $\mathbb{Z}/n$ is equivalent to $(n,\varphi(n))=1$.
The proof of the above can be found on my blog here.
$\endgroup$ $\begingroup$Suppose $G$ is a finite group of order $|G| = 255 = 3 \cdot 5 \cdot 17$. We will show that $G$ must be cyclic.
As you noticed, there must be only one Sylow subgroup $P$ of order $17$. Thus $P$ must be a normal subgroup. Since $17$ is prime the subgroup $P$ is cyclic, say $P = \langle a \rangle$. It is not difficult to see that every group of order $15$ is cyclic, so $G/P = \langle bP \rangle$ is also cyclic. Now $(bP)^{|b|} = P$, so $|bP| = 15 \mid |b|$.
If $|b| = 255$ we are done. Therefore we can assume $|b| = 15$. We will show that $ab = ba$ which implies that $ab$ has order $255$ since $15$ and $17$ are coprime. Because $P$ is normal, $bab^{-1} = a^i$ for some integer $i$. Using this fact we get $b^2ab^{-2} = ba^ib^{-1} = (bab^{-1})^i = a^{i^2}$ and by induction $b^kab^{-k} = a^{i^k}$ for all $k \geq 1$. Thus $a = b^{15}ab^{-15} = a^{i^{15}}$, which gives $i^{15} \equiv 1 \mod{17}$. By Fermat's little theorem $i^{16} \equiv 1 \mod{17}$ and since $15$ and $16$ are coprime, $i \equiv 1 \mod{17}$. Therefore $bab^{-1} = a$, proving the claim.
This exercise is a special case of a more general fact. Using the idea of this solution and the fact that in a group of squarefree order the Sylow subgroup corresponding to the largest prime is normal, you can prove that every group of order $n$ is cyclic when $n$ and $\varphi(n)$ are coprime ($\varphi$ is the totient function). This is done by Tibor Szele in the following short article (in German).
$\endgroup$ 1 $\begingroup$T. Szele, Uber die endlichen ordnungszahlen zu denen nur eine Gruppe gehirt, Commentarii Mathematici Helvetici, 20, 265-67, (1947).
$$|N|=255=3\cdot 5\cdot 17$$
By Sylow theorem, the $\,17-$Sylow sbgp. $\,P_{17}\,$ is normal in $\,N\,$, so taking any $\,5-$Sylow sbgp. $\,P_5\,$, we get a sbgp. $\,K:=P_5P_{17}\,$ of index $\,3\,$ in $\,N\,$, and since this last is the minimal prime dividing $\,|N|\,$ , we get that $\,K\triangleleft N\,$ .
It's easy to prove that any group of order $\,85\,$ is cyclic , and thus its automorphism group has order $\,\phi(5)\phi(17)=64\,$ , so the only homomorphism $\,C_3\to Aut(K)\,$ ( the first group is the cyclic one of order $\,3\,$) there is is the trivial one (why?), and since by the above it follows that $\,N\cong K\rtimes P_3\,$ (with, of course, $\,P_3\cong C_3\,$), we get this semidirect product is in fact a direct one of abelian groups and, thus, $\,N\,$ is abelian and, in fact, cyclic.
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